I saw a usage of the * operator in front of the new operator. What does it do?
int x= *(new int);
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11It leaks memory. Don't do it. – user2357112 Apr 03 '17 at 05:53
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5Also causes undefined behaviour by reading an uninitialized int – M.M Apr 03 '17 at 05:53
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@M.M More of a garbage value than UB, unless we know how `x` is used later ;) – CinCout Apr 03 '17 at 05:55
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Related question: http://stackoverflow.com/questions/35191252/what-does-this-reference-notation-signify – msc Apr 03 '17 at 05:58
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2I hope that was in a section titled **Things You Should Not Do**. – R Sahu Apr 03 '17 at 05:58
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4@CinCout the code reads from the uninitialized memory allocated by `new int` , in order to initialize x, which causes UB , [see here](http://stackoverflow.com/a/23415662/1505939) – M.M Apr 03 '17 at 06:00
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Also see [What happens if you dereference `new int`?](http://stackoverflow.com/q/25395297/1708801) – Shafik Yaghmour Apr 03 '17 at 15:49
2 Answers
4
new int
is allocating memory.
By using defering operator * you are getting the value present at that memory location which would be garbage.
#include <iostream>
void main()
{
int x = *(new int);
std::cout << x;
system("pause");
}
But this will cause the memory leak.
Sumeet
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The expression new int creates an object of type int in allocated memory and returns a pointer to said object. Because no initializer is specified, the object is default-initialized and therefore has an indeterminate value.
The expression int x = *(new int) initializes x with the value of the object referenced by the pointer returned by new int. However, accessing indeterminate values in this way is undefined behaviour. Since standard imposes no requirements on the behaviour of programs containing undefined behaviour, any program containing the expression int x = *(new int) could, in theory, do anything at all.
If your were to initialize the object created by the new expression (e.g. new int()), then x would have whatever value you initialized the original object with. However, since you didn't store the pointer to the original object, there is no way to destroy the object using delete, so you have a memory leak.
Joseph Thomson
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Technically, it just produces a reference to the new object. E.g. consider `int &r = *(new int);` – Apr 03 '17 at 08:40
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@Hurkyl Thanks. You are right. The undefined behaviour will only be invoked when the value of `*(new int)` is evaluated. I have updated my answer, so it should be technically correct now. – Joseph Thomson Apr 03 '17 at 14:18
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@Hurkyl, in that case you are forced to use `delete &r` once you don't need the allocated object anymore. Or you'll continue to have a memory leak. – Luis Colorado Apr 04 '17 at 15:00