CODEWITHSUNDEEP
c++function-pointers
nzomkxia
nzomkxia

Reputation: 1267

function call with function pointer in c++

Here's the code: 

#include<iostream>
using namespace std;
typedef struct ptrs
{
    int (*addptr)(int a, int b);
}mems;

int add(int a, int b)
{
    int result = a+b;
    return result;
}

int main()
{
    mems ptrtest;
    ptrtest.addptr = &add;
    int c = (*ptrtest.addptr)(3,4);
    //int c = ptrtest.addptr(3,4);
    cout << c << endl;
    return 0;
}

if I replace the code int c = (*ptrtest.addptr)(3,4); with it's next line(annotated now), the result will be the same, why is that?

Upvotes: 1

Views: 83

Answers (3)

IanPudney
IanPudney

Reputation: 6031

C++ will automatically cast a function name to a function pointer (and vise versa) if doing so will create correct syntax.

Upvotes: 0

Vaughn Cato
Vaughn Cato

Reputation: 64308

Functions and function pointers can be used interchangeably, presumably for convenience. In particular, section 5.2.2 of the C++11 standard specifies that a function call can occur using a function or a pointer to a function.

Upvotes: 1

user529758
user529758

Reputation:

Of course, int c = (*ptrtest.addptr)(3,4); is the base case. However, in C++ (and in C as well), if you use the call (()) operator on a function pointer, it will do the "dereferencing" automatically. Just like when assigned to a variable of function pointer type, the name of a function decays into a function pointer, i. e.

int (*fptr)() = some_func;

is just as valid as

int (*fptr)() = &some_func;

albeit the type of func is int ()(void).

Upvotes: 3

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