Arrays and pointers in C ++

The array and pointer in C ++ first look at two definitions (Declaration) int * aryp [10]; int (* pary) [10]; aryp is an array of 10 elements, the element type in the array is int * pary, which is a pointer to the array, the type of the specified array is int [10]. In this case, the Pary definition will clear a bit of typdef int int_ary_ten [10]; int_ary_ten * pary; c ++ is a strong type (strong typed) language. c ++ has a special set of types: the type of the array is not only related to the element type, it is also related to the number of elements. Int Aten [10]; int anine [9]; then the Aten type is different from the anine type. The Aten type is int [10], the anine type is int [9] instead of the int * const, although we can use it as an int * Const. This is different from Java. in Java, the two types are array arrays. They cannot be used as function parameters or return value types void Foo (INT Arg [10]). {} The ARG type is actually int * const, and it has nothing to do with that 10. It doesn't even matter if you don't write anything in. This is why some people say that the array name is a constant pointer. In addition to functions, arrays cannot be used as the typeid parameter typeid (Aten ). name () returns "int *". For this reason, typeid (Aten) = typeid (anine) returns trueaten. When anine is passed in, the type is changed to int * const (typeid does not record constant information, so cosnt is removed). Since the array is also a type, the * operator can define another type: pointer to the array. For example, the starting pary can define int (* pary) [10] = & Aten; and INT (* pary) [10] = & anine; is invalid. The compiler complained: cannot convert from 'int (*) [9] 'to 'int (*) [10]'