CODEWITHSUNDEEP
cmemory
isshinBD
isshinBD

Reputation: 33

Pointer and memory confusion in C

If I have the following line of code:

int* a[3];

Does that mean it is a pointer to the first item in an array of 3 integers (thus, giving it a size of 4 bytes)? Or does it mean it is an array of 3 pointers (giving it a size of 12 bytes)?

Upvotes: 2

Views: 92

Answers (3)

Peter - Reinstate Monica
Peter - Reinstate Monica

Reputation: 16112

In order to understand expressions and declarations you must know the operator precedences. There are plenty online, for example here.

On that page you can see that angle brackets have a higher priority than the dereferencing asterisk, which means that they are applied first. If you like you can bracket parts of the declaration with parentheses, just like in an expression:

int *(a[10]); // brackets are redundant and show operator precedence

Therefore, the variable a must be an array. The array element resulting from the indexing yields, after the dereferencing operator is applied, an int, so that the element must be a pointer to int.

Thus, the declaration declares an array of int pointers.

As an extra, you cannot declare a pointer to an array without parentheses or typedefs, just because the operator precedences are made for the much more common case above. A pointer to an array would look like this:

int (*a)[10]; // brackets are necessary to make "*" apply first

Here, the dereferencing asterisk is applied first; a therefore must be a pointer. The resulting type can be indexed, therefore must be an array, whose elements are declared ints. This makes a a pointer to an array of ints.

Upvotes: 1

abhiarora
abhiarora

Reputation: 10450

It is the array of pointers to int data type in C programming language. Therefore, each element of this array can hold an address of a variable of int data type.

 int x=5, y=9, z=8;
 int *a[3]; // Declare an array of pointers to int

 a[0] = &x;
 a[1] = &y;
 a[2] = &z;

However, the declaration like below will declare a pointer to an array of 3 int.

int (*a)[3];

Size of a is dependent on the platform for which you will compile your code. Please use sizeof operator for determining size in your platform.

 int *a[3];
 printf("Size of a: %zu\r\n", sizeof(a));

NOTE:

To print result of sizeof operator, use %zu if your compiler supports C99; otherwise, or if you want maximum portability, the best way to print a size_t value is to convert it to unsigned long and use %lu. You can read about it over here.

Upvotes: 4

Govind Parmar
Govind Parmar

Reputation: 21572

You can test this yourself with the following code:

#include <stdio.h>
int main()
{
        int *a[3];
        printf("%zu\n", sizeof(a));
        return 0;
}

The output is 12, which, on the machine I'm testing it on, and apparently yours too from what you wrote in your question, is 3*sizeof(int).

Type declaration precedence in C puts the "array" declaration [] one step above the "pointer to" declaration (*), so the declaration is:

Declare 'a' as an array of 3 pointers to type 'int'

You can use parentheses (()) to change the order in which the type declaration is processed:

int (*a)[3];

sizeof(a) gives it an output of 4, which on your machine is the size of a pointer.

As a side note, it's generally bad practice to assume that a pointer is always 4 bytes and that an int is always 4 bytes. This varies depending on architecture, compiler, CPU mode, etc.

Upvotes: 0

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