我怎么理解&*(x+i)？

Question

This is c++ code:

x[5]={1,2,3,4,5};
&x[i]==&*(x+i)==x+i;

as for &*(x+i) ,in my opinion, *(x+i) is a rvalue ,why use & . why?

Answer 1

This line &x[i]==&*(x+i)==x+i; shell show you 3 different ways on getting the address of an specific element of an array.

In your case there is an array x which stores 5 values. The first thing probaly looks kind of familiar to you. x[i] will get the i -th element of the x and by using the address-of-operator you will get the address of the i -th element of.

Let's go on to the third statement x+i . Since x is actually just a pointer pointing to the adress of the first element of the actual array in your storage you can get the adress of other elements by adding an offset to x . However since you're working with x ' address you wont get the value of the 'offsetted' element but its address. To get the value you need to derefence the address by using the magical * . So if &x[i] gets the adress of the i th element and x+i does the same, *(x+i) will do the same thing as x[i] . It will get the i th value of x .

No you know what the adress-of-operater & and the derefencing-operator * . Theirefore the second statement is fairly simple. (x+i) will get the address of the i th element. *(x+i) will get the value by dereferencing the address and &*(x+i) will get the adress of the derefenced address. So the second statement is the same thing as just writing x+i .

However the explained line will probably give you a compiler error since it should more look like this: (&x[i]) == (&*(x+i)) && (&x[i]) == (x+i);

Conclusion: All these things do the same thing yet slightly differently.

Answer 2

Lets suppose we have this piece of code

int i = 0;
int x[5]={1,2,3,4,5};

then this comparision is true:

&x[i] == &*(x+i)

and this comparision is also true for similar reasons.

&*(x+i) == x+i;