Related
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++.
Can I use a variable to define an array's size?
int test = 12;
int testarr[test];
Would this work, I don't want to change the size of the array after initialization. The int test's value isn't known at compile time.
From C99 it is allowed but only for the automatic variables.
this is illegal:
int test = 12;
int testarr[test]; // illegal - static storage variable
int foo(void)
{
int test = 12;
static int testarr[test]; // illegal - static storage variable
}
the only valid form is:
int foo(void)
{
int test = 12;
int testarr[test]; // legal - automatic storage variable
}
can i use an variable to define an arrays size?
This is called Variable Length Arrays (VLA).
Read Modern C then the C11 standard n1570 (and see this reference). VLAs are permitted in §6.7.6; also read the documentation of your C compiler (e.g. GCC).
But you don't want to overflow your call stack, typically limited to a megabyte on laptops (and OS specific).
So you may prefer C dynamic memory allocation with e.g. malloc(3) (and free ...) or calloc.
Beware of memory leaks.
You might be interested by tools such as valgrind. You need to learn to use your debugger (e.g. GDB).
Can I use an integer variable to define an arrays length?
Yes, that is what is called a variable length array and is part of the C standard since C99. Note that an implementation does not need to support it. Therefore you may prefer dynamic memory allocation. Take a look at here:
malloced array VS. variable-length-array
To cite the C standard:
"If the size is not present, the array type is an incomplete type. If the size is * instead of being an expression, the array type is a variable length array type of unspecified size, which can only be used in declarations or type names with function prototype scope;146) such arrays are nonetheless complete types. If the size is an integer constant expression and the element type has a known constant size,the array type is not a variable length array type; otherwise, the array type is a variable length array type. (Variable length arrays are a conditional feature that implementations need not support; see 6.10.8.3.)"
"146) Thus, * can be used only in function declarations that are not definitions (see 6.7.6.3)."
Source: C18, 6.7.6.2/4
Also note:
"Array objects declared with the _Thread_local, static, or extern storage-class specifier cannot have a variable length array (VLA) type."
Source: C18, 6.7.6.2/10
VLAs cannot be used:
at file scope.
when qualified with _Thread_local, static, or extern storage-class specifier.
if they have linkage.
Which means that they can only be used at function-scope and when of storage-class automatic, which is done by default when omitting any explicit specifier.
Feel free to ask for further clarification if you don't understand something.
Without knowing much of the context, wouldn't it be easier to just do this?
#define TEST 12 //to ensure this value will not change at all
int testarr[TEST];
Technically your method should work too but the value of test may change later on depending on the piece of code you written
So, when you try this on C, as in this [https://onlinegdb.com/rJzE8yzC8][1]
It successfully compiles and you can also update the value of the variable int test
However, after the update of test, the size of array does not change since arrays are static-defined.
For guarantee, I suggest you to use either const int or macro variables for doing this.
I am creating an array on stack as
static const int size = 10;
void foo() {
..
int array[size];
..
}
However, I get the compile error: "expression must have a constant value", even though size is a constant. I can use the macro
#define SIZE (10)
But I am wondering why size marked const causes compilation error.
In C language keyword const has nothing to do with constants. In C language, by definition the term "constant" refers to literal values and enum constants. This is what you have to use if you really need a constant: either use a literal value (define a macro to give your constant a name), or use a enum constant.
(Read here for more details: Shall I prefer constants over defines?)
Also, in C99 and later versions of the language it possible to use non-constant values as array sizes for local arrays. That means that your code should compile in modern C even though your size is not a constant. But you are apparently using an older compiler, so in your case
#define SIZE 10
is the right way to go.
The answer is in another stackoverflow question, HERE
it's because In C objects declared with the const modifier aren't true
constants. A better name for const would probably be readonly - what
it really means is that the compiler won't let you change it. And you
need true constants to initialize objects with static storage (I
suspect regs_to_read is global).
if you are on C99 your IDE compiler option may have a thing called variable-length array (VLA) enable it and you won't get compile error, efficiently without stressing your code though is with MALLOC or CALLOC.
static const int size = 10;
void foo() {
int* array;
array = (int *)malloc(size * sizeof(int));
}
Which one is better to use among the below statements in C?
static const int var = 5;
or
#define var 5
or
enum { var = 5 };
It depends on what you need the value for. You (and everyone else so far) omitted the third alternative:
static const int var = 5;
#define var 5
enum { var = 5 };
Ignoring issues about the choice of name, then:
If you need to pass a pointer around, you must use (1).
Since (2) is apparently an option, you don't need to pass pointers around.
Both (1) and (3) have a symbol in the debugger's symbol table - that makes debugging easier. It is more likely that (2) will not have a symbol, leaving you wondering what it is.
(1) cannot be used as a dimension for arrays at global scope; both (2) and (3) can.
(1) cannot be used as a dimension for static arrays at function scope; both (2) and (3) can.
Under C99, all of these can be used for local arrays. Technically, using (1) would imply the use of a VLA (variable-length array), though the dimension referenced by 'var' would of course be fixed at size 5.
(1) cannot be used in places like switch statements; both (2) and (3) can.
(1) cannot be used to initialize static variables; both (2) and (3) can.
(2) can change code that you didn't want changed because it is used by the preprocessor; both (1) and (3) will not have unexpected side-effects like that.
You can detect whether (2) has been set in the preprocessor; neither (1) nor (3) allows that.
So, in most contexts, prefer the 'enum' over the alternatives. Otherwise, the first and last bullet points are likely to be the controlling factors — and you have to think harder if you need to satisfy both at once.
If you were asking about C++, then you'd use option (1) — the static const — every time.
Generally speaking:
static const
Because it respects scope and is type-safe.
The only caveat I could see: if you want the variable to be possibly defined on the command line. There is still an alternative:
#ifdef VAR // Very bad name, not long enough, too general, etc..
static int const var = VAR;
#else
static int const var = 5; // default value
#endif
Whenever possible, instead of macros / ellipsis, use a type-safe alternative.
If you really NEED to go with a macro (for example, you want __FILE__ or __LINE__), then you'd better name your macro VERY carefully: in its naming convention Boost recommends all upper-case, beginning by the name of the project (here BOOST_), while perusing the library you will notice this is (generally) followed by the name of the particular area (library) then with a meaningful name.
It generally makes for lengthy names :)
In C, specifically? In C the correct answer is: use #define (or, if appropriate, enum)
While it is beneficial to have the scoping and typing properties of a const object, in reality const objects in C (as opposed to C++) are not true constants and therefore are usually useless in most practical cases.
So, in C the choice should be determined by how you plan to use your constant. For example, you can't use a const int object as a case label (while a macro will work). You can't use a const int object as a bit-field width (while a macro will work). In C89/90 you can't use a const object to specify an array size (while a macro will work). Even in C99 you can't use a const object to specify an array size when you need a non-VLA array.
If this is important for you then it will determine your choice. Most of the time, you'll have no choice but to use #define in C. And don't forget another alternative, that produces true constants in C - enum.
In C++ const objects are true constants, so in C++ it is almost always better to prefer the const variant (no need for explicit static in C++ though).
The difference between static const and #define is that the former uses the memory and the later does not use the memory for storage. Secondly, you cannot pass the address of an #define whereas you can pass the address of a static const. Actually it is depending on what circumstance we are under, we need to select one among these two. Both are at their best under different circumstances. Please don't assume that one is better than the other... :-)
If that would have been the case, Dennis Ritchie would have kept the best one alone... hahaha... :-)
In C #define is much more popular. You can use those values for declaring array sizes for example:
#define MAXLEN 5
void foo(void) {
int bar[MAXLEN];
}
ANSI C doesn't allow you to use static consts in this context as far as I know. In C++ you should avoid macros in these cases. You can write
const int maxlen = 5;
void foo() {
int bar[maxlen];
}
and even leave out static because internal linkage is implied by const already [in C++ only].
Another drawback of const in C is that you can't use the value in initializing another const.
static int const NUMBER_OF_FINGERS_PER_HAND = 5;
static int const NUMBER_OF_HANDS = 2;
// initializer element is not constant, this does not work.
static int const NUMBER_OF_FINGERS = NUMBER_OF_FINGERS_PER_HAND
* NUMBER_OF_HANDS;
Even this does not work with a const since the compiler does not see it as a constant:
static uint8_t const ARRAY_SIZE = 16;
static int8_t const lookup_table[ARRAY_SIZE] = {
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}; // ARRAY_SIZE not a constant!
I'd be happy to use typed const in these cases, otherwise...
If you can get away with it, static const has a lot of advantages. It obeys the normal scope principles, is visible in a debugger, and generally obeys the rules that variables obey.
However, at least in the original C standard, it isn't actually a constant. If you use #define var 5, you can write int foo[var]; as a declaration, but you can't do that (except as a compiler extension" with static const int var = 5;. This is not the case in C++, where the static const version can be used anywhere the #define version can, and I believe this is also the case with C99.
However, never name a #define constant with a lowercase name. It will override any possible use of that name until the end of the translation unit. Macro constants should be in what is effectively their own namespace, which is traditionally all capital letters, perhaps with a prefix.
#define var 5 will cause you trouble if you have things like mystruct.var.
For example,
struct mystruct {
int var;
};
#define var 5
int main() {
struct mystruct foo;
foo.var = 1;
return 0;
}
The preprocessor will replace it and the code won't compile. For this reason, traditional coding style suggest all constant #defines uses capital letters to avoid conflict.
It is ALWAYS preferable to use const, instead of #define. That's because const is treated by the compiler and #define by the preprocessor. It is like #define itself is not part of the code (roughly speaking).
Example:
#define PI 3.1416
The symbolic name PI may never be seen by compilers; it may be removed by the preprocessor before the source code even gets to a compiler. As a result, the name PI may not get entered into the symbol table. This can be confusing if you get an error during compilation involving the use of the constant, because the error message may refer to 3.1416, not PI. If PI were defined in a header file you didn’t write, you’d have no idea where that 3.1416 came from.
This problem can also crop up in a symbolic debugger, because, again, the name you’re programming with may not be in the symbol table.
Solution:
const double PI = 3.1416; //or static const...
I wrote quick test program to demonstrate one difference:
#include <stdio.h>
enum {ENUM_DEFINED=16};
enum {ENUM_DEFINED=32};
#define DEFINED_DEFINED 16
#define DEFINED_DEFINED 32
int main(int argc, char *argv[]) {
printf("%d, %d\n", DEFINED_DEFINED, ENUM_DEFINED);
return(0);
}
This compiles with these errors and warnings:
main.c:6:7: error: redefinition of enumerator 'ENUM_DEFINED'
enum {ENUM_DEFINED=32};
^
main.c:5:7: note: previous definition is here
enum {ENUM_DEFINED=16};
^
main.c:9:9: warning: 'DEFINED_DEFINED' macro redefined [-Wmacro-redefined]
#define DEFINED_DEFINED 32
^
main.c:8:9: note: previous definition is here
#define DEFINED_DEFINED 16
^
Note that enum gives an error when define gives a warning.
The definition
const int const_value = 5;
does not always define a constant value. Some compilers (for example tcc 0.9.26) just allocate memory identified with the name "const_value". Using the identifier "const_value" you can not modify this memory. But you still could modify the memory using another identifier:
const int const_value = 5;
int *mutable_value = (int*) &const_value;
*mutable_value = 3;
printf("%i", const_value); // The output may be 5 or 3, depending on the compiler.
This means the definition
#define CONST_VALUE 5
is the only way to define a constant value which can not be modified by any means.
Although the question was about integers, it's worth noting that #define and enums are useless if you need a constant structure or string. These are both usually passed to functions as pointers. (With strings it's required; with structures it's much more efficient.)
As for integers, if you're in an embedded environment with very limited memory, you might need to worry about where the constant is stored and how accesses to it are compiled. The compiler might add two consts at run time, but add two #defines at compile time. A #define constant may be converted into one or more MOV [immediate] instructions, which means the constant is effectively stored in program memory. A const constant will be stored in the .const section in data memory. In systems with a Harvard architecture, there could be differences in performance and memory usage, although they'd likely be small. They might matter for hard-core optimization of inner loops.
Don't think there's an answer for "which is always best" but, as Matthieu said
static const
is type safe. My biggest pet peeve with #define, though, is when debugging in Visual Studio you cannot watch the variable. It gives an error that the symbol cannot be found.
Incidentally, an alternative to #define, which provides proper scoping but behaves like a "real" constant, is "enum". For example:
enum {number_ten = 10;}
In many cases, it's useful to define enumerated types and create variables of those types; if that is done, debuggers may be able to display variables according to their enumeration name.
One important caveat with doing that, however: in C++, enumerated types have limited compatibility with integers. For example, by default, one cannot perform arithmetic upon them. I find that to be a curious default behavior for enums; while it would have been nice to have a "strict enum" type, given the desire to have C++ generally compatible with C, I would think the default behavior of an "enum" type should be interchangeable with integers.
A simple difference:
At pre-processing time, the constant is replaced with its value.
So you could not apply the dereference operator to a define, but you can apply the dereference operator to a variable.
As you would suppose, define is faster that static const.
For example, having:
#define mymax 100
you can not do printf("address of constant is %p",&mymax);.
But having
const int mymax_var=100
you can do printf("address of constant is %p",&mymax_var);.
To be more clear, the define is replaced by its value at the pre-processing stage, so we do not have any variable stored in the program. We have just the code from the text segment of the program where the define was used.
However, for static const we have a variable that is allocated somewhere. For gcc, static const are allocated in the text segment of the program.
Above, I wanted to tell about the reference operator so replace dereference with reference.
We looked at the produced assembler code on the MBF16X... Both variants result in the same code for arithmetic operations (ADD Immediate, for example).
So const int is preferred for the type check while #define is old style. Maybe it is compiler-specific. So check your produced assembler code.
I am not sure if I am right but in my opinion calling #defined value is much faster than calling any other normally declared variable (or const value).
It's because when program is running and it needs to use some normally declared variable it needs to jump to exact place in memory to get that variable.
In opposite when it use #defined value, the program don't need to jump to any allocated memory, it just takes the value. If #define myValue 7 and the program calling myValue, it behaves exactly the same as when it just calls 7.
static const int LOG_MAX = 31;
static int log_table[LOG_MAX];
This code is inside of a function in C. When I try to compile I get the error:
"main.c:19:16: error: storage size of 'log_table' isn't constant".
I don't understand this since LOG_MAX is const.
Just to clarify this is C code and I am using GCC.
In older C and C++ standards, the array bounds of an array had to be a constant literal evaluated at compile time. A const variable isn't necessary evaluated at compile time, it could be created in runtime as a local variable. Also, as pointed out in another answer, const should actually be regarded read-only rather than anything else.
In all C and C++ standards, static arrays must always have their size set using a constant literal. (Or to be picky, this applies to any variable with static storage duration)
In newer C standards (C99, C11) however, the code you posted is perfectly fine if you leave out the static keyword. It will then create a variable-length array (VLA), which may or may not be what you wanted to do.
I'm not sure about the latest C++11 standard, but as far as I know it does not support VLAs.
const does not mean constant in C but rather read-only. LOG_MAX is not a constant in your program.
Here are two ways to have a constant:
#define LOG_MAX 31
or
enum {
LOG_MAX = 31
};
This isn't valid C, even though the int is const and static.
I would recommend doing something like
#define LOG_MAX 31
static int log_table[LOG_MAX];
This fails since a const int variable is not considered a compile-time constant in C.
But since the array is going to be static, you can be sure that the "full" declaration is always available, i.e. it's never going to be referenced through a pointer. Thus, you can inline the size and use sizeof in all places where you wanted to use LOG_MAX:
static int log_table[32];
and then elsewhere in the same function:
const size_t log_max = sizeof log_table / sizeof *log_table;
This keeps the "magic constant" around, but only in one place and its purpose should be pretty clear given the way the log_table is used. This is all inside a single function, after all, it's not scary global data.
You should use the preprocessor :
#define LOG_MAX 31
static int log_table[LOG_MAX];