Hi all. Just working on a small virtual machine, and thought about using
vector iterators instead of pointer arithmetic. Question is, why does an
iterator plus any number out of range not generate a out_of_range exception?
Maybe this is a gcc issue?
I'm using gcc version 3.3.3 (cygwin special).
Here's the full sample code:
#include <iostream>
#include <vector>
#include <stdexcept>
using namespace std;
int main() {
vector<int> code;
code.push_back( 10L );
code.push_back( 20L );
vector<int>::it erator iter = code.begin();
try {
cout << *(iter + 5) << endl; // 0
cout << code.at( 10 ) << endl; // vector [] access out of range
} catch( out_of_range e ) {
cout << e.what() << endl;
}
return 0;
}
Thanks,
Mike 13 5080
"Mike Austin" <mi**@mike-austin.com> wrote in message
news:0L******** *************@b gtnsc04-news.ops.worldn et.att.net... Hi all. Just working on a small virtual machine, and thought about using vector iterators instead of pointer arithmetic. Question is, why does an iterator plus any number out of range not generate a out_of_range
exception?
Because that's the way the library is designed.
If you want range checking, use vector::at().
That's what it's for.
This follows the "don't pay for what you don't use"
principle of C++. ('at()' will necessarily add more
overhead which might be unacceptable for some
applications).
Maybe this is a gcc issue?
No.
I'm using gcc version 3.3.3 (cygwin special).
Here's the full sample code:
#include <iostream> #include <vector> #include <stdexcept> using namespace std;
int main() { vector<int> code; code.push_back( 10L ); code.push_back( 20L );
vector<int>::it erator iter = code.begin(); try { cout << *(iter + 5) << endl; // 0
I don't know what you mean by your comment "0",
but note that this statement produces 'undefined behavior'.
cout << code.at( 10 ) << endl; // vector [] access out of range } catch( out_of_range e ) { cout << e.what() << endl; }
return 0; }
If you don't use the protection of 'vector::at()' you can still protect
yourself by checking e.g. 'vector::size() ', 'vector::empty( )', comparing
against 'vector::end()' , etc.
-Mike
"Mike Wahler" <mk******@mkwah ler.net> wrote in message
news:s9******** *********@newsr ead3.news.pas.e arthlink.net... try { cout << *(iter + 5) << endl; // 0
Also note that you can use the possibly more intiutive
array notation with a vector:
iter[n]; /* but still no bounds checking */
-Mike
"Mike Austin" <mi**@mike-austin.com> wrote in message
news:0L******** *************@b gtnsc04-news.ops.worldn et.att.net... Hi all. Just working on a small virtual machine, and thought about using vector iterators instead of pointer arithmetic. Question is, why does an iterator plus any number out of range not generate a out_of_range exception? Maybe this is a gcc issue?
Vector iterators are often implemented as pointers, i.e. something like
template <class T>
class vector
{
public:
typedef T* iterator;
typedef const T* const_iterator;
So vector iterators don't throw exceptions for the same reasons that
ordinary pointers don't.
john
Mike Austin wrote: Hi all. Just working on a small virtual machine, and thought about using vector iterators instead of pointer arithmetic. Question is, why does an iterator plus any number out of range not generate a out_of_range exception?
It's never required to. It's undefined behavior to cause an iterator
to go outside the bounds of the array (with the exception of one past the
end). cout << *(iter + 5) << endl; // 0
Evaluating expression (iter + 5) is undefined beahvior.
cout << code.at( 10 ) << endl; // vector [] access out of range
This is OK. At specifically throws an exception for out of range.
"John Harrison" <jo************ *@hotmail.com> wrote in message
news:2t******** *****@uni-berlin.de... "Mike Austin" <mi**@mike-austin.com> wrote in message news:0L******** *************@b gtnsc04-news.ops.worldn et.att.net... Hi all. Just working on a small virtual machine, and thought about
using vector iterators instead of pointer arithmetic. Question is, why does
an iterator plus any number out of range not generate a out_of_range exception? Maybe this is a gcc issue?
Vector iterators are often implemented as pointers, i.e. something like
template <class T> class vector { public: typedef T* iterator; typedef const T* const_iterator;
So vector iterators don't throw exceptions for the same reasons that ordinary pointers don't.
That's possibly an 'incidental' reason for some implementations ,
but imo not "the" reason, which is to allow best possible performance
by not imposing bounds-check overhead.
-Mike
Mike Austin wrote: Hi all. Just working on a small virtual machine, and thought about using vector iterators instead of pointer arithmetic. Question is, why does an iterator plus any number out of range not generate a out_of_range exception? Maybe this is a gcc issue?
I'm using gcc version 3.3.3 (cygwin special).
Here's the full sample code:
#include <iostream> #include <vector> #include <stdexcept> using namespace std;
int main() { vector<int> code; code.push_back( 10L ); code.push_back( 20L );
vector<int>::it erator iter = code.begin(); try { cout << *(iter + 5) << endl; // 0 cout << code.at( 10 ) << endl; // vector [] access out of range } catch( out_of_range e ) { cout << e.what() << endl; }
return 0; }
May well be wrong, but I belive only the member function at() will throw
an out_of_range, so you need to use that to test, otherwise the compiler
has not idea what you are doing.
Adrian
"Mike Austin" <mi**@mike-austin.com> wrote in message
news:0L******** *************@b gtnsc04-news.ops.worldn et.att.net... Hi all. Just working on a small virtual machine, and thought about using vector iterators instead of pointer arithmetic. Question is, why does an iterator plus any number out of range not generate a out_of_range
exception? Maybe this is a gcc issue?
Thanks for everyone's reply. The reason I ask is that I want to point to a
index into vector, then read an offset by that. I could use code.at( ip +
offset ), but I wanted to do everything with iterators. I realize [] does
no bounds checking, but did not know that (iter + n) does not either.
"iter + 0" is the selector
"iter + 1" is the first argument
etc.
Thanks,
Mike
I'm using gcc version 3.3.3 (cygwin special).
Here's the full sample code:
#include <iostream> #include <vector> #include <stdexcept> using namespace std;
int main() { vector<int> code; code.push_back( 10L ); code.push_back( 20L );
vector<int>::it erator iter = code.begin(); try { cout << *(iter + 5) << endl; // 0 cout << code.at( 10 ) << endl; // vector [] access out of range } catch( out_of_range e ) { cout << e.what() << endl; }
return 0; }
Thanks, Mike
"Mike Wahler" <mk******@mkwah ler.net> wrote in message
news:s9******** *********@newsr ead3.news.pas.e arthlink.net... "Mike Austin" <mi**@mike-austin.com> wrote in message news:0L******** *************@b gtnsc04-news.ops.worldn et.att.net... Hi all. Just working on a small virtual machine, and thought about
using vector iterators instead of pointer arithmetic. Question is, why does
an iterator plus any number out of range not generate a out_of_range exception?
Because that's the way the library is designed. If you want range checking, use vector::at(). That's what it's for.
This follows the "don't pay for what you don't use" principle of C++. ('at()' will necessarily add more overhead which might be unacceptable for some applications).
Well, I don't like the way it works. I use vector so I don't have to worry
(as much) about my programs going awry.
I'd be delighted if you could help me write a "safe_itera tor" class. Here's
a start:
template <typename T>
class safe_iterator : public T::iterator {
Value operator ++() {
if( *this == container->end() ) // how to access "container" ?
throw out_of_range( "*** operator vector *(): out of range" );
}
};
Regards,
Mike
"Mike Wahler" <mk******@mkwah ler.net> wrote in message
news:s9******** *********@newsr ead3.news.pas.e arthlink.net... If you don't use the protection of 'vector::at()' you can still protect yourself by [..snip..] comparing against 'vector::end()' , etc.
Note that it is not obvious to compare against end(), because it may
be too late to avoid Undefined Behavior (UB):
iter += 5; // if result is > vect.end(), UB happens here!
if( iter >= vect.end() ) { ... /* but it's too late */ ... };
hth -Ivan
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