int num = atoi(argv[1]);
unsigned long times[num];
I have this code and I assumed it won't compile because I am trying to allocate the array using a value from a command line argument, which compiler doesn't know at the compile time.
But I compiled this code and it worked.
Can someone explain what is going on here??
Am I misunderstanding the basic concept of static allocation??
C99 allows to allocate an array with a var. This is called variable length arrays aka VLA
I don't have the C99 in my hand, the section is 6.7.5.2 in C99, and the following links are from the internet.
vla - wikipedia
c99 - wikipedia
be aware that vla is not supported in c++, more information here
Related
I have some concepts about the VLA and its behavior that I need to clarify.
AFIK since C99 it's possible to declare VLA into local scopes:
int main(int argc, char **argv)
{
// function 'main' scope
int size = 100;
int array[size];
return 0;
}
But it is forbidden in global scopes:
const int global_size = 100;
int global_array[global_size]; // forbidden in C99, allowed in C++
int main(int argc, char **argv)
{
int local_size = 100;
int local_array[local_size];
return 0;
}
The code above declares a VLA in C99 because the const modifier doesn't create a compile-time value. In C++ global_size is a compile-time value so, global_array doesn't become a VLA.
What I need to know is: Is my reasoning correct? Is the behaviour that I've described correct?
I also want to know: Why are the VLA not allowed in global scope? are they forbidden both in C and C++? Why is the behavior of arrays into global and local scope different?
Yes your reasoning is correct, that is how these different forms of array declarations and definitions are viewed by C and C++.
As others already stated, VLA with a veritable variable length (non-const) in global scope is difficult to make sense. What would the evaluation order be, e.g if the the length expression would refer to an object of a different compilation unit? C++ doesn't have VLA, but it has dynamic initialization of objects at file scope. And already this gives you quite a head ache, if you have to rely on evaluation order.
This leaves the small gap for C concerning length expressions that contain a const qualified object, which isn't allowed. This comes from the fact that such objects are not considered "integer constant expressions" by the C standard. This could perhaps change in future versions, but up to now the C committee didn't find it necessary to allow for such a thing: there are enum constants that play that role in C. Their only limitation is that they are limited to int in C, it would be nice to also have them size_t.
C++ doesn't support VLAs, period. The reason the second code snippet works in C++ is that the const keyword creates a compile-time constant in C++; in C, it doesn't.
C99 doesn't support VLAs outside of block scope, period, regardless of how you declare the size variable. Note that C2011 makes VLA support optional.
There is a difference between being forbidden and not being allowed. ;-)
The VLA feature is designed to allow the use of stack space for a local array, to avoid the use of malloc for heap allocation. It is mostly a speed optimization.
Now you want to use VLAs outside of functions. Why? There is not much to win speedwise in avoiding a single malloc call during program start. And what stack space are we supposed to use for variables with a static life time?
I think the fundamental reason is a global variable has linkage, its size has to be known at compile time. If not, how could one link the program?
local variables have no linkage and the VLAs are allocated on the stack, which grows dynamically as the program runs.
So, for global VLA's, one of the issues (there are many variants on the theme) can be shown here:
int size;
int a;
int v[size];
int b;
....
in another file:
extern int a;
extern int b;
The linker will have to know where a and be are in relation to each other at link time, or it won't be able to fix them up correctly at load-time.
Just curious, what actually happens if I define a zero-length array int array[0]; in code? GCC doesn't complain at all.
Sample Program
#include <stdio.h>
int main() {
int arr[0];
return 0;
}
Clarification
I'm actually trying to figure out if zero-length arrays initialised this way, instead of being pointed at like the variable length in Darhazer's comments, are optimised out or not.
This is because I have to release some code out into the wild, so I'm trying to figure out if I have to handle cases where the SIZE is defined as 0, which happens in some code with a statically defined int array[SIZE];
I was actually surprised that GCC does not complain, which led to my question. From the answers I've received, I believe the lack of a warning is largely due to supporting old code which has not been updated with the new [] syntax.
Because I was mainly wondering about the error, I am tagging Lundin's answer as correct (Nawaz's was first, but it wasn't as complete) -- the others were pointing out its actual use for tail-padded structures, while relevant, isn't exactly what I was looking for.
An array cannot have zero size.
ISO 9899:2011 6.7.6.2:
If the expression is a constant expression, it shall have a value greater than zero.
The above text is true both for a plain array (paragraph 1). For a VLA (variable length array), the behavior is undefined if the expression's value is less than or equal to zero (paragraph 5). This is normative text in the C standard. A compiler is not allowed to implement it differently.
gcc -std=c99 -pedantic gives a warning for the non-VLA case.
As per the standard, it is not allowed.
However it's been current practice in C compilers to treat those declarations as a flexible array member (FAM) declaration:
C99 6.7.2.1, §16: As a special case, the last element of a structure with more than one named member may have an incomplete array type; this is called a flexible array member.
The standard syntax of a FAM is:
struct Array {
size_t size;
int content[];
};
The idea is that you would then allocate it so:
void foo(size_t x) {
Array* array = malloc(sizeof(size_t) + x * sizeof(int));
array->size = x;
for (size_t i = 0; i != x; ++i) {
array->content[i] = 0;
}
}
You might also use it statically (gcc extension):
Array a = { 3, { 1, 2, 3 } };
This is also known as tail-padded structures (this term predates the publication of the C99 Standard) or struct hack (thanks to Joe Wreschnig for pointing it out).
However this syntax was standardized (and the effects guaranteed) only lately in C99. Before a constant size was necessary.
1 was the portable way to go, though it was rather strange.
0 was better at indicating intent, but not legal as far as the Standard was concerned and supported as an extension by some compilers (including gcc).
The tail padding practice, however, relies on the fact that storage is available (careful malloc) so is not suited to stack usage in general.
In Standard C and C++, zero-size array is not allowed..
If you're using GCC, compile it with -pedantic option. It will give warning, saying:
zero.c:3:6: warning: ISO C forbids zero-size array 'a' [-pedantic]
In case of C++, it gives similar warning.
It's totally illegal, and always has been, but a lot of compilers
neglect to signal the error. I'm not sure why you want to do this.
The one use I know of is to trigger a compile time error from a boolean:
char someCondition[ condition ];
If condition is a false, then I get a compile time error. Because
compilers do allow this, however, I've taken to using:
char someCondition[ 2 * condition - 1 ];
This gives a size of either 1 or -1, and I've never found a compiler
which would accept a size of -1.
Another use of zero-length arrays is for making variable-length object (pre-C99). Zero-length arrays are different from flexible arrays which have [] without 0.
Quoted from gcc doc:
Zero-length arrays are allowed in GNU C. They are very useful as the last element of a structure that is really a header for a variable-length object:
struct line {
int length;
char contents[0];
};
struct line *thisline = (struct line *)
malloc (sizeof (struct line) + this_length);
thisline->length = this_length;
In ISO C99, you would use a flexible array member, which is slightly different in syntax and semantics:
Flexible array members are written as contents[] without the 0.
Flexible array members have incomplete type, and so the sizeof operator may not be applied.
A real-world example is zero-length arrays of struct kdbus_item in kdbus.h (a Linux kernel module).
I'll add that there is a whole page of the online documentation of gcc on this argument.
Some quotes:
Zero-length arrays are allowed in GNU C.
In ISO C90, you would have to give contents a length of 1
and
GCC versions before 3.0 allowed zero-length arrays to be statically initialized, as if they were flexible arrays. In addition to those cases that were useful, it also allowed initializations in situations that would corrupt later data
so you could
int arr[0] = { 1 };
and boom :-)
Zero-size array declarations within structs would be useful if they were allowed, and if the semantics were such that (1) they would force alignment but otherwise not allocate any space, and (2) indexing the array would be considered defined behavior in the case where the resulting pointer would be within the same block of memory as the struct. Such behavior was never permitted by any C standard, but some older compilers allowed it before it became standard for compilers to allow incomplete array declarations with empty brackets.
The struct hack, as commonly implemented using an array of size 1, is dodgy and I don't think there's any requirement that compilers refrain from breaking it. For example, I would expect that if a compiler sees int a[1], it would be within its rights to regard a[i] as a[0]. If someone tries to work around the alignment issues of the struct hack via something like
typedef struct {
uint32_t size;
uint8_t data[4]; // Use four, to avoid having padding throw off the size of the struct
}
a compiler might get clever and assume the array size really is four:
; As written
foo = myStruct->data[i];
; As interpreted (assuming little-endian hardware)
foo = ((*(uint32_t*)myStruct->data) >> (i << 3)) & 0xFF;
Such an optimization might be reasonable, especially if myStruct->data could be loaded into a register in the same operation as myStruct->size. I know nothing in the standard that would forbid such optimization, though of course it would break any code which might expect to access stuff beyond the fourth element.
Definitely you can't have zero sized arrays by standard, but actually every most popular compiler gives you to do that. So I will try to explain why it can be bad
#include <cstdio>
int main() {
struct A {
A() {
printf("A()\n");
}
~A() {
printf("~A()\n");
}
int empty[0];
};
A vals[3];
}
I am like a human would expect such output:
A()
A()
A()
~A()
~A()
~A()
Clang prints this:
A()
~A()
GCC prints this:
A()
A()
A()
It is totally strange, so it is a good reason not to use empty arrays in C++ if you can.
Also there is extension in GNU C, which gives you to create zero length array in C, but as I understand it right, there should be at least one member in structure prior, or you will get very strange examples as above if you use C++.
Yesterday I was helping a friend w/ some C progamming and he used to declare arrays in this way:
/*includes*/
int i;
scanf("%d",&i);
int array[i];
I don't think it's a legit way but the compilers gave no error... So is it right/possible to declare an array or there could be any problem with that kind of declaration?
As stated in (How do compilers treat variable length arrays) in the C99 version of the C standard, variable length arrays are permitted. However, they are not permitted in any version of C++.
I am trying to create an array according to the size that the user inputs but it does not seem to be working for c programming. The following are my codes:
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int x, y;
scanf("%d", &x);
scanf("%d", &y);
double arr[x][y];
}
The compiler keeps returning an error of" Error: Expression must have a constant value. at the line double ... could anyone help point out the error?
You have two choices:
Either use a decent C compiler that supports C99 (or later) and variable-length arrays (I'd go with this approach, personally);
or if that is not possible, or the resulting array would be too large to fit in a block-scope variable (causing, for example, a stack overflow), you can use malloc(); however, you won't be able to create a true two-dimensional array using that approach, only a pointer-to-pointer, which may or may not be what you are looking for.
The code can work in C99 mode or when the compiler supports VLA(variable length arrays) as an extension (e.g, GCC supports VLA as GNU extesnion).
In C89, you have to use pointers with dynamic memory to simulate.
Older C standards don't provide support for arrays that do not have compile-time sizes:
int array[42];
char text[] = "Hello World";
int numbers = { 1, 2, 3, 4 };
(in the case of the latter two examples, the size is derived from the data)
You either need a newer compiler, to specify the -std=c99 if you are using GCC, or you need to allocate memory for the array yourself.
You're using a C89 (Visual Studio? Though that would give you an error on the declaration of arr next) or a C++ compiler. VLA's (Variable Length Arrays) are a C99 feature.
You need to allocate the memory in run-time . The compiler doesn't allow the declaration of arrays becuase it does not know the size of the array before hand . You need to use malloc() to allocate memory
why can we do this in c?
int n;
scanf("%d",&n);
int a[n];
I thought array is located memory during load time but seems like the above example works during runtime.
Do I misunderstand any thing? can you guys help?
Thanks,
I am no expert in C, but this could be a variable-length array as added by C99 and supported by GCC, for example. GCC allocates the memory for such array on stack, so that it gets automatically freed when you return from the function.
Variable-length arrays are not found in C89, but are a new feature in C99.
I thought array is *al*located memory during load time but seems like the above example works during run-time.
Yes, ordinary arrays like <datatype> <Array_Name> [<size>] is allocated memory during load time it is there in C89 and also existed in C99.
But in the code snippet int a[n]; is a Variable Length Array or VLA for short.VLA's in C99 are defined just like any other array, except that the length doesn’t need to be a compile-time constant.
A decent article on the need of VLAs can be found here :http://www.ddj.com/cpp/184401444 :)
Given how your code is written (specifically, that you have a statement), this must be code within a function.
While I'm not sure if this is strictly required in the spec, within a function, all auto (i.e. function level, not static) arrays are put on the stack. So regardless of whether you have a regular or VL array, the memory is allocated at runtime.
The memory for non-auto arrays is not handled at runtime so do no support VLA. If you try to compile the following code:
extern int size;
char buff1[size];
void doit(int x)
{
static int buff2[x];
int buff3[x];
}
On the compiler I tested this on (gcc 4.2.1), I got following errors:
moo.c:2: error: variably modified ‘buff1’ at file scope
moo.c: In function ‘doit’:
moo.c:6: error: storage size of ‘buff2’ isn’t constant