echo $PATH in system() give me a wrong output

Question

This is a piece of code found on Internet

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


int main(int argc, char* argv[])
{
    putenv("PATH=/nothinghere");
    //setenv("PATH","/nothinghere");
    system(argv[1]);
    return 0;
}

if I do

$./a.out "ls"
sh: 1: ls: not found

Of course But what if

$./a.out "echo $PATH"
/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games:/usr/local/games

It print the original $PATH !!

If we create a new shell then do the samethings

int main(int argc, char* argv[])
{
    putenv("PATH=/nothinghere");
    //setenv("PATH","/nothinghere");
    system("/bin/sh");
    return 0;
}

$./a.out
$ echo $PATH
/nothinghere
$ ls
/bin/sh: 2: ls: not found

Why? Is it kind of problem about fork or the implementation of echo?

Run `set -x` before your command, and it'll show you what it's doing. — Charles Duffy, Aug 16 '18 at 20:53
`fprintf(stderr, "argv[1]=%s\n", argv[1]);` would likewise have avoided this confusion. — Charles Duffy, Aug 16 '18 at 20:58

score 11 · Accepted Answer · answered Aug 16 '18 at 20:54

11

This is because you're using double quotes, telling your shell to replace $PATH with the value of the PATH variable before it even starts a.out.

The wrong value is thus being inserted not by the shell invoked by system(), but by the shell you're interactively typing commands at.

To fix it, change:

$ ./a.out "echo $PATH"

to:

$ ./a.out 'echo $PATH'

answered Aug 16 '18 at 20:54

Charles Duffy

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Oh... I see, thanks – Terrynini Aug 16 '18 at 21:04

echo $PATH in system() give me a wrong output

1 Answers1