GCC interprets uint8_t and uint16_t as signed? [duplicate] - linux

Question

This question already has answers here:
Format specifiers for uint8_t, uint16_t, ...?
(7 answers)
Closed 3 years ago.
My test code:
#include <cstdint>
#include <cstdio>
int main() {
const constexpr uint8_t x = 64;
printf("%u", x);
}
Here is how I compiled with GCC 8.2:
g++ -Wall test_format.cpp -o test_format -O3 -std=c++17 -Wformat-signedness
And here is GCC's output:
test_format.cpp: In function ‘int main()’:
test_format.cpp:6:9: warning: format ‘%u’ expects argument of type ‘unsigned int’, but argument 2 has type ‘int’ [-Wformat=]
printf("%u", x);
^~~~
Tho, if I try to print an uint32_t, it has no error/warning.
I wonder why GCC expects uint8_t to be signed int.
Thanks.

Answer 1

Default argument promotions are applied to operands of a variadic function. Under these, an expression of type unsigned char is promoted to int.

Answer 2

In C and C++ types narrower than int are always promoted to int. See Why must a short be converted to an int before arithmetic operations in C and C++?
And inside variadic functions default promotion also applies, which means you can't pass types narrower than int to vararg functions. So uint8_t must be printed with %d, not %u. But anyway you're printing it the wrong way. The correct way is to use PRIu8
printf("%" PRIu8 "\n", x);
Format specifiers for uint8_t, uint16_t, ...?
printing the uint8_t
Why is the format specifier for uint8_t and uint16_t the same (%u)?
How do I print uint32_t and uint16_t variables value?

Answer 3

To print a uint8_t variable with printf(), you should do something like the following:
#include <cinttypes>
#include <cstdio>
int print_u8(std::uint8_t x) {
return std::printf("%" PRIu8 "\n", x);
}
The <cinttypes> header includes printf and scanf format specifiers for all the <cstdint> types (and explicitly includes that header) that should be used for maximum portability.