CODEWITHSUNDEEP
cpointersreverse
Adonis geek
Adonis geek

Reputation: 33

How does this reverse() function work in the below code?

I found this code on the Internet which reverses strings in C using pointers. I understand the most of it but I don't understand the reverse() function: why it is using length / 2?

If I change the condition in second for loop to length in reverse() function, it also displays the same output.

// function to reverse the string s which is an array of some size
void reverse(char *s) {
    int length, c;
    char *begin = NULL, *end = NULL, temp;

    length = string_length(s);
    begin = s;
    end = s;

    for (c = 0; c < length - 1; c++)
        end++;

    for (c = 0; c < length / 2; c++) {
        temp = *end;
        *end = *begin;
        *begin = temp;

        begin++;
        end--;
    }
}

int string_length(char *pointer) {
    int c = 0;

    while (*(pointer + c) != '\0')
        c++;

    return c;
}

Upvotes: 0

Views: 129

Answers (3)

chqrlie
chqrlie

Reputation: 144780

Let's use a trivial comparison:

  • you have a ball in each hand, say a black one in the right hand and a white one the left hand.
  • if you swap the balls once, you have effectively reversed the situation.
  • but if you swap the balls twice, you are back in the original situation.
  • hence the number of swaps must be half the number of balls.

The same goes for the letters in the string, if you perform length swaps, you will get the original string. To get a reversed string, you must stop at half the length of the string.

Note also that the code is more complicated than required. Here is a simplified version:

// function to reverse the string s which is an array of some size
void reverse(char *s) {
    char *begin, *end;

    for (end = begin = s; *end; end++)
        continue;

    while (begin < end) {
        char temp = *begin;
        *begin++ = *--end;
        *end = temp;
    }
}

Upvotes: 2

schnaader
schnaader

Reputation: 49719

Have a look at what is done in the loop:

temp = *end;
*end = *begin;
*begin = temp;

This is some typical swap code that here swaps two characters *begin and *end of a string. The pointers iterate from the beginning and the end of the string and approach the middle.

Using the minimal example abcd, let's see how many characters we have to swap:

abcd (Original string)
dbca (Swapped 'a' and 'd') - iteration 1
dcba (Swapped 'b' and 'c') - finished after 2 iterations

So you only need to swap half of the characters (length / 2 iterations) because you change two of them with each operation. This also explains why you get the original string when changing to length - 1 iterations:

dcba (result from above after 2 iterations)
dbca (Swapped 'c' and 'b') - iteration 3
abcd (Swapped 'd' and 'a') - original string after 4 iterations

Upvotes: 0

klutt
klutt

Reputation: 31389

Because when c is 0, you will swap the elements at s[0] and s[length-1]. When c is 1, you will swap the elements at s[1] and s[length-2] and so on. If you used length instead of length/2 you would reverse the string and then reverse it back to the original.

Upvotes: 0

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