为什么使用`execl`而不是`system`会使我的程序无法正常工作？

Question

I'm trying to do basic IPC using pipes. I spent hours searching the internet, doing this and that, reading the API documentations, and ended up with the code below. But it does not work, as I quite expected. Just any help making my code 'work' would be many thanks.

<edit>
I've just found that using system instead of execl makes my program run perfectly as expected. So what is going wrong here when I use execl , while it doesn't happen with the system function?
</edit>

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

int main(void){
    int hInPipe[2];
    int hOutPipe[2];
    FILE *hInFile;
    FILE *hOutFile;
    char *s;

    pipe(hInPipe);
    pipe(hOutPipe);
    if(fork()){
        close(hInPipe[0]);
        close(hOutPipe[1]);
        hInFile=fdopen(hInPipe[1],"w");
        fprintf(hInFile,"2^100\n");
        fclose(hInFile);
        hOutFile=fdopen(hOutPipe[0],"r");
        fscanf(hOutFile,"%ms",&s);
        fclose(hOutFile);
        printf("%s\n",s);
        free(s);
    }else{
        dup2(hInPipe[0],STDIN_FILENO);
        dup2(hOutPipe[1],STDOUT_FILENO);
        close(hInPipe[0]);
        close(hInPipe[1]);
        close(hOutPipe[0]);
        close(hOutPipe[1]);

        system("bc -q");/*this works*/
        /*execl("bc","-q",NULL);*/ /*but this doesn't*/
    }
}

Answer 1

Read the fine man page. :)

execl(const char *path, const char *arg0, ... /*, (char *)0 */);

arg0 (aka argv[0], the name the program is told it was invoked under) is not the same argument as the path (the location of the executable for said program). Moreover, execl takes, as its first argument, a fully-qualified pathname.

Thus, you want:

execl("/usr/bin/bc", "bc", "-q", NULL);

...or, to search the PATH for bc rather than hardcoding a location:

execlp("bc", "bc", "-q", NULL);