Why does my running function only return "0"?

Question

I’m doing a function that gets a string (char str[10]) and she has to return 1 if to string contain the word "wing" and otherwise return 0.

To make it easier to understand what I tried to do, I’ll show you an example:

0 1 2 3 4 5 6 7 8 9

x s a s a t i \0

In this case, the function has to return 1 because the word "wing" is contained in the string received (positions 2, 3 and 4).

The code I made:

#include <stdio.h>
int vetorPalavra(char str[10]){
    int i, cont;
    for(i=0;i<10;i++){
        if((str[i] == 'a') && (str[i] == 's')){
            return 1;
        }
        else{
            return 0;
        }
    }
}
main(){
    fflush(stdin);
    char str[10] = {'x','d','m','a','s','a','h','k','p','g'};
    printf("%i", vetorPalavra(str));
}

In function main(), note that I have already entered the values of the vector. Then it is not necessary to insert anything when executing. It will already display the direct value (0 or 1). However, when executing it only shows the value 0, when it was to show 1. What I did wrong?

I’ve tried to use fflush(stdin), but it didn’t do any good. I thought about using an accountant but I don’t think this exercise would work.

Answer 1

Failed to compare with the letter a and walk on the index in each letter. The a in an element of array, to be valid the next element should be s, and the following must be a. The question code is comparing to the same element, which is impossible for it to be two different things at the same time, but it’s not even that intention, it has to compare with three elements in sequence at a time.

Another problem is that if he does not find at the first opportunity closes the search saying that he does not have, and this should only happen when he searches throughout the string, otherwise you wouldn’t even need a bow.

It’s all about interpreting the problem correctly and translating it into code.

This algorithm only works with strings formed correctly and with the expected size (can have a small variation), according to what is said in the statement. The code does not create a string formed correctly. To make it work with a array of characters that is not a string the search must end 2 characters before to not pick up memory junk to compare. As optimization the new code already does this.

Organizing and fixing other minor problems:

#include <stdio.h>

int vetorPalavra(char str[10]) {
    for (int i = 0; i < 7; i++) if (str[i] == 'a' && str[i + 1] == 's' && str[i + 2] == 'a' ) return 1;
    return 0;
}
int main() {
    printf("%d", vetorPalavra("xdmasahkp"));
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

Just out of curiosity whether the comparison should really be just with the beginning of string and not in all of it, the function could have only this line:

return str[0] == 'a' && str[1] == 's' && str[2] == 'a';