Printing string using Arrays of Pointers

Question

Printing string using Arrays of Pointers

Trying to revise for a coding exam and I came across this slide. Can any of you help me to understand how to print the strings from a pointer to array? Here is the code sample:

#include <stdio.h>

int main(void)
{
    size_t i;
    const char *suit[4] = { "Hearts", "Diammonds", "Clubs", "Spades" };

    printf("The strings in array const char*suit[4] is:\n\n");


    for (i = 0; i < 5; i++)
    {
        printf("const char *suit[%u] is: %s\n", i, suit[i]);
    }
}

When I print, I get this output:

The strings in array const char*suit[4] is:

const char *suit[0] is: Hearts
const char *suit[1] is: Diammonds
const char *suit[2] is: Clubs
const char *suit[3] is: Spades

However, it came with this error as well. "Exception thrown at 0x7AAD170C (ucrtbased.dll) in Pointer Practice.exe: 0xC0000005: Access violation reading location 0xCCCCCCCC."

Why is there an error? How do I print the same output, but without an error?

Thank you in advance guys!

c

asked on Stack Overflow Dec 8, 2020 by

Fermi Toh • edited Dec 8, 2020 by

Fermi Toh

4 Answers

A simple fix! You only have 4 elements in your array but are iterating 5 times. Trying to access suit[5] will throw a Segmentation Fault (on Linux) or an Access Violation (on Windows). These errors occur when your accessing memory you shouldn't be (if you tried to access 0xEABCC8, but you don't have access to it, you will get hit with one of these errors). Try changing your max bound to 4, instead of 5.
In C (and basically every other language) indexes start at 0 when indexing, but start at 1 when initalisation. Example:

const int foo[4] = { 6, 9, 3, 2}; // Creates an array with 4 elements.
foo[4] // Gets the fifth element since it start at zero,
foo[0] // thus this is the first element.

answered on Stack Overflow Dec 8, 2020 by

Milo Banks

The largest index is 3. Your loop is trying to access index 4 which does not exist..

Sigh.

answered on Stack Overflow Dec 8, 2020 by

Fermi Toh

As you found that, in your code, the loop which is printing suit array strings is accessing array beyond its size. To not to repeat this mistake again, follow this mechanism to get the size of an array:

size_t sz = sizeof (arr) / sizeof (arr[0]);

In your case, you should do:

    for (size_t i = 0; i < sizeof (suit) / sizeof (suit[0]); ++i)
    {
        printf("const char *suit[%zu] is: %s\n", i, suit[i]);
    }

With this, if you change the size of suit array (adding or removing array members), you don't need to make any changes in the loop printing the suit array members.

Additional:

You don't need to specify the dimension when initialising an array. When we omit the dimension, compiler computes it for us based on the size of the initialiser. So, you can do:

const char *suit[] = { "Hearts", "Diammonds", "Clubs", "Spades" };
                ^^
                 |
           omit the dimension

To check it, you can do

#include <stdio.h>

int main(void)
{
    const char *suit[] = { "Hearts", "Diammonds", "Clubs", "Spades" };

    printf ("Size of suit array : %zu\n", sizeof (suit) / sizeof (suit[0]));

    return 0;
}

Output:

Size of suit array : 4

answered on Stack Overflow Dec 8, 2020 by

H.S. • edited Dec 8, 2020 by

H.S.

As many have pointed out, you are trying to access an element in the array that is not there.

You also need to change the print line to the following, as you are printing the string itself, and you are already de referencing the pointer by suit[i].

for (i = 0; i < 4; i++)
{
    printf("const char *suit[%lu] is: %s\n", i, suit[i]);
}

answered on Stack Overflow Dec 8, 2020 by

geebert

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