Re: string vs. vector<char>

From:
Richard Herring <junk@[127.0.0.1]>
Newsgroups:
comp.lang.c++
Date:
Thu, 4 Jun 2009 11:10:47 +0100
Message-ID:
<aGg+shFn25JKFwI9@baesystems.com>
In message
<5b6da5bd-445f-43cf-a09c-519c3d506166@s16g2000vbp.googlegroups.com>,
thomas <FreshThomas@gmail.com> writes

On Jun 4, 5:07?pm, thomas <FreshTho...@gmail.com> wrote:

On Jun 4, 4:21?pm, Ian Collins <ian-n...@hotmail.com> wrote:

thomas wrote:

Hi,
? ? In my mind, string and vector<char> are equivalent in some sense.
? ?But I have found that the vector operation "pop_back()" is missing
for "string". I prefer the "string" to "vector<char>" when playing
strings. But I will need the "pop_back()" operation from time to
time.
? ? I know that "resize(str.size()-1)" is a workaround, but I'm not
sure if there is any efficiency loss compared with "pop_back()" etc.


I'd have thought the vector equivalent would be to erase the second from
last item, assuming your "string" is null terminated.


so you mean that a "string" is null terminated. If a "pop_back()"
operation exists, there's confusion that whether the last character or
'\0' is poped out.


No. In the STL container model, pop_back() doesn't return a value, it
simply decrements the effective position of end().

en.. I think it makes sense.


By the way, I am not sure how the "string" is designed and
implemented. But I think the iterator operations (string.begin(),
string.end(), etc.) should have been specially handled for strings. Or
how can a "string.end()" tell whether it's after the '\0' or right at
the position of '\0'?


It can't and doesn't. end() simply points to size() elements beyond
begin().

In std::string, '\0' is not a special character, and str[str.size()-1]
is not generally '\0' unless you deliberately make it so.

It's only the result of c_str() that is guaranteed to have a terminating
null added at offset size().

--
Richard Herring

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