将C ++数组转换为向量

Question

Here is the code I found on Leetcode . However I cannot make sense of the following two lines, especially *(&a + 1) . And the results show a copy of the array a . Could anyone give some explanation to this? Thanks!

    int a[5] = {0, 1, 2, 3, 4};
    vector<int> v4(a, *(&a + 1));

Answer 1

The examples that you normally face when constructing a vector from an array usually looks like the following:

int a[5] = {...};
vector<int> v4(a, a + 5 );
// or vector<int>(a, a+sizeof(a)/int); // automatically calculate num elems inside a

The example above simply shows you want to construct a vector using all the elements inside array "a". "a" in this example is just the address to the starting element of the array, then you add 5 to get the address of the last element. A more intuitive way to write this would be:

vector<int> v4(begin(a), end(a));

Unless of course you don't want all the elements of a in v4

Now the example you have given us is the shorthand of the first example where you don't need to explicitly state the size of the array. I'm assuming you are just confused with the second argument of the vector constructor. The second argument just returns the address of the last element of array "a". But how?

Well let's break it down: &a returns the address to the "a[]". Essentially a pointer to the array "a". Now if you add one to this address, you will get a pointer to the address "a[0]" + sizeof a[] which will point to the last element of address a. Compare this to the first example above. You had to add 5, why is this not the case here? Well you are working with different units. &a points to the starting address of a[] rather than the address of the first element of the array. So you are essentially moving the address in units of a[5] rather than units of int.

Then finally you have the dereference operator "*" to dereference the pointer and get the address of array[] for the last element.

a is type a[] and &a is type *a[] so the dereference is needed to make it the same type otherwise you get a compiler error.

TLDR: You are getting the address for the last element using different methods. The +1 operator behaves relative to the type you are dealing with. a + 1 is the same as starting address of "a" + sizeof 1 integer. &a + 1 is the same as starting address of a[] + size of a[].

Answer 2

You're getting confused by array decay. a is the array "as a whole". It decays into a pointer pointing to the first element in MOST contexts, but the operand of unary-& is one of the few that it does not. So &a gives you the address of the array as a whole, not the address of the first element. These are the same place, but have different types ( int (*)[5] vs int * ) and that different type means that pointer arithmetic is different.