直到给出换行符时，输出才显示为非睡眠状态 [英] Output not displayed with usleep until a line break is given

问题描述

我正在尝试在C语言中编写一个简单的打字机"效果，其中文本一次出现一个字母，并有一定延迟.这是我的功能:

I'm trying to program a simple "typewriter" effect in C, where text appears one letter at a time with a delay. Here's the function I have:

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

void typestring(const char *str, useconds_t delay)
{
    while (*str) {
        putchar(*(str++));
        usleep(delay);
    }
}

问题是直到显示\n，文本才真正出现.我在做什么错了?

The problem is the text doesn't actually appear until a \n is displayed. What am I doing wrong?

推荐答案

缓冲到stdout的输出.使用\n强制刷新.如果要更改此设置，则需要更改终端的设置(对于Linux，请在此处)或使用

The output to stdout is buffered. Using \n you are forcing a flush. If you want to change this, you will need to change the settings of the terminal (for Linux look here) or use

void typestring(const char *str, useconds_t delay)
{
    while (*str) {
        putchar(*(str++));
        fflush(stdout);
        usleep(delay);
    }
}

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