pointer to function syntax and usage

Question

I apologize if this was asked many times. I'm trying to understand why both of this works fine without any warnings or other visible issues (in Xcode):

int testFunctionAcceptingIntPointer(int * p) {         
    return *p = *p +5;
}

void test() {
    int testY = 7;
    typedef int (*MyPointerToFunction)(int*);
    // Both this (simply a function name):
    MyPointerToFunction functionPointer = testFunctionAcceptingIntPointer;
    // And this works (pointer to function):
    MyPointerToFunction functionPointer = &testFunctionAcceptingIntPointer;
    int y = functionPointer(&testY);
}

You define functionPointer twice in your test() you should receive an error for that. — SpaceDust__
– SpaceDust__, Commented Dec 18, 2012 at 15:55
MyPointerToFunction functionPointer = testFunctionAcceptingIntPointer; is syntactic sugar of the other version. — nhahtdh
– nhahtdh, Commented Dec 18, 2012 at 15:57

Daniel Fischer · Accepted Answer · 2012-12-18 15:56:45Z

5

The code works fine without warnings both ways because a function designator is converted to a function pointer

MyPointerToFunction functionPointer = testFunctionAcceptingIntPointer;

unless it is the operand of the address operator

MyPointerToFunction functionPointer = &testFunctionAcceptingIntPointer;

(or sizeof and _Alignof).

In the first assignment, you don't use &, so the automatic conversion is done, resulting in a function pointer of appropriate type, in the second, you explicitly take the address, resulting in a function pointer of the appropriate type.

answered Dec 18, 2012 at 15:56

Daniel Fischer

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1 Comment

Daniel Fischer Over a year ago

The fun thing about the automatic conversion to a pointer is that you could also call int y = ************functionPointer(&testY); if you like asterisks. After each dereference, it's converted to a pointer again ;)

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