11 Beginner The Hundred

The Integer That Prints a Letter

int8_t is not what it says on the tin

#include <cstdint>
#include <iostream>

int main() {
    int8_t a = 65;
    uint8_t b = 66;

    std::cout << "a     = " << a << "\n";
    std::cout << "b     = " << b << "\n";
    std::cout << "a + b = " << a + b << "\n";
}

Run it. What does it print?

Answer

a     = A
b     = B
a + b = 131

The single values come out as letters; only the sum is a number.

Why

On every major platform, int8_t is a typedef for signed char and uint8_t for unsigned char — the <cstdint> names change nothing about the type itself. Overload resolution sees a character type, so operator<< picks the character overload and prints the byte as ASCII: 65 is 'A', 66 is 'B'. The sum escapes because a + b promotes both operands to int before adding, and int gets the numeric overload. Input is broken the same way: std::cin >> a reads exactly one character, so typing 7 stores 55, the ASCII code of '7'; read into an int instead.

The fix

Promote to a real integer before printing:

std::cout << +a << "\n";                    // unary plus promotes to int — prints 65
std::cout << static_cast<int>(b) << "\n";   // says what it means — prints 66

Takeaway: int8_t and uint8_t are chars in disguise — promote (+n or a cast) before streaming them.

Try it: g++ -std=c++17 main.cpp -o demo && ./demo

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