Re: template

From:
Robert Bauck Hamar <roberth+news@ifi.uio.no>
Newsgroups:
comp.lang.c++
Date:
Mon, 09 Jul 2007 15:53:03 +0200
Message-ID:
<f6tejv$3e3$1@readme.uio.no>
yashwant pinge wrote:

On Jul 9, 6:00 pm, Robert Bauck Hamar <roberth+n...@ifi.uio.no> wrote:

yashwant pinge wrote:

#include<iostream>

using namespace std;

template<class T, int size=100>
class Array
{
        T array[size];
public:
        T& operator [](int index)
        {
                return array[index];
        }


Often you would also want a
const T& operator[](int) const
too.

        int length() const
        {
                return size;
        }
};

class Number
{
        float f;
public:
        Number(float ff=0.0f):f(ff)
        {
        }
        Number& operator =(const Number& n)
        {
                f=n.f;
                return *this;
        }


This operator=() would be written by the compiler if you hadn't written
it yourself.

        operator float const()


Why not
operator float() const
?

        {
                return f;
        }
        friend ostream& operator <<(ostream& os,const Number& x)
        {
                return os << x.f;
        }
};

template<class T,int size=2>
class Holder
{
        Array<T,size>*np;
public:
        Holder():np(0)
        {
        }

        T& operator [](int i)
        {
                if(!np) np = new Array<T,size>;
                return np ->operator[](i);
        }

        int length() const
        {
                return size;
        }

        ~Holder()
        {
                delete np;
        }


You should take care. The compiler will generate
Holder(const Holder& other) : np(other.np) {}
and
Holder& operator=(const Holder& other) { np = other.np; }

If one of these are invoked, you would delete np twice. In addition, the
assignment operator is bound to leak memory. Remember the rule of three:
A class with any of {destructor, assignment operator, copy constructor}
generally needs all 3.

};

int main()
{
        Holder<Number> h;


invokes Holder<Number>::Holder().

        for(int i=0;i<2;i++)
                h[i]=i; // why the number constructor
called ...?


h[i] calls Holder<Number>::operator[](int). The first time this
executes, !np is false, and an Array<Number, 2> is default constructed.
This creates an array of size 2 of Number, which constructs these two
numbers with Number::Number(0.0f), since this is the default constructor
of number.

Holder<Number>::operator[](int) then returns a reference to a Number
object. This number is assigned to, which means a call to
Number::operator=(const Number&), which is the only assignment operator
defined. This assignment operator can use a temporary Number object, and
this is constructed by first converting i to a float, and then using this
float as an argument to Number::Number(float). That is:

h[i] = i;

means

h.operator[](i).operator=(Number(float(i)))

        for(int j=0;j<2;j++)
                cout<<h[j]<<endl;

        return 0;
}

Why the number constructor is called during the execution of line
" h[i]=i;"


--
rbh


Breakpoint 1, main () at 6.cpp:72
72 Holder<Number> h;
(gdb) step
Holder (this=0xbfffe2b0) at 6.cpp:49
49 {
(gdb) step
main () at 6.cpp:74
74 for(int i=0;i<2;i++)
(gdb) step
75 h[i]=i;
(gdb) step
Number (this=0xbfffe2a8, ff=0) at 6.cpp:26
26 {
(gdb) step
27 }
(gdb) step
Holder<Number, 2>::operator[](int) (this=0xbfffe2b0, i=0) at 6.cpp:54

I could not understand the flow of the code that GDB is given me
why the num constructor is called before
"older<Number, 2>::operator[](int) (this=0xbfffe2b0, i=0)"


Because the standard says that the order of evaluation is unspecified. That
is, the compiler may choose in which order the arguments are calculated.

--
rbh

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