Function pointer with default arguments

Question

Please refer below code :

#include <iostream>
using namespace std;

int sum(int a , int b = 5)
{
return a + b;
}

int main()
{
int (*fp)(int ,int);//function pointer
fp = &sum;
//int result = fp(3,5);
int result = fp(3);
cout << result;

return 0;
}

I am not able to call function pointer with default argument. Is there any other way we need to declare function pointer with default argument?

Answer 1

It seems default parameters mean something only to us and the compiler. They play no part in the function signature, and thus are ignored when you assign the function to a function pointer or a std::function object.

Here are more details on the same:

https://stackoverflow.com/questions/2576411/function-pointers-with-default-parameters-in-c

https://stackoverflow.com/questions/10718994/can-i-bind-to-a-function-that-takes-default-arguments-and-then-call-it

You can maybe define a class that stores the previous state of the argument and in case no new argument is provided, you use the old argument.

Answer 2

/*Fill in the blanks to declare a function sum returning the sum of its arguments and declare a function pointer psum pointing to sum.*/

int sum(int a, int b) {
return a + b;
}
int (*psum)(int, int) = sum;

Answer 3

// suppose f1, f2, and f3 are declared

void (*funcs[3])() = {f1, f2, f3};

for (int ix = 0; ix <
; ix++) {

[ix]();

}

Answer 4

int sum(int a, int b) {
return a

b;
}

int (

psum)(int,

) =

;

Answer 5

declare a function sum returning the sum of its arguments and declare a function pointer psum pointing to sum.

int sum(int a, int b) {
return a + b;
}

int (*psum)(int,int) =sum;

Answer 6

int sum(int a, int b) {

return a
+
b;

}

int (
*
psum)(int,
int
) =
sum
;

Answer 7

void (*funcs[3])() = {f1, f2, f3};

for (int ix = 0; ix <
; ix++) {

[ix]();

}

Answer 8

void (*funcs[3])() = {f1, f2, f3};

for (int ix = 0; ix <
; ix++) {

[ix]();

}