No compilation error comparing integer with string?
c++, compiler-errors, compiler-warnings
Solution
The `<<` operator has a higher precedence than `<`, so this is parsed as
(cout << a) < " ";
You are not really comparing a string with an integer. Instead, you are comparing the return value of `ostream::operator<<`, which is `std::cout` itself, to the string literal. This isn't legal (in the sense that is has an unspecified result, and it is not meaningful) either, `clang` warns:
warning: result of comparison against a string literal is unspecified
The reason why it compiles is that up until C++11, `std::ostream` can be implicitly converted to `void *`. Also, the string literal of type `const char[2]` decays into a pointer of type `const char *`. So, the `<` operator now takes two pointers, which is permitted, although its result is not specified, because the two pointers don't point to the same object.
Problem
Probably like many, I typed this typo ``` int a = 0; cout << a < " "; //note the '<' ``` However, the MSVC++ compiler threw just a warning warning C4552: '<' : operator has no effect; expected operator with side-effect though I expected a compilation error. Is it indeed standard complaint code? Does any implicit type conversion or overloading happen which make the code valid? I am also confused whether `<` operator is comparing the string `" "` with an integer `a` or with the result of `cout << a` A related SO post is here.