Question

我们是否可以通过易于阅读的东西替换strcmp的循环，类似于C＃扩展方法？

我维护了传统的C ++，并想知道如何开始向更现代的C ++过渡，并且有大量的代码看起来像这样：

int numberOfPipe = 10;
char* collection[5] = { "pompe","pipe","turbine","pompe", "pipe" };

// Count the pipes in the collection

int nPipeFound = 0;
int nPipe = 5;

for (int idx = 0; idx < nPipe; idx++)
{
    if (strcmp(collection[idx], "pipe") == 0)
        nPipeFound++;
}

cout << nPipeFound << endl;

Answer 1

使用标准库：

使用std::count并使用std::string比较。

#include <algorithm>
#include <iostream>
#include <string>

int main() {
  char const * collection[] = { "pompe","pipe","turbine","pompe","pipe" };

  auto n_pipe_found = std::count( std::begin( collection ), std::end( collection ), std::string{"pipe"});

  std::cout << n_pipe_found << '\n';
}

使用std::count_if并编写一个谓词，为您进行C字符串比较。

#include <algorithm>
#include <cstring>
#include <iostream>

int main() {
  char const * collection[] = { "pompe","pipe","turbine","pompe","pipe" };

  auto n_pipe_found = std::count_if( std::begin( collection ), std::end( collection ),
                                     [](char const * a) { return std::strcmp(a,"pipe") == 0; } );

  std::cout << n_pipe_found << '\n';
}

您还可以使用[](std::string const& a) { return a == "pipe"; }这样的谓词，并再次使用std::string比较。

如果您需要对计数进行更精细的控制，请使用std::accumulate。

#include <numeric>
#include <iostream>
#include <string>

int main() {
  char const * collection[] = { "pompe","pipe","turbine","pompe","pipe" };

  auto n_pipe_found = std::accumulate( std::begin( collection ), std::end( collection ), int{0},
                                   [](int a, std::string const& b) { return a + (b == "pipe"); });

  std::cout << n_pipe_found << '\n';
}

替换循环以计算符合条件的数字元素

1 个答案: