[Solved] Compare two character strings in C [closed]

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Ok, just to put an end to this. Let’s think what string comparing is. First, let’s disregard NULL pointers, and assume that both strings are actually valid strings. Then we need no special check.

We will do this by simply comparing them element for element. Whenever we compare two elements at the same index, one of three things can happen:

  1. Both elements are '\0'. The strings are equal. Return true.
  2. The elements differ. Stop comparing and return false.
  3. The elements are equal, but not '\0'. Continue comparing.

Note that the case where “one string has ended” is included in case 2. If only one string has ended, then we will compare '\0' with something else, and thus they will differ.

To make the code very easy to write, you can check if the length is the same first. It makes it a little bit easier, since if they have different length, then they are different, and if they have the same length, you can loop from zero to that length. You could then write it very simple like this:

bool is_equal(char str1[], char str2[])
{
    if(strlen(str1) != strlen(str2))
        return false;

    for(int i=0; i<strlen(str1); i++) {
        if(str1[i] != str2[i])
            return false;
    }

    return true;
}

Most important thing here that differs from your code is the realization that as soon as you discover ANY difference, you can return false. The above code is very clear, but a bit inefficient. An obvious optimization is this:

bool is_equal(char str1[], char str2[])
{
    int len1 = strlen(str1);
    if(len1 != strlen(str2))
        return false;

    for(int i=0; i<len1; i++) {
        if(str1[i] != str2[i])
            return false;
    }

    return true;
}

This removes unnecessary length check in the loop, but it looks messier. Actually, I think I would prefer this:

    int len1 = strlen(str1);
    int len2 = strlen(str1);
    if(len1 != len2)
        return false;

because it looks cleaner. But it’s an extra line and an extra variable. But since we can combine the length check with the character check, we can get rid of that completely.

So how can we incorporate the length check in the character check? Well, like this. I want to point out that !str[i] and str[i] != '\0' is completely equivalent in this situation, and it’s a very common way to check end of strings.

bool is_equal(char str1[], char str2[])
{
    // Infinite loop, but it does not matter. It will end when it should.
    for(int i=0; ; i++) {
        // If the strings are different, they are different
        if(str1[i] != str2[i])
            return false;
        // At this stage we know that they are equal for all elements
        // checked so far, so if both strings ends here, then they are equal.
        if(!str1[i] && !str2[i])
            return true;
    }
}

This is not how I would have implemented it, and I omitted a few details. For instance, I would have used size_t instead of int for the index variable and const for the pointers. It’s intended to show how it works in a very simple way. If I implemented it, I would do something like this:

bool is_equal(const char *str1, const char *str2)
{
    while(true) {
        if(*str1 != *str2)
            return false;
        if(!*str1 && !*str2)
            return true;
        str1++;
        str2++;
    }
}

If you want to work with index variables or pointers is often just a matter of taste. Personally, I think pointers is a bit clearer in this situation. I would also consider adding these lines in the beginning, before the loop:

assert(str1);
assert(str2);

Those will make the program crash if you feed them a NULL pointer.

Demonstration:

https://onlinegdb.com/r1XvzLR3I

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solved Compare two character strings in C [closed]