Are are elements of an empty array NULL valued?-CodePudding

I tried this Armstrong program but found myself entangled in this empty array thing. Working of this program has been bothering me for a while now and still can't seem to figure it what's wrong here. Yea, so just wanted to ask what value are the elements of an empty or incomplete array assigned to? Is it the NULL character i.e. '\0'? I tried checking it on an online C compiler where this assertion seems true but GCC tells us the opposite. I tried this approach to Armstrong problem and here's my code:

#include <stdio.h>
#include <math.h>

int main()
{
    int num,i,cub,j;
    i = cub = 0;
    int sto[20];
    scanf("%d",&num);
    j = num;
    while(num != 0) 
    {
        sto[i] = num%10;
        num = num / 10;
        i  ;
    }
    i = 0;
    while(sto[i] != '\0')
    {
       cub  = pow(sto[i],3);
       i  ;
    }
    num = j;
    printf("cub: %d     num: %d\n\n",cub,num);
    if(j == cub)
      printf("The number is an Armstrong number");
    else 
      printf("The number is not an Armstrong number");

    return 0;
}

I know there is other approach to this problem but what I am looking for is the answer to the above mentioned question.

CodePudding user response：

The values in that array are going to be undefined. They might be zero, they might be whatever values happened to be there from when that portion of the stack was last used. The same principle is applicable to heap memory.

CodePudding user response：

so just wanted to ask what value are the elements of an empty or incomplete array assigned to? Is it the NULL character i.e. '\0'?

No, when declared in the body of a function as you've done it here (i.e. with no initializer), the contents are undefined. In practice, the values will be whatever random data happens to have last been in the stack at those locations. When an "automatic" array is declared as you've done here, the compiler simply inserts a stack increment to reserve space, but does nothing else. It could be zero, but probably not.

CodePudding user response：

You have declared an array in a block scope without storage specifier static

int sto[20];

such an array has automatic storage duration. That is it will not be alive after exiting the block where it is defined.

If you would declare the array in the file scope for example before the function main or with the storage specifier static in a block it would have static storage duration.

From the C Standard (6.7.9 Initialization)

10 If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate. If an object that has static or thread storage duration is not initialized explicitly, then:

— if it has pointer type, it is initialized to a null pointer;

— if it has arithmetic type, it is initialized to (positive or unsigned) zero;

— if it is an aggregate, every member is initialized (recursively) according to these rules, and any padding is initialized to zero bits;

— if it is a union, the first named member is initialized (recursively) according to these rules, and any padding is initialized to zero bits;

CodePudding user response：

C doesn't have an end marker, the Programmer's supposed to handle it. Refer to this link:End of array in C language

Suggestion: Add an end marker in your array, for instance, insert 'x' at the end of your array, and loop until you find the marker. Or just use a counter variable as in this code.

 #include <stdio.h>
    #include <math.h>
    
    int main()
    {
        int num,i,cub,j;
        i = cub = 0;
        int sto[20];
        scanf("%d",&num);
        j = num;
        while(num != 0) 
        {
            sto[i] = num%10;
            num = num / 10;
            i  ;
        }
        int counter = 0;
        while(counter < i)
        {
           cub  = pow(sto[counter],3);
           counter  ;
        }
        num = j;
        printf("cub: %d     num: %d\n\n",cub,num);
        if(j == cub)
          printf("The number is an Armstrong number");
        else 
          printf("The number is not an Armstrong number");
    
        return 0;
    }

I hope this answers your question.

CodePudding user response：

The initial values of sto are indeterminate - they could be pretty much anything, and they can be different each time you run that code.

If you want to make sure all the elements contain a specific value, you can either:

explicitly initialize all 20 elements;
explicitly initialize some elements, with the remainder implicitly initialized to 0;
declare the array with the static keyword to implicitly initialize all elements to 0 (although this changes how the array is allocated and stored);

If you want all elements to initially be 0, you could do one of the following:

int sto[20] = {0}; // explicitly initializes first element to 0, implicitly 
                   // initializes remaining elements to 0

static int sto[20]; // implicitly initializes all elements to 0, *but*
                    // changes how sto is stored

If you want all the elements to be initialized to something other than 0, then you'll either need to initialize all the elements explicitly:

int sto[20] = {-1, -1, -1, -1, -1, ... };

or use a loop:

for ( size_t i = 0; i < 20; i   )
  sto[i] = -1; // or whatever initial value