How to initialize an array with only -1 values [duplicate]

Question

Possible Duplicate:
How to initialize an array in C
initializing an array of ints

I wonder over the fastest/simplest way to initialize an int array to only contain -1 values. The array I need is 90 ints long so the straightforward way should be to initialize it like this:

int array[90]={-1, -1, -1, ...};

but I only want to use the array once so I want to be able to use it dynamically and be able to free it after using it in the program, so Im more looking for a fast way like calloc, but instead of zeros, -1 of course.

Answer 1

If you are using gcc then use designated initializer

int array[90] = { [ 0 ... 89 ] = -1}

int array[90],i;
for(i = 0; i < 90 ; arr[i++] = -1);

To do this dynamically , you will have to allocate using malloc then you only free the memory, otherwise freeing the memory which is not allocated by malloc , calloc or realloc is undefined behavior.

Use this:

int *array;
array=malloc(sizeof(int)*n);
for(i=0;i<n;array[i++]=-1);
// After use free this
free(array);

Answer 2

It is not possible to do it in Standard C at initialization without explicitly enumerating all initializers.

In GNU C you can use GNU C designated initializers

 int array[90] = {[0 ... sizeof array - 1] = -1};

after initialization:

   int i;

   for (i = 0; i < sizeof array / sizeof *array; i++)
   {
       array[i] = -1;
   }

Answer 3

It hurts to write this, but you could always use a macro

#define FILL(arr, val) \
for(int i_##arr = 0; i_##arr < sizeof arr / sizeof *arr; ++i_##arr) \
{ \
    arr[i_##arr] = val;\
}

Then in other code:

int array[90];
FILL(array, -1);

Answer 4

90 words isn't much memory. You're likely to use a good fraction of your time allocating/de-allocating the memory. Putting it on the stack is probably faster than dynamically creating the memory. I'd see if a for loop or Omkant's answer would work. If it turns out to really be the bottleneck, then you can start to optimize.

for (i = 0; i < 90; ++i) { array[i] = -1; }

Answer 5

memset( array, -1 , sizeof(array) ) ;

It can be used for initialising with 0 or -1

Answer 6

There is no simple way, calloc only initializes to 0.

you can do

int *array = malloc(sizeof(int)*size);
for (i=0;i<size;i++) array[i] = -1;

or

memset(array,-1,sizeof(int)*size);

You can use memset BUT it only works if you want to use the values "0" or "-1", otherwise it won't work as expected because memset sets the same value for all the bytes.