C++函数退出循环，退出代码为None

Question

这是我报名参加的Coursera算法课程中的一个问题。这不是为了询问问题的答案，但我根本没有解决它，因为我的代码(在函数中(运行到for循环行之前的行，然后它退出。这是代码：

#include <iostream>
#include <vector>
#include <algorithm>
using std::vector;
using namespace std;
double get_optimal_value(int capacity, vector<int> weights, vector<int> values) {
  double value = 0.0;
  vector<int> rates;
  int len = end(weights) - begin(weights);
  cout << "pre-loop (values loop)" << endl; //the last line the code executes
  for (int i = 0; i < len; i++){
    rates[i] = values[i] / weights[i];
    cout << values[i] / weights[i] << endl; 
  }
  std::sort(begin(rates), end(rates), greater<int>());
  return value;
}
int main() {
  int n;
  int capacity;
  std::cin >> n >> capacity;
  vector<int> values(n);
  vector<int> weights(n);
  for (int i = 0; i < n; i++) {
    std::cin >> values[i] >> weights[i];
  }
  double optimal_value = get_optimal_value(capacity, weights, values);
  std::cout.precision(10);
  std::cout << optimal_value << std::endl;
  return 0;
}

文件本身由讲师提供，我们只需要使用我们正在处理的函数，而不需要使用main函数。

Answer 1

明确使用push_back是一种解决方案；不过，就变体而言，我要注意的是，在这种情况下，您可以很容易地从现有的vector之一初始化新的vector(从int到double的类型转换工作得很好(，然后以大致相同的方式进行计算，而不是逐个构建元素：

double get_optimal_value(int capacity, const vector<int>& weights, const vector<int>& values) {  // Arguments made into const references to avoid pointless copies
  double value = 0.0;
  vector<double> rates(std::begin(values), std::end(values));  // Initialize to copy of values converted to double
  const auto len = weights.size(); // No need for iterator arithmetic; size tells you the size
  cout << "pre-loop (values loop)" << endl; //the last line the code executes
  for (size_t i = 0; i < len; i++){
    rates[i] /= weights[i]; // No need to use values anymore
    cout << rates[i] << endl; 
  }
  std::sort(begin(rates), end(rates), greater<int>());
  return value;
}

Answer 2

#include <iostream>
#include <vector>
#include <algorithm>
//using std::vector; // why this and you're using std in the next line
using namespace std;
double get_optimal_value(int capacity, vector<int> weights, vector<int> values) {//capacity is unused
    double value = 0.0, temp{}; 
    vector<double> rates;
    //int len = end(weights) - begin(weights); //using .size() instead
    cout << "pre-loop (values loop)" << endl; 
    for (int i = 0; i <weights.size(); i++) 
    {
        temp = static_cast<double>(values[i]) / static_cast<double>(weights[i]);
        rates.push_back(temp);
        cout << static_cast<double>(values[i]) / static_cast<double>(weights[i]) << endl; //to get the floating number because you're dividing
    }
    std::sort(begin(rates), end(rates), greater<double>());
    return value;
}
int main() {
    int n;
    int capacity;
    std::cin >> n >> capacity;
    vector<int> values(n);
    vector<int> weights(n);
    for (int i = 0; i < n; i++) 
    {
        std::cin >> values[i] >> weights[i];
    }
    double optimal_value = get_optimal_value(capacity, weights, values);
    std::cout.precision(10);
    std::cout << optimal_value << std::endl;
    return 0;
}

你不能像以前那样给向量加一个值，因为它的大小为null，你必须把它放大，这样你才能加值。我对你的代码(评论(做了一些更正

但是您仍然会得到0作为函数的返回，因为在get_optimal_value()的执行过程中值不会改变。

你必须更具体，这样我们才能帮助你