The difference between array name A and &a in C + + learning

Define an array in C + +:

int a[5]={1,2,3,4,5}

So what does a really mean? The first thought is that a represents the first element address of an array. So what if you ask what type A is? A lot of people are not sure, look at the following procedure, let the compiler tell us what kind of data A is:

Main.cpp as follows:

From the error message in the picture, you can see:

The type of a is an int [5] type. It is not possible to assign a pointer to an int (*) [5] type directly, so you need to take an address action on a &a the type that is given is the INT (*) [5] Type (pointer type that points to the int [5] type).

So why is a in the picture and can be directly assigned to an int * type variable p3? This is because the int [5] type is implicitly converted at assignment, an int [5] type of a is converted to a variable of type int *, stating the int[5] type and int * The type is very similar to the data type. And we know &a[0] The data type is int * Type, so it is generally said that the array name A is the address of the first element represented, in fact there is a certain difference, the specific difference is not sure.

Based on the above analysis, a and &a are actually two different data types, a is an int [5] type (like int *, which can be understood as a pointer to the first address of a 5-element array), and &a is an int (*) [5] Type ( Can be understood as pointing to a pointer to an array of 5 child elements), note the difference between the whole and the individual.

So if we perform +1 operations on a and &a respectively, the results are different.

#include <iostream> #include <string> #include <vector> #include "Dog.h" using namespace Std;int main ( int argc, const char * argv[]) {    int a[5]={0,1,2,3,4};    cout << "A:" << a << "<< a+1 <<endl;    cout << "&a:" << &a << "<< &a+1 <<endl;    return 0;}

The output is:

Can see a words +1 is actually added an int size 4 bytes, and &a+1 Plus is the entire array of memory size, which also reflects a and &a individual and the difference between the whole. This should be the result of a different type overloaded + number operator, which, like an int [5] Type overloaded + number operator, is the size of an int, and the int (*) [5] Type is the size of the 5 int type.

The difference between array name A and &a in C + + learning