C iterate through char array with a pointer

Question

I am very new in C and was wondering about how to get each element of an array using a pointer. Which is easy if and only if you know the size of the array. So let the code be:

#include <stdio.h>

int main (int argc, string argv[]) {
    char * text = "John Does Nothing";
    char text2[] = "John Does Nothing";

    int s_text = sizeof(text); // returns size of pointer. 8 in 64-bit machine
    int s_text2 = sizeof(text2); //returns 18. the seeked size.

    printf("first string: %s, size: %d\n second string: %s, size: %d\n", text, s_text, text2, s_text2);

    return 0;
}

Now I want to determine the size of text. to do this, I found out, that the String will end with a '\0' character. So I wrote the following function:

int getSize (char * s) {
    char * t; // first copy the pointer to not change the original
    int size = 0;

    for (t = s; s != '\0'; t++) {
        size++;
    }

    return size;
}

This function however does not work as the loop seems to not terminate.

So, is there a way to get the actual size of the chars the pointer points on?

This unfortunately doesn't change anything in the result. doesn't matter whether I use s != '\0' or *s != '\0' or t != '\0' or *t != '\0', it ends up still not terminating… — christopher westburry
– christopher westburry, Commented Jan 21, 2018 at 13:06
Your function doesn't use s, so what's the point in preserving "the original" value? — melpomene
– melpomene, Commented Jan 21, 2018 at 13:08

mdatsev · Accepted Answer · 2018-01-21 13:07:06Z

22

Instead of checking the pointer you have to check the current value. You can do it like this:

int getSize (char * s) {
    char * t; // first copy the pointer to not change the original
    int size = 0;

    for (t = s; *t != '\0'; t++) {
        size++;
    }

    return size;
}

Or more concisely:

int getSize (char * s) {
    char * t;    
    for (t = s; *t != '\0'; t++)
        ;
    return t - s;
}

answered Jan 21, 2018 at 13:07

mdatsev

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2 Comments

christopher westburry Over a year ago

Only problem with this one is, it returns always one char less. So adding +1 in the end works. Only if the string isn't empty to begin with. But thanks! Ima work around it now :D

David C. Rankin Over a year ago

Also, if you are using an empty loop, make it easier to detect by using empty-braces, e.g. for (t = s; *t; t++) {} (you at least moved the ';' to a line of its own -- that was good)

Vlad from Moscow · Accepted Answer · 2018-01-21 13:37:06Z

3

There is a typo in this for loop

for (t = s; s != '\0'; t++) {
            ^^^^^^^^^

I think you mean

for (t = s; *t != '\0'; t++) {
            ^^^^^^^^^

Nevertheless in general the function does not provide a value that is equivalent to the value returned by the operator sizeof even if you will count also the terminating zero. Instead it provides a value equivalent to the value returned by the standard function strlen.

For example compare the output of this code snippet

#include <string.h>
#include <stdio.h>

//...

char s[100] = "Hello christopher westburry";

printf( "sizeof( s ) = %zu\n", sizeof( s ) );
printf( "strlen( s ) = %zu\n", strlen( s ) + 1 );

So your function just calculates the length of a string.

It would be more correctly to define it the following way (using pointers)

size_t getSize ( const char * s ) 
{
    size_t size = 0;

    while ( *s++ ) ++size;

    return size;
}

edited Jan 21, 2018 at 13:37

answered Jan 21, 2018 at 13:11

Vlad from Moscow

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1 Comment

Chukwujiobi Canon Over a year ago

This should be the accepted answer. size_t getSize ( const char * s ) { size_t size = 0; while ( *s++ ) ++size; return size; } it works even with pointer to any other container with contiguous memory as long as *s is a pointer to the first element and the elements are stored in contiguous memory such that the end of the container produces false.

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