return type of a function

Question

I came across this code but I couldn't understand it:

void *operator new(unsigned int some_variable) -> Here new is a function but I can't understand what void *operator is doing and here is the complete code:

int buf[3];   

void *operator new(unsigned int uiSize)
{
    return (void *)buf;
}

"`operator new`" is the *name* of the function (even if it is called only with `new`). — Jarod42, Feb 03 '20 at 09:39
More detail there: [operator_new](https://en.cppreference.com/w/cpp/memory/new/operator_new). — Jarod42, Feb 03 '20 at 09:42
See also https://stackoverflow.com/questions/1885849/difference-between-new-operator-and-operator-new — Bob__, Feb 03 '20 at 09:56

score 1 · Accepted Answer · answered Feb 03 '20 at 10:05

1

This is Operator Overloading that defines an implementation for the operator new and returns Void Pointer

For more information about overloading in C++, you can check this link.

C++ Overloading

answered Feb 03 '20 at 10:05

Yomna Hesham

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return type of a function

1 Answers1