Error comparing strings and acting on the comparison

WindowsBurner · 10-21-2004, 09:51 AM

Hi,

How come when I do :

Code:

int main()                                                                                                                                     
{                                                                                                                                              
   char input[6];                                                                                                                              
   scanf("%s", input);                                                                                                                         
   if(input == "look")                                                                                                                         
     printf("You typed : look\n");                                                                                                             
   else if(input =="quit")                                                                                                                    
     exit(0);                                                                                                                                  
   else                                                                                                                                        
     printf("You did not type a choice.\n");                                                                                                   
                                                                                                                                               
   return 0;                                                                                                                                   
}

I always get the message : You did not type a choice.?
I tried typing both :

Code:

look

and

Code:

quit

plus a few other things, but to no avail.
I tried replacing the double quotes with single quotes and I tried replacing the double equals with one equals and it still didn't work, though I got errors( I knew I would, but I'm trying everything).

With single-quotes :

Code:

test.c:7:16: warning: multi-character character constant
test.c: In function `main':
test.c:7: warning: comparison between pointer and integer
test.c:9:21: warning: multi-character character constant
test.c:9: warning: comparison between pointer and integer

With one equals :

Code:

test.c:7:15: warning: multi-character character constant
test.c: In function `main':
test.c:7: incompatible types in assignment
test.c:9:20: warning: multi-character character constant
test.c:9: incompatible types in assignment

I'm completely friggin baffled, I've had this problem since I started programming and I've never been able to figure out why.

Any help would be greatly appreciated. Thanks!

- WB

mfeat · 10-21-2004, 10:13 AM

change:
if(input == "look")
to:
if(strcmp(input, "look") == 0)

WindowsBurner · 10-21-2004, 10:42 AM

Thanks a lot mfeat. Thats alot better than the workaround I came up with. I was using a while loop and getchar to get around it.
But now that I have two work arounds.... Can you or someone else tell me WHY my previous post does not work?

I know it doesn't but my big problem is WHY. I know most of you wouldn't care, but I've got a crazy curiosity and its gonna bug me till I find out. Dont think I'm just trying to get an easy answer to my problem, I've spent easily 100 hours trying to work on that problem and read easily 100 tutorials and 4 or 5 books. Plus at least 50 hours of testing different ways myself.

As you can probably tell I'm not the best programmer since I spent all that time without fixing my problem.

Thanks again mfeat for helping. And thanks to anyone who can answer my WHY.

- WB

itsme86 · 10-21-2004, 11:48 AM

String literals in C (e.g. "look") are stored somewhere in memory. When you use "look" in your code you're actually playing with a pointer to that stored memory. I don't know how familiar you are with pointers, but basically what you're doing is comparing 2 memory addresses when you do if(input == "look").

So it's not comparing the contents of the strings, but the memory address of the strings. So:

Code:

char *ptr1 = "foo";
char *ptr2 = "foo";

printf("%d\n", ptr1 == ptr2)

that will print 0 since there's actually 2 "foo"s in memory and each pointer points to its own, but:

Code:

char *ptr1 = "foo";
char *ptr2 = ptr1;

printf("%d\n", ptr1 == ptr2);

that will print 1 because the value of ptr1 is equal to the value of ptr2 (they point to the same address).

You can also look at it like this:

Code:

int a = 5;
int b = 5;

printf("%d\n", a == b);
printf("%d\n", &a == &b);

The first printf() will print 1 because the values of a and b are the same, but the second printf() will print 0 because the address of a and b are not the same.

So to sum up, strings in C are really just an address to the first character in the string. When you do if(string1 == string2) you're comparing the memory addresses that the two strings point to, not the contents of the string.

WindowsBurner · 10-21-2004, 12:37 PM

Thanks a friggin million itsme86, NOW I finally understand. *Whew* now my curiosity can relax.
Thanks again.

- WB