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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










9















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++







share|improve this question









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 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 & 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
















9















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++







share|improve this question









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 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 & 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














9












9








9


1






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++







share|improve this question









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++






share|improve this question









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.











share|improve this question









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.









share|improve this question




share|improve this question








edited 2 days ago









Guillaume Racicot

15.9k53770




15.9k53770






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




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 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 & 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













  • 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 & 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








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













2 Answers
2






active

oldest

votes


















11














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.






share|improve this answer























  • 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


















11














You can use auto:



auto fptr = &f;


It skips the need of a typedef and conserve a nice syntax.






share|improve this answer























  • 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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









11














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.






share|improve this answer























  • 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















11














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.






share|improve this answer























  • 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













11












11








11







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.






share|improve this answer













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.







share|improve this answer












share|improve this answer



share|improve this answer










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

















  • 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













11














You can use auto:



auto fptr = &f;


It skips the need of a typedef and conserve a nice syntax.






share|improve this answer























  • 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















11














You can use auto:



auto fptr = &f;


It skips the need of a typedef and conserve a nice syntax.






share|improve this answer























  • 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













11












11








11







You can use auto:



auto fptr = &f;


It skips the need of a typedef and conserve a nice syntax.






share|improve this answer













You can use auto:



auto fptr = &f;


It skips the need of a typedef and conserve a nice syntax.







share|improve this answer












share|improve this answer



share|improve this answer










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

















  • 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










emma brain is a new contributor. Be nice, and check out our Code of Conduct.









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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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