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declare as function pointer and initialize in the same line

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17

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

edited yesterday

atomsymbol

21859

asked Mar 28 at 14:46

emma brainemma brain

1986

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.

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  You could always stick with auto x = &the_function;'.
  
  – François Andrieux
  Mar 28 at 14:48
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 28 at 14:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You're missing the & before whatever. FP x = &whatever ;
  
  – dave
  Mar 28 at 14:49
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @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
  Mar 28 at 14:52
  
  

  
  
  
  

add a comment | 

17

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

edited yesterday

atomsymbol

21859

asked Mar 28 at 14:46

emma brainemma brain

1986

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.

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  You could always stick with auto x = &the_function;'.
  
  – François Andrieux
  Mar 28 at 14:48
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 28 at 14:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You're missing the & before whatever. FP x = &whatever ;
  
  – dave
  Mar 28 at 14:49
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @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
  Mar 28 at 14:52
  
  

  
  
  
  

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++

share|improve this question

edited yesterday

atomsymbol

21859

asked Mar 28 at 14:46

emma brainemma brain

1986

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.

// fundamental language construct 
type name = value; 

// for example 
int x = y;

typedef char(*FP)(unsigned);

// AFAIK not possible in C++
FP x = y;

FP x = [](unsigned k) -> char return char(k); ;

void whatever () 
typedef void (*FP) (void);
FP x = whatever;

share|improve this question

edited yesterday

atomsymbol

21859

asked Mar 28 at 14:46

emma brainemma brain

1986

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

edited yesterday

atomsymbol

21859

asked Mar 28 at 14:46

emma brainemma brain

1986

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 yesterday

atomsymbol

21859

edited yesterday

atomsymbol

21859

asked Mar 28 at 14:46

emma brainemma brain

1986

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 Mar 28 at 14:46

emma brainemma brain

1986

3 Answers
3

active

oldest

votes

12

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

answered Mar 28 at 14:53

n.m.n.m.

73.8k885172

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  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
  Mar 28 at 21:09
  

  
  
  
  

add a comment | 

13

You can use auto:

auto fptr = &f;

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

share|improve this answer

answered Mar 28 at 15:02

Guillaume RacicotGuillaume Racicot

16.1k53771

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  Mar 29 at 7:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  By the way, it also works without the &.
  
  – El Profesor
  yesterday
  

  
  
  
  

add a comment | 

0

The code:

typedef char(*FP)(int);
FP x = y;

fails to compile with current C++ compilers if y is a lambda expression capturing a variable.

// Compiles OK
FP x0 = [](int k) -> char return char(k); ;

// Fails to compile
int i = 123;
FP x1 = [=](int k) -> char return char(k); ;
FP x2 = [=](int k) -> char return char(k+i); ;
FP x3 = [&](int k) -> char return char(k+i); ;
FP x4 = [i](int k) -> char return char(k+i); ;
// error: cannot convert ‘main()::<lambda(int)>’ to ‘FP aka char (*)(int)’
// in initialization

The reason why it fails to compile is that the size of the right side of the assignment to x1...x4 is greater than size of FP.

For a C++ compiler to make assignments to x1...x4 be valid it would need to generate code at runtime. Current C++ compilers such as GCC and clang do not support this, mainly because it would cause memory leaks because C++ isn't a garbage collected language. Some garbage collected language implementations, such as earlier versions of the official Go compiler, do support such assignments by performing runtime code generation.

share|improve this answer

answered yesterday

atomsymbolatomsymbol

21859

add a comment | 

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12

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

answered Mar 28 at 14:53

n.m.n.m.

73.8k885172

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  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
  Mar 28 at 21:09
  

  
  
  
  

add a comment | 

12

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

answered Mar 28 at 14:53

n.m.n.m.

73.8k885172

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  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
  Mar 28 at 21:09
  

  
  
  
  

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.

share|improve this answer

answered Mar 28 at 14:53

n.m.n.m.

73.8k885172

 // typedef
 typedef char(*FP)(unsigned);
 FP x = y ;

 // no typedef
 char(*x)(unsigned) = y;

share|improve this answer

answered Mar 28 at 14:53

n.m.n.m.

73.8k885172

answered Mar 28 at 14:53

n.m.n.m.

73.8k885172

answered Mar 28 at 14:53

n.m.n.m.

73.8k885172

13

You can use auto:

auto fptr = &f;

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

share|improve this answer

answered Mar 28 at 15:02

Guillaume RacicotGuillaume Racicot

16.1k53771

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  Mar 29 at 7:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  By the way, it also works without the &.
  
  – El Profesor
  yesterday
  

  
  
  
  

add a comment | 

13

You can use auto:

auto fptr = &f;

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

share|improve this answer

answered Mar 28 at 15:02

Guillaume RacicotGuillaume Racicot

16.1k53771

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  Mar 29 at 7:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  By the way, it also works without the &.
  
  – El Profesor
  yesterday
  

  
  
  
  

add a comment | 

You can use auto:

auto fptr = &f;

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

share|improve this answer

answered Mar 28 at 15:02

Guillaume RacicotGuillaume Racicot

16.1k53771

auto fptr = &f;

share|improve this answer

answered Mar 28 at 15:02

Guillaume RacicotGuillaume Racicot

16.1k53771

answered Mar 28 at 15:02

Guillaume RacicotGuillaume Racicot

16.1k53771

answered Mar 28 at 15:02

Guillaume RacicotGuillaume Racicot

16.1k53771

0

The code:

typedef char(*FP)(int);
FP x = y;

fails to compile with current C++ compilers if y is a lambda expression capturing a variable.

// Compiles OK
FP x0 = [](int k) -> char return char(k); ;

// Fails to compile
int i = 123;
FP x1 = [=](int k) -> char return char(k); ;
FP x2 = [=](int k) -> char return char(k+i); ;
FP x3 = [&](int k) -> char return char(k+i); ;
FP x4 = [i](int k) -> char return char(k+i); ;
// error: cannot convert ‘main()::<lambda(int)>’ to ‘FP aka char (*)(int)’
// in initialization

The reason why it fails to compile is that the size of the right side of the assignment to x1...x4 is greater than size of FP.

For a C++ compiler to make assignments to x1...x4 be valid it would need to generate code at runtime. Current C++ compilers such as GCC and clang do not support this, mainly because it would cause memory leaks because C++ isn't a garbage collected language. Some garbage collected language implementations, such as earlier versions of the official Go compiler, do support such assignments by performing runtime code generation.

share|improve this answer

answered yesterday

atomsymbolatomsymbol

21859

add a comment | 

0

The code:

typedef char(*FP)(int);
FP x = y;

fails to compile with current C++ compilers if y is a lambda expression capturing a variable.

// Compiles OK
FP x0 = [](int k) -> char return char(k); ;

// Fails to compile
int i = 123;
FP x1 = [=](int k) -> char return char(k); ;
FP x2 = [=](int k) -> char return char(k+i); ;
FP x3 = [&](int k) -> char return char(k+i); ;
FP x4 = [i](int k) -> char return char(k+i); ;
// error: cannot convert ‘main()::<lambda(int)>’ to ‘FP aka char (*)(int)’
// in initialization

The reason why it fails to compile is that the size of the right side of the assignment to x1...x4 is greater than size of FP.

For a C++ compiler to make assignments to x1...x4 be valid it would need to generate code at runtime. Current C++ compilers such as GCC and clang do not support this, mainly because it would cause memory leaks because C++ isn't a garbage collected language. Some garbage collected language implementations, such as earlier versions of the official Go compiler, do support such assignments by performing runtime code generation.

share|improve this answer

answered yesterday

atomsymbolatomsymbol

21859

add a comment | 

The code:

typedef char(*FP)(int);
FP x = y;

fails to compile with current C++ compilers if y is a lambda expression capturing a variable.

// Compiles OK
FP x0 = [](int k) -> char return char(k); ;

// Fails to compile
int i = 123;
FP x1 = [=](int k) -> char return char(k); ;
FP x2 = [=](int k) -> char return char(k+i); ;
FP x3 = [&](int k) -> char return char(k+i); ;
FP x4 = [i](int k) -> char return char(k+i); ;
// error: cannot convert ‘main()::<lambda(int)>’ to ‘FP aka char (*)(int)’
// in initialization

The reason why it fails to compile is that the size of the right side of the assignment to x1...x4 is greater than size of FP.

For a C++ compiler to make assignments to x1...x4 be valid it would need to generate code at runtime. Current C++ compilers such as GCC and clang do not support this, mainly because it would cause memory leaks because C++ isn't a garbage collected language. Some garbage collected language implementations, such as earlier versions of the official Go compiler, do support such assignments by performing runtime code generation.

share|improve this answer

answered yesterday

atomsymbolatomsymbol

21859

typedef char(*FP)(int);
FP x = y;

// Compiles OK
FP x0 = [](int k) -> char return char(k); ;

// Fails to compile
int i = 123;
FP x1 = [=](int k) -> char return char(k); ;
FP x2 = [=](int k) -> char return char(k+i); ;
FP x3 = [&](int k) -> char return char(k+i); ;
FP x4 = [i](int k) -> char return char(k+i); ;
// error: cannot convert ‘main()::<lambda(int)>’ to ‘FP aka char (*)(int)’
// in initialization

share|improve this answer

answered yesterday

atomsymbolatomsymbol

21859

answered yesterday

atomsymbolatomsymbol

21859

answered yesterday

atomsymbolatomsymbol

21859