How to get array length for array of strings

Question

I want to find the number of elements in that array, but as far as I know I'm not allowed to use strlen or sizeof. strlen(array[0]) gives out 5 cause apple consists of 5 characters, but I need length to equal to 4, because the arrays contains 4 words. Any suggestions?

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

int main() {
    char array[10][100] = {"apple", "banana", "strawberry", "grapefruit"};
    int length = strlen(array[0]);
    printf("%d", length);

    return 0;
}

Iterate over the words until you hit a word with zero length (or you hit the end of the array). To avoid the need for strlen, simply test the first character of each word (0 => empty string). — jarmod
– jarmod, Commented Sep 30, 2022 at 14:26
char array[10][100]; has 10 elements. Those elements are arrays of 100 char. If you decide to initialize only 4 of them, then the number of initialized elements if 4 but the size is still 10. You don't need strlen or size; you initialized 4 of them so 4 of them are initialized. And the size remains 10. Don't compute anything at run time; this is entirely knowable at compile time, and doing any computation at runtime is wasted cycles. "As far as I know I'm not allowed to use ... sizeof". Well, that's just silly, since sizeof is the right operator if you don't want to hard-code 4 or 10 — William Pursell
– William Pursell, Commented Sep 30, 2022 at 14:31
@WilliamPursell... <pedantic>all 10 elements are initialized, there is no partial initialization in C</pedantic> — pmg
– pmg, Commented Sep 30, 2022 at 14:44
@pmg A valid point. 6 of the 10 elements are initialized to be all zero. I still don't understand the point of this question! — William Pursell
– William Pursell, Commented Sep 30, 2022 at 14:55

yano · Accepted Answer · 2022-09-30 15:04:47Z

2

You can search over array[i] until you find an empty string:

size_t arrayLength = 0;
for (size_t i=0; i<10; i++)
{
  if (array[i][0] != '\0')
  {
    arrayLength++;
  }
  else
  {
    // with brace initialization, there will be no other words in the
    // array, we're safe to break
    break;
  }
}

When you use a brace-initialization list like that with a specified size, it will initialize the array to the contents you provided, and 0 for everything else. The pointers in the first dimension will still be valid (ie, array[i] is not-NULL for all i [0, 10) ), but they point to empty strings.

Demonstration

edited Sep 30, 2022 at 15:04

answered Sep 30, 2022 at 14:16

yano

5,3852 gold badges29 silver badges41 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

pmg Over a year ago

ITYM if (array[i][0]) to check for empty string rather than NULL pointer. ... ideone.com/8Yf72w

yano Over a year ago

@pmg Absolutely correct. In fact my last answer was completely wrong,, the first dimension of pointers are valid for each array[i], but after the 4th word they don't point to anything. Whoops.

Weather Vane Over a year ago

They decay to point to the empty strings, just as with the assigned elements.

William Pursell · Accepted Answer · 2022-09-30 15:06:54Z

1

If you want to know the size of the array, the "correct" operator to use is sizeof. For example:

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

int 
main(void)
{
    char array[][100] = {"apple", "banana", "strawberry", "grapefruit"};
    printf("%zd\n", sizeof array / sizeof *array);

    return 0;
}

I suspect the real issue here is that the char array[10][100] ought to be replaced with char array[][100]. There is no need to do any run-time computation here.

answered Sep 30, 2022 at 15:06

William Pursell

214k49 gold badges279 silver badges317 bronze badges

Collectives™ on Stack Overflow

How to get array length for array of strings

2 Answers 2

3 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related