为什么在c ++中使用静态数组＆a和相同？

Question

int a[10];

Now here &a and a is same, I don't clearly understand logic why is it so. Can I say that value of a is same as &a and *(a) is a[0]. (But how a is not a pointer). In case of dynamic array things are clear to me.

int *da = new int[10];

value of da is base address, &da gives the address where pointer is stored, and we deference da to reach da[0].

Answer 1

They are not the same.

a is an int[10] type; an array .

&a is a pointer to an int[10] type.

In certain situations (passing to functions, when used in arithmetic expressions), a decays to an int* type. You can then use pointer arithmetic to reach other elements of the array. That could be what is causing your confusion.

da is an int* type.

The only thing that decays to itself is a function pointer .

Answer 2

well 'a' and '&a' are definitely not the same

int a[10] takes up a specific block from memory to allocate exactly 10 integers and this block is called a.

'&a' refers to the memory address of this memory block of 10 integers you made. it's like you have a hotel room (the array of 10 elements) in your hotel. And (your memory address) '&a' refers to your room number, your address.

now saying for example int *p = &a means I'm making an integer pointer 'p' that can only point to integers, and i'm giving it the memory address of 'a' (the array of integers) so it can point to it.

so :

int a[10] is your hotel room;

&a is your room number;

*p = &a is the key chain attached to your room number (ur keychain points to ur room number, your address);

this is why sometimes we use : int *da = new int[10]; to immediately allocate in memory a space for 10 integers and directly point a pointer (in this case da) to it.

Hope this helps.