array of array of strings in c program

Question

It seems I'm confused with how to use array of strings in c code. Here, what I wanted to do is making a master_array which includes all other array of strings in.

I made a code like below. And the output is

 arr1[1] = pear
 arr2[1] = lettuce
 arr3[1] = Smith
 master_array[1][2] = onion
 sizeof(arr1) = 4
 sizeof(master_array[0]) = 1

Except for the last one, everything was as expected. I expected sizeof(master_array[0]) = 4. But the result was 1.

Would someone point out what I am confused with?

#include <stdio.h>

int main(void)
{
   char *arr1[] = {"apple", "pear", "banana", "melon"};
   char *arr2[] = {"carrot", "lettuce", "onion", "spinach"};
   char *arr3[] = {"Tom", "Smith", "Dave", "John"};
   char **master_array[] = {arr1, arr2, arr3};

   printf("arr1[1] = %s\n", arr1[1]);
   printf("arr2[1] = %s\n", arr2[1]);
   printf("arr3[1] = %s\n", arr3[1]);
   printf("master_array[1][2] = %s\n", master_array[1][2]);

   printf("sizeof(arr1) = %ld\n", sizeof(arr1)/sizeof(char *));
   printf("sizeof(master_array[0]) = %ld\n", sizeof(master_array[0])/sizeof(char *));

   return 0;
}

I found one line missing. Now it has everything.

Cprogrammer
– Cprogrammer

2022-09-12 03:52:45 +00:00
Commented Sep 12, 2022 at 3:52 — Cprogrammer
– Cprogrammer, Commented Sep 12, 2022 at 3:52

Bill Lynch · Accepted Answer · 2022-09-12 03:53:58Z

1

arr1 has a type of char*[4].

However, master_array[0] has a type of char **, not char*[4].

answered Sep 12, 2022 at 3:53

Bill Lynch

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2 Comments

Cprogrammer Over a year ago

Does it mean I cannot get the number of elements of sub array using master_array?

ash Over a year ago

Correct, you cannot figure out the size of an array unless it is statically allocated, and you use the literal name of that variable. Because everywhere else, an array is just a pointer to the first element.

ash · Accepted Answer · 2022-09-12 04:01:37Z

1

This line misleads:

printf("sizeof(master_array[0]) = %ld\n", sizeof(master_array[0])/sizeof(char *));

Because it says it is printing sizeof master_array[0] but is isn't. It is printing size of master_array[0] / sizeof(char *).

It helps to realize that C doesn't really have strings - instead String functions expect a pointer to the first character of a string, and for the string to be terminated with the null character (byte value 0).

answered Sep 12, 2022 at 4:01

ash

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