Re: Pointer or reference?
On 14 Okt., 14:33, davee <da...@nomail.com> wrote:
Lets assume that I have an object 'MyObject' with various fields. Below
foo1 and foo2 modifies the fields in a MyObject object:
void foo1(MyObject * mobj){
int value = 55;
myobj->setValue(value);
...
...
}
void foo2(MyObject *& mobj){
int value = 55;
myobj->setValue(value);
...
...
}
int main(){
MyObject * myobj = new MyObject();
foo1(myobj);
std::cout << "getValue = " << myobj.getValue() << std::endl;
foo2(myobj);
std::cout << "getValue = " << myobj.getValue() << std::endl;
}
For the record: MyObject is not an object but a type, specifically a
class-type.
As I understand its only when calling foo2 that the changes will visible
in main after returning from the call. The reason is that foo1 works
on a copy of the pointer which disappears when the function returns.
On contrary foo2 works on the original pointer. Or am I mistaken ?
Have you tried?
You don't seem to differentiate between the pointer and the pointee.
In the first version you pass a pointer by value, in the second by
reference. But "mobj" which in in both cases refers to a pointer to a
MyObject object is not modified so it does not matter which function
you call. Both functions will however mutate the object pointed to --
visibly. So, the expected output is
getValue = 55 (as set in foo1)
getValue = 55 (as set in foo2)
See "pointer fun with Binky" (C++ version)
http://www.youtube.com/watch?v=i49_SNt4yfk
Cheers,
SG
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