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

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

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.

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.

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.

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.

You can do the whole static initializer thing as detailed above, but it can be a real bummer when your array size changes (when your array embiggens, if you don't add the appropriate extra initializers you get garbage).

memset gives you a runtime hit for doing the work, but no code size hit done right is immune to array size changes. I would use this solution in nearly all cases when the array was larger than, say, a few dozen elements.

If it was really important that the array was statically declared, I'd write a program to write the program for me and make it part of the build process.

Here is another way:

static void
unhandled_interrupt(struct trap_frame *frame, int irq, void *arg)
{
    //this code intentionally left blank
}

static struct irqtbl_s vector_tbl[XCHAL_NUM_INTERRUPTS] = {
    [0 ... XCHAL_NUM_INTERRUPTS-1] {unhandled_interrupt, NULL},
};

See:

C-Extensions

Designated inits

Then ask the question: When can one use C extensions?

The code sample above is in an embedded system and will never see the light from another compiler.

For initializing 'normal' data types (like int arrays), you can use the bracket notation, but it will zero the values after the last if there is still space in the array:

// put values 1-8, then two zeroes
int list[10] = {1,2,3,4,5,6,7,8};

A slightly tongue-in-cheek answer; write the statement

array = initial_value

in your favourite array-capable language (mine is Fortran, but there are many others), and link it to your C code. You'd probably want to wrap it up to be an external function.

If the array happens to be int or anything with the size of int or your mem-pattern's size fits exact times into an int (i.e. all zeroes or 0xA5A5A5A5), the best way is to use memset().

Otherwise call memcpy() in a loop moving the index.

There is a fast way to initialize array of any type with given value. It works very well with large arrays. Algorithm is as follows:

• initialize first element of the array (usual way)
• copy part which has been set into part which has not been set, doubling the size with each next copy operation

For 1 000 000 elements int array it is 4 times faster than regular loop initialization (i5, 2 cores, 2.3 GHz, 4GiB memory, 64 bits):

loop runtime 0.004248 [seconds]

memfill() runtime 0.001085 [seconds]

#include <stdio.h>
#include <time.h>
#include <string.h>
#define ARR_SIZE 1000000

void memfill(void *dest, size_t destsize, size_t elemsize) {
   char   *nextdest = (char *) dest + elemsize;
   size_t movesize, donesize = elemsize;

   destsize -= elemsize;
   while (destsize) {
      movesize = (donesize < destsize) ? donesize : destsize;
      memcpy(nextdest, dest, movesize);
      nextdest += movesize; destsize -= movesize; donesize += movesize;
   }
}    
int main() {
    clock_t timeStart;
    double  runTime;
    int     i, a[ARR_SIZE];

    timeStart = clock();
    for (i = 0; i < ARR_SIZE; i++)
        a[i] = 9;    
    runTime = (double)(clock() - timeStart) / (double)CLOCKS_PER_SEC;
    printf("loop runtime %f [seconds]\n",runTime);

    timeStart = clock();
    a[0] = 10;
    memfill(a, sizeof(a), sizeof(a[0]));
    runTime = (double)(clock() - timeStart) / (double)CLOCKS_PER_SEC;
    printf("memfill() runtime %f [seconds]\n",runTime);
    return 0;
}

You can use memset function.

void *memset(void *array, int value, unsigned sizeofarray);

Nobody has mentioned the index order to access the elements of the initialized array. My example code will give an illustrative example to it.

#include <iostream>

void PrintArray(int a[3][3])
{
    std::cout << "a11 = " << a[0][0] << "\t\t" << "a12 = " << a[0][1] << "\t\t" << "a13 = " << a[0][2] << std::endl;
    std::cout << "a21 = " << a[1][0] << "\t\t" << "a22 = " << a[1][1] << "\t\t" << "a23 = " << a[1][2] << std::endl;
    std::cout << "a31 = " << a[2][0] << "\t\t" << "a32 = " << a[2][1] << "\t\t" << "a33 = " << a[2][2] << std::endl;
    std::cout << std::endl;
}

int wmain(int argc, wchar_t * argv[])
{
    int a1[3][3] =  {   11,     12,     13,     // The most
                        21,     22,     23,     // basic
                        31,     32,     33  };  // format.

    int a2[][3] =   {   11,     12,     13,     // The first (outer) dimension
                        21,     22,     23,     // may be omitted. The compiler
                        31,     32,     33  };  // will automatically deduce it.

    int a3[3][3] =  {   {11,    12,     13},    // The elements of each
                        {21,    22,     23},    // second (inner) dimension
                        {31,    32,     33} };  // can be grouped together.

    int a4[][3] =   {   {11,    12,     13},    // Again, the first dimension
                        {21,    22,     23},    // can be omitted when the 
                        {31,    32,     33} };  // inner elements are grouped.

    PrintArray(a1);
    PrintArray(a2);
    PrintArray(a3);
    PrintArray(a4);

    // This part shows in which order the elements are stored in the memory.
    int * b = (int *) a1;   // The output is the same for the all four arrays.
    for (int i=0; i<9; i++)
    {
        std::cout << b[i] << '\t';
    }

    return 0;
}

The output is:

a11 = 11                a12 = 12                a13 = 13
a21 = 21                a22 = 22                a23 = 23
a31 = 31                a32 = 32                a33 = 33

a11 = 11                a12 = 12                a13 = 13
a21 = 21                a22 = 22                a23 = 23
a31 = 31                a32 = 32                a33 = 33

a11 = 11                a12 = 12                a13 = 13
a21 = 21                a22 = 22                a23 = 23
a31 = 31                a32 = 32                a33 = 33

a11 = 11                a12 = 12                a13 = 13
a21 = 21                a22 = 22                a23 = 23
a31 = 31                a32 = 32                a33 = 33

11      12      13      21      22      23      31      32      33

1. If your array is declared as static or is global, all the elements in the array already have default default value 0.
2. Some compilers set array's the default to 0 in debug mode.
3. It is easy to set default to 0 : int array[10] = {0};
4. However, for other values, you have use memset() or loop;

example: int array[10]; memset(array,-1, 10 *sizeof(int));

I see no requirements in the question, so the solution must be generic: initialization of an unspecified possibly multidimensional array built from unspecified possibly structure elements with an initial member value:

#include <string.h> 

void array_init( void *start, size_t element_size, size_t elements, void *initval ){
  memcpy(        start,              initval, element_size              );
  memcpy( (char*)start+element_size, start,   element_size*(elements-1) );
}

// testing
#include <stdio.h> 

struct s {
  int a;
  char b;
} array[2][3], init;

int main(){
  init = (struct s){.a = 3, .b = 'x'};
  array_init( array, sizeof(array[0][0]), 2*3, &init; );

  for( int i=0; i<2; i++ )
    for( int j=0; j<3; j++ )
      printf("array[%i][%i].a = %i .b = '%c'\n",i,j,array[i][j].a,array[i][j].b);
}

Result:

array[0][0].a = 3 .b = 'x'
array[0][1].a = 3 .b = 'x'
array[0][2].a = 3 .b = 'x'
array[1][0].a = 3 .b = 'x'
array[1][1].a = 3 .b = 'x'
array[1][2].a = 3 .b = 'x'

EDIT: start+element_size changed to (char*)start+element_size

Cutting through all the chatter, the short answer is that if you turn on optimization at compile time you won't do better than this:

int i,value=5,array[1000]; 
for(i=0;i<1000;i++) array[i]=value; 

Added bonus: the code is actually legible :)