why can't I initialize the array like this?

Question

#include<stdio.h>

main() {
  int a=5;
  int array[a]={0};
  printf("Success\n");
}

when i am executing the program it will through a error as

b.c: In function ‘main’:
b.c:8:1: error: variable-sized object may not be initialized
b.c:8:1: warning: excess elements in array initializer [enabled by default]
b.c:8:1: warning: (near initialization for ‘array’) [enabled by default]

In cc complier . but i can assign like this

 int array[5]={0};

If anyone correct me?

@Bhuvanesh Have you checked your code using the cc -E. In this variable is not having the value to be placed in a array size. — Karthikeyan.R.S, Dec 11 '14 at 07:36

Vlad from Moscow · Accepted Answer · 2014-12-11T11:44:52.217

4

This statement

int array[a]={0};

declares a Variable Length Array (VLA).

According to C Standard (6.7.9 Initialization)

3 The type of the entity to be initialized shall be an array of unknown size or a complete object type that is not a variable length array type.

The problem is that the compiler shall know the array size at compile time that to generate the code that initialize an array.

Consider an example

void f( size_t n )
{
    int a[n] = { 1, 2, 3, 4, 5 };
    //...
}

Here is a is a variable length array. Now as n can have any value then the number of initializers in the array definition can be greater than the size of the array. So this code breaks the Standard from another side because the number of initializers may not be greater than the number of elements of array. On the other hand if the number of initializers less than the number of elements of array then what to do in this case? Maybe the programmer did not mean that some elements shall be zero-initialized.

As for this declaration

int array[5]={0};

then there is no variable length array. The size of the array is known at compile time. So there is no problem

.

edited Dec 11 '14 at 11:44

answered Dec 11 '14 at 07:28

Vlad from Moscow

301,070
26
186
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but we can declare the array as array[a];, but the problem is while we initializing then only the error will occur – Bhuvanesh Dec 11 '14 at 07:30
@Bhuvanesh: The quote that Vlad mentions explicitly states that you can't initialize a variable length array. – Cornstalks Dec 11 '14 at 07:35
Well done for being the only person to answer the actual question. – Cornstalks Dec 11 '14 at 07:39
@Cornstalks I updated my post because at first I did not explain why it can not be done. – Vlad from Moscow Dec 11 '14 at 11:47
@Vlad - Thank you for your post its useful for me.. – Bhuvanesh Dec 11 '14 at 12:00

score 1 · Answer 2 · answered Dec 11 '14 at 07:14

1

"Variable-length automatic arrays are allowed in ISO C99, and as an extension GCC accepts them in C90 mode and in C++"

says here.

Compilers can differ in some situations. It's normal to have problems with variable-sized arrays. This should work if you really need to do this

#include<stdio.h>
#DEFINE A 5

 main()
 {
 int array[A]={0};
 printf("Success\n");
 }

answered Dec 11 '14 at 07:14

Emre Acar

920
9
24

Its same as writing int a[5] = {0} as preprocessor will be resolved at compile time itself. – anurag-jain Dec 11 '14 at 07:20
_This should work **if you really need to do this**_ – Emre Acar Dec 11 '14 at 07:21

score 0 · Answer 3 · answered Dec 11 '14 at 07:20

0

The elements field within square brackets [], representing the number of elements in the array, must be a constant expression, since arrays are blocks of static memory whose size must be determined at compile time, before the program runs.

answered Dec 11 '14 at 07:20

Akhilesh Jaiswal

227
2
14

@Akilesh Jaiswal - but both are local variables. this will work during the run time – Bhuvanesh Dec 11 '14 at 07:22
since arrays are blocks of static memory whose size must be determined at compile time, before the program runs. :) – Akhilesh Jaiswal Dec 11 '14 at 07:36
This is a VLA so the array size doesn't have to be a constant expression. – Lundin Dec 11 '14 at 08:52

why can't I initialize the array like this?

3 Answers3