Dynamic char array, size is fixed?

Question

I am trying to create a char array dynamically. However, do not matter which number I put as the desired value of the size array it just keep using 4 has the size. If I place 3,2 or even 8, 10 it does not change...

Part where I initialize it...

char *x;

x = (char *)calloc(10, sizeof(char));

if(x == NULL )
{
    return;
}

cout <<"Size_Vector = "<<sizeof(x)<<endl; //Keep making the size of the array 4..no matter what

score 2 · Answer 1 · edited May 23 '17 at 12:31

x is a pointer of type char *. sizeof(x) returns the size of the pointer instead of the array size.

Generally, there is no way to know the array size from a pointer type even if it's pointing to an array, because of array decaying.

char *x = malloc(sizeof(char)*10);
// This will give you 4
printf ("%d\n", sizeof(x));

But you can get the size of an array type.

char x[10];
// This will give you 10
printf ("%d\n", sizeof(x));

score 2 · Accepted Answer · answered Apr 18 '12 at 04:33

2

sizeof(x) returns the size of the pointer not the allocated memory or the size of array.

It always returns 4 size of an pointer on your envrionment is 4and it will be the same for all.

You will have to keep track of how much memory you allocated yourself & also ensure that you do not write beyond the bounds of that allocated memory.

answered Apr 18 '12 at 04:33

Alok Save

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What about if I want to create the array size 2? How I change the sizeof(char) in the calloc then? – SpcCode Apr 18 '12 at 04:35
1

@SpcCode: `x = (char *)calloc(2, sizeof(char));` that gives you memory allocation of 2 `char`. – Alok Save Apr 18 '12 at 04:38
Oh understood... The compiler was crashing because I was printing like this:for (i=0; i – SpcCode Apr 18 '12 at 04:44

score 0 · Answer 3 · edited Jul 24 '20 at 23:15

0

The malloc call returns a void pointer and when you say sizeof(pointer) you ask for the size of the pointer and not the chunk of memory that you just declared.So it gives you 4/8/... bytes depending on the m/c you are using.

Try this out :

char *cptr;

int *iptr;

printf("size of integer pointer is : %d\n",sizeof(iptr));


printf("size of character pointer is : %d\n",sizeof(cptr));

Note : malloc points to the address of the first element of the chunk of memory(ie 10 * 1 = 10 bytes) that you have dynamically obtained by calling it.

edited Jul 24 '20 at 23:15

trinaldi

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answered Apr 18 '12 at 05:07

Phoenix225

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Pointers aren't always 4 bytes. On a 64 bit system, they're 8. An on unusual systems, they could be any size. – Timothy Jones Apr 18 '12 at 05:16
Sorry i was referring to a 32 bit m/c out here. – Phoenix225 Apr 18 '12 at 05:22
but in any case the call to calloc or malloc will return a void pointer and when you do sizeof(pointer) you are always going to get the size which can be 4/8/ ...or whatever depending on the m/c – Phoenix225 Apr 18 '12 at 05:24
If you edit that in to your answer instead of saying "always 4 bytes" I'll give you an upvote :) – Timothy Jones Apr 18 '12 at 05:28

score 0 · Answer 4 · edited Apr 01 '20 at 12:55

This is a short piece of code i wrote for you in case you really want to find the size of the memory piece you reserved by calling calloc though its obvious that when you say 10 spaces of 1 byte each then you are going to get 10 bytes like the case here which says you want 10 * sizeof(char) :

#include<stdio.h>

#include<stdlib.h>

main(){

    char *ptr = NULL;

    int i = 0;

    int size = 0;

    int n = 0;

    printf("Enter the size of the dynamic array in bytes : ");

    scanf("%d",&n);

    ptr = (char*) calloc (n,sizeof(char));

    for(i=0;i<n;i++){

            size = size + sizeof(ptr[i]);

    }

    printf("Size is %d\n",size);

}

Dynamic char array, size is fixed?

4 Answers4