Checking if two char array elements are the same c

Question

I've got some code in my program that essentially does this

char array[3] = {'g', 'w', 'a'};
char array2[3] = {'h', 'o', 'd'};
int i; 
int b = 1;
for (i = 0; i<=2; i++){
    if (array[i] != array2[i]){
        b = 0;
    }
}
if (b == 1){
    printf("true");
}
else{
    printf("false");
}

}

When I run it it outputs true no matter what values are in the arrays, even if they are completely different. Why aren't they being compared properly?

Here's the actual code because people can't seem to reproduce my error.

#include <stdio.h>
#include <string.h>
int compare(char word1[], char word2[]){
    int i, k, same = 1, j, a, b;

    for (i =0; i<=79; i++){
        for (k=i; k <= 79; k++){
            if (word1[i] > word1[k]){
                char temp = word1[i];
                word1[i] = word1[k];
                word1[k] = temp;
            }
        }
    }


    for (a =0; a<=79; a++){
        for (b=a; b <= 79; b++){
            if (word2[a] > word2[b]){
                char temp = word2[a];
                word2[a] = word2[b];
                word2[b] = temp;
            }
        }
    }

    for (j =0; j <= 79; j++){
        putchar(word1[j]);
        putchar(word2[j]);
        putchar('\n');
        if (word1[j] != word2[j]){
            same = 0;
        }
    }
    if (same = 1){
        printf("Anagrams");
    }
    else{
        printf("Not Anagrams");
    }
}

int split(char array[]){
    int a, b, c, second = 0, count = 0;
    char word2[80] = "", word1[80] ="";
    for(a=0; a <= 79; a++){
        if (array[a] != 0){
            if (array[a] == ' '){
                second = 1;
                count = 0;
            }
            else{  //add to array
                if (second == 0){
                    word1[count] = array[a];
                    count ++;
                }
                else{
                    word2[count] = array[a];
                    count ++;
                }
            } 
        }
        else{
            break;
        }
    }
    compare(word1, word2);
    putchar('\n');
    return 0;
}
int main(){
    char temp, words[80] = "";
    int count = 0;
    while ((temp=getchar()) != EOF){
        if (temp == '\n'){
            split(words);
            memset(words, 0, 80);
            count = 0;
        }
        else{
            words[count] = temp;
            count ++;
        }
    }
    if (count > 0){
        split(words);
    }
    return 0;
}

@TimBiegeleisen The code is fine until and unless he use arrays as string . — ameyCU
– ameyCU, Commented Oct 9, 2015 at 4:07
@Jay We need to see code that reproduces the problem otherwise there's no point asking. The code you've shown us looks right. — Paul Boddington
– Paul Boddington, Commented Oct 9, 2015 at 4:08

R Sahu · Accepted Answer · 2015-10-09 04:15:32Z

7

I suspect the problem is here:

if (same = 1){

That should have been

if (same == 1){

Using -Wall in gcc reveals the problem quickly.

cc -Wall -std=c99     soc.c   -o soc
soc.c: In function ‘compare’:
soc.c:35:5: warning: suggest parentheses around assignment used as truth value [-Wparentheses]
     if (same = 1){

answered Oct 9, 2015 at 4:15

R Sahu

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3 Comments

Jeff H Over a year ago

Wow I am the dumbest guy ever. That solved it, thanks for the help.

R Sahu Over a year ago

@Jay, don't feel bad. You are not the first to make that mistake and certainly won't be the last.

dave Over a year ago

@Jay: people say that C is a risky language that lets you make these kinds of errors, but no one who uses C seriously compiles without -Wall (and probably even -Werror), you will do well to use them, especially if you are learning.

LearningCODE · Accepted Answer · 2015-10-09 04:19:39Z

3

logic error, replace if(same = 1) to if(same).

answered Oct 9, 2015 at 4:19

LearningCODE

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1 Comment

LearningCODE Over a year ago

@immibis Correct, but a lot of matured 'c' programmers tend to use if(same) just because you don't care what value `same' holds, as long as its not 0...

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