C ++ pointer Recognition

I started to understand the pointer from Java to C ++ and talked about my understanding of the pointer. the pointer is also a variable type, and the address of an object or variable is stored, let's start with the code: the code is the most direct:

# Include "stdafx. H "int _ tmain (INT argc, _ tchar * argv []) {int I = 100; // assign a value to an int * P = & I; // you can define a pointer and assign a value in this way. // The Value assignment method is int * PT; Pt = & I;
// * PT = & I cannot be defined like this. The value assignment method is defined like this. // The following is the test result printf ("I: % d \ n", I ); /* // The memory address of a variable pointed to. It can be seen that p = & I is a little different from the value assignment above. At first, I always thought the value assignment method was * P = & I; then * P = & I should be fine; or I may have poor understanding, */printf ("P: % d \ n", P ); printf ("& I: % d \ n", & I); printf ("* P: % d \ n", * P ); // The content or value of the memory address printf ("PT: % d \ n", pt); printf ("* PT: % d \ n", * PT ); /* the pointer memory address, indicating that the pointer also has its own memory address. The pointer is a type */printf ("& PT: % d \ n", & pt ); printf ("& P: % d \ n", & P); Return 0;

The above is my personal understanding of pointers.

Use Pointer:

Int * mypointer = new int;

In this case, the pointer points to an integer address. To access this value, you need to remove the reference from this pointer. The following code

*myPointer = 8;

　　

Int * myponiter = new int; // dynamically allocates the memory printf ("& myponiter: % d \ n", & myponiter); printf ("* myponiter: % d \ n ", * myponiter); * myponiter = 8; printf ("& myponiter: % d \ n", & myponiter); printf ("* myponiter 8: % d \ n ", * myponiter); Delete myponiter; // release memory

　　

The memory address of the pointer is not changed here, which refers to the content of the memory address changed;

The following code is an understanding of array pointers;

const size_t  size = 5;int int_arr[size] = {0,1,2,3,4};int *b = int_arr;int *e = int_arr+size;printf("b:%d\n",b);printf("b++:%d\n",++b);printf("int_arr0:%d\n",&int_arr[0]);//b = &&int_arr[0]printf("int_arr1:%d\n",&int_arr[1]);//++b=&int_arr[1]printf("e:%d\n",e);for (int *pbegin = int_arr,*end =int_arr+ size;pbegin != end;++pbegin){std::cout<<*pbegin<<endl;}