Quick C++ question
02-13-2008
, 12:37 PM
Quote:
In C++, what does the assignment operator '=' do for classes, assuming that I haven't overloaded it. For example, what would the following code do? I know it doesn't do what one would want it to do, but I can't remember exactly what goes on. I know Google is my friend, but all I got were a bunch of pages telling me how to overload the operator, which I already know. Thanks in advance to anyone who can help me out.
class X
{
...
}
.
.
.
X a;
X b;
a = b;
class X
{
...
}
.
.
.
X a;
X b;
a = b;
Code:
class X
{
int I;
char S[100];
}
.
.
.
X a;
X b;
a = b;
Code:
class X
{
char* S;
X() { S=new char[100]; }
~X() { delete [] S; }
}
.
.
.
X a;
X b;
a = b;
Juk
02-13-2008
, 12:44 PM
Another way to think of it is:
and
are the same and if memcpy wouldn't work (because of dynamically allocated memory, file handles, etc) then neither will the basic compiler generated assignment operator.
Juk
Code:
class X
{
...
}
.
.
.
X a;
X b;
a = b;
Code:
class X
{
...
}
.
.
.
X a;
X b;
memcpy((void*)a,(void*)b,sizeof(X));
Juk
02-13-2008
, 04:00 PM
Juk
02-13-2008
, 06:46 PM
Quote:
If I changed the function type to TestClass* and returned &testObj, that would cause trouble, correct? Because after the function returned, the memory location at which testObj used to reside could be overwritten by something else at that point.
Another thing to note is if you have a large data-type to return from a function it's usually better to use C++ object references to do it or else you end up making a temporary copy for every return (which gets placed on the stack). See here for an example:
http://www.functionx.com/cpp/example...nreference.htm
Quote:
I guess what's confusing me is the following. I thought when you passed arrays as parameters, they don't get copied for resource reasons. As a result, I thought the same reasoning applied to objects. So now I'm confused exactly when copies are made, and when they are not.
Juk
02-14-2008
, 09:25 PM
If the former, I don't understand why the 'a = b' wouldn't just copy the value of the S pointer from b to a. You now have char pointers in two different objects pointing to the same location, which is obviously "bad", but won't inherently make your program self-destruct, right?
02-14-2008
, 09:52 PM
02-14-2008
, 09:52 PM
Quote:
I guess what's confusing me is the following. I thought when you passed arrays as parameters, they don't get copied for resource reasons. As a result, I thought the same reasoning applied to objects. So now I'm confused exactly when copies are made, and when they are not.
02-14-2008
, 11:03 PM
Quote:
I think. It's been a while.
EDIT:
02-15-2008
, 08:24 AM
Quote:
just use vectors, imo
Juk
02-15-2008
, 08:32 AM
Very rarely you might actually want to do this, but then you could make it clear by using a single pointer declared as static for all classes to share. At least then if the pointer gets changed it gets changed for all classes.
Juk
02-15-2008
, 08:37 AM
Quote:
Ooh, do modern OS actually prevent this these days and just throw exceptions? as I say, it's been a while... we used to just be able to trash another running processes memory by accident with wrong pointer stuff IIRC - at least we were careful not to do this for that reason - Was Win98 days FWIW
There is some info about why Win95 memory protection wasn't that great here too:
http://everything2.com/index.pl?node_id=668981
but compared to Window 3.1 it was a big improvement.
Juk
02-15-2008
, 01:27 PM
Code:
#include <iostream>
#include <vector>
using std::vector; using std::cout; using std::endl;
int main()
{
vector<int> a,b;
a.push_back(1);
a.push_back(2);
a.push_back(3);
b=a;
a.clear();
a.push_back(4);
for (vector<int>::iterator i = a.begin(); i != a.end(); i++)
cout << *i << " ";
for (vector<int>::iterator i = b.begin(); i != b.end(); i++)
cout << *i << " ";
cout << endl;
return 0;
}
02-15-2008
, 02:28 PM
perator= does what you think it should do.
02-15-2008
, 02:53 PM
If you use a STL vector inside another class which doesn't have an assignment operator defined, then the compiler generated default assignment operator will not invoke the assignment operator of the SDL vector contained within it.
Juk
EDIT: See post below.
Last edited by jukofyork; 02-15-2008 at 03:05 PM.
02-15-2008
, 03:04 PM
Quote:
#include <cstdlib>
#include <string>
class foo
{
public:
int a, b, c;
std::string s;
};
int main()
{
foo right;
right.a = 1;
right.b = 2;
right.c = 3;
right.s = "hello foo";
foo left = right;
return 0;
}
...the statement 'foo left = right' resolves to the semantic equivilent of this:
left.a = right.a;
left.b = right.b;
left.c = right.c;
left.s = right.s;
#include <string>
class foo
{
public:
int a, b, c;
std::string s;
};
int main()
{
foo right;
right.a = 1;
right.b = 2;
right.c = 3;
right.s = "hello foo";
foo left = right;
return 0;
}
...the statement 'foo left = right' resolves to the semantic equivilent of this:
left.a = right.a;
left.b = right.b;
left.c = right.c;
left.s = right.s;
http://www.codeproject.com/KB/recipe...yingRules.aspx
Juk
EDIT: I am definitely wrong here. It seems that C++ does a "shallow memberwise copy" (see here) and I was assuming it only did a "bitwise copy". So the assignment operator will get called for each member, but the "shallow" part means that pointer assignment will only make a copy of the pointer as opposed to "deep" which would make a copy of the memory that the pointer pointed too.
Last edited by jukofyork; 02-15-2008 at 03:16 PM.