Assign different strings to char array using a variable

You should use char *lakenamer() and return your string (or strdup(your_string)) from it.

By the way, you're using printf ("%s", lakenamer(lakenum));, so you need the return value be char * but you're returning int instead.

Also you are declaring lname as a local variable within your function, but returning a pointer to it to your calling function. This isn't good, because once you return that memory address can get written over at any time.

You could instead have the caller allocate the memory, and pass its address to you, i.e.

char[20] lname;
lakenamer(lakenum, lname);

Then inside your function you can strcpy the result into lname. Then have lakenamer return void, or int if you would like to return a success or error status code.

You need to dynamically allocate the memory for the string inside the function if you want to use it outside the function.

char* lakenamer( int lakenum )
{
    char* lname = ( char* )malloc( 20 );
    if ( lakenum == 1 )
    {
        strncpy( lname, "Ontario", 20 );
    }
    else if ( lakenum == 2 )
    {
        strncpy( lname, "Erie", 20 );
    }
    else if ( lakenum == 3 )
    {
        strncpy( lname, "Huron", 20 );
    }
    else if ( lakenum == 4 )
    {
        strncpy( lname, "Muskoka", 20 );
    }
    else if ( lakenum == 5 )
    {
        strncpy( lname, "Simcoe", 20 );
    }
    printf( "%s", lname );
    return ( lname );
}

int main()
{
    int lakenum = 1;
    char* s = lakenamer( lakenum );
    printf( "%s", s );

    /*
    .
    .Do other stuff with s
    .
    */

    free( s ); // Very important !!
}

error: return makes integer from pointer without a cast

Ans: You are getting this error because you are returning address and collecting with int without cast.

error: function returns address of local variable

Ans: You are getting above error because you are returning local variable address from a function which wont be exist after returning to Main() function.

Please find the below code which solves above problem.

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

void lakenamer(int, char *);

main(int argc, char *argv[])
{
    char string[20]={0,};
    lakenamer(atoi(argv[1]), string);
    printf("\nIn Main fn:[%s]\n",string);
}

void lakenamer(int lakenum, char *str)
{
    char lname[20] = "0"; 
    //char lname[20] = {0,};

    if (lakenum == 1) {
        strncpy(lname, "Ontario", 20); //u can change lname to str
    } else if (lakenum == 2) {
        strncpy(lname, "Erie", 20);
    } else if (lakenum == 3) {
        strncpy(lname, "Huron", 20);
    } else if (lakenum == 4) {
        strncpy(lname, "Muskoka", 20);
    } else if (lakenum == 5) {
        strncpy(lname, "Simcoe", 20);
    }

    strcpy(str, lname);
    printf("%s",lname);
    //return (lname);//do not return any reference for local variable
}