Warning : overflow in implicit constant conversion

Question

In the following program, the line 5 does give overflow warning as expected, but surprisingly the line 4 doesn\'t give any warning in GCC: http://www.ideone.com/U0

Answer 1

In the general case of assigning an int value to a char object, the compiler doesn't know whether the int is out of range of the char.

Look at the actual warning more closely:

warning: overflow in implicit constant conversion

It is in this specific case, where a constant is being converted to char that the compiler is able to warn you. Likewise, if you changed the declaration of i to be const:

const int i = 256;

you will also get the warning, because the value being assigned to c2 is a constant expression.

Note also that the warning is somewhat misleading as the conversion does not technically "overflow." Arithmetic overflow yields undefined behavior in C++. A narrowing conversion (like int to char, if int has a larger range than char) yields some implementation-defined conversion.