Pointers and such...
Author
Yes, I know this is a very basic question, but I wanted to know more for clarification, since I see things done two ways.
If you do this:
int bob;
int *x; // x is a pointer to an integer
x = &bob // x points to address of bob
Is that the same as this:
int bob;
int *x = &bob / x points to address of bob
Or do those mean two different things?
And if you do this:
int tom;
int *x = &tom
Would *x hold the address of tom, and x hold the value of tom? Is that correct? And would changing x to 27 make tom 27, or changing tom to 100, make x 100? Any clarification on these things would be helpful.
While programming in C, what do the -> operators mean? I don''t really know what they are called, so I can''t look up information on them.
Thanks all.
I am not worthy of a sig!
int bob;
int * x;
x = &bob
is the same as
int bob;
int * x = &bob
it just saves a line of code
x holds the address of bob and *x holds the value of bob. Think of * as ''contents of'' when dereferencing a variable. Remember, x is a pointer, so it only holds an address.
D
int * x;
x = &bob
is the same as
int bob;
int * x = &bob
it just saves a line of code
x holds the address of bob and *x holds the value of bob. Think of * as ''contents of'' when dereferencing a variable. Remember, x is a pointer, so it only holds an address.
D
Drakkor is totally right, but just to clarify your second question:
int tom;
int *x = &tom
It might look, as you thought, like *x is holding the address of tom, but it''s not. In this instance you''re using the * to declare the pointer not to dereference it, so as usual x will hold the address and *x will hold the value.
The -> is called "pointer to member notation," it might have some other names too but that''s what I''ve seen it called. If you have a pointer to a struct or an object you use -> to simultaneously dereference it and access its fields and methods. So if class CObject has field foo and method bar:
CObject myObj= new CObject();
myObj->foo = myObj->bar();
instead of something like:
CObject myObj = new CObject();
(*myObj).foo = (*myObj).bar();
int tom;
int *x = &tom
It might look, as you thought, like *x is holding the address of tom, but it''s not. In this instance you''re using the * to declare the pointer not to dereference it, so as usual x will hold the address and *x will hold the value.
The -> is called "pointer to member notation," it might have some other names too but that''s what I''ve seen it called. If you have a pointer to a struct or an object you use -> to simultaneously dereference it and access its fields and methods. So if class CObject has field foo and method bar:
CObject myObj= new CObject();
myObj->foo = myObj->bar();
instead of something like:
CObject myObj = new CObject();
(*myObj).foo = (*myObj).bar();
Dobbs, woundn't it be:
CObject *myObj= new CObject;
myObj->foo = myObj->bar;
Wouldn't
CObject myObj= new CObject();
be for Java?
Edem Attiogbe
Edited by - KwamiMatrix on January 9, 2002 1:04:23 AM
CObject *myObj= new CObject;
myObj->foo = myObj->bar;
Wouldn't
CObject myObj= new CObject();
be for Java?
Edem Attiogbe
Edited by - KwamiMatrix on January 9, 2002 1:04:23 AM
This topic is closed to new replies.
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