无法在shell_exec函数中传递参数

Question

I am pretty new to using the shell commands.I am doing a debugger application in php.I need to check the result of the .exe files to check whether the logic of the code is right.So I tried a test program .

The C program is

include

#define MAX 128
int main(char c)
{
const int max=127;
char array[max]; // char array[10];
char string[MAX];
scanf("%c",&c);
array[0] = string[0] = c;
printf("%c %c
", array[0], string[0]);
return 0;
}

This is compiled as se.exe

PHP CODE

<?php 
define('STDIN',fopen("php://stdin","r"));
$cor1=1;
$op=shell_exec("se.exe H");
echo($op);  
if($op=="H H")
$cor1+=1;
echo $cor1;
if($cor1>1)
{echo "PASSED";}
else
{echo "FAILED";}
?>

This is not echoing any values.

Answer 1

I think the argument is successfully being passed. The c code not using the argument. It is doing scanf to get user input and not really using the argument that is being passed. You need to change the c code to use the argument.

int main(int argc, char *argv[]) { 
    /* ... 
    argv[0] will have the executable name as a char array (char *)
    argv[1] will have the first command line argument as a char array (char *)
    argv[1][0] will have the first character of the first parameter that you seem to be interested.
    ensure that you do proper null check or you will get core dump/application crash
    */ 
}