c++--functions that return array pointers

Basics: Arrays cannot be copied, functions cannot return arrays, only pointers or references to arrays are returned.

A method that defines a function that returns an array pointer , taking as an example a function that receives a parameter as a reference to an array containing 10 integral elements and a pointer to an array containing 10 integral elements.

Law one: (do not use aliases)

int(*getcopyarr (int(&arr) [Ten]))[Ten]{    int(*n) [Ten] = (int(*) [Ten])New int[Ten];  for(inti =0; I <Ten; i++)        (*N) [I] =Arr[i]; returnN;}//Note: The function returns an array allocated in the heap, and if it does not release memory after the array is obtained, it will cause a memory leak, which is only as an example .

Law II:

// tail return type (c++11 standard)auto getCopyArr2 (int(&arr) [ten]),int(*) [ ]{    return  Getarr (arr);}

Fahsarm

// Use type aliases (1) int arrt[];arrt* GETCOPYARR3 (int(&arr) [ten]) {     return Getarr (arr);}

Law IV:

// Use type aliases (2) (C++11 standard) using int [];arrt2* GETCOPYARR4 (int(&arr) [ten]) {    return Getarr (arr);}

A method that declares a function pointer to a function that returns an array pointer :

int (* (*GETCA) (int(&) [ten])) [ten= Getarr;
// A function pointer   int (*fun) (int, int) that resembles the         int  sum (int a, int b) function // .

Reference blog post and "C + + Primer 5th Edition"

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Finally, here are two examples of "refreshing the Brain":

// declares a pointer to an array of 10 elements, where each element is a function pointer, and the return value of the function is int, and the argument is int* int (* (*p) [ten]) (int *) // or int (* (*p[)) (int *) // Source

// Defines a pointer to an array of 4 elements, each of which is a pointer to a function whose reference returns a pointer to an array of 10 elements with a value of 10 elements int (* (*getca[4]) (int(&) [ten])) [ten];

c++--functions that return array pointers