Analysis of examples of C + + pointers

See a long time of C, basically is the basic grammar clearance, met a C + + face test, analysis, as a period of time punch.

Code inside the compiler,

1#include <iostream>2 3 using namespacestd;4 5 intMainintargcChar**argv)6 {7     Char*str[]={"Welcome"," to","Boyaa","Shenzhen"};8     Char**p=str+1;9 Tenstr[0]= (*p++) +1; Onestr[1]=* (p+1); Astr[2]=p[1]+3; -cout<<str[2]-str[1]<<Endl; -  thestr[3]=p[0]+ (str[2]-str[1]); -cout<<str[0]<<" "<<str[1]<<" "<<str[2]<<" "<<str[3]<<Endl; -     return 0; -}

The result of the code operation is:

30 (0x0)   0.366continue.

Here is your own simple analysis process, halfway may be a little error (people remind me) ———————— "

First, there are some initialization steps.

i) request an array of strings, which is the address of the string that is stored in the array:
Char *str[]={"Welcome", "to", "Boyaa", "Shenzhen"};
STR is here equivalent to a constant level two pointer.

II) Define a level two pointer char**p=str+1;
Assign P value &str[1]

III) OK, the play is coming back to the string array of Str just mentioned. It's all a series of operations on a string array
str[0]= (*p++) +1;
str[1]=* (p+1);
str[2]=p[1]+3;
cout<<str[2]-str[1]<<endl;

1) First (*p++) +1, wherein the *p++ priority is * (p++), and then first return the value of p to the expression, equivalent to *p,p++;
Then str[0]=*p+1 equivalent to str[1]+1 equivalent to str[1][2], that is, from the second letter to the beginning to the last
Notice that this time the string array changes and becomes
Char *str[]={"O", "to", "Boyaa", "Shenzhen"};

2) Okay, the second one.
str[1]=* (p+1), where * (p+1) equals * (&str[2]+1) equals * (&str[3]) equals str[3], which is
Char *str[]={"O", "Shenzhen", "Boyaa", "Shenzhen"};

3) The third str[2]=p[1]+3, where the p[1]+3 equivalent to str[3]+3 equivalent to "Shenzhen" +3 is equivalent to "Nzhen"
"Shenzhen" is not a string, it is an address. All string constants are addresses.
This time is the "Nzhen" address assigned to STR[3], so there are changes
Char *str[]={"O", "Shenzhen", "Nzhen", "Shenzhen"};

4) Then the final output is the first one.
cout<<str[2]-str[1]<<endl;
It looks like a two string subtraction, and the result is that the first address of the string is subtracted from the value "Shenzhen"-"nzhen" equals the address of S minus N.
The result is 3.

5) The fifth step is to assign value, str[3]=p[0]+ (Str[2]-str[1]);
That is str[3]= "Nzhen" +3= "en", which is an operation of an address
Update STR has
Char *str[]={"O", "Shenzhen", "Nzhen", "en"};

6) Finally the result output
cout<<str[0]<< "<<str[1]<<" "<<str[2]<<" "<<str[3]<<endl;
o shenzhen nzhen en

Analysis of examples of C + + pointers