Member Avatar for Farisnet

Hi...


I am a bit new to programming, and I'm trying to learn C++. I was wondering if there was a way to copy the contents of one variable into another... So that, for example:

int x=5;
int y;
(Some fancy code here involving x and y);
(y=5 but it is not related at all to x. If x is assigned a value of 10438478234 later in the code, y will still equal 5)

The reason is that if I use variable assingment or pointers, and x changes, then so will y... If you please, list a code or method in order to achieve this data transfer

:icon_smile: Thank you,

  • 5 Contributors
  • 8 Replies
  • 20K Views
  • 2 Days Discussion Span
  • Latest Post Latest Post by dougy83

Dani AI

Generated

In C++, built-in types (int, long long, double, etc.) use value semantics: assigning one variable to another copies the current value, and later changes do not affect the copy. Also note that = is assignment, while == tests equality.

long long a = 5;
long long b = a;      // copy current value of a
a = 10;               // b remains 5

const long long c = a; // snapshot at 10; c cannot change
a = 20;                // c is still 10

Pointers and references alias the same object, so changes are seen through all aliases. If you only want a snapshot of the pointee, copy the value, not the address:

int v = 42;
int* p = &v;         // alias
int snapshot = *p;   // independent copy of v’s value

For non-trivial types like std::string or std::vector, assignment also makes independent copies of the elements. If you write classes that own raw pointers, define proper copy behavior (rule of three/five) to avoid shallow copies.

Side note: the literal 10438478234 does not fit in a 32-bit int. Use a wider type (e.g., long long) and the LL suffix: 10438478234LL.

References:

Recommended Answers

All 8 Replies

Member Avatar for Salem

int *y = &x; If x changes, then will *y

Member Avatar for dougy83

If you want 2 variables to refer to the same data, you can also use references:

int x = 5;
   int &y = x;
   
   cout << x << "," << y << endl;
   x = 10;
   cout << x << "," << y << endl;
   y = 15;
   cout << x << "," << y << endl;

output:
5,5
10,10
15,15

Member Avatar for Farisnet

You missed my point. What I want is that if I change x, y is not affected.

Thank you!

Member Avatar for Nick Evan

Ok...

int x =3;
int y= 2;
cout << "x: " << x << " y: " << y << endl;
y=x;
cout << "x: " << x << " y: " << y << endl;
x=10;
cout << "x: " << x << " y: " << y << endl;

output:

x: 3 y: 2
x: 3 y: 3
x: 10 y: 3
Member Avatar for Farisnet

It does work like that?? But what about the "x=y" part, followed by "x=10"???? Doesn't that say "x has same value as y, and x is now 10" Doesn't that mean that y will also be 10?? (Please excuse my noobish questions, I'm just staring with this thing!!!)

Member Avatar for dougy83

You sound like a mathematician, not a programmer.

In programming, x = 5 means set x to 5. y = x means set x to the value of y (the old value of x is discarded).

Member Avatar for Yellowdog428

er...

look at it this way

when there is a = sign remember that it is an assignment operator not an "equal" sign

anything on the left is assigned to what is on the right

But what about the "x=y" part, followed by "x=10"???? Doesn't that say "x has same value as y, and x is now 10" Doesn't that mean that y will also be 10??

Looks like you read it wrong.
the only variable assignment he made was y = x

cheers

Member Avatar for dougy83

You sound like a mathematician, not a programmer.

In programming, x = 5 means set x to 5. y = x means set x to the value of y (the old value of x is discarded).

That should have been x = y. The complexity of the statement must have confounded me..

commented: Haha, next time I'll make ik less complex ;) +4
Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.