函数printf（）打印退格问题

Question

There are two programs and they get different results, but I don't understand why. Here is the first:

int main()
{
    printf("12345");
    fflush(stdout);
    printf("\b\b");
    fflush(stdout);
    return 0;
}

The result is: 123 . Then the second:

int main()
{
    printf("12345");
    fflush(stdout);
    sleep(1);
    printf("\b\b");
    fflush(stdout);
    return 0;
}

but the result is: 12345 .

Why does the sleep call make the second result different when I expect a "123" result? The code was running in the CLion. and My OS is macOS, if it matters.

Answer 1

Your code first displays "12345" and then displays "123". Without the sleep , the "12345" is displayed for too little time to be visible. With the sleep , you can see the "12345" for about a second before the "4" and "5" are erased.

Answer 2

Depending on the Terminal, '\\b' might "erase" a character or only move the cursor to the left. To have a foolproof solution, use:

#include <unistd.h>
#include <stdio.h>

int main(void)
{
    printf("12345");
    fflush(stdout);
    sleep(1);
    printf("\b\b  \b\b");
    fflush(stdout);
}

Answer 3

With the original code (called bs97 , compiled from bs97.c ), and with my command line prompt Osiris JL: , I get the output:

Osiris JL: bs97
123Osiris JL:

Note that the new prompt has overwritten the 45 . You can see that if you add putchar('\\n'); after the second fflush() . (I added void to int main(void) to avoid compilation warnings (errors) with my default compilation options.)

Osiris JL: make bs97 && bs97
    gcc -O3 -g -std=c11 -Wall -Wextra -Werror -Wmissing-prototypes -Wstrict-prototypes bs97.c -o bs97
12345
Osiris JL: 

This time, the 4 and the 5 are still visible.

Simply using backspace does not erase the content that was previously there; it positions the write position on screen back one place, that's all. If you want to erase what was there, you need to output printf("\\b \\b\\b \\b") to go back, output a space, and go back again — repeated twice. Or you could use printf("\\b\\b ") , but that would leave the write position after the two spaces; or you could use printf("\\b\\b \\b\\b") to leave the write position after the 3 . And there are other variations on this theme.

With the second program with the sleep, I get similar behaviour, except that the 12345 is on display while the program sleeps.

I'm testing on a Mac running macOS 10.14 Mojave, using a home-built GCC 8.2.0 or the clang compiler from XCode 10.0.