Question

Possible Duplicate:
nul terminating a int array

I’m trying to print out all elements in an array:

int numbers[100] = {10, 9, 0, 3, 4};
printArray(numbers); 

using this function:

void printArray(int array[]) {
    int i=0;
    while(array[i]!='\0') {
        printf("%d ", array[i]);
        i++;
    }
    printf("\n");
}

the problem is that of course C doesn’t differentiate between just another zero element in the array and the end of the array, after which it’s all 0 (also notated \0).

I’m aware that there’s no difference grammatically between 0 and \0 so I was looking for a way or hack to achieve this:

10 9 0 3 4

instead of this

10 9

The array could also look like this: {0, 0, 0, 0} so of course the output still needs to be 0 0 0 0.

Any ideas?

Answer 1

Don’t terminate an array with a value that could also be in the array.

You need to find a UNIQUE terminator.

Since you didn’t indicate any negative numbers in your array, I recommend terminating with -1:

int numbers[100] = {10, 9, 0, 3, 4, -1};

If that doesn’t work, consider: INT_MAX, or INT_MIN.

As a last resort, code a sequence of values that are guaranteed not to be in your array, such as: -1, -2, -3 which indicates the termination.

There is nothing “special” about terminating with 0 or \0. Terminate with whatever works for your case.

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If your array truly can hold ALL values in ANY order, then a terminator isn’t possible, and you will have to keep track of the length of the array.

From your example, this would look like:

int numbers[100] = {10, 9, 0, 3, 4};
int Count = 5;
int i;

for(i=0; i<Count; ++i)
{
    // do something with numbers[i]
}