Re: Why am I so stupid?
On Jun 4, 8:27 am, none <n...@none.none> wrote:
Every little thing I try to do using STL turns into a five
mile crawl through raw sewage.
I want to use the STL to lex/parse some text. This must have
been done at least 5.0e+75 times before, but I can't find the
solution in the FAQ or in any examples.
The STL is not a lexer/parser tool. It addresses a much lower
level. Parts of the STL can be used in parsing.
This is my utterly failed attempt:
std::string s = "5.0 This is a string";
std::istringstream i(s);
std::cout << "Before: \"" << s << "\"" << std::endl;
float x;
i >> x;
std::cout << "Extracted a float: " << x << std::endl;
std::cout << "After: \"" << s << "\"" << std::endl;
And this is the output produced:
Before: "5.0 This is a string"
Extracted a float: 5
After: "5.0 This is a string"
All that I want in this world is for that output to be,
instead, this:
Before: "5.0 This is a string"
Extracted a float: 5
After: " This is a string"
Why would you expect this? If you were using an ifstream, you
wouldn't expect characters extracted from the stream to be
removed from the file.
Then I can just continue parsing the remainder of the string.
I have tried the following, and all have failed:
1) Making the assignment "s = i.str();" after the line "i >> x;" in
hopes that "i.str()" would return only the un-extracted portion of the
string. It doesn't. It returns the entire string.
2) Calling "i.sync();" I don't know why I thought this would help. A
shot in the dark.
3) Calling "i.ignore();" a few times. No effect.
4) Changing the last line to:
std::cout << "After: \"" << i << "\"" << std::endl;
(displaying the value of "i" instead of "s". It displays a
hex number, probably a pointer I guess.
There is no << operator for an istream, so the compiler tries
any implicit conversions available. There's an implicit
conversion to void*; what you're seeing is a result of that.
But you're on the right track.
(
std::cout << "After: \"" << i.rdbuf() << "\"" << std::endl ;
will do what you are trying to do. But you don't want to do
what you are trying to do, see below.)
5) Changing the last line to:
std::cout << "After: \"" << i.str() << "\"" << std::endl;
I knew before I tried that it wouldn't help, but was desperate.
How do you get the >> operator to actually remove the
characters from a string?
You can't, because doing so would break too many expectations
and invariants. A stream is an access method into the
underlying sequence; an input stream should *never* modify the
underlying sequence, in any way.
The answer to your problem, of course, is to use istream for all
of your parsing. After extracting the float, continue
extracting using the same istream (istringstream, etc.). (FWIW:
I have an IteratorInputStream, in which the input sequence is
defined by two STL iterators---it's really simple to do. And
this streambuf has additional functions which return or set the
current iterators, in case part of the input should be parsed
using functions in <algorithm>. But in practice, such cases are
rare; algorithm doesn't contain much useful for parsing.)
--
James Kanze (GABI Software) email:james.kanze@gmail.com
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