Question

日志：

@myLaptop：〜/ test $ gcc -o ./exe ./main.c

@myLaptop：〜/ test $ ./exe

waitpid之前：：成功

dup2之后：：成功

...。（我的终端在这一行中等待一些内容……只能激活Ctrl + C）我认为waitpid等待一些事情。：（

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

#include <wait.h>

#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>

int main(){
    system("rm -rf ./stdinTest");
    system("ls -all / > ./stdinTest");

    char **argv = (char **)malloc(sizeof(char *) * 3);

    argv[0] = (char *)malloc(strlen("grep") + 1);
    strcpy(argv[0], "grep");

    argv[1] = (char *)malloc(strlen("bin") + 1);
    strcpy(argv[1], "bin");

    argv[2] = NULL;

    int fd = open("./stdinTest", O_RDONLY);
    pid_t cpid;

    if ((cpid = fork()) == 0){
        dup2(STDIN_FILENO, fd);
        perror("After dup2: ");
        execvp(argv[0], argv);
    }
    else{
        perror("Before waitpid: ");
        waitpid(cpid, NULL, 0);
        perror("After waitpid: ");
        close(fd);
    }
    return 0;
}

Answer 1

第dup2(STDIN_FILENO, fd);行应为dup2(fd, STDIN_FILENO);

请参阅http://man7.org/linux/man-pages/man2/dup.2.html

以下code可以工作：

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

#include <wait.h>

#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>

int main(){
    system("rm -rf ./stdinTest");
    system("ls -all / > ./stdinTest");

    char **argv = (char **)malloc(sizeof(char *) * 3);

    argv[0] = (char *)malloc(strlen("grep") + 1);
    strcpy(argv[0], "grep");

    argv[1] = (char *)malloc(strlen("bin") + 1);
    strcpy(argv[1], "bin");

    argv[2] = NULL;

    int fd = open("./stdinTest", O_RDONLY);
    pid_t cpid;

    if ((cpid = fork()) == 0){
        dup2(fd, STDIN_FILENO);
        perror("After dup2: ");
        execvp(argv[0], argv);
    }
    else{
        perror("Before waitpid: ");
        waitpid(cpid, NULL, 0);
        perror("After waitpid: ");
        close(fd);
    }
    return 0;
}

为什么我的“ dup2（STDIN_FILENO，fd）”无法将参数发送给grep？

1 个答案: