C++ Overflow when converting from integer to signed integer


When either a signed or unsigned integer is converted to a signed integer type, and its value is not representable in the destination type, the value produced is implementation-defined. Example:

// Suppose that on this implementation, the range of signed char is -128 to +127 and
// the range of unsigned char is 0 to 255
int x = 12345;
signed char sc = x;   // sc has an implementation-defined value
unsigned char uc = x; // uc is initialized to 57 (i.e., 12345 modulo 256)