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Can you explain surprising C++ parsing?
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\n";
else
std::cout << "false";
return 0;
}
the output is "true".
Using Microsoft Visual C++ version 6.0
2 Answers
- 1 decade agoFavorite 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.
- Anonymous1 decade ago
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.
