Why does setting function pointer as pointer and pointerpointer work?

Asked Apr 22 '13 at 07:17

Active Apr 22 '13 at 07:20

Viewed 71 times

One more time I got trapped by pointer logic. The following code uses a function typedef tOutFunction. It defines a global variable to this function pointer and a set method.

Why does

SetOutFunction(OutFunction);

and

SetOutFunction(&OutFunction);

both work? How could autoconversion work in this place. Or is one of it wrong and it works by accident?

#include <iostream>

typedef int (*tOutFunction)(const char*);

int OutFunction(const char* out)
{
  std::cout << "OutFunction:" << out << "\n";
  return 10;
}

tOutFunction out__ = 0;

void SetOutFunction(tOutFunction outFunc)
{
  out__ = outFunc;
}

int main()
{
  SetOutFunction(OutFunction);
  std::cout << out__("Without Ref") << "\n";

  SetOutFunction(&OutFunction);
  std::cout << out__("With Ref") << "\n";

  return 0;
}

Also tried it here: http://codepad.org/yfxj2QAo

asked Apr 22 '13 at 07:17

Totonga

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RandyGaul- How did you find it. I swear I searched for it for some minutes. – Totonga Apr 22 '13 at 07:21
Googled "address of function c", was the first search result : ] – RandyGaul Apr 22 '13 at 07:22
for more thought provocation, why does `SetOutFunction(*****OutFunction);` work? – newacct Apr 22 '13 at 08:53
newacct - You won't get me again because I read http://stackoverflow.com/a/6893288/101635 – Totonga Apr 22 '13 at 09:03

Why does setting function pointer as pointer and pointerpointer work?

0 Answers0