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iwongu · Guest

After I learned some Ruby syntax, I could write the code that print
one's tables in a line like the following.

(2..9).each { |i| (1..9).each { |j| puts "#{i}*#{j}=#{i * j}" } }

So, I thought it's good to have something like this in C++.
The written code is this.

----
using namespace std;
using namespace boost;
using namespace boost::lambda;
using namespace boost::mpl;

class range
{
public:
range() {
}
range(int s, int e) {
s_.push_back(s);
e_.push_back(e);
}

range& operator()(int s, int e) {
s_.push_back(s);
e_.push_back(e);

return *this;
}

template <class F>
range& operator()(F func) {
// now this range only support 2-depth loop.
// but it can be extended.
typedef typename if_c<F::arity == 2, do_2, do_1>::type do_type;

do_type()(s_, e_, func);

return *this;
}

struct do_2
{
void operator()(vector<int>& s_, vector<int>& e_, function<void
(int, int)> func) {
for (int i = s_[0]; i <= e_[0]; ++i)
for (int j = s_[1]; j <= e_[1]; ++j)
func(i, j);
}
};

struct do_1
{
void operator()(vector<int>& s_, vector<int>& e_, function<void
(int)> func) {
for (int i = s_[0]; i <= e_[0]; ++i)
func(i);
}
};

private:
vector<int> s_;
vector<int> e_;
};

With this, I could write the print one's tables code like the
following.

function<void (int, int)> f(cout << _1 << '*' << _2 << '=' << _1 * _2
<< '\n');
range(2, 9)(1, 9)( f );

or

range(2, 9)(1, 9)( function<void (int, int)>(cout << _1 << '*' << _2 <<
'=' << _1 * _2 << '\n' ) );

But What I really want to do is using lamda directly as the parameter
like the below.

range(2, 9)(1, 9)( cout << _1 << '*' << _2 << '=' << _1 * _2 << '\n' );

But I can't find something like ::arity with the lambda expression.

Is there any solution? or is there already some 'range' class like
this?

----
And one more question.
In the documentation of lambda, the 'protect' function is introduced
with sample code.

int x = 1, y = 10;
(_1 + protect(_1 + 2))(x)(y);

But, I can't compile this with g++ 3.2.3.
The following is successfully compiled and the result id 12.

(protect(_1 + 2))(x)(y);

----
I found that the 'function_traits' can not be used with 'function.' I
wish it'll be compatible. :-)

Thanks in advance.

iwongu

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Peter Dimov · Guest

iwongu wrote: