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
1 depending on architecture and compiler
std::cout << +value
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