Simple C array declaration / assignment question

Question

In higher level languages I would be able something similar to this example in C and it would be fine. However, when I compile this C example it complains bitterly. How can I assign new arrays to the array I declared?

int values[3];

if(1)
   values = {1,2,3};

printf("%i", values[0]);

Thanks.

Answer 1

You can only do multiple assignment of the array, when you declare the array:

int values[3] = {1,2,3};

After declaration, you'll have to assign each value individually, i.e.

if (1) 
{
  values[0] = 1;
  values[1] = 2;
  values[2] = 3;
}

Or you could use a loop, depending on what values you want to use.

if (1)
{
  for (i = 0 ; i < 3 ; i++)
  { 
    values[i] = i+1;
  }
}

Answer 2

In C99, using compound literals, you could do:

memcpy(values, (int[3]){1, 2, 3}, sizeof(int[3]));

or

int* values = (int[3]){1, 2, 3};

Answer 3

 //compile time initialization
 int values[3] = {1,2,3};

//run time assignment
 value[0] = 1;
 value[1] = 2;
 value[2] = 3;

Answer 4

you can declare static array with data to initialize from:

static int initvalues[3] = {1,2,3};
…
if(1)
    memmove(values,initvalues,sizeof(values));

Answer 5

#include<stdio.h>
#include<stdlib.h>
#include<stdarg.h>

int *setarray(int *ar,char *str)
{
    int offset,n,i=0;
    while (sscanf(str, " %d%n", &n, &offset)==1)
    {
        ar[i]=n;
        str+=offset;
        i+=1;
    }
    return ar;
}

int *setarray2(int *ar,int num,...)
{
   va_list valist;
   int i;
   va_start(valist, num);

   for (i = 0; i < num; i++) 
        ar[i] = va_arg(valist, int);
   va_end(valist);
   return ar;
}

int main()
{
    int *size=malloc(3*sizeof(int*)),i;
    setarray(size,"1 2 3");

    for(i=0;i<3;i++)
        printf("%d\n",size[i]);

    setarray2(size,3 ,4,5,6);
    for(i=0;i<3;i++)
        printf("%d\n",size[i]);

    return 0;
}