Why does int8_t and user input via cin shows strange result
c++, int, uint8t
Solution
`int8_t` is a typedef for an integer type with the required characteristics: pure 2's-complement representation, no padding bits, size of exactly 8 bits.
For most (perhaps all) compilers, that means it's going to be a typedef for `signed char`.(Because of a quirk in the definition of the term signed integer type, it cannot be a typedef for plain `char`, even if `char` happens to be signed).
The `>>` operator treats character types specially. Reading a character reads a single input character, not sequence of characters representing some integer value in decimal. So if the next input character is `'0'`, the value read will be the character value `'0'`, which is probably 48.
Since a `typedef` creates an alias for an existing type, not a new distinct type, there's no way for the `>>` operator to know that you want to treat `int8_t` as an integer type rather than as a character type.
The problem is that in most implementations there is no 8-bit integer type that's not a character type.
The only workaround is to read into an `int` variable and then convert to `int8_t` (with range checks if you need them).
Incidentally, `int8_t` is a signed type; the corresponding unsigned type is `uint8_t`, which has a range of 0..255.
(One more consideration: if `CHAR_BIT > 8`, which is permitted by the standard, then neither `int8_t` nor `uint8_t` will be defined at all.)
Problem
A tiny piece of code drives me crazy but hopefully you can prevent me from jumping out of the window. Look here: ``` #include <iostream> #include <cstdint> int main() { int8_t i = 65; int8_t j; std::cout << "i = " << i << std::endl; // the 'A' is ok, same as uchar std::cout << "Now type in a value for j (use 65 again): " << std::endl; std::cin >> j; std::cout << "j = " << j << std::endl; if (i != j) std::cout << "What is going on here?????" << std::endl; else std::cout << "Everything ok." << std::endl; return 0; } ``` If I use `int` instead of `int8_t` everything ok. I need this as 8-bit `unsigned integers`, not bigger. And btw. with `unsigned char` it's the same behaviour - of course - as with `int8_t`. Anyone with a hint?