Why is there no error flagged for wrong dynamic allocation of array in C?

Question

#include<stdio.h>
void main(){
    int n = 0;
    scanf("%d",&n);
    int a[n];
    printf("Number of elements in array is %ld\n",sizeof(a)/sizeof(int));
}

I have shown above a wrong method for dynamically allocating array in C. The correct way to allocate dynamically is to use malloc. In the current declaration of the array (owing to the syntax of C), the variable "a" should allocated in the compile-time but not in the run-time. So it should be in data-segment. But anything allocated dynamically is in heap. I have 2 questions:

The variable "n" must be a constant so that "a" is allocated during compile time in data segment and not during run-time. This program compiles with no error. Why is there no error flagged or at least warning at the end of compile time?
I executed this code and gave a value for n, but found that the number of elements allocated for array "a" is exactly n. How is this happening? Also where is the array allocated?

I am using gcc compiler 5.4.0 in ubuntu 16.04.

few things first 1. void main() -> int main() ; 2 printf is "%ld" not "%d"; 3. return statement in main as per point 1. — asio_guy, Nov 11 '17 at 06:43
Note that if you used C89 mode, you couldn't even define `int a[1]` at that position - in C89 all variable definitions had to go to the beginning of the block. — Antti Haapala -- Слава Україні, Nov 11 '17 at 07:47

user2736738 · Answer 1 · 2017-11-11T07:57:33.587

3

The answer to 1 will be:-

No error is flagged because your compiler understands C99 code. This is known as a VLA (variable length array). And it has been supported since C99. (VLA is a C99 specific language feature). The mode is gnu11 (by default).

Internally here also it uses some allocation (using alloca etc.).

Your compiler by default runs in C99 mode. Check for the compiler's options to switch to strict C89 or another mode. You will get an error.

The answer to the second question is quite obvious as you are specifying the array size to be n not n-1 or n+1. So it allocates that much of memory for you.

edited Nov 11 '17 at 07:57

answered Nov 11 '17 at 06:41

user2736738

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The mode is gnu11 (by default). – Antti Haapala -- Слава Україні Nov 11 '17 at 07:49
@AnttiHaapala.: Now I remember..yes it is.. – user2736738 Nov 11 '17 at 07:55
@psmears.: Thanks for the contribution and help. – user2736738 Nov 11 '17 at 07:56
@coderredoc Thanks for your answer. No, even in C89 mode there is no error flagged. You did not understand my second question. As per the syntax of the declaration of array "a", it should be allocated in data segment. But it is allowing us to allocate dynamically. So where is array "a" present? in the data segment or in heap?? My guess is in heap. – varsh Nov 11 '17 at 08:08
@varshaneya.:normally stack. It uses `alloca` mostly. – user2736738 Nov 11 '17 at 08:12
@varshaneya there is no *error*. The C language allows the use of variable length arrays, why should it be an error?! if you really want an error use `-std=c89 -pedantic-errors`. But why? – Antti Haapala -- Слава Україні Nov 11 '17 at 08:19
@AnttiHaapala Yes that is true. I did not use -pedantic-errors flag thought I used -std=c89. Nothing its just a doubt. We are taught not allocate arrays dynamically like that but to use only malloc. – varsh Nov 11 '17 at 08:48

Why is there no error flagged for wrong dynamic allocation of array in C?

1 Answers1