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pointer_variable=(typecasting
datatype*)malloc(sizeof(datatype));
This is the syntax for malloc?Please explain this,how it
work with an example?
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ya, this is the above syntax for malloc function........
ya i will clearly explain ,
let us take a small block of coding , here my aim is to get 'n' numbers and print the 'n' numbers ......
#include<alloc.h>
void main()
{
int n ,*pointer;
clrscr();
printf("enter the number of elements u r going to enter :");
scanf("%d",&n);
pointer=(int *)malloc(n*sizeof(int));
the above statement states that : , this function is requesting the OPERATING SYSTEM to allocate 'n' amount of memory of a data type integer. and since the return format of the malloc function is an address , so we are type casting as (int*)before malloc , and the returned starting address will be stored in the pointer variable (pointer) ..
this 'pointer' will have the starting address of the allocated memory dynamically...
that's all..
for(int i=0;i<n;i++)
{
scanf("%d",(pointer+i));
}
for(i=0;i<n;i++)
printf("%d\n",*(pointer+i));
getch();
}
thank u
| Is This Answer Correct ? | 6 Yes | 1 No |
Answer / pramod
I'll add few more lines to answer above just to make it a
bit more understandable.
Malloc allocates consequetive blocks of memory of requested
size.Then it returns the starting address of this block ,
the address returned is of type void , since it is a void
pointer so we need to typecast it to one that we requested
so typecasting is done( This headache has been removed in
C++ by using new as new automatically does this typecasting)
| Is This Answer Correct ? | 2 Yes | 0 No |
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