CODEWITHSUNDEEP
arrayscpointers
It's probable
It's probable

Reputation: 137

Array name and pointer

I was reading this answer- https://stackoverflow.com/a/1644292/10352155

After reading it I have some doubts which go as follows:

It is said in the answer that, as soon as that array expression "decays" to a pointer, all you have is a pointer to a single element.

So does it means that when the array expression is getting converted to a pointer, an extra memory space(other than the memory spaces already allocated for the array elements) of required bytes is allocated to store the value of the pointer expression(which is none other than the base address of the array)?

If this is so, then how is it different from the language B where it was mentioned that a separate memory space is allocated for a pointer expression?

And if it is not so and no extra memory is being allocated, then how is the pointer expression storeing the value of the base address?

Upvotes: 1

Views: 89

Answers (2)

anastaciu
anastaciu

Reputation: 23832

Getting the obvious out of the way, the array is always in the same memory location, it does not change, and you can't move it around nor convert it, once declared it will occupy the same memory space during its entire existance and it will serve the same purpose, to store elements of the declared type. No extra memory is needed, the only memory used is for the array elements themselves.

The name of the array is not an element, it is simply for reference and to ease the programming process, it is compiled away, as C does not have reflection it is not needed, the compiler will only need the array address and that's what is kept, it will also know the number of elements and the size of the elements, so you can use sizeof to render the total memory space occupied by it.

The decay process usually happens on assignment or when you pass the array as parameter, what you end up having is a new pointer to the existing array, and the pointer will contain the address of the initial element of the array, that's all.

That pointer has to be stored, and it is, accordingly, depending on the situation, the system knows the size of the pointer, it is implementation defined.

If you use manual memory allocation then the pointer will also have to be stored as well as the memory space for the array, in the heap for the latter.

This also means that you can no longer know the size of the array using sizeof as you can with the array itself, you will be getting the size of the pointer instead.

Upvotes: 1

Eric Postpischil
Eric Postpischil

Reputation: 224546

The conversion is done in the process of evaluating an expression. The original array is not changed. For example, in:

int x = 3;
int y = 4;
int z = x*y + 7;

the product of x and y is calculated to be 12, but that is just used in the arithmetic. Neither x nor y is changed.

Similarly, in:

int a[] = { 0, 1, 2, 3 };
int *p = a + 2;

the address of the first element of a is calculated by the compiler (based on its knowledge of where a is stored), and a is not changed by this.

Upvotes: 0

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