comparing the ending of the strings

Question

I am writing a program to compare different strings. Specifically chemical elements that end with an OH. I have to return -1 if the string ends with OH. However, my program doesn't work. Where am I wrong?

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

int hydroxide(char *string);

int main() {
    char *string;
    printf("Enter String:");
    gets(string);
    printf("%d", hydroxide(string));
}

int hydroxide(char *string) {
    string = strrchr(string, 'O');
    if (string != NULL)
        return (strcmp(string, "OH"));
    return (-1);
 }

Answer 1

For starters the logic of the function is wrong.

Usually such a function should return 1 (or a positive value) that corresponds to the logical true or 0 that corresponds to the logical false when it answers to a question like "yes or no".

This call

strcmp(string, "OH")

returns 0 if two strings are equal. Otherwise the function can return any positive or negative value depending on whether the first string is greater or less than the second string.

Apart from this the function parameter should have the qualifier const because the passed string is not changed within the function.

You did not reserve a memory where you are going to read a string. The declared pointer

char *string;

is uninitialized and has an indeterminate value. Thus this call

gets(string);

invokes undefined behavior.

Take into account that the function gets is an unsafe function and is not supported by the C Standard. Instead you should use the standard C function fgets.

And it will be much better if the function will be more generic. That is when it can check any supplied suffix of a string. Always try to write more generic functions. In this case they can be reusable.

Below there is a demonstrative program that shows how the function can be defined.

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

int hydroxide( const char *s, const char *suffix )
{
    size_t n1 = strlen( s );
    size_t n2 = strlen( suffix );

    return !( n1 < n2 ) && strcmp( s + n1 - n2, suffix )  == 0;
 }

int main(void) 
{
    enum { N = 100 };
    char s[N];

    while ( 1 )
    {
        printf( "Enter a String (empty string - exit): " );

        if ( fgets( s, N, stdin ) == NULL || s[0] == '\n' ) break;

        s[ strcspn( s, "\n" ) ] = '\0';

        printf( "%s\n", hydroxide( s, "OH" ) ? "true" : "false" );
    }

    return 0;
}

The program output might look like

Enter a String (empty string - exit): brogrammerOH
true
Enter a String (empty string - exit): 

Answer 2

Your function is too complicated and it returns 0 if the suffix is OH. A simpler way is to compute the length of the string and if this length is at least 2, compare the last 2 characters with 'O' and 'H':

int hydroxide(const char *string) {
    size_t len = strlen(string);
    if (len >= 2 && string[len - 2] == 'O'  && string[len - 1] == 'H')
        return -1;
    else
        return 0;
}

Furthermore, the main function has undefined behavior: string is an uninitialized char pointer: passing it to gets() will cause undefined behavior when gets() tries to write bytes to it. Note also that gets() is obsolete and has been removed from the latest version of the C Standard because there is no way to prevent a buffer overflow for sufficiently long input strings. Use fgets() instead and remove the trailing newline, if any:

Here is a modified version:

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

int hydroxide(const char *string) {
    size_t len = strlen(string);
    if (len >= 2 && string[len - 2] == 'O'  && string[len - 1] == 'H')
        return -1;
    else
        return 0;
}

int main() {
    char buf[80];
    printf("Enter String: ");
    if (fgets(buf, sizeof buf, stdin)) {
        buf[strcspn(buf, "\n")] = '\0';  // strip the newline if any
        printf("%d\n", hydroxide(string));
    }
    return 0;
}

Answer 3

Get the length of the string and check the last two characters.

int len = strlen(string);
      if(string[len-1] == 'H' && string[len-2] =='O')
          return -1;