Why is std::cout not printing the correct value for my int8_t number?

Question

I have something like:

int8_t value;
value = -27;

std::cout << value << std::endl;

When I run my program I get a wrong random

Answer 1

This is because int8_t is synonymous to signed char.

So the value will be shown as a char value.

To force int display you could use

std::cout << (int) 'a' << std::endl;

This will work, as long as you don't require special formatting, e.g.

std::cout << std::hex << (int) 'a' << std::endl;

In that case you'll get artifacts from the widened size, especially if the char value is negative (you'd get FFFFFFFF or FFFF¹ for (int)(int8_t)-1 instead of FF)

Edit see also this very readable writeup that goes into more detail and offers more strategies to 'deal' with this: http://blog.mezeske.com/?p=170

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_{¹ depending on architecture and compiler}