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How to make a variable always equal to the result of some calculations?
The Next CEO of Stack OverflowHow to detect unsigned integer multiply overflow?The Definitive C++ Book Guide and ListIs the sizeof(some pointer) always equal to four?Using scanf() in C++ programs is faster than using cin?Pretty-print C++ STL containersImage Processing: Algorithm Improvement for 'Coca-Cola Can' RecognitionReplacing a 32-bit loop counter with 64-bit introduces crazy performance deviationsCreate istream and ostream objects in C++Deoptimizing a program for the pipeline in Intel Sandybridge-family CPUsSetting two variables *always* equal?
In math, if z = x+y/2
, then z
will always change whenever we replace the value of x
and y
. Can we do that in programming without having to specifically updating z
whenever we change the value of x
and y
?
I mean something like that won't work, right?
int x;
int y;
int zx + y;
cin >> x;
cin >> y;
cout << z;
c++ c++11
New contributor
|
show 1 more comment
In math, if z = x+y/2
, then z
will always change whenever we replace the value of x
and y
. Can we do that in programming without having to specifically updating z
whenever we change the value of x
and y
?
I mean something like that won't work, right?
int x;
int y;
int zx + y;
cin >> x;
cin >> y;
cout << z;
c++ c++11
New contributor
1
Why do you want to do that?
– Robert Andrzejuk
12 hours ago
1
Right, that won't work. That's a spreadsheet thing.
– Pete Becker
12 hours ago
2
@RobertAndrzejuk Because it'd be very useful. For example, if you write a game and have something like networth(cash+the worth of all you own). You will have to call the function of networth everytime one of those update. That would be very annoying and error prone if you forgot to call the function somewhere.
– Nay Wunna Zaw
12 hours ago
5
@NayWunnaZaw In my experience this is why getters and setters are encouraged over direct variable access. If you always wanted networth to be updated, you could retrieve the value using getNetWorth() which would itself call updateNetWorth() before returning the value... or just calculate it before returning it.
– Onyz
11 hours ago
3
That's called a "function."
– ApproachingDarknessFish
6 hours ago
|
show 1 more comment
In math, if z = x+y/2
, then z
will always change whenever we replace the value of x
and y
. Can we do that in programming without having to specifically updating z
whenever we change the value of x
and y
?
I mean something like that won't work, right?
int x;
int y;
int zx + y;
cin >> x;
cin >> y;
cout << z;
c++ c++11
New contributor
In math, if z = x+y/2
, then z
will always change whenever we replace the value of x
and y
. Can we do that in programming without having to specifically updating z
whenever we change the value of x
and y
?
I mean something like that won't work, right?
int x;
int y;
int zx + y;
cin >> x;
cin >> y;
cout << z;
c++ c++11
c++ c++11
New contributor
New contributor
edited 7 hours ago
Nay Wunna Zaw
New contributor
asked 12 hours ago
Nay Wunna ZawNay Wunna Zaw
8810
8810
New contributor
New contributor
1
Why do you want to do that?
– Robert Andrzejuk
12 hours ago
1
Right, that won't work. That's a spreadsheet thing.
– Pete Becker
12 hours ago
2
@RobertAndrzejuk Because it'd be very useful. For example, if you write a game and have something like networth(cash+the worth of all you own). You will have to call the function of networth everytime one of those update. That would be very annoying and error prone if you forgot to call the function somewhere.
– Nay Wunna Zaw
12 hours ago
5
@NayWunnaZaw In my experience this is why getters and setters are encouraged over direct variable access. If you always wanted networth to be updated, you could retrieve the value using getNetWorth() which would itself call updateNetWorth() before returning the value... or just calculate it before returning it.
– Onyz
11 hours ago
3
That's called a "function."
– ApproachingDarknessFish
6 hours ago
|
show 1 more comment
1
Why do you want to do that?
– Robert Andrzejuk
12 hours ago
1
Right, that won't work. That's a spreadsheet thing.
– Pete Becker
12 hours ago
2
@RobertAndrzejuk Because it'd be very useful. For example, if you write a game and have something like networth(cash+the worth of all you own). You will have to call the function of networth everytime one of those update. That would be very annoying and error prone if you forgot to call the function somewhere.
– Nay Wunna Zaw
12 hours ago
5
@NayWunnaZaw In my experience this is why getters and setters are encouraged over direct variable access. If you always wanted networth to be updated, you could retrieve the value using getNetWorth() which would itself call updateNetWorth() before returning the value... or just calculate it before returning it.
– Onyz
11 hours ago
3
That's called a "function."
– ApproachingDarknessFish
6 hours ago
1
1
Why do you want to do that?
– Robert Andrzejuk
12 hours ago
Why do you want to do that?
– Robert Andrzejuk
12 hours ago
1
1
Right, that won't work. That's a spreadsheet thing.
– Pete Becker
12 hours ago
Right, that won't work. That's a spreadsheet thing.
– Pete Becker
12 hours ago
2
2
@RobertAndrzejuk Because it'd be very useful. For example, if you write a game and have something like networth(cash+the worth of all you own). You will have to call the function of networth everytime one of those update. That would be very annoying and error prone if you forgot to call the function somewhere.
– Nay Wunna Zaw
12 hours ago
@RobertAndrzejuk Because it'd be very useful. For example, if you write a game and have something like networth(cash+the worth of all you own). You will have to call the function of networth everytime one of those update. That would be very annoying and error prone if you forgot to call the function somewhere.
– Nay Wunna Zaw
12 hours ago
5
5
@NayWunnaZaw In my experience this is why getters and setters are encouraged over direct variable access. If you always wanted networth to be updated, you could retrieve the value using getNetWorth() which would itself call updateNetWorth() before returning the value... or just calculate it before returning it.
– Onyz
11 hours ago
@NayWunnaZaw In my experience this is why getters and setters are encouraged over direct variable access. If you always wanted networth to be updated, you could retrieve the value using getNetWorth() which would itself call updateNetWorth() before returning the value... or just calculate it before returning it.
– Onyz
11 hours ago
3
3
That's called a "function."
– ApproachingDarknessFish
6 hours ago
That's called a "function."
– ApproachingDarknessFish
6 hours ago
|
show 1 more comment
7 Answers
7
active
oldest
votes
You can get close to this with by using a lambda in C++. Generally, when you set a variable like
int x;
int y;
int zx + y;
z
will only be the result of x + y
at that time. You'd have to do z = x + y;
every time you change x
or y
to keep it update.
If you use a lambda though, you can have it capture what objects it should refer to, and what calculation should be done, and then every time you access the lambda it will give you the result at that point in time. That looks like
int x;
int y;
auto z = [&]() return x + y; ;
cin >> x;
cin >> y;
cout << z();
and now z()
will have the correct value instead of the uninitialized garbage that the original code had.
If the computation is very expensive you can even add some caching to the lambda to make sure you aren't running the computation when you don't need to. That would look like
auto z = [&]() static auto cache_x = x;
static auto cache_y = y;
static auto cache_result = x + y;
if (x == cache_x && y == cache_y)
return cache_result;
else
cache_x = x;
cache_y = y;
cache_result = x + y;
return cache_result;
;
2
Shouldn't the caches have a type?
– Aconcagua
12 hours ago
1
Also, at that point you should not repeat the cache code for each kind of lambda. A class encapsulation is pretty much required there.
– Max Langhof
12 hours ago
And a class can be used to avoid thez()
syntax as well: coliru.stacked-crooked.com/a/3c3dafe9856c28b8
– Mooing Duck
7 hours ago
@MooingDuck That's awesome! I didn't think it was possible. It deserves to be an answer on its own!
– Fabio Turati
6 hours ago
@FabioTurati See Aconcagua's answer
– NathanOliver
5 hours ago
|
show 1 more comment
The closest you probably can get is to create a functor:
#include <iostream>
int main()
int x;
int y;
auto z = [&x, &y] return x + y; ; // a lambda capturing x and y
while(true)
std::cin >> x;
std::cin >> y;
std::cout << z() << "n";
add a comment |
You mean something like this:
class Z
int& x;
int& y;
public:
Z(int& x, int& y) : x(x), y(y)
operator int() return x + y;
;
The class delays calculation of the result until casted as int. As cast operator is not explicit, Z
can be used whenever an int is required. As there's an overload of operator<<
for int, you can use it with e. g. std::cout
directly:
int x, y;
Z z(x, y);
std::cin >> x >> y;
if(std::cin) // otherwise, IO error! (e. g. bad user input)
std::cout << z << std::endl;
add a comment |
- You create a function for that.
- You call the function with the appropriate arguments when you need the value.
int z(int x, int y)
return (x + y);
int x;
int y;
// This does ot work
// int zx + y;
cin >> x;
cin >> y;
cout << z(x, y);
1
yes, i mean the only way to get z updated is to call that function everytime we change x and y values?
– Nay Wunna Zaw
12 hours ago
@NayWunnaZaw, yes, that is correct.
– R Sahu
12 hours ago
1
@NayWunnaZaw, you can avoid the repeated use ofx
andy
by using a lambda function, as shown by Nathan but you still have to make the call.
– R Sahu
12 hours ago
3
@NayWunnaZaw To complete the issue: Even with my solution, there is a function call (to the cast operator!), just that it is implicit and not required explicitly...
– Aconcagua
12 hours ago
add a comment |
You can define the following lambda z
which always returns the current value of x+y
because x
and y
are captured by reference:
DEMO
int x;
int y;
const auto z = [&x, &y]() return x+y; ;
1
Don't use parentheses around return values –return
is not a function, and in worst case, the parentheses can even create a dangling reference (with return type beingdecltype(auto)
).
– Aconcagua
12 hours ago
@Aconcagua thx! you are right. I edited my answer.
– Hiroki
12 hours ago
Why downvoted ? At least I posted my answer faster than Ted with my tested code.
– Hiroki
5 hours ago
add a comment |
There are two chief techniques:
Deferred calculation - instead of
z
being a simple variable, make it a function which calculates the value on demand (see other answers for examples). This can be source-code transparent ifz
is some proxy object with implicit conversion to the required type.Explicit notification of changes. This requires
x
andy
to be observable types; when either changes value, thenz
updates itself (and notifies its observers if applicable).
The first version is usually preferred, but the second may be more appropriate if you need z
to be an observable type.
add a comment |
You can get what you're asking for by using macros:
int x, y;
#define z (x + y)
/* use x, y, z */
#undef z
The #undef
is for a little sanity. For more sanity, don't use macros at all, and go with one of the other answers, and deal with the extra verbosity.
Although a class with a custom operator int
would work in a lot of cases ... hmm.
add a comment |
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can get close to this with by using a lambda in C++. Generally, when you set a variable like
int x;
int y;
int zx + y;
z
will only be the result of x + y
at that time. You'd have to do z = x + y;
every time you change x
or y
to keep it update.
If you use a lambda though, you can have it capture what objects it should refer to, and what calculation should be done, and then every time you access the lambda it will give you the result at that point in time. That looks like
int x;
int y;
auto z = [&]() return x + y; ;
cin >> x;
cin >> y;
cout << z();
and now z()
will have the correct value instead of the uninitialized garbage that the original code had.
If the computation is very expensive you can even add some caching to the lambda to make sure you aren't running the computation when you don't need to. That would look like
auto z = [&]() static auto cache_x = x;
static auto cache_y = y;
static auto cache_result = x + y;
if (x == cache_x && y == cache_y)
return cache_result;
else
cache_x = x;
cache_y = y;
cache_result = x + y;
return cache_result;
;
2
Shouldn't the caches have a type?
– Aconcagua
12 hours ago
1
Also, at that point you should not repeat the cache code for each kind of lambda. A class encapsulation is pretty much required there.
– Max Langhof
12 hours ago
And a class can be used to avoid thez()
syntax as well: coliru.stacked-crooked.com/a/3c3dafe9856c28b8
– Mooing Duck
7 hours ago
@MooingDuck That's awesome! I didn't think it was possible. It deserves to be an answer on its own!
– Fabio Turati
6 hours ago
@FabioTurati See Aconcagua's answer
– NathanOliver
5 hours ago
|
show 1 more comment
You can get close to this with by using a lambda in C++. Generally, when you set a variable like
int x;
int y;
int zx + y;
z
will only be the result of x + y
at that time. You'd have to do z = x + y;
every time you change x
or y
to keep it update.
If you use a lambda though, you can have it capture what objects it should refer to, and what calculation should be done, and then every time you access the lambda it will give you the result at that point in time. That looks like
int x;
int y;
auto z = [&]() return x + y; ;
cin >> x;
cin >> y;
cout << z();
and now z()
will have the correct value instead of the uninitialized garbage that the original code had.
If the computation is very expensive you can even add some caching to the lambda to make sure you aren't running the computation when you don't need to. That would look like
auto z = [&]() static auto cache_x = x;
static auto cache_y = y;
static auto cache_result = x + y;
if (x == cache_x && y == cache_y)
return cache_result;
else
cache_x = x;
cache_y = y;
cache_result = x + y;
return cache_result;
;
2
Shouldn't the caches have a type?
– Aconcagua
12 hours ago
1
Also, at that point you should not repeat the cache code for each kind of lambda. A class encapsulation is pretty much required there.
– Max Langhof
12 hours ago
And a class can be used to avoid thez()
syntax as well: coliru.stacked-crooked.com/a/3c3dafe9856c28b8
– Mooing Duck
7 hours ago
@MooingDuck That's awesome! I didn't think it was possible. It deserves to be an answer on its own!
– Fabio Turati
6 hours ago
@FabioTurati See Aconcagua's answer
– NathanOliver
5 hours ago
|
show 1 more comment
You can get close to this with by using a lambda in C++. Generally, when you set a variable like
int x;
int y;
int zx + y;
z
will only be the result of x + y
at that time. You'd have to do z = x + y;
every time you change x
or y
to keep it update.
If you use a lambda though, you can have it capture what objects it should refer to, and what calculation should be done, and then every time you access the lambda it will give you the result at that point in time. That looks like
int x;
int y;
auto z = [&]() return x + y; ;
cin >> x;
cin >> y;
cout << z();
and now z()
will have the correct value instead of the uninitialized garbage that the original code had.
If the computation is very expensive you can even add some caching to the lambda to make sure you aren't running the computation when you don't need to. That would look like
auto z = [&]() static auto cache_x = x;
static auto cache_y = y;
static auto cache_result = x + y;
if (x == cache_x && y == cache_y)
return cache_result;
else
cache_x = x;
cache_y = y;
cache_result = x + y;
return cache_result;
;
You can get close to this with by using a lambda in C++. Generally, when you set a variable like
int x;
int y;
int zx + y;
z
will only be the result of x + y
at that time. You'd have to do z = x + y;
every time you change x
or y
to keep it update.
If you use a lambda though, you can have it capture what objects it should refer to, and what calculation should be done, and then every time you access the lambda it will give you the result at that point in time. That looks like
int x;
int y;
auto z = [&]() return x + y; ;
cin >> x;
cin >> y;
cout << z();
and now z()
will have the correct value instead of the uninitialized garbage that the original code had.
If the computation is very expensive you can even add some caching to the lambda to make sure you aren't running the computation when you don't need to. That would look like
auto z = [&]() static auto cache_x = x;
static auto cache_y = y;
static auto cache_result = x + y;
if (x == cache_x && y == cache_y)
return cache_result;
else
cache_x = x;
cache_y = y;
cache_result = x + y;
return cache_result;
;
edited 2 hours ago
isanae
2,53711437
2,53711437
answered 12 hours ago
NathanOliverNathanOliver
97.1k16137214
97.1k16137214
2
Shouldn't the caches have a type?
– Aconcagua
12 hours ago
1
Also, at that point you should not repeat the cache code for each kind of lambda. A class encapsulation is pretty much required there.
– Max Langhof
12 hours ago
And a class can be used to avoid thez()
syntax as well: coliru.stacked-crooked.com/a/3c3dafe9856c28b8
– Mooing Duck
7 hours ago
@MooingDuck That's awesome! I didn't think it was possible. It deserves to be an answer on its own!
– Fabio Turati
6 hours ago
@FabioTurati See Aconcagua's answer
– NathanOliver
5 hours ago
|
show 1 more comment
2
Shouldn't the caches have a type?
– Aconcagua
12 hours ago
1
Also, at that point you should not repeat the cache code for each kind of lambda. A class encapsulation is pretty much required there.
– Max Langhof
12 hours ago
And a class can be used to avoid thez()
syntax as well: coliru.stacked-crooked.com/a/3c3dafe9856c28b8
– Mooing Duck
7 hours ago
@MooingDuck That's awesome! I didn't think it was possible. It deserves to be an answer on its own!
– Fabio Turati
6 hours ago
@FabioTurati See Aconcagua's answer
– NathanOliver
5 hours ago
2
2
Shouldn't the caches have a type?
– Aconcagua
12 hours ago
Shouldn't the caches have a type?
– Aconcagua
12 hours ago
1
1
Also, at that point you should not repeat the cache code for each kind of lambda. A class encapsulation is pretty much required there.
– Max Langhof
12 hours ago
Also, at that point you should not repeat the cache code for each kind of lambda. A class encapsulation is pretty much required there.
– Max Langhof
12 hours ago
And a class can be used to avoid the
z()
syntax as well: coliru.stacked-crooked.com/a/3c3dafe9856c28b8– Mooing Duck
7 hours ago
And a class can be used to avoid the
z()
syntax as well: coliru.stacked-crooked.com/a/3c3dafe9856c28b8– Mooing Duck
7 hours ago
@MooingDuck That's awesome! I didn't think it was possible. It deserves to be an answer on its own!
– Fabio Turati
6 hours ago
@MooingDuck That's awesome! I didn't think it was possible. It deserves to be an answer on its own!
– Fabio Turati
6 hours ago
@FabioTurati See Aconcagua's answer
– NathanOliver
5 hours ago
@FabioTurati See Aconcagua's answer
– NathanOliver
5 hours ago
|
show 1 more comment
The closest you probably can get is to create a functor:
#include <iostream>
int main()
int x;
int y;
auto z = [&x, &y] return x + y; ; // a lambda capturing x and y
while(true)
std::cin >> x;
std::cin >> y;
std::cout << z() << "n";
add a comment |
The closest you probably can get is to create a functor:
#include <iostream>
int main()
int x;
int y;
auto z = [&x, &y] return x + y; ; // a lambda capturing x and y
while(true)
std::cin >> x;
std::cin >> y;
std::cout << z() << "n";
add a comment |
The closest you probably can get is to create a functor:
#include <iostream>
int main()
int x;
int y;
auto z = [&x, &y] return x + y; ; // a lambda capturing x and y
while(true)
std::cin >> x;
std::cin >> y;
std::cout << z() << "n";
The closest you probably can get is to create a functor:
#include <iostream>
int main()
int x;
int y;
auto z = [&x, &y] return x + y; ; // a lambda capturing x and y
while(true)
std::cin >> x;
std::cin >> y;
std::cout << z() << "n";
answered 12 hours ago
Ted LyngmoTed Lyngmo
3,5802522
3,5802522
add a comment |
add a comment |
You mean something like this:
class Z
int& x;
int& y;
public:
Z(int& x, int& y) : x(x), y(y)
operator int() return x + y;
;
The class delays calculation of the result until casted as int. As cast operator is not explicit, Z
can be used whenever an int is required. As there's an overload of operator<<
for int, you can use it with e. g. std::cout
directly:
int x, y;
Z z(x, y);
std::cin >> x >> y;
if(std::cin) // otherwise, IO error! (e. g. bad user input)
std::cout << z << std::endl;
add a comment |
You mean something like this:
class Z
int& x;
int& y;
public:
Z(int& x, int& y) : x(x), y(y)
operator int() return x + y;
;
The class delays calculation of the result until casted as int. As cast operator is not explicit, Z
can be used whenever an int is required. As there's an overload of operator<<
for int, you can use it with e. g. std::cout
directly:
int x, y;
Z z(x, y);
std::cin >> x >> y;
if(std::cin) // otherwise, IO error! (e. g. bad user input)
std::cout << z << std::endl;
add a comment |
You mean something like this:
class Z
int& x;
int& y;
public:
Z(int& x, int& y) : x(x), y(y)
operator int() return x + y;
;
The class delays calculation of the result until casted as int. As cast operator is not explicit, Z
can be used whenever an int is required. As there's an overload of operator<<
for int, you can use it with e. g. std::cout
directly:
int x, y;
Z z(x, y);
std::cin >> x >> y;
if(std::cin) // otherwise, IO error! (e. g. bad user input)
std::cout << z << std::endl;
You mean something like this:
class Z
int& x;
int& y;
public:
Z(int& x, int& y) : x(x), y(y)
operator int() return x + y;
;
The class delays calculation of the result until casted as int. As cast operator is not explicit, Z
can be used whenever an int is required. As there's an overload of operator<<
for int, you can use it with e. g. std::cout
directly:
int x, y;
Z z(x, y);
std::cin >> x >> y;
if(std::cin) // otherwise, IO error! (e. g. bad user input)
std::cout << z << std::endl;
answered 12 hours ago
AconcaguaAconcagua
12.9k32144
12.9k32144
add a comment |
add a comment |
- You create a function for that.
- You call the function with the appropriate arguments when you need the value.
int z(int x, int y)
return (x + y);
int x;
int y;
// This does ot work
// int zx + y;
cin >> x;
cin >> y;
cout << z(x, y);
1
yes, i mean the only way to get z updated is to call that function everytime we change x and y values?
– Nay Wunna Zaw
12 hours ago
@NayWunnaZaw, yes, that is correct.
– R Sahu
12 hours ago
1
@NayWunnaZaw, you can avoid the repeated use ofx
andy
by using a lambda function, as shown by Nathan but you still have to make the call.
– R Sahu
12 hours ago
3
@NayWunnaZaw To complete the issue: Even with my solution, there is a function call (to the cast operator!), just that it is implicit and not required explicitly...
– Aconcagua
12 hours ago
add a comment |
- You create a function for that.
- You call the function with the appropriate arguments when you need the value.
int z(int x, int y)
return (x + y);
int x;
int y;
// This does ot work
// int zx + y;
cin >> x;
cin >> y;
cout << z(x, y);
1
yes, i mean the only way to get z updated is to call that function everytime we change x and y values?
– Nay Wunna Zaw
12 hours ago
@NayWunnaZaw, yes, that is correct.
– R Sahu
12 hours ago
1
@NayWunnaZaw, you can avoid the repeated use ofx
andy
by using a lambda function, as shown by Nathan but you still have to make the call.
– R Sahu
12 hours ago
3
@NayWunnaZaw To complete the issue: Even with my solution, there is a function call (to the cast operator!), just that it is implicit and not required explicitly...
– Aconcagua
12 hours ago
add a comment |
- You create a function for that.
- You call the function with the appropriate arguments when you need the value.
int z(int x, int y)
return (x + y);
int x;
int y;
// This does ot work
// int zx + y;
cin >> x;
cin >> y;
cout << z(x, y);
- You create a function for that.
- You call the function with the appropriate arguments when you need the value.
int z(int x, int y)
return (x + y);
int x;
int y;
// This does ot work
// int zx + y;
cin >> x;
cin >> y;
cout << z(x, y);
answered 12 hours ago
R SahuR Sahu
170k1294193
170k1294193
1
yes, i mean the only way to get z updated is to call that function everytime we change x and y values?
– Nay Wunna Zaw
12 hours ago
@NayWunnaZaw, yes, that is correct.
– R Sahu
12 hours ago
1
@NayWunnaZaw, you can avoid the repeated use ofx
andy
by using a lambda function, as shown by Nathan but you still have to make the call.
– R Sahu
12 hours ago
3
@NayWunnaZaw To complete the issue: Even with my solution, there is a function call (to the cast operator!), just that it is implicit and not required explicitly...
– Aconcagua
12 hours ago
add a comment |
1
yes, i mean the only way to get z updated is to call that function everytime we change x and y values?
– Nay Wunna Zaw
12 hours ago
@NayWunnaZaw, yes, that is correct.
– R Sahu
12 hours ago
1
@NayWunnaZaw, you can avoid the repeated use ofx
andy
by using a lambda function, as shown by Nathan but you still have to make the call.
– R Sahu
12 hours ago
3
@NayWunnaZaw To complete the issue: Even with my solution, there is a function call (to the cast operator!), just that it is implicit and not required explicitly...
– Aconcagua
12 hours ago
1
1
yes, i mean the only way to get z updated is to call that function everytime we change x and y values?
– Nay Wunna Zaw
12 hours ago
yes, i mean the only way to get z updated is to call that function everytime we change x and y values?
– Nay Wunna Zaw
12 hours ago
@NayWunnaZaw, yes, that is correct.
– R Sahu
12 hours ago
@NayWunnaZaw, yes, that is correct.
– R Sahu
12 hours ago
1
1
@NayWunnaZaw, you can avoid the repeated use of
x
and y
by using a lambda function, as shown by Nathan but you still have to make the call.– R Sahu
12 hours ago
@NayWunnaZaw, you can avoid the repeated use of
x
and y
by using a lambda function, as shown by Nathan but you still have to make the call.– R Sahu
12 hours ago
3
3
@NayWunnaZaw To complete the issue: Even with my solution, there is a function call (to the cast operator!), just that it is implicit and not required explicitly...
– Aconcagua
12 hours ago
@NayWunnaZaw To complete the issue: Even with my solution, there is a function call (to the cast operator!), just that it is implicit and not required explicitly...
– Aconcagua
12 hours ago
add a comment |
You can define the following lambda z
which always returns the current value of x+y
because x
and y
are captured by reference:
DEMO
int x;
int y;
const auto z = [&x, &y]() return x+y; ;
1
Don't use parentheses around return values –return
is not a function, and in worst case, the parentheses can even create a dangling reference (with return type beingdecltype(auto)
).
– Aconcagua
12 hours ago
@Aconcagua thx! you are right. I edited my answer.
– Hiroki
12 hours ago
Why downvoted ? At least I posted my answer faster than Ted with my tested code.
– Hiroki
5 hours ago
add a comment |
You can define the following lambda z
which always returns the current value of x+y
because x
and y
are captured by reference:
DEMO
int x;
int y;
const auto z = [&x, &y]() return x+y; ;
1
Don't use parentheses around return values –return
is not a function, and in worst case, the parentheses can even create a dangling reference (with return type beingdecltype(auto)
).
– Aconcagua
12 hours ago
@Aconcagua thx! you are right. I edited my answer.
– Hiroki
12 hours ago
Why downvoted ? At least I posted my answer faster than Ted with my tested code.
– Hiroki
5 hours ago
add a comment |
You can define the following lambda z
which always returns the current value of x+y
because x
and y
are captured by reference:
DEMO
int x;
int y;
const auto z = [&x, &y]() return x+y; ;
You can define the following lambda z
which always returns the current value of x+y
because x
and y
are captured by reference:
DEMO
int x;
int y;
const auto z = [&x, &y]() return x+y; ;
edited 12 hours ago
answered 12 hours ago
HirokiHiroki
2,1253320
2,1253320
1
Don't use parentheses around return values –return
is not a function, and in worst case, the parentheses can even create a dangling reference (with return type beingdecltype(auto)
).
– Aconcagua
12 hours ago
@Aconcagua thx! you are right. I edited my answer.
– Hiroki
12 hours ago
Why downvoted ? At least I posted my answer faster than Ted with my tested code.
– Hiroki
5 hours ago
add a comment |
1
Don't use parentheses around return values –return
is not a function, and in worst case, the parentheses can even create a dangling reference (with return type beingdecltype(auto)
).
– Aconcagua
12 hours ago
@Aconcagua thx! you are right. I edited my answer.
– Hiroki
12 hours ago
Why downvoted ? At least I posted my answer faster than Ted with my tested code.
– Hiroki
5 hours ago
1
1
Don't use parentheses around return values –
return
is not a function, and in worst case, the parentheses can even create a dangling reference (with return type being decltype(auto)
).– Aconcagua
12 hours ago
Don't use parentheses around return values –
return
is not a function, and in worst case, the parentheses can even create a dangling reference (with return type being decltype(auto)
).– Aconcagua
12 hours ago
@Aconcagua thx! you are right. I edited my answer.
– Hiroki
12 hours ago
@Aconcagua thx! you are right. I edited my answer.
– Hiroki
12 hours ago
Why downvoted ? At least I posted my answer faster than Ted with my tested code.
– Hiroki
5 hours ago
Why downvoted ? At least I posted my answer faster than Ted with my tested code.
– Hiroki
5 hours ago
add a comment |
There are two chief techniques:
Deferred calculation - instead of
z
being a simple variable, make it a function which calculates the value on demand (see other answers for examples). This can be source-code transparent ifz
is some proxy object with implicit conversion to the required type.Explicit notification of changes. This requires
x
andy
to be observable types; when either changes value, thenz
updates itself (and notifies its observers if applicable).
The first version is usually preferred, but the second may be more appropriate if you need z
to be an observable type.
add a comment |
There are two chief techniques:
Deferred calculation - instead of
z
being a simple variable, make it a function which calculates the value on demand (see other answers for examples). This can be source-code transparent ifz
is some proxy object with implicit conversion to the required type.Explicit notification of changes. This requires
x
andy
to be observable types; when either changes value, thenz
updates itself (and notifies its observers if applicable).
The first version is usually preferred, but the second may be more appropriate if you need z
to be an observable type.
add a comment |
There are two chief techniques:
Deferred calculation - instead of
z
being a simple variable, make it a function which calculates the value on demand (see other answers for examples). This can be source-code transparent ifz
is some proxy object with implicit conversion to the required type.Explicit notification of changes. This requires
x
andy
to be observable types; when either changes value, thenz
updates itself (and notifies its observers if applicable).
The first version is usually preferred, but the second may be more appropriate if you need z
to be an observable type.
There are two chief techniques:
Deferred calculation - instead of
z
being a simple variable, make it a function which calculates the value on demand (see other answers for examples). This can be source-code transparent ifz
is some proxy object with implicit conversion to the required type.Explicit notification of changes. This requires
x
andy
to be observable types; when either changes value, thenz
updates itself (and notifies its observers if applicable).
The first version is usually preferred, but the second may be more appropriate if you need z
to be an observable type.
answered 9 hours ago
Toby SpeightToby Speight
17.2k134367
17.2k134367
add a comment |
add a comment |
You can get what you're asking for by using macros:
int x, y;
#define z (x + y)
/* use x, y, z */
#undef z
The #undef
is for a little sanity. For more sanity, don't use macros at all, and go with one of the other answers, and deal with the extra verbosity.
Although a class with a custom operator int
would work in a lot of cases ... hmm.
add a comment |
You can get what you're asking for by using macros:
int x, y;
#define z (x + y)
/* use x, y, z */
#undef z
The #undef
is for a little sanity. For more sanity, don't use macros at all, and go with one of the other answers, and deal with the extra verbosity.
Although a class with a custom operator int
would work in a lot of cases ... hmm.
add a comment |
You can get what you're asking for by using macros:
int x, y;
#define z (x + y)
/* use x, y, z */
#undef z
The #undef
is for a little sanity. For more sanity, don't use macros at all, and go with one of the other answers, and deal with the extra verbosity.
Although a class with a custom operator int
would work in a lot of cases ... hmm.
You can get what you're asking for by using macros:
int x, y;
#define z (x + y)
/* use x, y, z */
#undef z
The #undef
is for a little sanity. For more sanity, don't use macros at all, and go with one of the other answers, and deal with the extra verbosity.
Although a class with a custom operator int
would work in a lot of cases ... hmm.
answered 2 hours ago
o11co11c
10.9k43154
10.9k43154
add a comment |
add a comment |
Nay Wunna Zaw is a new contributor. Be nice, and check out our Code of Conduct.
Nay Wunna Zaw is a new contributor. Be nice, and check out our Code of Conduct.
Nay Wunna Zaw is a new contributor. Be nice, and check out our Code of Conduct.
Nay Wunna Zaw is a new contributor. Be nice, and check out our Code of Conduct.
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1
Why do you want to do that?
– Robert Andrzejuk
12 hours ago
1
Right, that won't work. That's a spreadsheet thing.
– Pete Becker
12 hours ago
2
@RobertAndrzejuk Because it'd be very useful. For example, if you write a game and have something like networth(cash+the worth of all you own). You will have to call the function of networth everytime one of those update. That would be very annoying and error prone if you forgot to call the function somewhere.
– Nay Wunna Zaw
12 hours ago
5
@NayWunnaZaw In my experience this is why getters and setters are encouraged over direct variable access. If you always wanted networth to be updated, you could retrieve the value using getNetWorth() which would itself call updateNetWorth() before returning the value... or just calculate it before returning it.
– Onyz
11 hours ago
3
That's called a "function."
– ApproachingDarknessFish
6 hours ago