Advertisement
Advertisement


How to initialize all members of an array to the same value?


Question

I have a large array in C (not C++ if that makes a difference). I want to initialize all members of the same value.

I could swear I once knew a simple way to do this. I could use memset() in my case, but isn't there a way to do this that is built right into the C syntax?

2020/04/19
1
991
4/19/2020 1:47:43 PM

Accepted Answer

Unless that value is 0 (in which case you can omit some part of the initializer and the corresponding elements will be initialized to 0), there's no easy way.

Don't overlook the obvious solution, though:

int myArray[10] = { 5, 5, 5, 5, 5, 5, 5, 5, 5, 5 };

Elements with missing values will be initialized to 0:

int myArray[10] = { 1, 2 }; // initialize to 1,2,0,0,0...

So this will initialize all elements to 0:

int myArray[10] = { 0 }; // all elements 0

In C++, an empty initialization list will also initialize every element to 0. This is not allowed with C:

int myArray[10] = {}; // all elements 0 in C++

Remember that objects with static storage duration will initialize to 0 if no initializer is specified:

static int myArray[10]; // all elements 0

And that "0" doesn't necessarily mean "all-bits-zero", so using the above is better and more portable than memset(). (Floating point values will be initialized to +0, pointers to null value, etc.)

2017/05/23
1268
5/23/2017 11:33:24 AM

If your compiler is GCC you can use following syntax:

int array[1024] = {[0 ... 1023] = 5};

Check out detailed description: http://gcc.gnu.org/onlinedocs/gcc-4.1.2/gcc/Designated-Inits.html

2009/10/13

For statically initializing a large array with the same value, without multiple copy-paste, you can use macros:

#define VAL_1X     42
#define VAL_2X     VAL_1X,  VAL_1X
#define VAL_4X     VAL_2X,  VAL_2X
#define VAL_8X     VAL_4X,  VAL_4X
#define VAL_16X    VAL_8X,  VAL_8X
#define VAL_32X    VAL_16X, VAL_16X
#define VAL_64X    VAL_32X, VAL_32X

int myArray[53] = { VAL_32X, VAL_16X, VAL_4X, VAL_1X };

If you need to change the value, you have to do the replacement at only one place.

Edit: possible useful extensions

(courtesy of Jonathan Leffler)

You can easily generalize this with:

#define VAL_1(X) X
#define VAL_2(X) VAL_1(X), VAL_1(X)
/* etc. */

A variant can be created using:

#define STRUCTVAL_1(...) { __VA_ARGS__ }
#define STRUCTVAL_2(...) STRUCTVAL_1(__VA_ARGS__), STRUCTVAL_1(__VA_ARGS__)
/*etc */ 

that works with structures or compound arrays.

#define STRUCTVAL_48(...) STRUCTVAL_32(__VA_ARGS__), STRUCTVAL_16(__VA_ARGS__)

struct Pair { char key[16]; char val[32]; };
struct Pair p_data[] = { STRUCTVAL_48("Key", "Value") };
int a_data[][4] = { STRUCTVAL_48(12, 19, 23, 37) };

macro names are negotiable.

2017/05/23

If you want to ensure that every member of the array is explicitly initialized, just omit the dimension from the declaration:

int myArray[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };

The compiler will deduce the dimension from the initializer list. Unfortunately, for multidimensional arrays only the outermost dimension may be omitted:

int myPoints[][3] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9} };

is OK, but

int myPoints[][] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9} };

is not.

2008/10/14

I saw some code that used this syntax:

char* array[] = 
{
    [0] = "Hello",
    [1] = "World"
};   

Where it becomes particularly useful is if you're making an array that uses enums as the index:

enum
{
    ERR_OK,
    ERR_FAIL,
    ERR_MEMORY
};

#define _ITEM(x) [x] = #x

char* array[] = 
{
    _ITEM(ERR_OK),
    _ITEM(ERR_FAIL),
    _ITEM(ERR_MEMORY)
};   

This keeps things in order, even if you happen to write some of the enum-values out of order.

More about this technique can be found here and here.

2012/03/22

int i;
for (i = 0; i < ARRAY_SIZE; ++i)
{
  myArray[i] = VALUE;
}

I think this is better than

int myArray[10] = { 5, 5, 5, 5, 5, 5, 5, 5, 5, 5...

incase the size of the array changes.

2008/10/14

Source: https://stackoverflow.com/questions/201101
Licensed under: CC-BY-SA with attribution
Not affiliated with: Stack Overflow
Email: [email protected]