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digitalmars.D.learn - Need help with basic functional programming

reply "Eric" <eric makechip.com> writes:
I have been writing several lexers and parsers. The grammars I 
need to
parse are really complex, and consequently I didn't feel 
confident about
the code quality, especially in the lexers.  So I decided to jump 
on the functional progamming bandwagon to see if that would help. 
  It definitely
does help, there are fewer lines of code, and I feel better about 
the code
quality.  I started at the high level, and had the input buffer 
return a
range of characters, and the lexer return a range of tokens.  But 
when I got
down to the lower levels of building up tokens, I ran into a 
problem:

First I started with this which worked:

private void getNumber(MCInputStreamRange buf)
{
     while (!buf.empty())
     {
         p++;
         buf.popFront();
         if (buf.front() <= '0' || buf.front() >= '9') break;
         *p = buf.front();
     }
     curTok.kind = Token_t.NUMBER;
     curTok.image = cast(string) cbuffer[0 .. (p - 
cbuffer.ptr)].dup;
}

I thought I could improve this like so:

private void getNumber(MCInputStreamRange buf)
{
     auto s = buf.until("a <= '0' || a >= '9'");
     curTok.kind = Token_t.NUMBER;
     curTok.image = to!string(s);
}

The problem is that "until" seems to not stop at the end of the 
number,
and instead continues until the end of the buffer.  Am I doing 
something
wrong here?  Also, what is the fastest way to convert a range to 
a string?

Thanks,

Eric
Jul 22 2014
next sibling parent reply "bearophile" <bearophileHUGS lycos.com> writes:
Eric:


     while (!buf.empty())
     {
         p++;
         buf.popFront();


Those () can be omitted, if you mind the noise (but you can also 
keep them).



         if (buf.front() <= '0' || buf.front() >= '9') break;


std.ascii.isDigit helps.



     curTok.image = cast(string) cbuffer[0 .. (p - 
 cbuffer.ptr)].dup;


If you want a string, then idup is better. Try to minimize the 
number of casts in your code.



     auto s = buf.until("a <= '0' || a >= '9'");


Perhaps you need a ! after the until, or a !q{a <= '0' || a >= 
'9'}.



 Also, what is the fastest way to convert a range to a string?


The "text" function is the simplest.

Bye,
bearophile
Jul 22 2014
next sibling parent "Eric" <eric makechip.com> writes:
On Tuesday, 22 July 2014 at 17:09:29 UTC, bearophile wrote:

 Eric:


    while (!buf.empty())
    {
        p++;
        buf.popFront();


 Those () can be omitted, if you mind the noise (but you can 
 also keep them).



        if (buf.front() <= '0' || buf.front() >= '9') break;


 std.ascii.isDigit helps.



    curTok.image = cast(string) cbuffer[0 .. (p - 
 cbuffer.ptr)].dup;


 If you want a string, then idup is better. Try to minimize the 
 number of casts in your code.



    auto s = buf.until("a <= '0' || a >= '9'");


 Perhaps you need a ! after the until, or a !q{a <= '0' || a >= 
 '9'}.



 Also, what is the fastest way to convert a range to a string?


 The "text" function is the simplest.

 Bye,
 bearophile


Thanks!  All very good suggestions...

-Eric
Jul 22 2014
prev sibling parent "Marc =?UTF-8?B?U2Now7x0eiI=?= <schuetzm gmx.net> writes:
On Tuesday, 22 July 2014 at 17:09:29 UTC, bearophile wrote:

 Eric:


    while (!buf.empty())
    {
        p++;
        buf.popFront();


 Those () can be omitted, if you mind the noise (but you can 
 also keep them).


Actually, the ones behind `empty` and `front` are wrong, because 
these are defined to be properties. They just happen to work 
currently.
Jul 22 2014
prev sibling parent reply "anonymous" <anonymous example.com> writes:
On Tuesday, 22 July 2014 at 16:50:47 UTC, Eric wrote:

 private void getNumber(MCInputStreamRange buf)
 {
     auto s = buf.until("a <= '0' || a >= '9'");
     curTok.kind = Token_t.NUMBER;
     curTok.image = to!string(s);
 }

 The problem is that "until" seems to not stop at the end of the 
 number,
 and instead continues until the end of the buffer.  Am I doing 
 something
 wrong here?


You've forgotten the exclamation mark: buf.until!(...)
Without it, the string is not the predicate, but the sentinel
value. I.e. the range stops when it sees the characters "a <= '0'
|| a >= '9'".

By the way, do you really mean to stop on '0' and '9'? Do you
perhaps mean "a < '0' || a > '9'"?


  Also, what is the fastest way to convert a range to a string?


The fastest to type is probably text(r) (or r.text). The fastest
for me to come up with is r.to!string, which does exactly the
same. I don't know about run time, but text/to!string is
hopefully fine.
Jul 22 2014
parent "Eric" <eric makechip.com> writes:
 By the way, do you really mean to stop on '0' and '9'? Do you
 perhaps mean "a < '0' || a > '9'"?


Yes, my bad...
Jul 22 2014