为什么exevp之前的printf无法运行? [英] Why is printf before exevp not running?

问题描述

我得到"hi！"的输出.为什么这还不打印东西"?

I get an output of "hi!". Why is this not also printing "something"?

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

int main(int argc, char** argv) {
    char* program_name = "echo";
    char* args[]= {program_name,"hi!",NULL};

    printf("something");
    execvp(program_name,args);
    return 0;
}

我知道我不是先创建子进程.如果我取出execvp行，它会按预期工作.奇怪的. (注意:回声"是指 https://en.wikipedia.org/wiki/Echo_(命令))

I know I'm not creating a child process first. If I take out the execvp line, it works as expected. Weird. (Note: "echo" refers to https://en.wikipedia.org/wiki/Echo_(command))

推荐答案

该字符串位于io缓冲区中-因此请拉链并刷新该缓冲区

The string is in the io buffer - so pull the chain and flush that buffer

即添加

fflush(stdout)

在printf之后(或将\n添加到printf)

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