Loop through an array with pointers?

Question

I am trying to learn pointers which is a real PITA when you don't quite fully understand it.

I am trying to print all the elements in the array but only the first element is printed.

#include <stdio.h>

int count(const int* numbers, int size)
{
    for(; numbers < size; numbers++)
    {
        printf("%d", *numbers);
    }
}

int main(void)
{

    int numbers[] = {3, 4, 6, 3, 46};

    int result = count(numbers, 5);

    printf("%d\n", result);

    return 0;
}

The loop in the count function does not seem to work properly since it is only looping through one time but I can't understand why.

My bad. I haven't really got there yet. I am focusing on getting the loop to work first. — user3032809
– user3032809, Commented Nov 26, 2013 at 9:57
Stupid question (just curious): are you running a non x86 architecture? — jcm
– jcm, Commented Nov 26, 2013 at 10:11

glglgl · Accepted Answer · 2013-11-26 10:14:45Z

4

The comparison numbers < size is wrong.

You have 3 options:

Have a separate index variable:

int count;
for(count = 0; count < size; count++)
{
    printf("%d", numbers[count]);
}

Have a separate "cursor" pointer:

const int * const end = numbers + size;
for(; numbers < end; numbers++)
{
    printf("%d", *numbers);
}

Decrement size:

for(; size != 0; size--, numbers++)
{
    printf("%d", *numbers);
}

edited Nov 26, 2013 at 10:14

answered Nov 26, 2013 at 9:55

glglgl

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1 Comment

unwind Over a year ago

Well put. I'd make it int * const end = orig + size; to make sure it's not accidentally modified, too.

MOHAMED · Accepted Answer · 2013-11-26 09:55:28Z

2

the count() function should be

int count(const int* numbers, int size)
{
    for(; size > 0; numbers++, size--)
    {
        printf("%d", *numbers);
    }
}

answered Nov 26, 2013 at 9:55

MOHAMED

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Comments

Eregrith · Accepted Answer · 2013-11-26 10:12:33Z

2

Because numbers is a pointer, it contains an address. size is an int and equals 5. The address is not 5, nor is it even near 5.

You might save the start address and calculate the offset:

int count(const int* numbers, int size)
{
  const int* start = numbers;
  for(; (int)(numbers - start) < size; numbers++)
  {
    printf("%d\n", *numbers);
  }
}

edited Nov 26, 2013 at 10:12

answered Nov 26, 2013 at 9:54

Eregrith

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4 Comments

user3032809 Over a year ago

I do not really understand how const int* start = numbers; differs from numbers, since they are equal? :S Can I not type for(*numbers < size; numbers++)? Oh, no I can't, since then I am comparing size with the values in the array, not the index...

Eregrith Over a year ago

numbers is the address of your array cell, *numbers the value in the cell pointed to. size is apparently the number of cells you have in your array, so you need to know how many cells you've printed. start differs from number since your for loop does numbers++ after each iteration.

Eregrith Over a year ago

It's the same as saying "From 121 Smith Road, tell me the names of people living in the 5 next houses". You won't check the address of the house against 5, you will remember having started at 121, so upon arriving at 127 you will do the calculation in your head (127-121 = 6) and realise you've done the 5 asked for so you're finished

Eregrith Over a year ago

Yes, on ideone. The only thing misssing is the return in count() but since OP sends size to that method, I don't know what he wants returned. I only addressed the condition problem.

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