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[英]How to convert a C++ STL vector iterator to a vector reverse iterator?
[英]How to keep track of current and previous iterator on a c++ vector?
我有一個向量容器,我想使用當前迭代器對上一個迭代器對其內容的值進行減法運算,我們將不勝感激
vector<MyClass>::iterator itPrevious = my_vec.begin();
vector<MyClass>::iterator itCurrent = itPrevious;
if (itCurrent != my_vec.end())
{
for (++itCurrent; itCurrent != my_vec.end(); ++itCurrent)
{
// do something with itPrevious and itCurrent
itPrevious = itCurrent;
}
}
if(v.size() < 2)
return;
auto curr = v.begin();
auto next = curr;
++next;
do
{
whatever(*next - *curr );
curr = next++;
} while( next != v.end() )
給定您要的向量,可以使用迭代器算法:
#include <vector>
#include <iostream>
int main() {
std::vector<int> v{ 1, 2, 3, 4 };
for ( auto i = v.begin(); i != v.end(); ++i ) {
if ( i != v.begin() )
*i = *i - *(i-1);
}
for ( auto i : v )
std::cout << i << std::endl;
}
std::vector
的迭代器是RandomAccessIterator
,因此您可以對其執行整數運算。 因此,只要從begin() + 1
迭代, begin() + 1
不需要單獨的“當前”和“上一個”指針:
vector<Foo> myVec = ...;
if(myVec.size() > 1) {
for(vector<Foo>::iterator iter = myVec.begin()+1; iter != myVec.end(); iter++) {
Foo current = *iter;
Foo previous = *(iter - 1);
Foo subtraction = current - previous;
...
}
}
當然,如果向量中的元素少於兩個,則不能減去當前元素和上一個元素。 如果您知道輸入向量將始終至少包含兩個元素,則大小檢查可能是多余的,但是為了安全起見,我將其包括在內。
替代:
for (auto previous = v.begin(), current = previous + 1, end = v.end();
previous != end && current != end;
++previous, ++current)
{
std::cout << *current << " - " << *previous << " = " << *current - *previous << std::endl;
}
即有三種或多或少的優雅方法來解決您的問題。
並在for_each
迭代中使用它
template<class T>
struct substractor {
substractor() : last(nullptr) {}
void operator()(T& item) const
{
if(last != nullptr)
*last -= item;
last = &item;
}
mutable T* last;
};
...
vector<int> v = {3, 2, 1};
for_each(v.begin(), v.end(), substractor<int>());
某種形式的成對原位轉換
template<typename It, typename Op>
void pair_transform(It begin, It end, Op op){
while(begin != end)
{
It next = std::next(begin);
if(next == end) break;
*begin = op(*begin, *next);
++begin;
}
}
...
vector<int> w = {3, 2, 1};
pair_transform(w.begin(), w.end(), std::minus<int>());
恕我直言最好的一個:)簡潔,標准,無其他地方才能理解此代碼。
vector<int> z = {3, 2, 1};
std::transform(z.begin(), z.end() - 1, z.begin() + 1, z.begin(),
std::minus<int>());
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