Re: ostream output - indenting

From:

Christopher <cpisz@austin.rr.com>

Newsgroups:

comp.lang.c++

Date:

Mon, 28 Jul 2008 11:21:22 -0700 (PDT)

Message-ID:

<275b99b6-bf2d-46d0-b5bf-3f61ff1ee8c8@y38g2000hsy.googlegroups.com>

On Jul 28, 1:10 pm, Joe Greer <jgr...@doubletake.com> wrote:

Christopher <cp...@austin.rr.com> wrote in news:dfc7ee09-8cdf-4e7d-b8f0-
69531c079...@34g2000hsf.googlegroups.com:

How would I go about indenenting for each level of recursion, if I am
trying to output the contents of a class which contains its own type?

where AttributeGroupMap is
typdef std::map<std::string, AttributeGroup *> AttributeGroupMap;

const std::string AttributeGroup::ToString() const
{
   std::stringstream ss;

   ss << m_name << std::endl;

   for(AttributeGroupMap::const_iterator it =
m_attributeGroups.begin();
        it != m_attributeGroups.end(); ++it)
   {
      // need everything from here indented for each level of
recursion
      ss << (it->second)->ToString() << std::endl;
   }

   return ss.str();
}

I thought if changing the parameters to something like
const std::string AttributeGroup::ToString(unsigned indent = 0) const

and then
ss << /** do something to indent here */ (it->second)-

ToString(indent + 3) << std::endl;

but still don't know how to indent the entire thing.

Indenting is pretty simple, simply prepend std::string(indent, ' ') to
the string. If I were you though, I would separate the recursive outpu=

from the ToString() method. A method named ToString() is too general
for such a specific use. I would start with something like:

// since we return by value, there is no need to return a const string
std::string AttributeGroup::ToString() const
{
   return m_name; // assumes m_name is a string

}

std::string AttributeGroup::RecursiveAttributeDump(int lvl) const
{
   std::string ret = std::string(lvl * 3, ' ') + ToString() + '\n';

   for (AttributeGroupMap::const_iterator it = m_attributeGroups.be=

gin

(),
      it != m_AttributeGroups.end(), ++it)
   {
      RecursiveAttributeDump(lvl+1);
   }
   return ret;

}

This is all untried but it should work.

joe

}

mmm
works if there is one data member to output, but I failed to show in
my concept code there that there are multiple lines of output, as
there will be multiple data members in the object. Name was just one.
Sorry I left that bit out.