CODEWITHSUNDEEP
carraysstringchar
nevzatseferoglu
nevzatseferoglu

Reputation: 1186

Confusing output of two different Char Array

Problem : Although I declared two char strings , whose contents are the same , Outputs are different.

#include <stdio.h>

int main(void)
{
    /* Initialization of two different array that We deal with */

    char arr1[10]={'0','1','2','3','4','5','6','7','8','9'};
    char arr2[10]="0123456789";

    /* Initialization End */

    for(int i = 0 ; i < 11 ; ++i)
    {
        printf("arr1[%d] is %c \t\t",i,arr1[i]);
        printf("arr2[%d] is %c\n",i,arr2[i]);

        if(arr1[i]=='\0')
            printf("%d . character is \\0 of arr1 \n",i);

        if(arr2[i]=='\0')
            printf("%d . character is \\0 of arr2 \n",i);
    }

    return 0;
}

Expectation : I expected that both if statements are going to be true for any kind of value of 'i'.

Output : It is an output that I got it.

arr1[0] is 0        arr2[0] is 0
arr1[1] is 1        arr2[1] is 1
arr1[2] is 2        arr2[2] is 2
arr1[3] is 3        arr2[3] is 3
arr1[4] is 4        arr2[4] is 4
arr1[5] is 5        arr2[5] is 5
arr1[6] is 6        arr2[6] is 6
arr1[7] is 7        arr2[7] is 7
arr1[8] is 8        arr2[8] is 8
arr1[9] is 9        arr2[9] is 9
arr1[10] is 0       arr2[10] is 
10 . character is \0 of arr2

Upvotes: 3

Views: 277

Answers (3)

hrishi007
hrishi007

Reputation: 133

when a character array is initialized with a double quoted string and array size is not specified, compiler automatically allocates one extra space for string terminator ‘\0’

Ref

Upvotes: 0

Morten Jensen
Morten Jensen

Reputation: 5936

There are a few small things wrong with your code and the assumptions you seem to make about it.

1. These two declarations are not the same

char arr1[10]={'0','1','2','3','4','5','6','7','8','9'};
char arr2[10]="0123456789";

The second line is equal to this:

char arr2[10]={'0','1','2','3','4','5','6','7','8','9', 0x00};

... which defines an array containing 11 elements. Check out implicit zero-termination for string literals.

EDIT: I'm getting quite a lot of down-votes for this point specifically. Please see Lundin's comment below, which clarifies the issue.

2. Your for-loop iterates over 11 elements

for(i=0 ; i<11 ;++i)

The loop above goes through i = 0..10, which is 11 elements.... but you only wanted to compare the first 10 right?

You could change your loop to only compare the first ten elements [for(i = 0; i < 10; ++i)] and that would make your program work as you expect.

Because of what it seems you are assuming, I would recommend reading up on strings in C, array-indices and undefined behavior.

Upvotes: 3

Lundin
Lundin

Reputation: 213513

Both cases invoke undefined behavior by accessing the array out of bounds. You cannot access index 10 of an array with items allocated from index 0 to 9. Therefore you need to change the loop to i<10 or anything might happen. It just happened to be different values printed - because you have no guarantees of what will be printed for the byte at index 10.

In both examples, there is no null terminator, so they are equivalent. Due to a subtle, weird rule in the C language (C17 6.7.9/14 emphasis mine):

An array of character type may be initialized by a character string literal or UTF−8 string literal, optionally enclosed in braces. Successive bytes of the string literal (including the terminating null character if there is room or if the array is of unknown size) initialize the elements of the array.

Normally when trying to store too many initializes inside an array, we get a compiler error. But not in this very specific case with a string literal initializer, which is a "language bug" of sorts. Change to char arr2[9]="0123456789"; and it won't compile. Change to char arr2[11]="0123456789"; and it will work just fine, even when iterating over 11 elements.

Upvotes: 6

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