Different Between the operator '*' and '&'

[PIEBALDconsult]

* is backward compatible with C. :-D Edit: I was off by one character, I meant to indicate that C doesn't support references.

modified on Thursday, March 5, 2009 9:56 AM

[Cedric Moonen]

You can use pointers in C too (thus using the '*' is valid in C).

Cédric Moonen Software developer
Charting control [v1.5] OpenGL game tutorial in C++

[PIEBALDconsult]

Ah, I'm not awake yet, coffee just finished dripping. I must have meant it the other way around. I only dabbled in C++ a little, and not recently.

[jeansea]

Is all the position the '*' use can replace by '&'? :doh:

[ky_rerun]

in a declaration ie int *X means a a pointer to x named *; You cannot define a reference with initializing it so int &X //doesn't work however --- int Y; int &x = Y; does Y and X now refer to the same place in memory changing the value of one will change the value of the other. When you use a & in a function prototype we are saying I want my parameter to access memory at the location of the variable the caller specified; When you declare a pointer in the prototype you are going to explicitly pass in a pointer that pointer can however be reassigned but once reassinged the caller will no longer be able to read changes to what was passed in I wrote a little example. // CppTest.cpp : Defines the entry point for the console application. // #include "stdafx.h" void NoSideEffect(int *T) { int G = 5; T = &G; } void SideEffect(int &T) { int G = 5; T = G; } void PointerSideEffect(int *T) { int G = 10; *T = G; } int _tmain(int argc, _TCHAR* argv[]) { char buffer[2]; int X = 2; NoSideEffect(&X); printf("X is Now %d \n",X); SideEffect(X); printf("X is Now %d \n",X); PointerSideEffect(&X); printf("X is Now %d \n",X); gets(buffer); } this will print out X is Now 2 X is Now 5 X is Now 10

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a programmer traped in a thugs body

[Maximilien]

using references is safer, you do not have to check for NULL pointers.

This signature was proudly tested on animals.

[jeansea]

void NoSideEffect(int *T) { int G = 5; T = &G; } int X = 2; NoSideEffect(&X); printf("X is Now %d \n",X); X is Now 2 don't change why?could you explain more detail ;P

[ky_rerun]

This is because c++ accurately always passes by value When you pass a pointer you are passing the value of the pointer which is a memory address. when the compiler creates a stack frame the value of the memory address is put on the stack as a local variable when you change the value of that variable you point it to a new memory location and no longer maps to memory address of the variable in the caller. So any change will not affect the caller since you are changing the location associated with another variable in this case another local variable which is allso on the stack.

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a programmer traped in a thugs body

[bulg]

I can't believe there're people who know C and not C++ still! :)

[PIEBALDconsult]

No?

[PIEBALDconsult]

C++ is all hype; it's not necessary for most real work. :-D

[_Superman_]

That's right. You cannot pass a null to a reference like you can for a pointer.

«_Superman_» I love work. It gives me something to do between weekends.

[_Superman_]

It is more neat sometimes to use &. Take this simple example of a multiply function. Using *

double multiply(double *x, double *y)
{
return *x * *y;
}

double d = multiply(&a, &b);

Using &

double multiply(double &x, double &y)
{
return x * y;
}

double d = multiply(a, b);

«_Superman_» I love work. It gives me something to do between weekends.