Dynamic memory allocation from the heap

Question

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

int main ()
{
    int a,n;
    int * ptr_data;


    printf ("Enter amount: ");
    scanf ("%d",&a);
    int arr[a];
    ptr_data = (int*) malloc ( sizeof(int) );
    for ( n=0; n<a; n++ )
    {
        printf ("Enter number #%d: ",n);
        scanf ("%d",&ptr_data[n]);
    }

    for ( n=0; n<a; n++ )
        arr[n] = ptr_data[n]; //values in heap are copied to an array.
    printf("%d\n",sizeof(ptr_data));
    free (ptr_data);
    printf ("Output: ");
    for ( n=0; n<a; n++ )
        printf("%d\n",arr[n]);
    return 0;
}

the sizeof function prints the size of pointer as 4(integer's size). but i run a loop to get the values to be stored in the heap (scanf,.), so how does this subsequent pointee pointer mapping happens after first time? as in, first time i explicitly code that the ptr_data to contain a size of 4. but after that when i want to store another integer this pointer shows another random memory refernce(this mapping explanation is what all i want.)

score 1 · Accepted Answer · answered Dec 04 '13 at 16:12

What you have in your code is an Undefined behavior.

 ptr_data = (int*) malloc ( sizeof(int) );

Allocates memory equivalent to size of a single integer on heap and ptr_data points to this allocated memory. Further,

for ( n=0; n<a; n++ )
{
    printf ("Enter number #%d: ",n);
    scanf ("%d",&ptr_data[n]);
}

Writes beyond the bounds of the allocate memory:

ptr_data[n];

is equivalent to:

*(ptr_data + n);

Note that writing beyond the bounds of memory which belongs to you results in Undefined behavior in C and C++ both. With undefined behavior the program doesn't need to behave in any specific way.It many work correctly or crash or show any random behavior, simple the behavior is undefined.

but its working. so you say that the compiler allocates the memory automatically? This is one beautiful explanation btw! Thanks! — bks4line, Dec 04 '13 at 16:18
@bks4line: A [good read](http://stackoverflow.com/questions/2397984/undefined-unspecified-and-implementation-defined-behavior).It is important to understand these concepts if you are working on C and C++. — Alok Save, Dec 04 '13 at 16:21

trifork · Answer 2 · 2013-12-05T01:21:00.930

0

The line ptr_data = (int*) malloc ( sizeof(int) ) allocates memory for only one integer pointer. You're trying to assign more than one value into ptr_data. If you want to do this, then:

ptr_data = malloc (sizeof(int) * a);

This will give the pointer room for int a integers.

edited Dec 05 '13 at 01:21

answered Dec 04 '13 at 16:34

trifork

57
1
10

You accidentally casted the return value of malloc. Don't do that. – Dec 04 '13 at 17:10
I changed it. But why? What's the problem? – trifork Dec 05 '13 at 01:21
2

Casting has no benefits, but it can hide errors and it decreases readability. – Dec 05 '13 at 05:37

Dynamic memory allocation from the heap

2 Answers2