Template operators

[Alex Tibbles]

quick:

struct A
{
template< typename T >
operator[](size_t index)
{
return vec_[index];
}
std::vector< int > vec_;
};

int main()
{
A a;
float f = a[1]; // doesn't specify template parameter
or
float f = a.operator()< float >(1); // gcc 3.3.2 Debian gives error
"operator[] not defined"
or what?
}

Any clues?

alex

[Patrik Stellmann]

Alex Tibbles wrote:

quick:

struct A
{
template< typename T >
operator[](size_t index)
{
return vec_[index];
}
std::vector< int > vec_; you don't even use the template parameter! I guess you mean

template< typename T >
T operator[](size_t index)
{
return vec_[index];
}
};

int main()
{
A a;
float f = a[1]; // doesn't specify template parameter not necessary since your code doesn't use the template parameter and by
default the result is int so everythings working... or
float f = a.operator()< float >(1); // gcc 3.3.2 Debian gives error

should be
a.operator[]<float>(1);
?

[tom_usenet]

On Fri, 23 Jan 2004 12:34:19 -0000, "Alex Tibbles"
<al**********@yahoo.co.uk> wrote:

quick:

struct A
{
template< typename T >
operator[](size_t index)
Did you mean:
T operator[](size_t index)
?
Or
int operator[](size_t index)
perhaps?
{
return vec_[index];
}
std::vector< int > vec_;
};

int main()
{
A a;
float f = a[1]; // doesn't specify template parameter
or
float f = a.operator()< float >(1); // gcc 3.3.2 Debian gives error
"operator[] not defined"
or what?
}

Any clues?

Ignoring the typos and missing headers,

float f = a.operator[]<float>(1);

should compile (it does on my compilers other than GCC). But why make
the operator a template anyway? It means you have to call it with
explicitly specified template parameters...

Tom

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