How do I properly compare strings in C?

Question

You can't (usefully) compare strings using != or ==, you need to use strcmp:

while (strcmp(check,input) != 0)

The reason for this is because != and == will only compare the base addresses of those strings. Not the contents of the strings themselves.

Ok a few things: gets is unsafe and should be replaced with fgets(input, sizeof(input), stdin) so that you don't get a buffer overflow.

Next, to compare strings, you must use strcmp, where a return value of 0 indicates that the two strings match. Using the equality operators (ie. !=) compares the address of the two strings, as opposed to the individual chars inside them.

And also note that, while in this example it won't cause a problem, fgets stores the newline character, '\n' in the buffers also; gets() does not. If you compared the user input from fgets() to a string literal such as "abc" it would never match (unless the buffer was too small so that the '\n' wouldn't fit in it).

Use strcmp.

This is in string.h library, and is very popular. strcmp return 0 if the strings are equal. See this for an better explanation of what strcmp returns.

Basically, you have to do:

while (strcmp(check,input) != 0)

or

while (!strcmp(check,input))

or

while (strcmp(check,input))

You can check this, a tutorial on strcmp.

You can't compare arrays directly like this

array1==array2

You should compare them char-by-char; for this you can use a function and return a boolean (True:1, False:0) value. Then you can use it in the test condition of the while loop.

Try this:

#include <stdio.h>
int checker(char input[],char check[]);
int main()
{
    char input[40];
    char check[40];
    int i=0;
    printf("Hello!\nPlease enter a word or character:\n");
    scanf("%s",input);
    printf("I will now repeat this until you type it back to me.\n");
    scanf("%s",check);

    while (!checker(input,check))
    {
        printf("%s\n", input);
        scanf("%s",check);
    }

    printf("Good bye!");

    return 0;
}

int checker(char input[],char check[])
{
    int i,result=1;
    for(i=0; input[i]!='\0' || check[i]!='\0'; i++) {
        if(input[i] != check[i]) {
            result=0;
            break;
        }
    }
    return result;
}

Whenever you are trying to compare the strings, compare them with respect to each character. For this you can use built in string function called strcmp(input1,input2); and you should use the header file called #include<string.h>

Try this code:

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

int main() 
{ 
    char s[]="STACKOVERFLOW";
    char s1[200];
    printf("Enter the string to be checked\n");//enter the input string
    scanf("%s",s1);
    if(strcmp(s,s1)==0)//compare both the strings  
    {
        printf("Both the Strings match\n"); 
    } 
    else
    {
        printf("Entered String does not match\n");  
    } 
    system("pause");  
} 

Welcome to the concept of the pointer. Generations of beginning programmers have found the concept elusive, but if you wish to grow into a competent programmer, you must eventually master this concept — and moreover, you are already asking the right question. That's good.

Is it clear to you what an address is? See this diagram:

----------     ----------
| 0x4000 |     | 0x4004 |
|    1   |     |    7   |
----------     ----------

In the diagram, the integer 1 is stored in memory at address 0x4000. Why at an address? Because memory is large and can store many integers, just as a city is large and can house many families. Each integer is stored at a memory location, as each family resides in a house. Each memory location is identified by an address, as each house is identified by an address.

The two boxes in the diagram represent two distinct memory locations. You can think of them as if they were houses. The integer 1 resides in the memory location at address 0x4000 (think, "4000 Elm St."). The integer 7 resides in the memory location at address 0x4004 (think, "4004 Elm St.").

You thought that your program was comparing the 1 to the 7, but it wasn't. It was comparing the 0x4000 to the 0x4004. So what happens when you have this situation?

----------     ----------
| 0x4000 |     | 0x4004 |
|    1   |     |    1   |
----------     ----------

The two integers are the same but the addresses differ. Your program compares the addresses.

You need to use strcmp() and you need to #include <string.h>

The != and == operators only compare the base addresses of those strings. Not the contents of the strings

while (strcmp(check, input))

Example code:

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

int main()
{
    char input[40];
    char check[40] = "end\n"; //dont forget to check for \n

    while ( strcmp(check, input) ) //strcmp returns 0 if equal
    {
        printf("Please enter a name: \n");
        fgets(input, sizeof(input), stdin);
        printf("My name is: %s\n", input);
    }

    printf("Good bye!");
    return 0;
}

Note1: gets() is unsafe. Use fgets() instead

Note2: When using fgets() you need to check for '\n' new line charecter too