Can you define the size of an array at runtime in C

Question

New to C, thanks a lot for help.

Is it possible to define an array in C without either specifying its size or initializing it.

For example, can I prompt a u

Answer 1

Above given answers are correct but there is one correction, the function malloc() reserve a block of memory of specified size and return a pointer of type void* which can be casted into pointer of any form. Syntax: ptr = (cast-type*) malloc(byte-size)

#include<stdio.h>
#include<cstdlib>
int main(int argc,char* argv[]){
int *arraySize,length;
scanf("%d",&length);
arraySize = (int*)malloc(length*sizeof(int));
for(int i=0;i<length;i++)
    arraySize[i] = i*2;
for(int i=0;i<length;i++)
    printf("arrayAt[%d]=%d\n",i,arraySize[i]);
free(arraySize);
}

Answer 2

Yes, absolutely. C99 introduced the VLA or Variable Length Array. Some simple code would be like such:

#include <stdio.h>

int main (void) {

    int arraysize;
    printf("How bid do you want your array to be?\n");
    scanf("%d",&arraysize);
    int ar[arraysize];  
    return 0;
}

Answer 3

Well, you can dynamically allocate the size:

#include <stdio.h>

int main(int argc, char *argv[])
{
  int *array;
  int cnt;
  int i;

  /* In the real world, you should do a lot more error checking than this */
  printf("enter the amount\n");
  scanf("%d", &cnt);

  array = malloc(cnt * sizeof(int));

  /* do stuff with it */
  for(i=0; i < cnt; i++)
    array[i] = 10*i;

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

  free(array);

  return 0;
}

Answer 4

If all you need is a data structure where in you can change its size dynamically then the best option you can go for is a linked list. You can add data to the list dynamically allocating memory for it and this would be much easier!!

Answer 5

Perhaps something like this:

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

/* An arbitrary starting size. 
   Should be close to what you expect to use, but not really that important */
#define INIT_ARRAY_SIZE 8

int array_size = INIT_ARRAY_SIZE;
int array_index = 0;
array = malloc(array_size * sizeof(int));

void array_push(int value) {
  array[array_index] = value;
  array_index++;
  if(array_index >= array_size) {
    array_size *= 2;
    array = realloc(array, array_size * sizeof(int));
  }
}

int main(int argc, char *argv[]) {
  int shouldBreak = 0;
  int val;
  while (!shouldBreak) {
    scanf("%d", &val);
    shouldBreak = (val == 0);
    array_push(val);
  }
}

This will prompt for numbers and store them in a array, as you asked. It will terminated when passed given a 0.

You create an accessor function array_push for adding to your array, you call realloc from with this function when you run out space. You double the amount of allocated space each time. At most you'll allocate double the memory you need, at worst you will call realloc log n times, where is n is final intended array size.

You may also want to check for failure after calling malloc and realloc. I have not done this above.

Answer 6

Arrays, by definition, are fixed-size memory structures. You want a vector. Since Standard C doesn't define vectors, you could try looking for a library, or hand-rolling your own.

You need to do dynamic allocation: You want a pointer to a memory address of yet-unkown size. Read up on malloc and realloc.