C + + Auto is a confusing situation when initializing with multidimensional arrays

First look at the following procedure:

intarr[1][Ten] = {0,1,2,3,4,5,6,7,8,9}; for(Auto *p = arr;p! = End (arr); p++) {cout<<"P is"<< typeid (P). Name () <<Endl; cout<<"*p is"<< typeid (*p). Name () <<Endl;} for(Auto *Q:arr) {cout<<"Q is"<< typeid (q). Name () <<Endl; cout<<"*q is"<< typeid (*q). Name () <<Endl;} for(Auto &R:arr) {cout<<"R is"<< typeID (R). Name () <<Endl;}

The result of the program running is

P is int (*) [10]

*p is int [10]

Q is int *

*Q is int

R is int [10]

In C + +, operations on arrays are usually performed on pointers. When you initialize a pointer with an array name, you get a pointer to the first element of the array.

The essence of a multidimensional array is an array of elements. such as int a[3][4] is an array of three elements whose elements are arrays of four integral elements. Therefore, auto *p = A,p is a pointer to the first element of array A, which is a pointer to an array with four integral elements, and *p is the first element of a, which is an array of four integral elements.

In range for, however, auto *q = arr is a pointer to the first element of an array element (an array with four elements) . So q is a pointer to an integral type element. If the array has more dimensions, such as the first element of int a2[3][4][5],a2 is an array with four elements, all four elements are arrays of five integral elements. In this case, the type of p in the for (auto *P:A2) is int (*) [5].

To prevent errors caused by this difference, you can use a reference in range for. For Auto &r:arr, R is a reference to an array element with the same type as auto *p = arr, *p.

C + + auto is a confusing situation when initializing with multidimensional arrays