Creating integer variable of a defined size.

Question

I want to define an integer variable in C/C++ such that my integer can store 10 bytes of data or may be a x bytes of data as defined by me in the program. for now..! I tried the

int *ptr;
ptr = (int *)malloc(10);

code. Now if I'm finding the sizeof ptr, it is showing as 4 and not 10. Why?

Answer 1

Because the size of the pointer is 4. Try something like:

typedef struct
{
    int a[10];
} big_int_t;

big_int_t x;

printf("%d\n", sizeof(x));

Note also that an int is typically not 1 byte in size, so this will probably print 20 or 40, depending on your platform.

Answer 2

Because on you machine, size of a pointer is 4 byte. Please note that type of the variable ptr is int *. You cannot get complete allocated size by sizeof operator if you malloc or new the memory, because sizeof is a compile time operator, meaning that at compile time the value is evaluated.

Answer 3

Integers in C++ are of a fixed size. Do you mean an array of integers? As for sizeof, the way you are using it, it tells you that your pointer is four bytes in size. It doesn't tell you the size of a dynamically allocated block.

Answer 4

It is showing 4 bytes because a pointer on your platform is 4 bytes. The block of memory the pointer addresses may be of any arbitrary size, in your case it is 10 bytes. You need to create a data structure if you need to track that:

struct VariableInteger
{
    int *ptr;
    size_t size;
};

Also, using an int type for your ptr variable doesn't mean the language will allow you to do arithmetic operations on anything of a size different than the size of int on your platform.

Answer 5

Few or no compilers support 10-byte integer arithmetic. If you want to use integers bigger than the values specified in <limits.h>, you'll need to either find a library with support for big integers or make your own class which defines the mathematical operators.

Answer 6

C and C++ compilers implement several sizes of integer (typically 1, 2, 4, and 8 bytes {8, 16, 32, and 64 bits}), but without some helper code to preform arithmetic operations you can't really make arbitrary sized integers.

The declarations you did:

int *ptr;
ptr = (int *)malloc(10);

Made what is probably a broken array of integers. Broken because unless you are on a system where (10 % sizeof(int) ) == 0) then you have extra bytes at the end which can't be used to store an entire integer.

There are several big number Class libraries you should be able to locate for C++ which do implement many of the operations you may want preform on your 10 byte (80 bit) integers. With C you would have to do operation as function calls because it lacks operator overloading.

Your sizeof(ptr) evaluated to 4 because you are using a machine that uses 4 byte pointers (a 32 bit system). sizeof tells you nothing about the size of the data that a pointer points to. The only place where this should get tricky is when you use sizeof on an array's name which is different from using it on a pointer. I mention this because arrays names and pointers share so many similarities.

Answer 7

I believe what you're looking for is known as "Arbitrary-precision arithmetic". It allows you to have numbers of any size and any number of decimals. Instead of using fixed-size assembly level math functions, these libraries are coded to do math how one would do them on paper.

Here's a link to a list of arbitrary-precision arithmetic libraries in a few different languages, compliments of Wikipedia: link.