How to initialize an array in c with incrementing values at compile time?

Question

For example I want to initialize an array of 5000 integers - and I want to initialize it like a[] = {1,2,....4999} I dont want to write 5000 integers while initializing the array at compile time.

Why do you think you need to do this rather than just initialising at run-time ? — Paul R, Sep 17 '13 at 09:40
Rather than an array, you can just use `i+1` instead of `a[i]` — David Heffernan, Sep 17 '13 at 09:44
@PaulR One obvious reason for doing so would be storing the array in a const. For example when implementing a lookup table stored in non-volatile memory. — Lundin, Sep 17 '13 at 10:02
@Lundin: yes, I think we were just discussing this very point in the comments to my answer below. — Paul R, Sep 17 '13 at 10:03

Paul R · Answer 1 · 2013-09-17T09:52:03.080

4

You could do it with a bunch of macros - its evil and it's ugly and I can't see why you'd want to, but here you go:

#define V10(x) (x), (x)+1, (x)+2, (x)+3, (x)+4, (x)+5, (x)+6, (x)+7, (x)+8, (x)+9

#define V100(x) V10(x), V10(x+10), V10(x+20), V10(x+30), V10(x+40), \
                V10(x+50), V10(x+60), V10(x+70), V10(x+80), V10(x+90)

#define V1000(x) V100(x), V100(x+100), V100(x+200), V100(x+300), V100(x+400), \
                 V100(x+500), V100(x+600), V100(x+700), V100(x+800), V100(x+900)

const int a[5000] = { V1000(1), \
                      V1000(1001), \
                      V1000(2001), \
                      V1000(3001), \
                      V1000(4001) };

edited Sep 17 '13 at 09:52

answered Sep 17 '13 at 09:45

Paul R

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@Neil: quite - I was giving the OP the benefit of the doubt though - maybe there is a genuine requirement (unlikely I know), e.g. it's some embedded thing that has a real-time constraint on startup time... – Paul R Sep 17 '13 at 09:48
@PaulR - Exactly. The system reboots quite a lot and it needs to be up as soon as possible. Apart from this, why do something at run time, which can easily be done at compile time. – Sumit Nathany Sep 17 '13 at 09:56
@Sumit: well if you can avoid a copy (e.g. if it's a const array and it is loaded directly from a const segment) then there might be a valid case, but otherwise you're just trading an initialisation loop for a copy loop and performance will not be very different. – Paul R Sep 17 '13 at 09:59

score 3 · Answer 2 · answered Sep 17 '13 at 09:46

For C, your best bet might be to use an external script to generate the C code, like:

$ echo "unsigned int myarray[] = { $(seq -s ", " 0 9) };"
unsigned int myarray[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };

You can pipe that into a dedicated source file and include that in your code.

(In C++ you have better options thanks to template magic.)

score 2 · Answer 3 · answered Sep 17 '13 at 09:41

2

You can't initialize the array directly with the incrementing values in C. You will have to either initialize it by specifying all the values or using a loop at runtime.

answered Sep 17 '13 at 09:41

HAL

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score 0 · Answer 4 · answered Sep 17 '13 at 09:55

0

Initializing the array directly with the incrementing values in C is not possible, although it may be possible using linker scripts as mentioned by others. Best is to use a for loop... for(i = 0, i < 5000, i++) a[i] = i + 1;

answered Sep 17 '13 at 09:55

Trilok M

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score 0 · Answer 5 · answered Sep 17 '13 at 09:59

For smaller arrays, you could do like this:

#include <stdio.h>

#define OFFSET (9 - 1)  // start at line 9, -1 for 1 indexing

int main()
{
  int array[] = 
  {
    __LINE__ - OFFSET,
    __LINE__ - OFFSET,
    __LINE__ - OFFSET,
    __LINE__ - OFFSET,
    __LINE__ - OFFSET
  };

  for(int i=0; i<sizeof(array)/sizeof(*array); i++)
  {
    printf("%d ", array[i]);
  }
}

But of course with 5000 integers, this method will be quite tedious. The best option is to generate the C source code from an external script.

(For example, every beginner C programmer should be capable of writing a program that prints `{1, 2, 3.... 5000}` to a text file.) — Lundin, Sep 17 '13 at 09:59

score 0 · Answer 6 · answered Sep 17 '13 at 10:26

0

Use an external assembly file to initialize the data and compile it with your C code. You may need to specify the .data segment directive.

answered Sep 17 '13 at 10:26

TwoCode

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score -1 · Answer 7 · answered Sep 17 '13 at 09:44

-1

Why not do something like this:

int size = 5000;
int arr[size];
int i = 1;
for(i=1; i <= size; i++)
{
    arr[i - 1] = i; // Initializing each element seperately
}

answered Sep 17 '13 at 09:44

Armand

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2

And what about the buffer overrun? Note that `sizeof(arr)` is at least 10000 and likely more. – David Heffernan Sep 17 '13 at 09:45
@hal sorry missed that one – Armand Sep 17 '13 at 09:45
2

sizeof(arr) = sizeof(int) * 5000. You probably want sizeof(arr)/sizeof(arr[0]) – Neil Sep 17 '13 at 09:45
1

The question asks for initialization at compile time. run time initialisation can always be done using some loop. – Sumit Nathany Sep 17 '13 at 09:54
1

No better. Now you've got a VLA. In any case, I suspect that the asker already knows how to run a for loop to initialize an array. – David Heffernan Sep 17 '13 at 10:01

How to initialize an array in c with incrementing values at compile time?

7 Answers7