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
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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 2 days ago
Guillaume Racicot
15.9k53770
asked 2 days ago
emma brainemma brain
1083
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 2 days ago
Guillaume Racicot
15.9k53770
asked 2 days ago
emma brainemma brain
1083
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
2 days 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
2 days ago
You're missing the&beforewhatever.FP x = &whatever ;
– dave
2 days 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
2 days 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 2 days ago
Guillaume Racicot
15.9k53770
asked 2 days ago
emma brainemma brain
1083
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 2 days ago
Guillaume Racicot
15.9k53770
asked 2 days ago
emma brainemma brain
1083
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 2 days ago
Guillaume Racicot
15.9k53770
asked 2 days ago
emma brainemma brain
1083
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 2 days ago
Guillaume Racicot
15.9k53770
edited 2 days ago
Guillaume Racicot
15.9k53770
edited 2 days ago
Guillaume Racicot
15.9k53770
15.9k53770
asked 2 days ago
emma brainemma brain
1083
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 2 days ago
emma brainemma brain
1083
asked 2 days ago
emma brainemma brain
1083
1083
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
2 days 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
2 days ago
You're missing the&beforewhatever.FP x = &whatever ;
– dave
2 days 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
2 days ago
add a comment |
3
You could always stick withauto x = &the_function;'.
– François Andrieux
2 days 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
2 days ago
You're missing the&beforewhatever.FP x = &whatever ;
– dave
2 days 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
2 days ago
3
3
You could always stick with
auto x = &the_function;'.– François Andrieux
2 days ago
You could always stick with
auto x = &the_function;'.– François Andrieux
2 days 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
2 days 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
2 days ago
You're missing the
& before whatever. FP x = &whatever ;– dave
2 days ago
You're missing the
& before whatever. FP x = &whatever ;– dave
2 days 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
2 days 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
2 days 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 2 days 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
2 days ago
add a comment |
You can use auto:
auto fptr = &f;
It skips the need of a typedef and conserve a nice syntax.
answered 2 days ago
Guillaume RacicotGuillaume Racicot
15.9k53770
I like this answer a bit more than the accepted one. It is more C++11-ish even though it is not clear that OP has or wants a C++11 solution. But hey, C++11 is not the "all new exotic" standard any more.
– TobiMcNamobi
yesterday
add a comment |
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2 Answers
2
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 2 days 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
2 days 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 2 days 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
2 days 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 2 days 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 2 days ago
n.m.n.m.
73.7k885172
answered 2 days ago
n.m.n.m.
73.7k885172
answered 2 days ago
n.m.n.m.
73.7k885172
answered 2 days 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
2 days 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
2 days 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
2 days 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
2 days ago
add a comment |
You can use auto:
auto fptr = &f;
It skips the need of a typedef and conserve a nice syntax.
answered 2 days ago
Guillaume RacicotGuillaume Racicot
15.9k53770
I like this answer a bit more than the accepted one. It is more C++11-ish even though it is not clear that OP has or wants a C++11 solution. But hey, C++11 is not the "all new exotic" standard any more.
– TobiMcNamobi
yesterday
add a comment |
You can use auto:
auto fptr = &f;
It skips the need of a typedef and conserve a nice syntax.
answered 2 days ago
Guillaume RacicotGuillaume Racicot
15.9k53770
I like this answer a bit more than the accepted one. It is more C++11-ish even though it is not clear that OP has or wants a C++11 solution. But hey, C++11 is not the "all new exotic" standard any more.
– TobiMcNamobi
yesterday
add a comment |
You can use auto:
auto fptr = &f;
It skips the need of a typedef and conserve a nice syntax.
answered 2 days ago
Guillaume RacicotGuillaume Racicot
15.9k53770
You can use auto:
auto fptr = &f;
It skips the need of a typedef and conserve a nice syntax.
answered 2 days ago
Guillaume RacicotGuillaume Racicot
15.9k53770
answered 2 days ago
Guillaume RacicotGuillaume Racicot
15.9k53770
answered 2 days ago
Guillaume RacicotGuillaume Racicot
15.9k53770
answered 2 days ago
Guillaume RacicotGuillaume Racicot
15.9k53770
15.9k53770
I like this answer a bit more than the accepted one. It is more C++11-ish even though it is not clear that OP has or wants a C++11 solution. But hey, C++11 is not the "all new exotic" standard any more.
– TobiMcNamobi
yesterday
add a comment |
I like this answer a bit more than the accepted one. It is more C++11-ish even though it is not clear that OP has or wants a C++11 solution. But hey, C++11 is not the "all new exotic" standard any more.
– TobiMcNamobi
yesterday
I like this answer a bit more than the accepted one. It is more C++11-ish even though it is not clear that OP has or wants a C++11 solution. But hey, C++11 is not the "all new exotic" standard any more.
– TobiMcNamobi
yesterday
I like this answer a bit more than the accepted one. It is more C++11-ish even though it is not clear that OP has or wants a C++11 solution. But hey, C++11 is not the "all new exotic" standard any more.
– TobiMcNamobi
yesterday
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
2 days 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
2 days ago
You're missing the
&beforewhatever.FP x = &whatever ;– dave
2 days 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
2 days ago