Question

我试图在C ++的循环中操纵0初始化向量的值。但是，当我打印出矢量的值时，它只显示＆＃34; myFunction＆＃34;中的合理值。除此之外，价值观没有改变。这是我的例子草图：

vector<float> meanDistances(numDescriptors);  

for (int i = 0; i < numDescriptors, i++)  
{
    myFunction(arg1, arg2,...,meanDistances[i]));
    cout << "meanDistance OUTSIDE myFunction: " << meanDistances[i] << endl;
}

这就是＆＃34; myFunction＆＃34;看起来像：

void myFunction(arg1, arg2, ..., float & meanDistance)
{
    meanDistance = someFloatNumber;
    cout << "meanDistance INSIDE myFunction: " << meanDistance << endl;
}

如何将矢量的单个元素作为函数的引用传递？

编辑：
这里有一些代码

boost::thread_group threadGroup;
for (int i = 0; i < numDescs ; i++ )
{
    int threadIdx = i % numThreads;

    vecHists[i].convertTo(vecHists[i], CV_32F);
    threadGroup.create_thread(boost::bind( &myFunction, vecHists[i],vecHists[i],vecDistanceMats[threadIdx], meanDistances[i]));



    sleep(1);

    if(threadIdx == numThreads-1 || i == numDescs-1)
    {
        threadGroup.join_all();
        for (int j = 0; j < numThreads ; j++)
        {
            sumDistanceMats += vecDistanceMats[j];
        }
    }
}
for (int i = 0; i < numDescs ; i++ )
{
  cout << "meanDistances OUTSIDE myFunction after all threads are joined again: " << meanDistances[i] << endl;
}

EDIT2：
事实证明，这个函数的问题在于boost::threadGroup的功能。如果我通过1x1 dim opencv Mats vector<float> meanDistances的向量交换vector<cv::Mat> meanDistances向量，一切正常。
当然，我也相应地修改了myFunction：

void myFunction(arg1, arg2, ..., Mat& matMeanDistance)
{
    meanDistance.at<float>(0,0) = someFloatNumber;
    cout << "meanDistance INSIDE myFunction: " << matMeanDistance.at<float>(0,0) << endl;
}

所以某种方式boost::threadGroup处理float的向量不同于cv::Mat的向量

Answer 1

下面你将找到一个例子，说明如何通过引用（使用 range-for ）或通过lambda辅助的std::transform算法修改向量中的所有元素。

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

// utility function to print to screen
template<typename T>
void show(const T& v) {
  for(const auto & vi : v) {
    std::cout << vi << " ";
  }
  std::cout << std::endl;
}

// function used to modify by reference
void fun(double arg, double& vi) {
  vi += arg;
}

int main() {

  // initialize ten elements to 0.0
  std::vector<double> v(10, 0.0);
  std::cout << "Initialized as: " << std::endl;
  show(v);

  // modification using references in a loop
  for(auto & vi : v) {
    fun(2.0, vi);
  }
  std::cout << "Modified to: " << std::endl;
  show(v);

  // modification using an algorithm and a lambda (this one modifies by value)
  std::transform(v.begin(), v.end(), v.begin(),
                 [](double vi) { return vi + 3; });
  std::cout << "Modified using std::transform() and lambda" << std::endl;
  show(v);
}

编译并运行：

$ g++ example.cpp -std=c++14 -Wall -Wextra
$ ./a.out
Initialized as:
0 0 0 0 0 0 0 0 0 0
Modified to:
2 2 2 2 2 2 2 2 2 2
Modified using std::transform() and lambda
5 5 5 5 5 5 5 5 5 5

如何在C ++中将向量的一个元素作为引用传递给函数？

1 个答案: