Overflow of integer calculation in C and C++

Question

Please check the following snippet:

  unsigned char a = 100;
  unsigned char b = 100;
  unsigned char c = 100;
  unsigned short x = a * b + c;

I expected this would overflow and the calculation would be done in 8-bit type unsigned char (not in 16-bit unsigned short), exceeding the value range of unsigned char. But it doesn't.

Why does the calculation not overflow in C and C++?

Because the integral types get promoted to `int` before the calculation? http://en.cppreference.com/w/cpp/language/operator_arithmetic — UnholySheep, Mar 13 '18 at 09:14

Tamás Zahola · Accepted Answer · 2018-03-13T10:05:42.160

[...] arithmetic operators do not accept types smaller than int as arguments, and integral promotions are automatically applied after lvalue-to-rvalue conversion, if applicable.

http://en.cppreference.com/w/cpp/language/implicit_conversion#Integral_promotion

So your code behaves like this:

unsigned char a = (unsigned char)100;
unsigned char b = (unsigned char)100;
unsigned char c = (unsigned char)100;
unsigned short x = (unsigned short)((int)a * (int)b + (int)c);

Overflow of integer calculation in C and C++

1 Answers1