Signed/unsigned comparisons

Question

I\'m trying to understand why the following code doesn\'t issue a warning at the indicated place.

//from limits.h
#define UINT_MAX 0xffffffff /* maximum unsi         


        

Answer 1

Why signed/unsigned warnings are important and programmers must pay heed to them, is demonstrated by the following example.

Guess the output of this code?

#include 

int main() {
        int i = -1;
        unsigned int j = 1;
        if ( i < j ) 
            std::cout << " i is less than j";
        else
            std::cout << " i is greater than j";

        return 0;
}

Output:

i is greater than j

Surprised? Online Demo : http://www.ideone.com/5iCxY

Bottomline: in comparison, if one operand is unsigned, then the other operand is implicitly converted into unsigned if its type is signed!