CODEWITHSUNDEEP
c++pointerscopy
mahmood
mahmood

Reputation: 24685

making a copy for a variable

Consider this piece of code:

int *a, *b;
a = foo();
if (a)
   b = a;
a = bar();

The problem is, when a updates by calling bar(), b also updates. However I want to make a backup by using b = a. What is the problem then?

Upvotes: 2

Views: 29549

Answers (4)

Branko Dimitrijevic
Branko Dimitrijevic

Reputation: 52107

The...

int *a;
// ...
a = foo();

...would not even compile if the function was declared as int foo();, so I'll assume it's declared as int* foo();. Ditto for int* bar();.

Don't return a raw pointer like this. It might not be clear whether caller should delete it or not. The good documentation can only partially resolve this problem (human beings have a tendency of ignoring the documentation from time to time). Why not just copy the integer value, or at least return a reference to integer in case it really needs to be shared?

That being said, in your particular case, foo() and bar() apparently use the same int "object" internally, so a continues to point to the same object as b even after bar() has modified it.

Upvotes: 0

Zéychin
Zéychin

Reputation: 4205

int *a, *b;
b = new int;

a = foo();
if (a)
   *b = *a;
a = bar();
...
delete(b);

(The value at address a is assigned to the value at address b.)

What you are doing right now is making a and b point to the same place in memory. Then, if the value in a or b is updated, they pointers both point to the new value.

By the way, unless bar() returns a pointer, you probably want *a = foo() and *a = bar().

Upvotes: 2

TJ C.
TJ C.

Reputation: 11

if a == b, then bar() is changing the value at a, not assigning a new a. This makes the most sense memory wise. If what you care about is the actual integer, not the address, instead of assigning b = a assign *b = *a.

Upvotes: 0

talnicolas
talnicolas

Reputation: 14053

You could just back up the value of a:

int *a, b;
a = foo();
if (a)
   b = *a;
a = bar();

or if you want to keep b as a pointer:

...
   *b = *a;
...

Upvotes: 1

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