About the find() is STL algorithm

[Draw]

Hi All,

Just a thought, about the find() algorithm in the C++ STL. I read that
the find algorithm can take a range of iterators. If it does not find
the element it is looking for in that range it returns the iterator to
the last element in the range, not to the last element in the
container, to the last element in the range. That being said, how can
we tell if find() has been successful in finding the element we need?
Its easy when we are searching the whole container, we can always the
container's end() method, but when we are searching only a subset of
all the elements in a container, what then? Do we have a better way
than checking the element that find returns again?(sounds inefficient,
repetitive).

Would appreciate your thoughts on this matter.

Thanks, Draw

 

[Alf P. Steinbach]

* Draw:

Just a thought, about the find() algorithm in the C++ STL. I read that
the find algorithm can take a range of iterators. If it does not find
the element it is looking for in that range it returns the iterator to
the last element in the range, not to the last element in the
container, to the last element in the range.

A range of elements is specified as an iterator A to the first element,
and an iterator B one past the last element.

If std::find fails it returns B, in all cases.

Compare the result iterator to B.

 

[Draw]

* Draw:

A range of elements is specified as an iterator A to the first element,
and an iterator B one past the last element.

If std::find fails it returns B, in all cases.

Compare the result iterator to B.

--

Thanks, but I don't think I got an answer for my question. Are you
saying that I have to always give the start and end as the range? What
if I don't want to, what if I want to search only the first 10 elements
of the vector?
I went ahead and tried to do that, the code is right after I stop
blabbering , and find() does indeed return the last element of the
range if it doesn't find anything(not that of the container), but what
if the last element of the range does satisfy the condition find() is
looking for, in that case, it would still return the last element in
the range. Thats ambiguous, isn't it?
And what you say above proves it, it does return the last element in
the range you provide, but you are able to work around the situation I
mention because your last element happens to equal the iterator
*.end()(indeed, one past the last element).

//[START]

#include<iostream>
#include<algorithm>
#include<vector>


int main(void){
unsigned int index;
vector<int> intvec(20,3);
vector<int>::iterator it,ret;


for(it = intvec.begin(),index=0;it != intvec.end();++it,++index){
*it =+ index;
cout << "vector Element value equated to subscript "<<*it<<endl;
}
intvec[15] = 45;


cout << "Finding the element with value equal to 45 \n";
it = intvec.begin();
ret = find(it,it+10,45);


if(ret == intvec.end()){
cout << "No such value in the vector.\n";
}
else{
cout << "Element value " << *ret << endl;
}
return(0);
}

//[END]

 

[Markus Schoder]

Thanks, but I don't think I got an answer for my question. Are you
saying that I have to always give the start and end as the range? What
if I don't want to, what if I want to search only the first 10 elements
of the vector?
I went ahead and tried to do that, the code is right after I stop
blabbering , and find() does indeed return the last element of the
range if it doesn't find anything(not that of the container), but what
if the last element of the range does satisfy the condition find() is
looking for, in that case, it would still return the last element in
the range. Thats ambiguous, isn't it?

The "one past the last element" concept is what you seem to be missing.

The range is specified by giving an iterator A to the first element and
an iterator B to _one past_ the last element in the range i.e. iterator
B itself does not point to an element within the range. Iterator B will
therefore never be returned if the element is found.

 

[Draw]

The "one past the last element" concept is what you seem to be missing.

The range is specified by giving an iterator A to the first element and
an iterator B to _one past_ the last element in the range i.e. iterator
B itself does not point to an element within the range. Iterator B will
therefore never be returned if the element is found.

Thanks Guys! Now I see. Thanks for clarifying that.

 

[Jack Saalweachter]

Thanks, but I don't think I got an answer for my question. Are you
saying that I have to always give the start and end as the range? What
if I don't want to, what if I want to search only the first 10 elements
of the vector?
I went ahead and tried to do that, the code is right after I stop
blabbering , and find() does indeed return the last element of the
range if it doesn't find anything(not that of the container), but what
if the last element of the range does satisfy the condition find() is
looking for, in that case, it would still return the last element in
the range. Thats ambiguous, isn't it?
And what you say above proves it, it does return the last element in
the range you provide, but you are able to work around the situation I
mention because your last element happens to equal the iterator
*.end()(indeed, one past the last element).

See the changed code.

//[START]

#include<iostream>
#include<algorithm>
#include<vector>


int main(void){
unsigned int index;
vector<int> intvec(20,3);
vector<int>::iterator it,ret;


for(it = intvec.begin(),index=0;it != intvec.end();++it,++index){
*it =+ index;
cout << "vector Element value equated to subscript "<<*it<<endl;
}
intvec[15] = 45;


cout << "Finding the element with value equal to 45 \n";
it = intvec.begin();
ret = find(it,it+10,45);

// if(ret == intvec.end()){
if (ret == (it+10)) {
// If find does not find the value in [it, it+10), it
// will return it+10, which is one element past the last
// element in [it, it+10)

cout << "No such value in the vector.\n";
}
else{
cout << "Element value " << *ret << endl;
}
return(0);
}

//[END]