Pointer calculation:
Resolving reference symbol (* &: can be canceled)
* A number of bytes to take the contents
Subtraction between two pointers: the unit number of the phase difference between the two addresses;
1 unit = sizeof (address pointed)
Then the concept of the previous chapter, said pointer, let's look at some examples
int n[5] = { 5, 3, 1, 4, 2 };
int * p = n * P1;
p1 = &n[3];
int** pp = &p;
cout << (int*)(p1 - n) << endl; //00000003
p1 point n [3] addresses, representative of n n [0] to address
Therefore p1 - n = & n [3] - & n [0] = 12 / sizeof (int *) = 3;
There convert int * type, so the final result is 00000003
cout << n[2]+++**pp<<endl; //6
Rear jerk, to calculate n [2] + ** pp; n [2] is equal to 1, ** p points p, p and point n [0];
Therefore n [2] + ** pp = n [2] + n [0] = 5 + 1 = 6; and n [2] ++, n [2] = 2;
cout << (*p1 == *p + 1) << endl; //0
p1 point n [3], so * p1 = n [3] = 4, p points to n [0], so * p = n [0] = 5, 6 is equal to 1 plus
So the result is 0
cout << (&p == &p1) << endl; //0
p1 point n [3] addresses, n-point P [0] of the address, are not equal, the result is 0
cout << (int)p1 - (int)&n[0] << endl;
p1 point n [3] of the address, (int) p1 - (int) & n [0] = (int) & n [3] - (int) & n [0] = n [3] to the n [0] distance , equal to 12
cout << *p + 1 << endl;//6
point p n [0] address, the * p + 1 = n [0] +1 = 6;
cout << p - n << endl;//0
point p n [0] of the address, but also point to the address n-n [0], so that pn = 0;
cout << *p - (*p1)++ << endl;//1
point p n [0] address, p1 point n [3] of the address, the home ++, so * p- * p1 = n [0] - n [3] = 1; p1 points to the n [4];
cout << **pp << endl;//5
pp point p, p and point n [0], so ** pp = n [0] = 5
cout << *p1***pp << endl;//25
pp point p, p and point n [0], so ** pp = n [0] = 5, p1 points to the n [4] = 5, so the result is 25;
2
int n[5] = { 1, 2, 3, 4, 5 };
int* pn[4] = { n, &n[2], n, &n[0] };
int ** ppn = pn; // ppn point pn [0]
cout << **ppn + 1 << endl;//2
ppn point pn [0], ** ppn = n [0] = 1; 2 result;
cout << ppn - &pn[2] << endl;//-2
ppn - &pn[2] = &pn[0] - &pn[2] = -2;
cout << pn[2] - n << endl;//0
pn [2] points to address n, n, n also points to the address, so the result is 0;
cout << *pn[2] + 2 << endl;//3
pn [2] = & n [0], so * pn [2] + 2 = 1+ 2 = 3;
cout << *pn - n << endl;
also points to address n pn, so the result is 0;
cout << (short*)pn - (short*)(&pn[3]) << endl;
pn point pn [0] of the address, (short *) pn - (short *) (& pn [3]) = (short *) (& pn [0]) - (short *) (& pn [3]) = - 12; divided by sizeof (short *) = -12/2 = -6;
cout << (short*)((char*)pn[1] - (char*)&n[3]) << endl;
pn[1] = &n[2] ;(short*)((char*)pn[1] - (char*)&n[3]) = (short*)((char*)&n[2] - (char*)&n[3]) = (short*)-4（转16进制） = fffffffc
char* pch = "hello world!";
cout << sizeof(pch) << endl;//4
Representative pch "hello world!", the length is generally any type pointer 4
cout << strlen(pch) << endl;//12
"Hello world!" There are 12 characters
cout << sizeof(*pch) << endl;//1
* Pch points to the first element of h; char length type 1
cout << *pch + 1<< endl;
* Pch points to the first element of h; * pch + 1 = 'h' +1;
cout << pch << endl;//"hello world!"
Print "hello world!"
cout << (* pch) ++ << endl; // error: Constants can not ++
cout << *&pch << endl;//pch//"hello world!"
Print "hello world!"
cout << (char*)(pch + 1) - pch << endl;//1
pch itself is a char *, not changed
3
char* pch[3] = { "monday", "apple", "hello" };
char** ppch = pch;
cout << *ppch << endl;
PPCH point pch [0], so ppch = pch [0] = monday;
cout << **ppch + 1<< endl;
PPCH point pch [0], so ** ppch = 'm' + 1 = 78;
cout << (int*)(pch - &pch[1]) << endl;
pch point pch [0] (int *) (pch - & pch [1]) = (int *) (& pch [0] - & pch [1]) = -1 (rotation FFFFFFFF in hexadecimal)
cout << pch[2] << endl;
Print hello
cout << pch[2] - *ppch << endl;
pch[2] - *ppch = &'h' - &'m' =16;