Question

Given some unknown input, how do you tell which variables are being substituted into a (s)printf statement?

printf("%s %s", "a", "b");     // both used
printf("%s",    "a", "b");     // only the first one used
printf( %1$s %1$s , "a", "b"); //      "        "
printf( %s %1$s , "a", "b");    //     "        "
printf( %1$s %s %1$s , "a", "b");   // "        "
printf( %2$s , "a", "b");      // only the second one used.

Checking the resultant string for the presence of the first or second variables won t help, because they could have the same value.

In my own situation, there is only ever 2 variables that could be substituted, and I need to know whether the second one is used or not.

Answer 1

Here is a bit hackish function to do it:

function printf_and_return_min_args_required(/* var args */) {
    $old_track = ini_set( track_errors ,  1 );
    $args = func_get_args();
    $args2 = array();
    $required = 100;    
    foreach ($args as $arg) {
        $args2[] = $arg;
        if (@call_user_func_array("printf", $args2)) {
            $required = count($args2) - 1;
            break;
        }
    }
    ini_set( track_errors , $old_track);
    return $required;
}

For sprintf() you might have to return array($result, $required) instead if you don t want the $result printed by the function.

Answer 2

Parse the format string yourself and see how many args it expects.

EDIT: A psuedocode example (no error checking):

bool arg1used = false;
bool arg2used = false;
int unspecifiedscount = 0;
for (int i = 0; i < s.length; i++) {
    if s[i] !=  %  continue;
    switch s[i+1] {
        case  % :
            i++;
            break;
        case  s :
            if unspecifiedscount == 0 arg1used = true;
            if unspecifiedscount == 1 arg2used = true;
            break;
        case  1 :
            if s[i+2] ==  $  && s[i+3] ==  s  {
                arg1used = true;
                i+=3;
                true;
            }
            break;
        case  2 :
            if s[i+2] ==  $  && s[i+3] ==  s  {
                arg2used = true;
                i+=3;
                true;
            }
            break;
    }
}

Answer 3

I ve come up with a kludgey way which works in this situation, but would be interested to hear better solutions which are more generic.

if (($result = @sprintf($input, "a")) === false) {
    // need two arguments
    $result = sprintf($input, "a", "b");
} else {
    // only needed one argument
}

Basically, it just tries it with one argument, and if that didn t work, then you know it needs two.

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