Why `static_cast(uint32_t)` works unexpected?

Asked Oct 29 '18 at 22:18

Active Oct 29 '18 at 22:18

Viewed 63 times

The following code does what's expected

uint32_t a = 1, b = 2;
std::cout << static_cast<int64_t>(a) - b << '\n';

Prints -1. But if I change int64_t to int everything breaks:

std::cout << static_cast<int>(a) - b << '\n';

Prints 4294967295.

Does anybody know what's the trick?

asked Oct 29 '18 at 22:18

InFamous X

2

Mixing signed and unsigned types in arithmetic raises red flags in general. – chris Oct 29 '18 at 22:21
2

Both code snippets do what's expected. – Eljay Oct 29 '18 at 22:21
[operator_arithmetic#Conversions](https://en.cppreference.com/w/cpp/language/operator_arithmetic#Conversions) – Jarod42 Oct 29 '18 at 22:42

0 Answers0