I have:
struct A
{
int index;
A(): index(0) {}
}
std::vector<A> ManyAs(10, A());
And wants to do:
for (int i = 0, size = ManyAs.size(); i < size; ++i)
{
ManyAs[i].index = i;
}
I want to do this with std algotrithm, maybe std::for_each?
How to do it, thanks!
I would do the following way
struct A
{
int index;
A(): index(0) {}
A & operator =( int i )
{
index = i;
return ( *this );
}
};
std::iota( ManyAs.begin(), ManyAs.end(), 0 );
In this particular case, I'd leave the code as it is. Standard library algorithms are best used when you want to apply the same operation to all elements of the range. But here, you don't actually apply the same operation, as you assign a different number to each element's index. So a counter-based for loop seems to be the most natural solution here.
However, if you really want to use a standard algorithm, you can use a stateful functor:
struct Functor
{
size_t index;
Functor() : index(0) {}
void operator() (A &a) { a.index = index++; }
};
std::for_each(ManyAs.begin(), ManyAs.end(), Functor());
for_each is for the same operation, I got it. So maybe a generate or transform? Is there any more suitable algorithm? By the way, your functor is still a good point, thanks!Here are two approaces:
struct my_functor
{
my_functor()
: i(0)
{
}
void operator () (A & a)
{
a.index = i++;
}
int i;
};
void foo();
{
//old c++ style
std::for_each(ManyAs.begin(), ManyAs.end(), my_functor());
}
second:
//c++11
int i = 0;
std::for_each(ManyAs.begin(), ManyAs.end(), [&](A & a){ a.index = i++; });
static int i = 0; inside the lambda itself.
i will never get set to 0 if the code that uses the lambda is reused.Somehow I found my ideal answer:
std::for_each(ManyAs.begin(), ManyAs.end(),
[&](A& a)
{
int offset=(&a) - (&ManyAs.front());
a.index = offset;
});
Actually it's quite similar to Raxvan's answer, but everything is local, which is better imo.