[SOLVED] The scanf function in C language uses a width specifier for char

The scanf function in C language uses a width specifier for char

There is a code where I enter ABCDEFGH and press enter, and the final result is HGF. When I use debug mode to observe variables. When executing the first sentence input, x='A'. After the next step, x='D', y='C'. After the next step, x='H', y='G', z='F'。

#include <stdio.h>

int main() {
    char x, y, z;
    scanf("%2c", &x);
    scanf("%3c", &y);
    scanf("%4c", &z);
    printf("%c%c%c", x, y, z);
    return 0;
}

I am currently very confused as to why this is happening. As far as I know, "%3c" means to read in 3 characters, but only store the first one and discard the last two. I'm not sure if there's a problem with this code. Can you explain why the output is like this, regardless of whether the code was written incorrectly or not?

Solution

You tell scanf that is should read two or more characters from the input, and then store them into a one-character variable. That leads to undefined behavior.

From this scanf (and family) reference regarding the c format:

If a width specifier is used, matches exactly width characters (the argument must be a pointer to an array with sufficient room).

_{[Emphasis mine]}

As a somewhat related tip, you should always check what scanf returns.