digitalmars.D.learn - Need help with basic functional programming
- Eric (44/44) Jul 22 2014 I have been writing several lexers and parsers. The grammars I
- bearophile (11/20) Jul 22 2014 Those () can be omitted, if you mind the noise (but you can also
- Eric (3/23) Jul 22 2014 Thanks! All very good suggestions...
- "Marc =?UTF-8?B?U2Now7x0eiI=?= <schuetzm gmx.net> (4/11) Jul 22 2014 Actually, the ones behind `empty` and `front` are wrong, because
- anonymous (11/23) Jul 22 2014 You've forgotten the exclamation mark: buf.until!(...)
- Eric (1/3) Jul 22 2014 Yes, my bad...
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
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
On Tuesday, 22 July 2014 at 17:09:29 UTC, bearophile wrote:Eric:Thanks! All very good suggestions... -Ericwhile (!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
On Tuesday, 22 July 2014 at 17:09:29 UTC, bearophile wrote:Eric:Actually, the ones behind `empty` and `front` are wrong, because these are defined to be properties. They just happen to work currently.while (!buf.empty()) { p++; buf.popFront();Those () can be omitted, if you mind the noise (but you can also keep them).
Jul 22 2014
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
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