[FIXED] How to convert char* argv[1] to int and print it without warning in C

Issue

I want to convert the argv[1] to an int. But I get this warning:

xorcipher.c:7:9: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int *’ [-Wformat=]

And the printf display me

-799362156

if I type

./xorcipher 4

How to correct this?

Here is my code:

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

int main(int argc,char* argv[])
{
    int key_length = atoi(argv[1]);
    printf("key_length = %d", &key_length);
    return(0);
}

Solution

You are passing the address of key_length, where as the error states, printf expects just the value. Try this:

printf("key_length = %d", key_length);

See this tutorial on C format specifiers.

Answered By - Nik

Answer Checked By - Senaida (PHPFixing Volunteer)