Thread: help with simple question

Code:

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

void search_array (char array[], char type)
{
    if (strcmp ("nop", array) == 0){
        type = 'R';
        }else if (strcmp ("syscall", array) == 0){
            type = 'I';
    }
    printf("%c\n", type);
}

int main (void)
{
    char type;
    char array[] = "syscall";
    search_array (array, type);
    printf ("%c\n", type);
    
}

I don't get why when I try to print value of type from the main it prints nothing, but when i print it from the function it prints the correct value.

Because "type in search_array" and "type in main" are not the same variable and consequently don't have the same value.

Which means you have to pass "type" by reference from main() to search_array().

cheers, all fixed now

Originally Posted by agentsmith

cheers, all fixed now

Hai
Here is your modified code

Code:

#include "stdafx.h"
#include "stdio.h"
// Change function.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>

void search_array (char array[], char *type)
{
    if (strcmp ("nop", array) == 0){
        *type = 'R';
        }else if (strcmp ("syscall", array) == 0){
            *type = 'I';
    }
    printf("%c\n", *type);
}

int main (void)
{
    char type;
    char array[] = "syscall";
    search_array (array, &type);
    printf ("%c\n", type);
	return 0;
    
}

You are expecting modified value of local variable from called function..to achieve this
you should go for either pass by address or pass by reference..but you are trying to use
pass by value..both variables(Function & main) stored at different addresses..hence you
will get garbage value when you print that...