If-directive macro comparison

Question

Why is the #if condition in the following code fulfilled:

#include <iostream> #define VALUE foo      int main() {     #if VALUE == bar     std::cout << "WORKS!" << std::endl; #endif // VALUE } 

Answer 1

The page on cppreference.com states:

After all macro expansion and evaluation of defined and __has_include (since C++17) expressions, any identifier which is not a boolean literal is replaced with the number ​0​ (this includes identifiers that are lexically keywords, but not alternative tokens like and).

So VALUE is first replaced with foo, and then both foo and bar are replaced with 0.

Answer 2

This is because neither foo nor bar have been given any definition or value - so they are the same (i.e. replaced with a "0" value). Compilers will give warnings about this.

The MSVC compiler (Visual Studio 2019) gives the following:

warning C4668: 'foo' is not defined as a preprocessor macro, replacing with '0' for '#if/#elif'
warning C4668: 'bar' is not defined as a preprocessor macro, replacing with '0' for '#if/#elif'

So VALUE is given the value '0' (default for foo) and bar also has '0', so VALUE == bar evaluates to "TRUE."

Similarly, clang-cl gives the following:

warning : 'foo' is not defined, evaluates to 0 [-Wundef]
warning : 'bar' is not defined, evaluates to 0 [-Wundef]