Problems with if_then_else_return in boost::lambda library

7 Aug 2003

      Hi,

If I have a vector of strings

       vector<string> v;

       v.push_back("One");

       v.push_back("Two");

       v.push_back("Three");

       v.push_back("Four");

       v.push_back("Five");

and define arithmetic action plus like this:

namespace boost {

namespace lambda {

        struct plain_return_type_2<arithmetic_action<plus_action>,
std::string, std::string> {

                typedef std::string type;

        };

}

}

then the following code works just fine:

cout << accumulate(v.begin()+1, v.end(), v.front(), _1 + string(", ") +
_2) << endl;

and the following also works wonderfully:

        int i = 1;

        cout << accumulate(v.begin()+1, v.end(), v.front(), _1 +
bind(constructor<string>(), var(i)++, ' ') + _2) << endl;

Now, if I want to use if_then_else_return as shown below, it fails
miserably, producing a huge error message:

cout << accumulate(v.begin(), v.end(), string(), if_then_else_return(_1
== string(), 

_1,

_1 + bind(constructor<string>(), var(i)++, ' ') + _2)) << endl;

Now for the questions:

What is the proper way to use if_then_else_return? Why do I need to
define a separate arithmetic action for strings, even though they are a
common library type? Is there a way to say the most intuitive:

cout << accumulate(v.begin()+1, v.end(), v.front(), _1 + ", " + _2) <<
endl;

Is there way to do more intuitive:

cout << accumulate(v.begin()+1, v.end(), v.front(), _1 + string(i++, '
') + _2) << endl;

Thank you for your help.

Robert Ioffe

[Non-text portions of this message have been removed]

Problems with if_then_else_return in boost::lambda library

Ioffe, Robert