Dynamically assigning numbers to an array in C

Question

Hello,

I want to do this:

char a[5];

//run some code here
//then

a[]={0,1,2,3,4};  //**compiler doesn't like it

but I don't want to do this:

a[0]=0;
a[1]=1;
a[2]=2;
a[3]=3;
a[4]=4;

Do you know any way to populate an array at run time with numbers, not a string (i.e., not a ="hello"), in one go, instead of defining every element individually?

Thanks,

Raed

Answer 1

No, there is no way of doing that. You need to use a loop, or assign each value individually.

Answer 2

Have you tried this:

char a[] = {0, 1, 2, 3, 4};

The compiler will automatically set this to be an array of 5, initialized with each element corresponding to their value and index.

Edit: Looking at it again, I realized what you were looking for. In short the compiler will not accept this.

Hope this helps, Best regards, Tom.

Answer 3

You could do memcpy(a, "\0\1\2\3\4", 5);, but it's bad practice and coding style.

Short of that no you can't.

Answer 4

sprintf( a, "%d%d%d%d%d", 0,1,2,3,4 );

Changed my answer.

Answer 5

Well, you could set up a template to be copied into it later:

#include <stdio.h>
#include <string.h>

static char a_src[] = {0,10,20,30,40};
int main() {
    char a[5];

    memcpy (a, a_src, sizeof(a_src));
    printf ("%d\n", a[3]);
    return 0;
}

Outputs 30 when run.

But this is still techically sourcing the data at compile-time. If you really want to do it at runtime (with calculated values), you need to do it element by element.

Answer 6

char a[5];

//run some code here
//then

static const char a_01234[sizeof(a)] = {0,1,2,3,4};
memcpy(a, a_01234, sizeof(a));

Answer 7

You could use a pointer and reference it as an array.

char * a;
char b[5] = {0,1,2,3,4};
char c[5] = {5,6,7,8,9};

//run some code here
//then

a = b;  // "Populate" the "array"

// Then reference a using array notation
printf ("%d\n", a[3]);  // Print the number 3

// run some more code

a = c;  // "Populate the "array" with some new values
printf ("%d\n", a[3]);  // Print the number 8

Answer 8

Using C99 compound literals (supported by GCC but not MSVC),

char a[5];

//run some code here
//then

memcpy(a, (char[]){0,1,2,3,4}, sizeof(a));