C language FAQ list

 

6.13

 

Http://c-faq.com/aryptr/ptrtoarray.html

 

Q: How can I define a pointer to an array?

 

A: Generally, when talking about pointers to arrays, it actually refers to a pointer to the first element of an array. For example:

 

Int arr [3] = {1, 2, 3 };

Int * PTR = arr;

Printf ("* PTR = % d/N", * PTR); // * PTR = 1

 

Strictly speaking, the PTR here is actually a pointer to the first element 1 of the array arr, rather than an array pointer, that is, PTR is an int * type pointer, we can perform algebraic operations on PTR, such as auto-increment:

 

PTR ++;

Printf ("* PTR = % d/N", * PTR); // * PTR = 2

 

When the array name arr is assigned to PTR:

PTR = arr;

The function is equivalent to the following sentence:

PTR = & arr [0] l

 

If you want the PTR to point to ARR [1], it should be as follows:

PTR = & arr [1];

 

Okay, now we will discuss the pointer to the array instead of the pointer to a single element of the array. For example, we need to define a pointer to an int-type array with a size of 3, it should be defined as follows:

 

INT (* ptrarr) [3];

 

When ptrarr ++ is used, the distance ptrarr spans is 3 rather than 1. Generally, the true array pointer is used only when multiple arrays are operated:

 

INT (* ptrarr) [3];

Int arr [2] [3] = {3, 4, 5}, {6, 7, 8 }};

Ptrarr = arr;

Printf ("(* ptrarr) [0] = % d, (* ptrarr) [1] = % d/N", (* ptrarr) [0], (* ptrarr) [1]); // (* ptrarr) [0] = 3, (* ptrarr) [1] = 4

 

Ptrarr ++; // The distance between ptrarr and ptrarr is 3!

Printf ("(* ptrarr) [0] = % d, (* ptrarr) [1] = % d/N", (* ptrarr) [0], (* ptrarr) [1]); // (* ptrarr) [0] = 6, (* ptrarr) [1] = 7