/*Copy the line a token at a time into the output*/
copy(istream_it erator<string>( iss),
istream_iterato r<string>(),
ostream_iterato r<string>(oss, " "));
What's the meaning of this statement? Thanks!
14  2097 
"dover" <do*****@close. com> wrote in message
news:g1******** *************@b gtnsc04-news.ops.worldn et.att.net... /*Copy the line a token at a time into the output*/
copy(istream_it erator<string>( iss),
istream_iterato r<string>(),
ostream_iterato r<string>(oss, " "));
What's the meaning of this statement? Thanks!
Again as your previous post the comment gives you the answer. copy algorithm
takes three parameters - the first two are beginning and one past the end
iterators of the source range and the third parameter is the beginning
iterator of the destination range.
So your source range is a istringstream (strings can behave as external
devices to streams.) Your destination is an ostringstream with space as the
delimiter.
So if input stream contains "This is a test" then destination string will
also contain "This is a test".
-Sharad
dover wrote: /*Copy the line a token at a time into the output*/
copy(istream_it erator<string>( iss),
istream_iterato r<string>(),
ostream_iterato r<string>(oss, " "));
What's the meaning of this statement? Thanks!
It copies the line a token at a time into the output.
dover wrote: /*Copy the line a token at a time into the output*/
copy(istream_it erator<string>( iss),
istream_iterato r<string>(),
ostream_iterato r<string>(oss, " "));
What's the meaning of this statement? Thanks!
Others have answered this question. I'd like to pose a bonus question
to the group: what's the meaning of this statement?
vector<string> vec( istream_iterato r<string>(iss) ,
istream_iterato r<string>() );
--
Russell Hanneken eu*******@cbobk .pbz
Use ROT13 to decode my email address.
"Russell Hanneken" <me@privacy.net > wrote in message
news:_j******** *********@newsr ead2.news.pas.e arthlink.net... dover wrote: /*Copy the line a token at a time into the output*/
copy(istream_it erator<string>( iss),
istream_iterato r<string>(),
ostream_iterato r<string>(oss, " "));
What's the meaning of this statement? Thanks!
Others have answered this question. I'd like to pose a bonus question
to the group: what's the meaning of this statement?
vector<string> vec( istream_iterato r<string>(iss) ,
istream_iterato r<string>() );
C++'s most vexing parse! It's a function declaration and not that of a
vector. To get correct results put another set of parenthesis -
vector<string> vec( (istream_iterat or<string>(iss) ),
istream_iterato r<string>() );
Seems you have read Effective STL ;-)
-Sharad
"Russell Hanneken" <me@privacy.net > wrote in message
news:_j******** *********@newsr ead2.news.pas.e arthlink.net... dover wrote: /*Copy the line a token at a time into the output*/
copy(istream_it erator<string>( iss),
istream_iterato r<string>(),
ostream_iterato r<string>(oss, " "));
What's the meaning of this statement? Thanks!
Others have answered this question. I'd like to pose a bonus question
to the group: what's the meaning of this statement?
vector<string> vec( istream_iterato r<string>(iss) ,
istream_iterato r<string>() );
C++'s most vexing parse! It's a function declaration and not that of a
vector. To get correct results put another set of parenthesis -
vector<string> vec( (istream_iterat or<string>(iss) ),
istream_iterato r<string>() );
Seems you have read Effective STL ;-)
-Sharad
Sharad Kala wrote: "Russell Hanneken" <me@privacy.net > wrote in message
news:_j******** *********@newsr ead2.news.pas.e arthlink.net...
I'd like to pose a bonus question
to the group: what's the meaning of this statement?
vector<string> vec( istream_iterato r<string>(iss) ,
istream_iterato r<string>() );
C++'s most vexing parse! It's a function declaration and not that of a
vector. To get correct results put another set of parenthesis -
vector<string> vec( (istream_iterat or<string>(iss) ),
istream_iterato r<string>() );
Seems you have read Effective STL ;-)
Aw, you got it right. That's no fun. :^)
I haven't actually read _Effective STL_, but I did read "Guru of the
Week" #75, which covers the issue: http://www.gotw.ca/gotw/075.htm
If anyone's confused about why the statement above declares a function
rather than a vector, I recommend the article.
--
Russell Hanneken eu*******@cbobk .pbz
Use ROT13 to decode my email address.
"Sharad Kala" <no************ ******@yahoo.co m> wrote in message news:2l******** ****@uni-berlin.de...
"Russell Hanneken" <me@privacy.net > wrote in message
news:_j******** *********@newsr ead2.news.pas.e arthlink.net... dover wrote: /*Copy the line a token at a time into the output*/
copy(istream_it erator<string>( iss),
istream_iterato r<string>(),
ostream_iterato r<string>(oss, " "));
What's the meaning of this statement? Thanks!
Others have answered this question. I'd like to pose a bonus question
to the group: what's the meaning of this statement?
vector<string> vec( istream_iterato r<string>(iss) ,
istream_iterato r<string>() );
C++'s most vexing parse! It's a function declaration and not that of a vector.
Is it really a function _declaration_? : iss is not type, but variable (istringstream iss).
To get correct results put another set of parenthesis -
vector<string> vec( (istream_iterat or<string>(iss) ),
istream_iterato r<string>() );
[snip]
--
Alex Vinokur http://mathforum.org/library/view/10978.html http://sourceforge.net/users/alexvn
"Alex Vinokur" <al****@big.foo t.com> wrote in message
news:2l******** ****@uni-berlin.de...
"Sharad Kala" <no************ ******@yahoo.co m> wrote in message
news:2l******** ****@uni-berlin.de...
"Russell Hanneken" <me@privacy.net > wrote in message
news:_j******** *********@newsr ead2.news.pas.e arthlink.net...
Is it really a function _declaration_? : iss is not type, but variable
(istringstream iss).
Yes, it is.
void foo( int d);
void foo (int (d)); // same as above; parens around d are ignored
With this knowledge, in vector<string> vec( istream_iterato r<string>(iss) ,
istream_iterato r<string>() );
the first parameter is of type istream_iterato r<string>. The parentheses
around iss are superfluous and are ignored. I hope you have got the second
parameter right by yourself :-)
-Sharad
"Sharad Kala" <no************ ******@yahoo.co m> wrote in message news:2l******** ****@uni-berlin.de...
[snip]
void foo( int d);
void foo (int (d)); // same as above; parens around d are ignored
d is of int type
With this knowledge, in vector<string> vec( istream_iterato r<string>(iss) ,
istream_iterato r<string>() );
the first parameter is of type istream_iterato r<string>. The parentheses
around iss are superfluous and are ignored.
[snip]
But iss is not of istream_iterato r<string> type; iss is of istringstream type.
--
Alex Vinokur http://mathforum.org/library/view/10978.html http://sourceforge.net/users/alexvn This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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