how to count string array elements in C?

Question

I want my loop to repeat only as many times, as many are strings in "fruits" array.

Tried it first by using "while" loop, but I couldn't make it to work properly (now it is commented), because (as debugger showed) it segfaulted on 4th iteration. This method should work in array of chars or integers but does not work in array of strings. Why is it like this?

Then I tried to use "for" loop and wanted it to stop when reaching total number of string elements in array, stored in "count" variable. However, I could not find a good method to count number of strings in array. Is that even possible? "sizeof" operator doesn't seem like a good solution here.

Thanks in advance for response.

#include <stdio.h>

int main()
{
    char *fruits[]={"Apple", "Grapefruit", "Banana"};
    int i=0;
    int count=sizeof(*fruits);


    char **bitter=&fruits[1];

    printf("Bitter fruit is: %s", *bitter);

    puts(" ");

    printf("All fruits are: ");

    for (;i<count;i++);
    {
        printf("%s ",*(fruits+i));
    }

    /*
        while ( fruits[i] != '\0')
        {
        printf("%s ",*(fruits+i));
        }

    stuff above failed. why?
    */


    return 0;
}

Is the stuff with using for loop and counting array string elements even possible? — hardpenguin
– hardpenguin, Commented Aug 26, 2012 at 9:35

eyelash · Accepted Answer · 2012-08-26 09:33:59Z

4

An easy way would be to append a NULL to your list:

char *fruits[] = {"Apple", "Grapefruit", "Banana", NULL};

Then you can print all fruits like this:

int i = 0;
while (fruits[i] != NULL) {
    printf ("%s ",*(fruits+i));
    i++;
}

As to why your original while loop failed:

your i doesn't get incremented
you are checking if fruits[i] != '\0' where fruits[i] is a pointer to char and '\0' is a char equal to 0. So you are essentially checking if fruits[i] points to 0 but for the first three iterations this is not the case as fruits[i] points to the first character of the respective fruit and in the fourth iteration fruits[i] points to a memory location that doesn't belong to your program.

edited Aug 26, 2012 at 9:33

answered Aug 26, 2012 at 9:19

eyelash

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2 Comments

Paul R Over a year ago

Good, but it's probably cleaner to write fruits[i] rather than *(fruits+i).

hardpenguin Over a year ago

Thanks @eyelash. @PaulR *(fruits+i) is there just because I am recently trying to get familiar with pointers.

Varun · Accepted Answer · 2017-03-21 12:38:08Z

0

#include <stdio.h>
int main()
{
    char *fruits[]={"Apple", "Grapefruit", "Banana"};
    int i=0;
    int count=sizeof(fruits)/sizeof(fruits[0]);
    char **bitter=&fruits[1];

    printf("Bitter fruit is: %s\n", *bitter);
    printf("All fruits are:\n");
    for(i=0;i<count;i++)
    {
        printf("%s\n",fruits[i]);
    }
    return 0;
}

answered Mar 21, 2017 at 12:38

Varun

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1 Comment

NoCake Over a year ago

This doesn't work because (sizeof(fruits[0])) is smaller than sizeof(fruits[1]), etc... So there is no uniform size. Your approach would work with an array of fixed size elements, such as int numbers[], etc...

Mr'Black · Accepted Answer · 2016-08-04 13:49:32Z

-1

You can also consider using the strlen() from header string.h :

#include <stdio.h>
#include <string.h>
int main()
{
    char *myarray[] = {"Mere", "Pere", "Gutui"};
    int i = 0;
    int lungime = strlen(*myarray);

    while (i < (lungime - 1)){
            printf("Fruit no. %d is %s\n", i, *(myarray + i));
            i++;
    }
    return 0;
}

answered Aug 4, 2016 at 13:49

Mr'Black

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1 Comment

user694733 Over a year ago

-1 This is completely wrong. You are using length of the string "Mere" as a length of myarray. strlen does not return length of an array. It returns length of nul terminated string.

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