Can you explain surprising C++ parsing?

Question

Much to my surprise, an incorrect copy and paste seems to be legal C++.  

Instead of writing 
if (my_func())
I accidentally wrote
if (bool my_func())

It seems to compare the function address.  Can someone please explain this?  I didn't think a function declaration could appear in an expression.

#include <iostream>

bool my_func()
{
return false;
}

int main()
{
	if (bool my_func())
		std::cout << "true&#92;n";
	else
		std::cout << "false";
	return 0;
}

the output is "true".
Using Microsoft Visual C++ version 6.0

soa · Accepted Answer

I don't think the code you showed is legal, since it didn't compile with gcc 4.2.0. Also it failed to compile with llvm.

http://llvm.org/demo/index.cgi

Probably it's a bug in microsoft's compiler.

Anonymous · Answer

Wow - that is interesting. I think what is happening is that the expression 'bool my_func()' is being evaluated to "true" because the function is working properly - in other words, it seems like, by including the function in the 'if' condition, it is basically checking to see if the return from it is valid, which - even though it is 'false' - is valid.

Can you explain surprising C++ parsing?

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