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How to call a function with default parameter through a pointer to function that is the return of another function?
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I have a function Mult that takes two integers and returns the product of its parameters. And a function Double that takes an integer and returns a pointer to function that returns an integer and takes two integer parameters like Mult.
Mult's second parameter isdefaultSo when I callDouble,Doublereturns the address ofMultthus I can pass only one argument.
But It doesn't work with pointer to function:
int Mult(int x, int y = 2) // y is default
return x * y;
using pFn = int(*)(int, int);
pFn Double(int x)
return Mult;
int main(int argc, char* argv[])
pFn func = Double(0);
cout << func(7, 4) << endl; // ok
//cout << func(7) << endl; // error: Too few arguments
cout << Mult(4) << endl; // ok. the second argument is default
Above if I call Mult with a single argument it works fine because the second argument is default but calling it through the pointer func it fails. func is pointer to function that takes two integers and returns an int.
c++ function-pointers default-arguments
add a comment |
I have a function Mult that takes two integers and returns the product of its parameters. And a function Double that takes an integer and returns a pointer to function that returns an integer and takes two integer parameters like Mult.
Mult's second parameter isdefaultSo when I callDouble,Doublereturns the address ofMultthus I can pass only one argument.
But It doesn't work with pointer to function:
int Mult(int x, int y = 2) // y is default
return x * y;
using pFn = int(*)(int, int);
pFn Double(int x)
return Mult;
int main(int argc, char* argv[])
pFn func = Double(0);
cout << func(7, 4) << endl; // ok
//cout << func(7) << endl; // error: Too few arguments
cout << Mult(4) << endl; // ok. the second argument is default
Above if I call Mult with a single argument it works fine because the second argument is default but calling it through the pointer func it fails. func is pointer to function that takes two integers and returns an int.
c++ function-pointers default-arguments
1
What is the point ofDoubletaking an integer parameter that it doesn't use?
– scohe001
24 mins ago
1
Similar: Howto: c++ Function Pointer with default values
– TrebledJ
24 mins ago
add a comment |
I have a function Mult that takes two integers and returns the product of its parameters. And a function Double that takes an integer and returns a pointer to function that returns an integer and takes two integer parameters like Mult.
Mult's second parameter isdefaultSo when I callDouble,Doublereturns the address ofMultthus I can pass only one argument.
But It doesn't work with pointer to function:
int Mult(int x, int y = 2) // y is default
return x * y;
using pFn = int(*)(int, int);
pFn Double(int x)
return Mult;
int main(int argc, char* argv[])
pFn func = Double(0);
cout << func(7, 4) << endl; // ok
//cout << func(7) << endl; // error: Too few arguments
cout << Mult(4) << endl; // ok. the second argument is default
Above if I call Mult with a single argument it works fine because the second argument is default but calling it through the pointer func it fails. func is pointer to function that takes two integers and returns an int.
c++ function-pointers default-arguments
I have a function Mult that takes two integers and returns the product of its parameters. And a function Double that takes an integer and returns a pointer to function that returns an integer and takes two integer parameters like Mult.
Mult's second parameter isdefaultSo when I callDouble,Doublereturns the address ofMultthus I can pass only one argument.
But It doesn't work with pointer to function:
int Mult(int x, int y = 2) // y is default
return x * y;
using pFn = int(*)(int, int);
pFn Double(int x)
return Mult;
int main(int argc, char* argv[])
pFn func = Double(0);
cout << func(7, 4) << endl; // ok
//cout << func(7) << endl; // error: Too few arguments
cout << Mult(4) << endl; // ok. the second argument is default
Above if I call Mult with a single argument it works fine because the second argument is default but calling it through the pointer func it fails. func is pointer to function that takes two integers and returns an int.
c++ function-pointers default-arguments
c++ function-pointers default-arguments
edited 20 mins ago
ShadowRanger
64.1k661100
64.1k661100
asked 27 mins ago
Syfu_HSyfu_H
1506
1506
1
What is the point ofDoubletaking an integer parameter that it doesn't use?
– scohe001
24 mins ago
1
Similar: Howto: c++ Function Pointer with default values
– TrebledJ
24 mins ago
add a comment |
1
What is the point ofDoubletaking an integer parameter that it doesn't use?
– scohe001
24 mins ago
1
Similar: Howto: c++ Function Pointer with default values
– TrebledJ
24 mins ago
1
1
What is the point of
Double taking an integer parameter that it doesn't use?– scohe001
24 mins ago
What is the point of
Double taking an integer parameter that it doesn't use?– scohe001
24 mins ago
1
1
Similar: Howto: c++ Function Pointer with default values
– TrebledJ
24 mins ago
Similar: Howto: c++ Function Pointer with default values
– TrebledJ
24 mins ago
add a comment |
2 Answers
2
active
oldest
votes
Defaulted arguments are a bit of C++ syntactic sugar; when calling the function directly with insufficient arguments, the compiler inserts the default as if the caller had passed it explicitly, so the function is still called with the full complement of arguments (Mult(4) is compiled into the same code as Mult(4, 2) in this case).
The default isn't actually part of the function type though, so you can't use the default for an indirect call; the syntactic sugar breaks down there, since as soon as you are calling through a pointer, the information about the defaults is lost.
add a comment |
For the "why not" I refer you to this answer. If you want to somehow keep the ability to use a default, you need to provide something more than a function pointer, eg a lamdba will do:
#include <iostream>
int Mult(int x, int y = 2) // y is default
return x * y;
auto Double()
return [](int x,int y=2) return Mult(x,y); ;
int main(int argc, char* argv[])
auto func = Double();
std::cout << func(7, 4) << 'n'; // ok
std::cout << func(7) << 'n'; // ok
std::cout << Mult(4) << 'n'; // ok
Live demo
Take the example with some grain of salt. It is merely meant to demonstrate that it is possible to do something like this. The usefulness of this approach is rather limited. You have to repeat the default parameters in Double to make it work. This is however a general problem with default parameters. They dont really like to be forwarded or be nested. Consider for example
int Double(int x,int y = ?!? ) // How to make y default here ?!?
return Mult(x,y); // just call Mult and return result
Note that this involves repeating the default explicitly insideDoublewhen defining thelambda, which limits the utility significantly.
– ShadowRanger
6 mins ago
@ShadowRanger yes, added a note
– user463035818
2 mins ago
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
oldest
votes
active
oldest
votes
Defaulted arguments are a bit of C++ syntactic sugar; when calling the function directly with insufficient arguments, the compiler inserts the default as if the caller had passed it explicitly, so the function is still called with the full complement of arguments (Mult(4) is compiled into the same code as Mult(4, 2) in this case).
The default isn't actually part of the function type though, so you can't use the default for an indirect call; the syntactic sugar breaks down there, since as soon as you are calling through a pointer, the information about the defaults is lost.
add a comment |
Defaulted arguments are a bit of C++ syntactic sugar; when calling the function directly with insufficient arguments, the compiler inserts the default as if the caller had passed it explicitly, so the function is still called with the full complement of arguments (Mult(4) is compiled into the same code as Mult(4, 2) in this case).
The default isn't actually part of the function type though, so you can't use the default for an indirect call; the syntactic sugar breaks down there, since as soon as you are calling through a pointer, the information about the defaults is lost.
add a comment |
Defaulted arguments are a bit of C++ syntactic sugar; when calling the function directly with insufficient arguments, the compiler inserts the default as if the caller had passed it explicitly, so the function is still called with the full complement of arguments (Mult(4) is compiled into the same code as Mult(4, 2) in this case).
The default isn't actually part of the function type though, so you can't use the default for an indirect call; the syntactic sugar breaks down there, since as soon as you are calling through a pointer, the information about the defaults is lost.
Defaulted arguments are a bit of C++ syntactic sugar; when calling the function directly with insufficient arguments, the compiler inserts the default as if the caller had passed it explicitly, so the function is still called with the full complement of arguments (Mult(4) is compiled into the same code as Mult(4, 2) in this case).
The default isn't actually part of the function type though, so you can't use the default for an indirect call; the syntactic sugar breaks down there, since as soon as you are calling through a pointer, the information about the defaults is lost.
answered 23 mins ago
ShadowRangerShadowRanger
64.1k661100
64.1k661100
add a comment |
add a comment |
For the "why not" I refer you to this answer. If you want to somehow keep the ability to use a default, you need to provide something more than a function pointer, eg a lamdba will do:
#include <iostream>
int Mult(int x, int y = 2) // y is default
return x * y;
auto Double()
return [](int x,int y=2) return Mult(x,y); ;
int main(int argc, char* argv[])
auto func = Double();
std::cout << func(7, 4) << 'n'; // ok
std::cout << func(7) << 'n'; // ok
std::cout << Mult(4) << 'n'; // ok
Live demo
Take the example with some grain of salt. It is merely meant to demonstrate that it is possible to do something like this. The usefulness of this approach is rather limited. You have to repeat the default parameters in Double to make it work. This is however a general problem with default parameters. They dont really like to be forwarded or be nested. Consider for example
int Double(int x,int y = ?!? ) // How to make y default here ?!?
return Mult(x,y); // just call Mult and return result
Note that this involves repeating the default explicitly insideDoublewhen defining thelambda, which limits the utility significantly.
– ShadowRanger
6 mins ago
@ShadowRanger yes, added a note
– user463035818
2 mins ago
add a comment |
For the "why not" I refer you to this answer. If you want to somehow keep the ability to use a default, you need to provide something more than a function pointer, eg a lamdba will do:
#include <iostream>
int Mult(int x, int y = 2) // y is default
return x * y;
auto Double()
return [](int x,int y=2) return Mult(x,y); ;
int main(int argc, char* argv[])
auto func = Double();
std::cout << func(7, 4) << 'n'; // ok
std::cout << func(7) << 'n'; // ok
std::cout << Mult(4) << 'n'; // ok
Live demo
Take the example with some grain of salt. It is merely meant to demonstrate that it is possible to do something like this. The usefulness of this approach is rather limited. You have to repeat the default parameters in Double to make it work. This is however a general problem with default parameters. They dont really like to be forwarded or be nested. Consider for example
int Double(int x,int y = ?!? ) // How to make y default here ?!?
return Mult(x,y); // just call Mult and return result
Note that this involves repeating the default explicitly insideDoublewhen defining thelambda, which limits the utility significantly.
– ShadowRanger
6 mins ago
@ShadowRanger yes, added a note
– user463035818
2 mins ago
add a comment |
For the "why not" I refer you to this answer. If you want to somehow keep the ability to use a default, you need to provide something more than a function pointer, eg a lamdba will do:
#include <iostream>
int Mult(int x, int y = 2) // y is default
return x * y;
auto Double()
return [](int x,int y=2) return Mult(x,y); ;
int main(int argc, char* argv[])
auto func = Double();
std::cout << func(7, 4) << 'n'; // ok
std::cout << func(7) << 'n'; // ok
std::cout << Mult(4) << 'n'; // ok
Live demo
Take the example with some grain of salt. It is merely meant to demonstrate that it is possible to do something like this. The usefulness of this approach is rather limited. You have to repeat the default parameters in Double to make it work. This is however a general problem with default parameters. They dont really like to be forwarded or be nested. Consider for example
int Double(int x,int y = ?!? ) // How to make y default here ?!?
return Mult(x,y); // just call Mult and return result
For the "why not" I refer you to this answer. If you want to somehow keep the ability to use a default, you need to provide something more than a function pointer, eg a lamdba will do:
#include <iostream>
int Mult(int x, int y = 2) // y is default
return x * y;
auto Double()
return [](int x,int y=2) return Mult(x,y); ;
int main(int argc, char* argv[])
auto func = Double();
std::cout << func(7, 4) << 'n'; // ok
std::cout << func(7) << 'n'; // ok
std::cout << Mult(4) << 'n'; // ok
Live demo
Take the example with some grain of salt. It is merely meant to demonstrate that it is possible to do something like this. The usefulness of this approach is rather limited. You have to repeat the default parameters in Double to make it work. This is however a general problem with default parameters. They dont really like to be forwarded or be nested. Consider for example
int Double(int x,int y = ?!? ) // How to make y default here ?!?
return Mult(x,y); // just call Mult and return result
edited 2 mins ago
answered 10 mins ago
user463035818user463035818
19.3k42971
19.3k42971
Note that this involves repeating the default explicitly insideDoublewhen defining thelambda, which limits the utility significantly.
– ShadowRanger
6 mins ago
@ShadowRanger yes, added a note
– user463035818
2 mins ago
add a comment |
Note that this involves repeating the default explicitly insideDoublewhen defining thelambda, which limits the utility significantly.
– ShadowRanger
6 mins ago
@ShadowRanger yes, added a note
– user463035818
2 mins ago
Note that this involves repeating the default explicitly inside
Double when defining the lambda, which limits the utility significantly.– ShadowRanger
6 mins ago
Note that this involves repeating the default explicitly inside
Double when defining the lambda, which limits the utility significantly.– ShadowRanger
6 mins ago
@ShadowRanger yes, added a note
– user463035818
2 mins ago
@ShadowRanger yes, added a note
– user463035818
2 mins ago
add a comment |
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1
What is the point of
Doubletaking an integer parameter that it doesn't use?– scohe001
24 mins ago
1
Similar: Howto: c++ Function Pointer with default values
– TrebledJ
24 mins ago