Thread: Character string

Originally Posted by FourAngels

The array is a character array, so that is why I believed that there is a conversion. I thought that the '0' + value was the result of beginning at '0' and adding value number of bytes to '0'. If it was the other way around than you would have value (a 64 bit number) plus '0' which is some other result.

Ah. The thing is that when you have an expression like A + B, A and B are converted to a common type in what is known as the usual arithmetic conversions. So, it is not that the type of the expression is that of A: it could be the type of B instead, or sometimes neither. In this case, both operands (n % 10) and '0' are already of type int, so no further conversion is necessary. But, as the type of the destination is char (since s is a pointer to char, s[i++] is a char), there is a conversion from int to char, but it is in the C = D expression rather than A + B.