这个问题很愚蠢,但我需要以一种非常有效的方式来做——它将在我的代码中再次执行.我有一个返回向量的函数,我必须将返回的值逐个元素添加到另一个向量中.很简单:
The question is quite dumb, but I need to do it in a very efficient way - it will be performed over an over again in my code. I have a function that returns a vector, and I have to add the returned values to another vector, element by element. Quite simple:
vector<double> result;
vector<double> result_temp
for(int i=0; i< 10; i++) result_temp.push_back(i);
result += result_temp //I would like to do something like that.
for(int i =0; i< result_temp.size();i++)result[i] += result_temp[i]; //this give me segfault
我想做的数学运算是
u[i] = u[i] + v[i] 对于所有的 i
u[i] = u[i] + v[i] for all i
可以做什么?
谢谢
添加了一个简单的初始化,因为这不是重点.结果应该如何初始化?
added a simple initialization, as that is not the point. How should result be initialized?
如果您尝试将一个 vector
附加到另一个,您可以使用以下类似的方法.这些来自我的一个实用程序库 - std::vector
的两个 operator+=
重载:一个将单个元素附加到 vector
,另一个附加整个vector
:
If you are trying to append one vector
to another, you can use something like the following. These are from one of my utilities libraries--two operator+=
overloads for std::vector
: one appends a single element to the vector
, the other appends an entire vector
:
template <typename T>
std::vector<T>& operator+=(std::vector<T>& a, const std::vector<T>& b)
{
a.insert(a.end(), b.begin(), b.end());
return a;
}
template <typename T>
std::vector<T>& operator+=(std::vector<T>& aVector, const T& aObject)
{
aVector.push_back(aObject);
return aVector;
}
如果您正在尝试执行求和(即,创建一个新的 vector
包含其他两个 vector
的元素之和),您可以使用一些东西像下面这样:
If you are trying to perform a summation (that is, create a new vector
containing the sums of the elements of two other vector
s), you can use something like the following:
#include <algorithm>
#include <functional>
template <typename T>
std::vector<T> operator+(const std::vector<T>& a, const std::vector<T>& b)
{
assert(a.size() == b.size());
std::vector<T> result;
result.reserve(a.size());
std::transform(a.begin(), a.end(), b.begin(),
std::back_inserter(result), std::plus<T>());
return result;
}
您可以类似地实现 operator+=
重载.
You could similarly implement an operator+=
overload.
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