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From: horstman@mathcs.sjsu.edu (Cay Horstmann)
Newsgroups: comp.lang.c++
Subject: Re: Incorrect handling of destructors in Turbo C++
Keywords: Turbo C++, destructors
Message-ID: <1991Mar25.231208.29589@mathcs.sjsu.edu>
Date: 25 Mar 91 23:12:08 GMT
References: <546@humu.NOSC.Mil>
Organization: San Jose State University - Math/CS Dept.
Lines: 69


In article <546@humu.NOSC.Mil> you write:
>There seems to be a bug in the code that Turbo C++ 1.00 generates.
>In the following program the destructor routine should only be called
>once.
>
>------------
># include <stdio.h>
>
>class A {
>  protected:
>    int i;
>  public:
>    A();
>    ~A();
>};
>
>A::A()
>{
>  printf("Constructor called\n");
>}
>
>A::~A()
>{
>  printf("Destructor called\n");
>}
>
>void sub(A a)
>{
>  printf("in sub\n");
>}
>
>void main()
>{
>  A a;
>
>  printf("before call\n");
>  sub(a);
>  printf("after call\n");
>}
>------------
>
>When this program is executed:
>
>------------
>Constructor called
>before call
>in sub
>after call
>Destructor called
>Destructor called
>------------
>

You are wrong.

You didn't supply a copy constructor for A (i.e. no A(const A&)). That
made the argument in sub constructed with a bitwise copy but destroyed
with the ~A(). The only funny thing is that Turbo routinely translates
a function call sub(a) into A temp = a; sub( temp ) and changes 
void sub( A a ) into void sub( A& a ). It beats me why they do it, but
they do, and that is why the destructor is called both times after
sub exits.

I ran into some cases when the destructor didn't get called enough, but
I have a hard time reproducing it in a small and short program. If anyone 
has one, let me know!

Cay