C beginner - arrays and pointers

Question

here's a simple and short code I've been trying to run:

#include <stdio.h>

int const SIZE = 5;

void a(int *arr);

int main(){
    int arr[5] = {1,2,3,4,5};
    a(arr);
    return 0;
}

void a(int *arr){
    int *i;
    for (i=arr; i<&a[5]; i++)
            printf("%d",*arr[i]);
}

and i get the following errors/warnings:

main.c: In function ‘main’:
main.c:15: error: variable-sized object may not be initialized
main.c:15: warning: excess elements in array initializer
main.c:15: warning: (near initialization for ‘arr’)
main.c:15: warning: excess elements in array initializer
main.c:15: warning: (near initialization for ‘arr’)
main.c:15: warning: excess elements in array initializer
main.c:15: warning: (near initialization for ‘arr’)
main.c:15: warning: excess elements in array initializer
main.c:15: warning: (near initialization for ‘arr’)
main.c:15: warning: excess elements in array initializer
main.c:15: warning: (near initialization for ‘arr’)
main.c: In function ‘a’:
main.c:22: error: subscripted value is neither array nor pointer
main.c:23: error: array subscript is not an integer

the warning are all associated with the initialization of the array: if I put '5' instead of 'SIZE' its ok, why? The errors in a I don't get at all. I'm passing a pointer as an arguments, where's the problem? thatnks!

you declare a constant and then don't use it. Please clean up your code... — Mitch Wheat, Aug 07 '11 at 12:52

score 5 · Accepted Answer · answered Aug 07 '11 at 12:51

5

Firstly, that should be:

for (i=arr; i<&arr[5]; i++)
               ^^^

But secondly, i is not an index, it's a pointer. So your print statement should be:

printf("%d",*i);

answered Aug 07 '11 at 12:51

Oliver Charlesworth

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Also arr only has 5 elements, so it should be &arr[4], but since there's a constant, it should be used instead (or a size passed in). And i is a pointer to the index, so referencing it should be i, not arr[i] – beatgammit Aug 07 '11 at 12:54
1

@tjameson: `&arr[5]` is ok, as the comparison is `<`, not `<=`. – Oliver Charlesworth Aug 07 '11 at 12:56
No, arr[5] is out of the bounds of the array. You're taking the address of the next block (at the sixth element, which doesn't exist). – beatgammit Aug 07 '11 at 12:58
2

@tjameson: It is legal C. The address of one-past-the-end is well-defined. – Oliver Charlesworth Aug 07 '11 at 12:58
Taking the address of memory that you don't necessarily own? This sounds dangerous. And it sounds like the OP is not understanding C, so perhaps we should guide him/her to the best practice? – beatgammit Aug 07 '11 at 13:00
1

@tjameson: It is well-defined by the standard. – Oliver Charlesworth Aug 07 '11 at 13:01
http://stackoverflow.com/questions/988158/take-the-address-of-a-one-past-the-end-array-element-via-subscript-legal-by-the – Mitch Wheat Aug 07 '11 at 13:02
Ok I've made a little change and now it works. I know i can used indexing but I am trying to learn pointers. Anyway, how come when I initialize the array using arr[SIZE} it doesn't work, but when I write arr[5] it does?? thanks – yotamoo Aug 07 '11 at 13:13
@yotamoo: In C, the size of an array must be a *constant expression*. `const int SIZE = 5` is not a constant expression; `const` really just means "read-only". – Oliver Charlesworth Aug 07 '11 at 13:17
So what is the alternative? I want to have a fixed SIZE for all arrays in this module, how do I do that? – yotamoo Aug 07 '11 at 13:21
@yotamoo: You could use a `#define` instead. – Oliver Charlesworth Aug 07 '11 at 13:22
but #define is not type-safe, so I still have to cast it to int inside the function... – yotamoo Aug 07 '11 at 13:28
@yotmaoo: C is quite limited in this regard. – Oliver Charlesworth Aug 07 '11 at 13:32
your end-condition solution is overdesigned for this question, your answer not for the difference-part of the question – user411313 Aug 07 '11 at 14:38

Mitch Wheat · Answer 2 · 2011-08-07T13:04:21.750

1

Your code should presumably be:

void a(int *arr)
{
    int *i;
    for (i=arr; i < &arr[5]; i++)
            printf("%d",*i);
}

but why don't you just write the more understandable:

void PrintArray(int *arr, int size)
{
    int i;
    for (i=0; i < size; i++)
        printf("%d", arr[i]);
}

edited Aug 07 '11 at 13:04

answered Aug 07 '11 at 12:54

Mitch Wheat

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Again, there are only 5 elements, so arr[5] will get the 6th element, which is an index out of bounds error. – beatgammit Aug 07 '11 at 12:57
see oli's comment "The address of one-past-the-end is well-defined. " – Mitch Wheat Aug 07 '11 at 13:00

beatgammit · Answer 3 · 2011-08-07T13:03:18.547

0

Try this:

void a(int *arr, int size);

int main(){
    int arr[5] = {1,2,3,4,5};
    a(arr, 5);
    return 0;
}

void a(int *arr, int size){
    int i;
    for (i=0; i< size; i++)
        printf("%d", arr[i]);
}

edited Aug 07 '11 at 13:03

answered Aug 07 '11 at 12:56

beatgammit

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score 0 · Answer 4 · answered Aug 07 '11 at 12:56

0

Arrays in C are indexed from zero. So the last element is arr[4], not arr[5]

answered Aug 07 '11 at 12:56

Géal

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score 0 · Answer 5 · answered Aug 07 '11 at 15:08

In C89 array-dimensions must known to compile-time, the content of an variable is'nt known at compile-time (also it is const), a define can known at compile-time. a strictly example show like this:

#include <stdio.h>

#define MY_ARRAY_SIZE 5  /* "SIZE" is a bad name for a #define */

typedef int MY_TYPE[MY_ARRAY_SIZE]; /* if possible, use your OWN types for your usecases */

void my_function(int *arr); /* "a" is a bad name for a function */

int main(){
    MY_TYPE arr = {1,2,3,4,5};
    my_function(arr);
    return 0;
}

void my_function(MY_TYPE arr){
    int i;
    for (i=0; i<sizeof(MY_TYPE)/sizeof*arr; i++)
            printf("%d",arr[i]);
}

C beginner - arrays and pointers

5 Answers5