Comparing Characters to Strings

Question

I'm to write a code in C to let the user enter the first and last characters of the word "fantastic". If he is correct for the two answers it should print "Well done", if he gets one wrong then it should print "one of your answers is incorrect" if he gets both incorrect i tell him to try again later.

Below is the code I tried which doesn't allow me to enter the second character and also gets the answer wrong.

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

int main()
{
    char word[] = "fantastic";
    char in1, in2;
    printf("Please enter the first letter of the word \'fantastic\': ");
    scanf("%c", &in1);
    printf("\nPlease enter the last letter of the word \'fantastic\': ");
    scanf("%c", &in2);
    if (in1 == word[0], in2 == word[8]) {
        printf("\nWell Done!\n");
    }
    else if (in1 == word[0], in2 != word[8], in1 != word[0], in2 == word[8]) {
        printf("\nOne of your answers is incorrect!\n");
    }
    else {
        printf("\nTry again next time!\n");
    }
    return 0;
}

You have to use logical operators like `&&` (and) or `||` (or) instead of the comma operator in the conditions of your `if`-statements. — Swordfish, Nov 21 '18 at 20:17
What would be the acted upon condition in `if (in1 == word[0], in2 == word[8])`? `in2 == word[8]` only? — Fiddling Bits, Nov 21 '18 at 20:18
@Swordfish thanks, but I would be much happier if I get a sample code of it — 3D0N, Nov 21 '18 at 20:23
Sorry, but you cannot write code at random trying to guess the syntax until it compiles. the commas aren't doing what you expect — Jean-François Fabre, Nov 21 '18 at 20:29
I'm tempted to close as dupe of https://stackoverflow.com/questions/52550/what-does-the-comma-operator-do — Jean-François Fabre, Nov 21 '18 at 20:30
It isn't necessary to define this: `char word[] = "fantastic";`. Instead of `word[0]` and `word[8]` you can use `'f'` and `'c'`. — Fiddling Bits, Nov 21 '18 at 20:33
worse than a dupe: a "fix my code" question. or composite duplicate. Since the scanf issue is also a well known dupe — Jean-François Fabre, Nov 21 '18 at 20:38
but I've always answered stuff like this in the C section when I was "younger". So... — Jean-François Fabre, Nov 21 '18 at 20:39

Swordfish · Answer 1 · 2018-11-21T20:51:52.073

code I tried which doesn't allow me to enter the second character

After

scanf("%c", &in1);

The newline ('\n') you entered after the character is still in the input buffer and will get consumed by

scanf("%c", &in2);

so you have no opportunity to enter another one. You can change both to

scanf(" %c" // ...
//     ^

to skip leading whitespace.

also gets the answer wrong

The next problems are with your if-statements: If you use the comma-operator between the tests for equality, the result for the whole expression will only be the rightmost subexpression:

in1 == word[0], in2 == word[8]

evaluates to in2 == word[8]. The part before the comma has no effect on it.

When you have two boolean expressions like tests for equality you have to link them with a logical and (&&) to make the whole expression true when both sides of && are true:

in1 == word[0] && in2 == word[8]

Similarly you should use a logical or (||) if you want the whole expression be true when at least one side of || is true.

So your code should look like that:

if (in1 == word[0] && in2 == word[8]) {
    printf("\nWell Done!\n");
}
else if (in1 == word[0] && in2 != word[8] || in1 != word[0] && in2 == word[8]) {
    printf("\nOne of your answers is incorrect!\n");
}
else {
    printf("\nTry again next time!\n");
}

As suggested by Fiddling Bits you don't have to define an array for the word if you have no other use for it other than checking for two letters:

char const first = 'f';
char const last  = 'c';

if (in1 == first && in2 == last) {
    printf("\nWell Done!\n");
}
else if (in1 == first && in2 != last || in1 != first && in2 == last) {
    printf("\nOne of your answers is incorrect!\n");
}
else printf("\nTry again next time!\n");

Less comparisons and logic is needed if you swap the order:

if (in1 == first && in2 == last) {  // both correct
    printf("\nWell Done!\n");
}
else if (in1 != first && in2 != last) {  // both wrong
    printf("\nTry again next time!\n");
}
else printf("\nOne of your answers is incorrect!\n");

thanks so much, I've tried all your comments and I think its working... — 3D0N, Nov 21 '18 at 20:51

score 0 · Answer 2 · answered Nov 21 '18 at 20:57

I tried your suggestions and it's working out Thanks so much..... Below is the final code which is working:

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

int main()
{
char word[] = "fantastic";
char in1, in2;
printf("Please enter the first letter of the word \'fantastic\':");
scanf(" %c", &in1);
printf("Please enter the last letter of the word \'fantastic\':");
scanf(" %c", &in2);
if (in1 == word[0] && in2 == word[8]) {
printf("\nWell Done!\n");
}
else if (in1 == word[0] && in2 != word[8] || in1 != word[0] && in2 == word[8]) {
printf("\nOne of your answers is incorrect!\n");
}
else {
printf("\nTry again next time!\n");
}
return 0;
}

Comparing Characters to Strings

2 Answers2