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Tyche
Joined: 13 May 2005
Posts: 176
Location: Ohio, USA
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 Posted: Wed Apr 19, 2006 3:24 pm Post subject: STL assign |
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| Code: |
#include <iostream>
#include <string>
using namespace std;
int main(int argc, char** argv) {
string s1;
string s2("stl sucks");
string::const_iterator p = s2.begin();
p++;
s1.assign(p, s2.end() - 1);
cout << s1 << endl;
return 0;
}
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Why in the hell doesn't this compile?
No that's not really the question.
It's more like why doesn't assign() accept const_iterator? |
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Gromble
Joined: 08 Oct 2005
Posts: 23
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| Posted: Wed Apr 19, 2006 5:44 pm Post subject: Re: STL assign |
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According to the documentation I have, string::assign() takes string::size_type arguments, not string::iterator or string::const_iterator arguments. So your assign() options for s2 to s1 are...
s1.assign(s2); // copy entire string
s1.assign(s2, <length>); // copy first <length> chars
s1.assign(s2, <index>, <length>); // copy <length> chars starting at <index> |
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Tyche
Joined: 13 May 2005
Posts: 176
Location: Ohio, USA
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| Posted: Wed Apr 19, 2006 6:49 pm Post subject: |
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This works:
| Code: |
#include <iostream>
#include <string>
using namespace std;
int main(int argc, char** argv) {
string s1;
string s2("stl sucks");
string::iterator p = s2.begin();
const string s3("anally const correct");
string::const_iterator p2 = s3.begin();
p++; p2++;
s1.assign(p, s2.end() - 1);
cout << s1 << endl;
s1.assign(p2, s3.end() - 1);
cout << s1 << endl;
return 0;
}
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I figured it out. begin() and end() return const_iterators only if the string is const.
So while this is okay:
assign (const InputIterator, const InputIterator)
assign (InputIterator, InputIterator)
Neither of these is legal:
assign (const InputIterator, InputIterator)
assign (InputIterator, const InputIterator) |
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Kaz
Joined: 05 Jun 2005
Posts: 24
Location: Hampshire, UK
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| Posted: Thu Apr 20, 2006 3:17 pm Post subject: |
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Ah, here's why. According to my copy of TC++PL, the function for string assignment with iterators is:
| Code: |
template <class In>
basic_string& assign(In first, In last)
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What happened is that one of your arguments was an iterator, the other a const_iterator. Thus, there was no match to this function, because the two types passed in must be the same. However, iterator is convertible to const_iterator, so you can have:
| Code: |
string x("foobar!");
string::iterator begin = x.begin();
string::const_iterator end = x.end();
string y;
y.assign(string::const_iterator(begin), end);
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Edit/Addendum: In retrospect, they could have used different template types for each argument. The reason they didn't, of course, was to make it difficult for the user to try and assign a range from two disparate objects. However, the technology now exists to check at compile-time whether one type is convertible to another, so this could be improved. |
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