Exchanging two strings using pointers prints strange output - c++

Question

I wrote a program to swap two strings using pointers.
char *names[]={"Sachin","Kapil","Ajay","Sunil","Anil"};
cout<<"String II is ";cout<<names[1];
cout<<"\nString IV is ";cout<<names[3];
char *t;
t=names[1];
names[1]=names[3];
names[3]=t;
cout<<"\nString II is ";cout<<names[1];
cout<<"\nString IV is ";cout<<names[3];
In the first line I used an array of char pointers to store the address of the 1st characters of the strings.
When I cout names[1] and names[3]:
The entire string was printed,but shouldn't the address of the character pointer alone be printed?
Also,how is the program printing the entire string instead of printing the first character alone?
After I compiled I got a warning from line 1 saying:
warning: deprecated conversion from string constant to 'char*' [-Wwrite-strings]|
The program runs as intended,but I would love to know how it is working.I came across an implementation where they used cout.write(names[i],len) to print the entire string(which makes sense),but to my surprise the above method works too (even without any sort of increase in the pointer value to print the next character).
Any tips and suggestions on the issues listed above would be appreciated.Thank you!

Answer 1

First up - why you don't just get the address output.
When you 'insert' any char* into a stream such as std::cout<<s; where s is of type char* the insertion logic assumes s points to the first character of a 'C' String. A 'C' String is a series of characters ending in a nul ('\0' character).
So the insertion logic outputs that series of characters.
You'll get a different result if you use std::cout<<static_cast<void*>(c); because the insertion of a void* pointer is treated differently.
The << operator on streams is overloaded to do different things depending on what you're inserting.
Now why you get that warning. You're uncovering some of the unhappy legacy of C++ particularly baggage inherited from C.
In the bad old days of early C without the const modifier there was no way of indicating that string literals (e.g. "Sachin") couldn't be modified at run-time and on different platforms doing so had different effects ranging from harmlessy amending the data, throwing an access exception to modifying all literal references to that string value in the code (there may be several).
So when C++ came along and introduced const (soon back adopted by C) it was quite rightly seen fit to make the type of string literals in the code const char[]. However some compilers (when you've got the wrong settings!) still allow you to assign literals to char[] (or char*) for backward compatibility.
Unless you're maintaining ancient code and the senior programmer has instructed you accordingly never leave that option enabled.

Answer 2

The stream library recognises the char* type and treats it as a null terminated c-style string which is simply an array of chars terminated by null.
To get it to print addresses, type cast the string to void*.
i.e.
cout << "String II is "; cout << (void *)names[1];
Will now output:
String II is 0x400a0b
As to why the first character is not printed only, it's because the stream library treats it as null terminated C string. But if you really only wanted the first character printed you could just as easily done this:
cout << "String II is "; cout << *names[1];
or:
cout << "String II is "; cout << names[1][0];
Manipulating pointers like you are is OK, but manipulation what they contain is not when declared like you have them. So you should really declared them const. This also removes the compilation warnings.
So the code would become.
const char *names[]={"Sachin","Kapil","Ajay","Sunil","Anil"};
cout << "\nString II is "; cout << (void*)names[1];
cout << "\nString IV is "; cout << (void*)names[3];
const char *t;
t = names[1];
names[1] = names[3];
names[3] = t;
cout << "\nString II is "; cout <<(void*)names[1];
cout << "\nString IV is "; cout <<(void*)names[3];