C++, weird behavior about copying char arrays by using pointers

Question

I got this code from a textbook:

#include <iostream>
using namespace std;
int main(){
    char str1[]="hello,world!", str2[20], *p1, *p2;
    p1=str1; p2=str2;
    /*
    for(;*p1!='\0';p1++,p2++){
      cout<<"p1="<<*p1<<endl;
      *p2=*p1;cout<<"p2="<<*p2<<endl;
    }
    *p2='\0';
    p1=str1; p2=str2;
    */   
    cout<<"p1="<<p1<<endl;
    cout<< "p2="<<p2<<endl;  

    return 0;
}

I ran this code, it will output p1=hello,world!p2= which I can understand.

But if I uncomment the for loop, the output shows here I got confused, why after the for loop, why it shows p1= instead of showing p1=hello,world!, and for pointer p2, even after the assignment in the for loop, it still shows p2=?

But after I uncomment p1=str1; p2=str2; this line, the output is p1=hello,world!, p2=hello,world!, why it works like that?

And what's the reason for writing this line *p2='\0';, it doesn't matter that this line is commented out or not, the previous outputs don't change.

can anyone tell me how the char pointer here is working?

Answer 1

The loop modifies p1 so that it points to the null terminator at the end of the string. That's the definition of an empty string. p2 likewise points to a null terminator at the end of a string.

If you reset p1 and p2 to their original values you can see the strings as they are.

Answer 2

Here is the output I see from VS2010 running that code with the commented parts uncommented:

p1=h
p2=h
p1=e
p2=e
p1=l
p2=l
p1=l
p2=l
p1=o
p2=o
p1=,
p2=,
p1=w
p2=w
p1=o
p2=o
p1=r
p2=r
p1=l
p2=l
p1=d
p2=d
p1=!
p2=!
p1=hello,world!
p2=hello,world!

That's pretty much what I would have expected! Basically this code is copying the contents of str1 into the (uninitialised) char array str2 via direct pointer manipulation, by copying each character from str1 into str2 one at a time.

To answer your last question, the reason for

*p2='\0';

is so that the second string that is being "created" by the for loop will be correctly null terminated. Without that line, it will just be a char array that cannot be treated like a 'C' string.

Overall this is a pretty contrived / non robust example though, as it won't work once we exceed 20 characters in length for the first string, due to str2[] being declared to be only 20 chars in size.

Answer 3

The code is for copying str1 to str2.

In C++, '\0' is used to end a string. When you try to print a char pointer (say ptr), the compiler prints the string starting from *ptr (the character pointed to by the pointer). When the compiler finds '\0', it stops printing.

In the beginning, p1 points to the first char of str1 and p2 points to the first char of str2. If you print them without doing anything else, the compiler will print both the strings out completely. So the output will be p1=hello,world!p2=.

The for loop makes p1 and p2 advance through str1 and str2. At the end, p1 points to the \0 at the end of the str1 and p2 points to the '\0' at the end of str2. So if you print p1 or p2 directly after the for loop ends, the compiler will immediately find '\0' and stop printing. So, you get the output p1=p2=.

Uncommenting p1=str1; p2=str2; will make both strings point to the first characters again, so printing them now will cause the whole string to be printed. So you get the output p1=hello,world!p2=hello,world! (because str1 got copied to str2 in the for loop).

The *p2 = '\0' is just for ending str2 with '\0'. If your code works without that line, it means that the compiler initialized all the characters of str2 to '\0' automatically. However, the compiler isn't guaranteed to do that, so you should always terminate strings with '\0' in your programs.

Answer 4

A c++ string is a char * under the hood, p1 and p2 are both pointing to the same string, as they are incremented they go through the characters of the string "*p2='\0';" sets the string to the null character it has no effect on the program because it is being reset anyway in the line after.