C integer overflow behaviour when assigning to larger-width integers

Question

If I execute the following code in C:

#include 

uint16_t a = 4000;
uint16_t b = 8000;

int32_t c = a - b;

printf(\"%d\", c);

Answer 1

The issue is actually somewhat complicated. Operands of arithmetic expressions are converted using specific rules that you can see in Section 3.2.1.5 of the Standard (C89). In your case, the answer depends on what the type uint16_t is. If it is smaller than int, say short int, then the operands are converted to int and you get -4000, but on a 16-bit system, uint16_t could be unsigned int and conversion to a signed type would not happen automatically.