小编Exc*_*bur的帖子

我需要一个针对以下情况的建议 - 我几个小时都无法弄明白:如何通过多个seq.容器大小相同(这里:两个向量)的简单方法？

int main() {
  int size = 3;
  std::vector<int> v1{ 1, 2, 3 }, v2{ 6, 4, 2 };

  // old-fashioned - ok
  for (int i = 0; i < size; i++) {
    std::cout << v1[i] << " " << v2[i] << std::endl;
  }

  // would like to do the same as above with auto range-for loop
  // something like this - which would be fine for ONE vector.
  // But this does not work. Do I need …

我目前陷入困境,试图为以下列表理解问题找到一个很好的解决方案:

在两个列表中很容易找到具有相同索引的相等值,例如

>>> vec1 = [3,2,1,4,5,6,7]
>>> vec2 = [1,2,3,3,5,6,9]
>>> [a for a, b in zip(vec1, vec2) if a == b]
[2,5,6]

但是,我只需要列表中出现这些匹配的索引,而不是值本身.使用上面的例子,我想要的输出是:[1,4,5]

我修好了但我只能想到一个"多线"的解决方案.有谁知道我怎么能以更Pythonic的方式找到索引？

一小段代码让我发疯,但希望你可以阻止我跳出窗外.看这里:

#include <iostream>
#include <cstdint>

int main()
{
    int8_t i = 65;
    int8_t j;

    std::cout << "i = " << i << std::endl; // the 'A' is ok, same as uchar

    std::cout << "Now type in a value for j (use 65 again): " << std::endl;
    std::cin >> j;
    std::cout << "j = " << j << std::endl;

    if (i != j) 
        std::cout << "What is going on here?????" << std::endl;
    else 
        std::cout << "Everything ok." << std::endl;

    return 0; …