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declare as function pointer and initialize in the same line
The Next CEO of Stack OverflowWhat are the differences between a pointer variable and a reference variable in C++?What is a smart pointer and when should I use one?Returning unique_ptr from functionsWhy is reading lines from stdin much slower in C++ than Python?Image Processing: Algorithm Improvement for 'Coca-Cola Can' RecognitionWhy should I use a pointer rather than the object itself?Replacing a 32-bit loop counter with 64-bit introduces crazy performance deviationsFunction pointer, which can point to every thing?Has a std::byte pointer the same aliasing implications as char*?declare and define function pointer variable in one line
In C++ how do we do the following
// fundamental language construct
type name = value;
// for example
int x = y;
with function pointers?
typedef (char)(*FP)(unsigned);
// AFAIK not possible in C++
FP x = y ;
I can use lambdas
FP x = []( unsigned k) -> char return char(k);
But I do not know how to do this without lambda. Any ideas? This is not an answer. we know this works:
void whatever ()
typedef void (*FP) (void);
FP x = whatever ;
The question is if one can do this in one line in C++. Like one can do it in one line in C++ with every other type kind.
c++
edited 13 hours ago
Guillaume Racicot
15.8k53569
asked 16 hours ago
emma brainemma brain
963
New contributor
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
In C++ how do we do the following
// fundamental language construct
type name = value;
// for example
int x = y;
with function pointers?
typedef (char)(*FP)(unsigned);
// AFAIK not possible in C++
FP x = y ;
I can use lambdas
FP x = []( unsigned k) -> char return char(k);
But I do not know how to do this without lambda. Any ideas? This is not an answer. we know this works:
void whatever ()
typedef void (*FP) (void);
FP x = whatever ;
The question is if one can do this in one line in C++. Like one can do it in one line in C++ with every other type kind.
c++
edited 13 hours ago
Guillaume Racicot
15.8k53569
asked 16 hours ago
emma brainemma brain
963
New contributor
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
You could always stick withauto x = &the_function;'.
– François Andrieux
16 hours ago
1
The name of a function pointer variable appears between the return type and the arguments It won't look liketype name = value;.
– François Andrieux
15 hours ago
You're missing the&beforewhatever.FP x = &whatever ;
– dave
15 hours ago
@dave: That's the same for functions, you don't need the ampersand. @emma: Why should that not be possible in C++? This should work fine (except for the(char), which should becharin yourtypedef)
– andreee
15 hours ago
add a comment |
In C++ how do we do the following
// fundamental language construct
type name = value;
// for example
int x = y;
with function pointers?
typedef (char)(*FP)(unsigned);
// AFAIK not possible in C++
FP x = y ;
I can use lambdas
FP x = []( unsigned k) -> char return char(k);
But I do not know how to do this without lambda. Any ideas? This is not an answer. we know this works:
void whatever ()
typedef void (*FP) (void);
FP x = whatever ;
The question is if one can do this in one line in C++. Like one can do it in one line in C++ with every other type kind.
c++
edited 13 hours ago
Guillaume Racicot
15.8k53569
asked 16 hours ago
emma brainemma brain
963
New contributor
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In C++ how do we do the following
// fundamental language construct
type name = value;
// for example
int x = y;
with function pointers?
typedef (char)(*FP)(unsigned);
// AFAIK not possible in C++
FP x = y ;
I can use lambdas
FP x = []( unsigned k) -> char return char(k);
But I do not know how to do this without lambda. Any ideas? This is not an answer. we know this works:
void whatever ()
typedef void (*FP) (void);
FP x = whatever ;
The question is if one can do this in one line in C++. Like one can do it in one line in C++ with every other type kind.
c++
c++
edited 13 hours ago
Guillaume Racicot
15.8k53569
asked 16 hours ago
emma brainemma brain
963
New contributor
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 hours ago
Guillaume Racicot
15.8k53569
asked 16 hours ago
emma brainemma brain
963
New contributor
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 hours ago
Guillaume Racicot
15.8k53569
edited 13 hours ago
Guillaume Racicot
15.8k53569
edited 13 hours ago
Guillaume Racicot
15.8k53569
15.8k53569
asked 16 hours ago
emma brainemma brain
963
New contributor
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 16 hours ago
emma brainemma brain
963
asked 16 hours ago
emma brainemma brain
963
963
New contributor
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
You could always stick withauto x = &the_function;'.
– François Andrieux
16 hours ago
1
The name of a function pointer variable appears between the return type and the arguments It won't look liketype name = value;.
– François Andrieux
15 hours ago
You're missing the&beforewhatever.FP x = &whatever ;
– dave
15 hours ago
@dave: That's the same for functions, you don't need the ampersand. @emma: Why should that not be possible in C++? This should work fine (except for the(char), which should becharin yourtypedef)
– andreee
15 hours ago
add a comment |
3
You could always stick withauto x = &the_function;'.
– François Andrieux
16 hours ago
1
The name of a function pointer variable appears between the return type and the arguments It won't look liketype name = value;.
– François Andrieux
15 hours ago
You're missing the&beforewhatever.FP x = &whatever ;
– dave
15 hours ago
@dave: That's the same for functions, you don't need the ampersand. @emma: Why should that not be possible in C++? This should work fine (except for the(char), which should becharin yourtypedef)
– andreee
15 hours ago
3
3
You could always stick with
auto x = &the_function;'.– François Andrieux
16 hours ago
You could always stick with
auto x = &the_function;'.– François Andrieux
16 hours ago
1
1
The name of a function pointer variable appears between the return type and the arguments It won't look like
type name = value;.– François Andrieux
15 hours ago
The name of a function pointer variable appears between the return type and the arguments It won't look like
type name = value;.– François Andrieux
15 hours ago
You're missing the
& before whatever. FP x = &whatever ;– dave
15 hours ago
You're missing the
& before whatever. FP x = &whatever ;– dave
15 hours ago
@dave: That's the same for functions, you don't need the ampersand. @emma: Why should that not be possible in C++? This should work fine (except for the
(char), which should be char in your typedef)– andreee
15 hours ago
@dave: That's the same for functions, you don't need the ampersand. @emma: Why should that not be possible in C++? This should work fine (except for the
(char), which should be char in your typedef)– andreee
15 hours ago
add a comment |
2 Answers
2
active
oldest
votes
Whenever you can write a typedef, you can also write a variable declaration with no typedef, with almost identical syntax.
Example:
// typedef
typedef char(*FP)(unsigned);
FP x = y ;
// no typedef
char(*x)(unsigned) = y;
Remove the typedef keyword, and you have a variable declaration. Slap an initialisation on it if you want.
answered 15 hours ago
n.m.n.m.
73.7k885172
Works just as well as in C, and can also be used for array types:char (*img)[width][3] = malloc(height*sizeof(*img));While this syntax is pretty useless in C++ (no array types with runtime sizes), it works well in C and sports the same counterintuitive syntax.
– cmaster
9 hours ago
add a comment |
You can use auto:
auto fptr = &f;
It skips the need of a typedef and conserve a nice syntax.
answered 15 hours ago
Guillaume RacicotGuillaume Racicot
15.8k53569
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Whenever you can write a typedef, you can also write a variable declaration with no typedef, with almost identical syntax.
Example:
// typedef
typedef char(*FP)(unsigned);
FP x = y ;
// no typedef
char(*x)(unsigned) = y;
Remove the typedef keyword, and you have a variable declaration. Slap an initialisation on it if you want.
answered 15 hours ago
n.m.n.m.
73.7k885172
Works just as well as in C, and can also be used for array types:char (*img)[width][3] = malloc(height*sizeof(*img));While this syntax is pretty useless in C++ (no array types with runtime sizes), it works well in C and sports the same counterintuitive syntax.
– cmaster
9 hours ago
add a comment |
Whenever you can write a typedef, you can also write a variable declaration with no typedef, with almost identical syntax.
Example:
// typedef
typedef char(*FP)(unsigned);
FP x = y ;
// no typedef
char(*x)(unsigned) = y;
Remove the typedef keyword, and you have a variable declaration. Slap an initialisation on it if you want.
answered 15 hours ago
n.m.n.m.
73.7k885172
Works just as well as in C, and can also be used for array types:char (*img)[width][3] = malloc(height*sizeof(*img));While this syntax is pretty useless in C++ (no array types with runtime sizes), it works well in C and sports the same counterintuitive syntax.
– cmaster
9 hours ago
add a comment |
Whenever you can write a typedef, you can also write a variable declaration with no typedef, with almost identical syntax.
Example:
// typedef
typedef char(*FP)(unsigned);
FP x = y ;
// no typedef
char(*x)(unsigned) = y;
Remove the typedef keyword, and you have a variable declaration. Slap an initialisation on it if you want.
answered 15 hours ago
n.m.n.m.
73.7k885172
Whenever you can write a typedef, you can also write a variable declaration with no typedef, with almost identical syntax.
Example:
// typedef
typedef char(*FP)(unsigned);
FP x = y ;
// no typedef
char(*x)(unsigned) = y;
Remove the typedef keyword, and you have a variable declaration. Slap an initialisation on it if you want.
answered 15 hours ago
n.m.n.m.
73.7k885172
answered 15 hours ago
n.m.n.m.
73.7k885172
answered 15 hours ago
n.m.n.m.
73.7k885172
answered 15 hours ago
n.m.n.m.
73.7k885172
73.7k885172
Works just as well as in C, and can also be used for array types:char (*img)[width][3] = malloc(height*sizeof(*img));While this syntax is pretty useless in C++ (no array types with runtime sizes), it works well in C and sports the same counterintuitive syntax.
– cmaster
9 hours ago
add a comment |
Works just as well as in C, and can also be used for array types:char (*img)[width][3] = malloc(height*sizeof(*img));While this syntax is pretty useless in C++ (no array types with runtime sizes), it works well in C and sports the same counterintuitive syntax.
– cmaster
9 hours ago
Works just as well as in C, and can also be used for array types:
char (*img)[width][3] = malloc(height*sizeof(*img)); While this syntax is pretty useless in C++ (no array types with runtime sizes), it works well in C and sports the same counterintuitive syntax.– cmaster
9 hours ago
Works just as well as in C, and can also be used for array types:
char (*img)[width][3] = malloc(height*sizeof(*img)); While this syntax is pretty useless in C++ (no array types with runtime sizes), it works well in C and sports the same counterintuitive syntax.– cmaster
9 hours ago
add a comment |
You can use auto:
auto fptr = &f;
It skips the need of a typedef and conserve a nice syntax.
answered 15 hours ago
Guillaume RacicotGuillaume Racicot
15.8k53569
add a comment |
You can use auto:
auto fptr = &f;
It skips the need of a typedef and conserve a nice syntax.
answered 15 hours ago
Guillaume RacicotGuillaume Racicot
15.8k53569
add a comment |
You can use auto:
auto fptr = &f;
It skips the need of a typedef and conserve a nice syntax.
answered 15 hours ago
Guillaume RacicotGuillaume Racicot
15.8k53569
You can use auto:
auto fptr = &f;
It skips the need of a typedef and conserve a nice syntax.
answered 15 hours ago
Guillaume RacicotGuillaume Racicot
15.8k53569
answered 15 hours ago
Guillaume RacicotGuillaume Racicot
15.8k53569
answered 15 hours ago
Guillaume RacicotGuillaume Racicot
15.8k53569
answered 15 hours ago
Guillaume RacicotGuillaume Racicot
15.8k53569
15.8k53569
add a comment |
add a comment |
emma brain is a new contributor. Be nice, and check out our Code of Conduct.
emma brain is a new contributor. Be nice, and check out our Code of Conduct.
emma brain is a new contributor. Be nice, and check out our Code of Conduct.
emma brain is a new contributor. Be nice, and check out our Code of Conduct.
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3
You could always stick with
auto x = &the_function;'.– François Andrieux
16 hours ago
1
The name of a function pointer variable appears between the return type and the arguments It won't look like
type name = value;.– François Andrieux
15 hours ago
You're missing the
&beforewhatever.FP x = &whatever ;– dave
15 hours ago
@dave: That's the same for functions, you don't need the ampersand. @emma: Why should that not be possible in C++? This should work fine (except for the
(char), which should becharin yourtypedef)– andreee
15 hours ago