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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
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  • 9
    They aren't the same. "0123456789" automatically includes the null-terminator, but your manually constructed string doesn't. Commented Feb 4, 2019 at 10:46
  • 3
    array[10] is the 11th element in the array. Commented Feb 4, 2019 at 10:47
  • 6
    @SeptemberSKY Contents are NOT the same :) Listen to @Blaze - the string literal has an automatically-appended 0-terminator. E.g. the 11th element of arr2 at idx 10 is 0x00, 0, '\0'. You are wrongly declaring arr2 to hold 10 characters, but your initializer-string contains 11 (including the implicit zero-terminator) Commented Feb 4, 2019 at 10:51
  • 3
    In arr1[10] you are invoking Undefined Behaviour (reading beyond the array size). Probably the '0' you are reading is the first element in the second array arr2, which is likely being stored just after arr1 Commented Feb 4, 2019 at 10:54
  • 4
    @MortenJensen Yes the string literal contains a null terminator but the array char[10] has no room to store it. Making the examples equivalent. Sander is correct. Commented Feb 4, 2019 at 11:42

3 Answers 3

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6

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.

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2 Comments

@Victor This is actually almost a FAQ. I too learned this from SO, just 4 years ago: stackoverflow.com/questions/31296727/…
Re “But not in this very specific case with a string literal initializer,…”: The rule prohibiting too many initializers does apply to string literals. The 6.7.9 14 text quoted in the answer does not tell us that a string literal is permitted to break this rule or that the rule does not apply to them. Rather, it tells us that a string literal provides its explicit characters as initializers and provides its terminating null character if it fits. If it does not fit, it is not considered to be an initializer, and the rule in 6.7.9 2 is satisfied (if the explicit characters fit).
3

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.

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13 Comments

You are right that the string literal takes 11 characters, but it is used to initialize an array holding only 10 bytes. The terminator is lost here.
The first point is not an issue. The contents of arr1 and arr2 will be the same (arr2 only contains the first 10 characters from the string literal, not the null terminator). The real issue is the undefined behavior from point 2.
but it does : the OP assumes that the contents of the two arrays are the same, and that assumption is correct.
Neither are well-defined and your point 1) is wrong. char arr2[10]="0123456789"; is equal to char arr2[10]={'0','1','2','3','4','5','6','7','8','9'} without null terminator. This because of a "language bug" in C that allows a string literal with a size exactly matching the one specified to drop the null termination.
your belief that such initialization is undefined behavior is wrong : "An array of character type may be initialized by a character string literal, optionally enclosed in braces. Successive characters of the character 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."
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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

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