Pointer Query
Hi,
I have declared:
Code:char c[] = "Good Morning"; char *pc = &c[2];The result I get is "od Morning"Code:cout << "pc is " << pc << endl;
Why does this happen? Can anyone explain how and why this happens?
And what the assignment statement for the pointer is doing.
What I expected...
I expected the output to be the address of the 3rd element of the array.
I was wrong.
Thanks!
You're using a reference on the character array ch (not the character pointer pch). So it is taking a pointer to the character array, and since pointers point to the first element of an array (i'm almost sure that's right) it will also point to everything after the index of 2. Try removing the & and see if it copies just the one character.
EDIT: you could always try something like this too
Code:char ch[] = "Good Morning"; char ch2 = ch[2]; char *pch = &ch2;
Last edited by scwizzo; 10-30-2008 at 10:33 AM.
If I remove the & address operaotr it will not compile
thanks
Because a string in C is an pointer to an array of characters, terminated with a 0 ('\0').
In the case of C++, it still supports "old" C style strings, and for cout the << operator that takes a char-pointer treats it as a string. You can print the value of the pointer by casting it to a non-character pointer - reinterpret_cast<void *>(pc)
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Mats
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Uh-oh. Careful with your wording there.Originally Posted by scwizzo
Since C++ actually has references (and you are not using or creating a reference), that wording is incorrect...
If you want the third element it would be
strings are zero-terminated so when you create a pointer to the third [3] letter it continues to get the string until the '\0' character is found.Code:char pc = c[2]; char* pc = &c[2];//This gets the string until '\0' is encountered.
edit: wow im slow...
pc outputs the way it does thanks to operator << for c-strings, where a c-string is char *. To print the one character, dereference pc.
Ok, Great. Thanks everyone who replied.
I understand that what you said Raigne is what happens. I just thought that the assignment statement worked like this:
Whereby, the pointer takes the value of the address of the the 3rd element of the array.Code:char *pc = &(c[2]);
But what you are saying happens is that:
The pointer to the third element continues to take all the elements of the string that it is part of until it encounters a null terminating charcter.
Is my understanding right?
thanks again
Actually,
This is correct.Whereby, the pointer takes the value of the address of the the 3rd element of the array.
What happens is when you pass a char* pointer to cout, it actually interprets that as that it is a string, and thus prints every character until it finds the '\0' char.
So basically, from the position from the pointer, it goes one step forward, print, forward, etc.
This is a left-over from C, since arrays cannot be passed by value or something else, so a pointer is instead passed. A pointer to the first element in the array. The functions would then interpret that pointer passed as a string and begin printing every character until the '\0' is found.
Cast with a static_cast, not a reinterpret_cast. Object*->void* is an implicit conversion that you want explicit, so static_cast is definitely the way to go.
All the buzzt!
CornedBee
"There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
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