Discover string inside string without string library. h?

Question

My function needs to check if one string is contained in another and speak where the first place where the letter was equal occurred, but is always returning the size of the first string

What’s wrong with my implementation?

my code:

void str_conteudo( char str1[], char str2[])
{  
int i=0;
int y;

while(str1[i] != str2[i] )
{
    if (str2[i] != str1[i])
    {
        y=i;
    }
    i++;

}

if(y>0)
printf("Indice do primeiro caractere que contem a string 2: %d ", y);
else
printf("A string 2 nao esta contida na string 1. \n");

}

Answer 1

In its implementation it has the logic that goes through the letters of the main sentence, but the logic that sees if each letter is equal to the letters of the second string is missing.

Trying to make the most of your logic can do so:

void str_conteudo( char str1[], char str2[]) {
    int i=0,j, k;
    int y = -1; //y começa a -1, pois estava sem valor de inicio

    while(str1[i] != '\0') { // enquanto a string principal não termina
        if (str2[0] == str1[i]) { //se a letra em que vai é igual à 1a letra da substring
            //percorrer enquanto ambas as letras são iguais e até ao fim da substring
            for (j=i+1, k=1; str1[j]!='\0' && str2[k]!='\0' && str1[j] == str2[k];++j,k++);

            if (str2[k] == '\0'){ //se a segunda foi até ao fim então achou
                y = i;
                break;
            }
        }
        i++;
    }

    if(y>=0) // >= 0 e não > 0
        printf("Indice do primeiro caractere que contem a string 2: %d \n", y);
    else
        printf("A string 2 nao esta contida na string 1. \n");
}

Testing

str_conteudo ("Frase de teste","de"); //6 
str_conteudo ("Frase de teste","teste"); //9
str_conteudo ("Frase de teste","Frase"); //0
str_conteudo ("Frase de teste","cuFra"); // não contem
str_conteudo ("Frase de teste","testem"); // não contem

See these tests in Ideone

The biggest difference to the code you had is in what was added inside the if in the while.

It’s never a good idea to write or read content. They should only return the result, giving freedom to those who call to do what they want with this result. Besides if that has in the while nor would it be necessary to change a few more things.

Taking these points that I mentioned I was switching function to something like:

int str_conteudo( char str1[], char str2[]) {
    int i=0, j, k;
    for(i = 0; str1[i] != '\0'; ++i) {
        for (j = i,k = 0;str1[j]!= '\0' && str2[k] != '\0' && str1[j] == str2[k]; ++j,k++);
        if (str2[k] == '\0'){
            return i;
        }
    }
    return -1;
}

Of course I could have switched the iteration and access to pointer syntax that would be shorter, but I chose to keep it with index to make the solution comparable and similar to yours.

Now the tests have to show the value obtained by the function, which returns -1 whenever you can’t find the substring:

printf("%d\n", str_conteudo ("Frase de teste","de")); //6
printf("%d\n", str_conteudo ("Frase de teste","teste")); //9
printf("%d\n", str_conteudo ("Frase de teste","Frase")); //0
printf("%d\n", str_conteudo ("Frase de teste","cuFra")); //-1
printf("%d\n", str_conteudo ("Frase de teste","testem")); //-1

See these tests also on Ideone

Just for reference, I leave the same example of top implemented with pointer notation, although exactly the same solution:

int str_conteudo( char *str1, char *str2) {
    char *curr1 = str1, *s1, *s2;
    while(*curr1){
        for(s1 = curr1, s2 = str2; *s1 && *s2 && *s1 == *s2; ++s1, ++s2);
        if (!*s2) return curr1 - str1;
        curr1++;
    }
    return -1;
}

As a final note, the logic you are implementing is that the native function strstr provides.