Look at my code:
void modify_element(BigType &x)
{
/* not shown */
}
BigType modify_element_copy(BigType x)
{
modify_element(x);
return x;
}
void modify_container(std::vector<BigType> &v)
{
std::transform(v.begin(), v.end(), v.begin(), modify_element_copy);
}
Can I replace the call to std::tranform with a call to std::for_each? My
documentation says that for_each must not modify the elements. Here is
modify_container using std::for_each rather than std::transform:
void modify_container(std::vector<BigType> &v)
{
std::for_each(v.begin(), v.end(), modify_element);
}
Any help is appreciated.
11  3887 
"Jason Heyes" <ja********@optusnet.com.au> wrote in message
news:42**********************@news.optusnet.com.au ... Look at my code:
void modify_element(BigType &x)
{
/* not shown */
}
BigType modify_element_copy(BigType x)
{
modify_element(x);
return x;
}
void modify_container(std::vector<BigType> &v)
{
std::transform(v.begin(), v.end(), v.begin(), modify_element_copy);
}
Can I replace the call to std::tranform with a call to std::for_each? My
documentation says that for_each must not modify the elements.
Here is
modify_container using std::for_each rather than std::transform:
void modify_container(std::vector<BigType> &v)
{
std::for_each(v.begin(), v.end(), modify_element);
}
Yes, 'for_each()' should work fine. Just remember not to try
using a modifying predicate with a const vector or via
const_iterators (well, the compiler should complain if you
do anyway).
-Mike
"Mike Wahler" <mk******@mkwahler.net> wrote in message
news:Qe****************@newsread3.news.pas.earthli nk.net...
Yes, 'for_each()' should work fine. Just remember not to try
using a modifying predicate with a const vector or via
const_iterators (well, the compiler should complain if you
do anyway).
-Mike
Ok thanks for that.
The for_each algorithm does not modify the elements of the sequence. It
calls Function F for each element in the range [First, Last) and
returns the input parameter F. This function does not modify any
elements in the sequence.
The for_each algorithm does not modify the elements of the sequence. It
calls Function F for each element in the range [First, Last) and
returns the input parameter F. This function does not modify any
elements in the sequence. See "The Difference between for_each and
transform", C/C++ Users Journal, February 2001, Klaus Kreft &
Angelika Langer. http://www.langer.camelot.de/Article...Transform.html
"Matthew Schaefer" <ma**************@yahoo.com> wrote in message
news:11*********************@f14g2000cwb.googlegro ups.com... The for_each algorithm does not modify the elements of the sequence. It
calls Function F for each element in the range [First, Last) and
returns the input parameter F. This function does not modify any
elements in the sequence. See "The Difference between for_each and
transform", C/C++ Users Journal, February 2001, Klaus Kreft &
Angelika Langer.
http://www.langer.camelot.de/Article...Transform.html
I read the article and it contradicts what you just said. It states that
for_each "permits side effects including modification of elements from the
input sequence." You said "this function does not modify any elements in the
sequence." What is right?
In message <11**********************@o13g2000cwo.googlegroups .com>,
Matthew Schaefer <ma**************@yahoo.com> writes The for_each algorithm does not modify the elements of the sequence. It
calls Function F for each element in the range [First, Last) and
returns the input parameter F. This function does not modify any
elements in the sequence.
The *algorithm* does not, of itself, modify any elements of the
sequence, and it discards the result of function F. That doesn't prevent
the function F from modifying its argument if you pass iterators of an
appropriate type.
--
Richard Herring
"Richard Herring" <ju**@[127.0.0.1]> wrote in message
news:vQ**************@baesystems.com... In message <11**********************@o13g2000cwo.googlegroups .com>,
Matthew Schaefer <ma**************@yahoo.com> writesThe for_each algorithm does not modify the elements of the sequence. It
calls Function F for each element in the range [First, Last) and
returns the input parameter F. This function does not modify any
elements in the sequence.
The *algorithm* does not, of itself, modify any elements of the sequence,
and it discards the result of function F. That doesn't prevent the
function F from modifying its argument if you pass iterators of an
appropriate type.
The result of F eventually comes back as a return value, I believe.
In message <42**********************@news.optusnet.com.au>, Jason Heyes
<ja********@optusnet.com.au> writes "Richard Herring" <ju**@[127.0.0.1]> wrote in message
news:vQ**************@baesystems.com... In message <11**********************@o13g2000cwo.googlegroups .com>,
Matthew Schaefer <ma**************@yahoo.com> writesThe for_each algorithm does not modify the elements of the sequence. It
calls Function F for each element in the range [First, Last) and
returns the input parameter F. This function does not modify any
elements in the sequence.
The *algorithm* does not, of itself, modify any elements of the sequence,
and it discards the result of function F. That doesn't prevent the
function F from modifying its argument if you pass iterators of an
appropriate type.
The result of F eventually comes back as a return value, I believe.
No. A copy of F itself (the functor or function pointer) is returned:
template <class InputIterator, class Function>
Function for_each (InputIterator first, InputIterator last, Function f)
{
while (first != last) f(*first++);
return f;
}
--
Richard Herring
"Matthew Schaefer" <ma**************@yahoo.com> wrote in message
news:11*********************@f14g2000cwb.googlegro ups.com... The for_each algorithm does not modify the elements of the sequence.
It calls the user supplied predicate function once for
each element, passing it a reference to the element.
The function can modify the referred to object if it likes.
It calls Function F for each element in the range [First, Last) and
returns the input parameter F.
This function does not modify any
elements in the sequence. See "The Difference between for_each and
transform", C/C++ Users Journal, February 2001, Klaus Kreft &
Angelika Langer.
http://www.langer.camelot.de/Article...ForEachTransfo
rm.html
Read it again. :-)
-Mike
"Richard Herring" <ju**@[127.0.0.1]> wrote in message
news:Cp**************@baesystems.com... In message <42**********************@news.optusnet.com.au>, Jason Heyes
<ja********@optusnet.com.au> writes"Richard Herring" <ju**@[127.0.0.1]> wrote in message
news:vQ**************@baesystems.com... In message <11**********************@o13g2000cwo.googlegroups .com>,
Matthew Schaefer <ma**************@yahoo.com> writes
The for_each algorithm does not modify the elements of the sequence. It
calls Function F for each element in the range [First, Last) and
returns the input parameter F. This function does not modify any
elements in the sequence.
The *algorithm* does not, of itself, modify any elements of the
sequence, and it discards the result of function F. That doesn't prevent the
function F from modifying its argument if you pass iterators of an
appropriate type.
The result of F eventually comes back as a return value, I believe.
No. A copy of F itself (the functor or function pointer) is returned:
template <class InputIterator, class Function>
Function for_each (InputIterator first, InputIterator last, Function f)
{
while (first != last) f(*first++);
return f;
}
Jason: Also note that if f() does return a value, it is
ignored. (So there's no point in f() having a return type
other than 'void', unless one plans to use it in other
contexts where the return value would be available).
-Mike
"Richard Herring" <ju**@[127.0.0.1]> wrote in message
news:Cp**************@baesystems.com... In message <42**********************@news.optusnet.com.au>, Jason Heyes
<ja********@optusnet.com.au> writesThe result of F eventually comes back as a return value, I believe.
No. A copy of F itself (the functor or function pointer) is returned:
template <class InputIterator, class Function>
Function for_each (InputIterator first, InputIterator last, Function f)
{
while (first != last) f(*first++);
return f;
}
Oh yea. Quite right. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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