CODEWITHSUNDEEP
c++pointers
Pink F
Pink F

Reputation: 21

using pointer with char variable

assume

char a='s';

now i want to store the address of variable a in char pointer p

char *p=&a;

if i use cout << p; then i get the value s, not the address!

can anyone explain the reason why is it so?

Upvotes: 2

Views: 525

Answers (5)

Gaim
Gaim

Reputation: 6844

Calling cout << p is type safe ( C++ objects and operators ) so it uses a operator for char *. This operator serves to print the strings saved in the char pointer. It takes the address and prints everything from start to the first occurrence of byte zero. If you are using MS windows then it is probably only this one char because this system zeros the memory so zero byte is soon. If you want to print the address then try cout << (void *)p;

Upvotes: 1

binW
binW

Reputation: 13692

As many answers above tell you, << operator is overloaded to handle char* as C type string i.e NULL terminated array of characters. In your case when you print p, it gets treated as C string and all the characters till first NULL character are printed. If you only get s printed then there is a NULL character immediately next to your variable a in memory. But this behavior may change depending on the memory contents. I guess you are declaring a as a global variable because space for global variables is usually initialized to all zeros.

Also it will be an interesting experiment for you to declare two character variables and then cout there pointer the same way as you have above. My guess is that pointer to first variable will print both of them.

Upvotes: 0

MartinStettner
MartinStettner

Reputation: 29164

In C a char* is typically used to point to the beginning of a (null-terminated) string. In order to make C++ IO streams (such as cout) compatible with this habit, there's an overloaded verion of operator<<() for handling char pointers. This operator just prints out the null-terminated string starting at the given address.

If you want to print out the 'value' of the pointer (note that it is not part of the C/C++ standard what a pointer 'looks like'), cast the pointer to void* such as

std::cout << static_cast<void*>(charpointer);

Upvotes: 0

kennytm
kennytm

Reputation: 523294

The basic_ostream's << operator is specialized for const char* arguments*, to output it like a C string**. See the answer to Why does cout print char arrays differently from other arrays? for more detail.

Basically, you could cast p to void* when printing to get the address.

cout << static_cast<void*>(p);

*: Also const charT* where charT is the character type adopted by the basic_ostream, and const unsigned char* and const signed char* if the charT == char. For example, wcout would treat a const wchar_t* as a (wide) C string, but cout would treat it as a pointer.

**: This is overiddable by the character traits used by that basic_ostream. See C++ standard §[ostream.inserters.character]/4 for the detailed algorithm.

Upvotes: 8

ur.
ur.

Reputation: 2947

In C/C++ a pointer to char normally is used as a pointer to a null-terminated string. The << operator handles this common use case.

Upvotes: -1

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