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 value of <E5> outputted to the screen, but when I run the program in gdb and use p value it prints out -27, which is the correct value. Does anyone have any ideas?

Answer 1

Because int8_t is the same as signed char, and char is not treated as a number by the stream. Cast into e.g. int16_t

std::cout << static_cast<int16_t>(value) << std::endl;

and you'll get the correct result.

Answer 2

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}

Answer 3

Most probably int8_t is

typedef char int8_t

Therefore when you use stream out "value" the underlying type (a char) is printed.

One solution to get a "integer number" printed is to type cast value before streaming the int8_t:

std::cout << static_cast<int>(value) << std::endl;

Answer 4

It looks like it is printing out the value as a character - If you use 'char value;' instead, it prints the same thing. int8_t is from the C standard library, so it may be that cout is not prepared for it(or it is just typedefd to char).