C ++没有访问if语句中的方法

时间:2015-06-14 17:10:22

标签: c++ if-statement strcmp

当我在这里添加\ n时,if语句块现在执行:

#include <stdio.h>
#include <unistd.h>
#include <cstring>
#include <stdlib.h>
#include <iostream>
#include <sys/wait.h>

char* getlistOfProcesses(const char* cmd) 
{
    FILE* pipe = popen(cmd, "r");
    if (!pipe) return (char*)"ERROR";
    char buffer[128];
    char *result = new char[1024];
    while(!feof(pipe)) {
        if(fgets(buffer, 128, pipe) != NULL)
            strcat(result, buffer);
    }
    pclose(pipe);
    return result;
}

int main(int argc, char **argv)
{   
    int P2C[2];
    int C2P[2];
    pipe(P2C);
    pipe(C2P);
    pid_t cPid = fork();
    char cmd[50];
    char* listOfProcesses = new char[1024];

    if (cPid == 0)
    {
        close(P2C[1]); 
        close(C2P[0]); 
        read(P2C[0], cmd, 50);
        if(strcmp(cmd,"LISTALL") == 0)
        {
            printf("Executing the command: %s", cmd);
            write(C2P[1], getlistOfProcesses("ps -ax -o pid,cmd"), 1024);
            close(P2C[0]);
            close(C2P[1]);
        }
    }
    else if (cPid > 0)
    {
        close(C2P[1]); 
        close(P2C[0]); 
        write(P2C[1], "LISTALL", 50);
        wait(NULL);
        read(C2P[0], listOfProcesses,1024);
        printf("%s\n",listOfProcesses); 
        close(C2P[0]);
        close(P2C[1]);
    }
    ...
    return 0;
}

在添加之前,它没有点击getListofProcesses()方法。原始代码如下所示:

{{1}}

我不确定这是因为我的管道完成方式还是被访问的方式而且我没有注意到,我试图用方法顶部附近的printf进行检查。

1 个答案:

答案 0 :(得分:2)

您需要在格式字符串中包含换行符,或者在printf语句后调用fflush(stdout)。有关详细信息,请参阅Why does printf not flush after the call unless a newline is in the format string?