Re: Return versus Side-Effect
In article
<fcb00168-7ca9-43ac-9d83-ca439e26aa77@s8g2000prg.googlegroups.com>,
jeungster@gmail.com wrote:
1.) I can write the function prototype like this:
void doManipulation( const std::string & input, std::string &
output ); // (uses side-effect to return ouput string)
2.) Or, I can write the function prototype like this:
std::string doManipulation( const std::string & input ); // (returns
the output string directly)
But how do you make the two functions do equivalent things? Their
semantics differ in subtle ways.
In general, to make (1) equivalent to (2), it would have to be written
like (this is pseudocode to show the issues):
void doManipulation(const std::string& input, std::string& output)
{
try
{
// Are input and output the same string object?
if (&input == &output)
// do something magical
if (WAYTOOBIG <= output.capacity())
// time for another trick
// do the work
}
catch (...)
{
// perform more magic
throw;
}
}
Compared with the much simpler but sometimes more expensive:
std::string doManipulation(const std::string& input)
{
// do the work
}
SInce it is easier to take correct code and make it faster when
necessary than to take faster but subtly buggy code and make it correct,
I usually use (2) unless I know (and can back up by measurement) that I
can't afford it.
--
Nevin ":-)" Liber <mailto:nevin@eviloverlord.com> 773 961-1620
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