Question

我正在完成一项任务，我很难完成任务。想法是编写一个执行一个程序的程序if.c，如果成功，则执行第二个程序。我应该抑制第一个程序的标准输出，并取消第二个程序的标准输出。我在多次测试中收到错误消息。例如：“。/ if echo no then echo yes”返回“echo：write error：Bad file descriptor”。我试过在网上找到我做错了但没有运气。

这是我的代码：

#include <fcntl.h>
#include <sys/wait.h>
#include <stdio.h>
#include "tlpi_hdr.h"

int main(int argc, char *argv[])
{
    if(argc < 4){
        fprintf(stderr,"Incorrect number of arguments.\n");
        exit(EXIT_FAILURE);
    }

    int thenArg = 0;
    char then[4];
    strcpy(then,"then");
    for(int x=1; x<argc; x++){
        if(strncmp(argv[x], then, 4) == 0) thenArg = x;
    }

    if(thenArg == 0){
        fprintf(stderr,"No 'then' argument found.\n");
        exit(EXIT_FAILURE);
    }

    int save_out = dup(STDOUT_FILENO);
    if(save_out == -1){
        fprintf(stderr,"Error in dup(STDOUT_FILENO)\n");
        exit(EXIT_FAILURE);
    }

    int devNull = open("/dev/null",0);
    if(devNull == -1){
        fprintf(stderr,"Error in open('/dev/null',0)\n");
        exit(EXIT_FAILURE);
    }

    int dup2Result = dup2(devNull, STDOUT_FILENO);
    if(dup2Result == -1) {
        fprintf(stderr,"Error in dup2(devNull, STDOUT_FILENO)\n");
        exit(EXIT_FAILURE);
    }

    int program1argLocation = 1;
    int program2argLocation = thenArg + 1;
    int program1argCount = thenArg-1;
    int program2argCount = argc-(program2argLocation);
    char *program1args[program1argCount+1];
    char *program2args[program2argCount+1];

    for(int i=0; i<program1argCount; i++){
        program1args[i]=argv[program1argLocation + i];
    }
    program1args[program1argCount] = NULL;
    for(int i=0; i<program2argCount; i++){
        program2args[i]=argv[program2argLocation + i];
    }
    program2args[program2argCount] = NULL;

    pid_t pid = fork();
    int child_status;
    switch (pid) {
    case -1:
        fprintf(stderr,"Fork failed\n");
        exit(EXIT_FAILURE);

    case 0: //child
        //child will run program 1
        if(execvp(program1args[0],&program1args[0]) == -1){
            fprintf(stderr,"Program 1 Failed.\n");
            exit(EXIT_FAILURE);
        }  

    default: //parent
        //parent will run program2
        pid = wait(&child_status);

        if(WEXITSTATUS(child_status) == 0){
            dup2(save_out, STDOUT_FILENO);

            int prog2status = execvp(program2args[0],&program2args[0]);
            if(prog2status == -1) {
                fprintf(stderr,"Program 2 failed.\n");
                exit(EXIT_FAILURE);
            }  
        }  
    }  

}

Answer 1

您的错误在这里：

int devNull = open("/dev/null",0);

要将devNull用作STDOUT_FILENO，必须打开它才能写入：

int devNull = open("/dev/null", O_WRONLY);

在C中，如何使用dup2将STDOUT_FILENO重定向到/ dev / null，然后再重定向回其原始值？

1 个答案: