C: dynamic declare array size

Question

I have the following declaration inside a function

int f[20000]

I want the number 20000 to be dynamic, How can i declare such array in code?

To be more specific, I have the following code to calculate PI.

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

#define BITS 2000
int a=10000,b,c=BITS*7/2,d,e,f[BITS*7/2+1],g; 

int main() 
{   
    for(;b-c;) 
        f[b++]=a/5; 
    for(;d=0,g=c*2;c-=14,printf("%.4d",e+d/a),e=d%a) 
        for(b=c;d+=f[b]*a,f[b]=d%--g,d/=g--,--b;d*=b);     
    //getchar();
    return 0;
}

I changed to

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

int main(int argc, char *argv[]) 
{   
    //
    // .... omit some lines here
    // read bits from user input at runtime
    // say precision = 200
    //
    int a=10000,b,c=precision *7/2,d,e,f[precision *7/2+1],g; 
    for(;b-c;) 
        f[b++]=a/5; 
    for(;d=0,g=c*2;c-=14,printf("%.4d",e+d/a),e=d%a) 
        for(b=c;d+=f[b]*a,f[b]=d%--g,d/=g--,--b;d*=b);     
    //getchar();
    return 0;
}

It doesn't work, I googled then changed to

int a=10000,b,c=precision *7/2,d,e,g; 
int *f=calloc(precision *7/2+1, sizeof(int));

It still doesn't work, I mean the program doesn't crash, the value it calculated is not correct. What's wrong? Thank you.

What is the value in precision. Where are you reading the user value for precision. — sumithdev
– sumithdev, Commented Mar 18, 2015 at 20:08
a 20000 * sizeof int array, (80000 bytes) on the stack is a very bad idea. probably resulting in an overflow of the stack. suggest: int arraySize = 20000; int pf = malloc( arraySizesizeof int); (then check that 'pf' does not contain NULL) (be sure to use: 'free(pf);' when done with the array — user3629249
– user3629249, Commented Mar 18, 2015 at 20:10
this code does not cleanly compile. a couple of the reasons are the parameters argc and argv are not used. suggest using 'int main(void) — user3629249
– user3629249, Commented Mar 18, 2015 at 20:12
the ''7/2' is performing an integer divide, so the result will always be '3' not 3.5' — user3629249
– user3629249, Commented Mar 18, 2015 at 20:14

Sourav Ghosh · Accepted Answer · 2015-03-18 20:49:17Z

3

There are two ways to achieve what you want.

use dynamic memory allocation. malloc()/calloc()
use variable-length array (in c99)

That said, as pointed out by @user3386109, the problem in your second code snippet is use of uninitiated variable b. You may want to explicitly initialize the local variables before using their value.

answered Mar 18, 2015 at 20:49

Sourav Ghosh

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mattm · Accepted Answer · 2015-03-18 20:07:40Z

3

You get a dynamically sized array by allocating on the heap using malloc (or calloc).

Replace

int f[20000];

with

int *f = (int *) malloc(20000 * sizeof(int) );

answered Mar 18, 2015 at 20:07

mattm

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Comments

user3386109 · Accepted Answer · 2015-03-18 20:15:08Z

1

The difference is that global variables are guaranteed to be initialized to 0 (unless initialized to some other value). But local variables are garbage unless you initialize them. So the problem is that variable b starts out as garbage in the second snippet.

In the original code:

int a=10000,b;
int main(void)
{
}

a will start with the value 10000 because you initialized it, and b will start as 0 because it is an uninitialized global variable.

In the changed code:

int main(void)
{
    int a=10000,b;
}

a will start with the value 10000 because you initialized it, and b will start as some random value (e.g. 0x5315fe) because it is an uninitialized local variable.

edited Mar 18, 2015 at 20:15

answered Mar 18, 2015 at 20:09

user3386109

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2 Comments

user3629249 Over a year ago

the question was NOT 'what is wrong with my code>', (of which there are lots of things wrong). the question was 'how to dynamically size an array?'

user3386109 Over a year ago

The question itself had two perfectly good ways to dynamically allocate an array, but from the OP's point-of-view neither appeared to work. So then the question does become, "Why doesn't this code work?"

Roger Alcindor · Accepted Answer · 2015-03-18 20:21:00Z

-1

Replace int f[2000]

with

int *f = new int[2000];

then use the array as f[0] = 1, f[1] = 2 etc...

when finished free up memory with delete [] f;

the array size could be allocated by a variable

eg. int x = 2000; f = new int[x];

answered Mar 18, 2015 at 20:21

Roger Alcindor

1

1 Comment

Sourav Ghosh Over a year ago

This question is tagged c and AFAIK, new is not c.

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