Question

string line= "";
vector<string> tokens;
//populate the vector 'tokens'

//loop through vector 'tokens' until it becomes empty
{
    //extract out 'line' containing the first element of the vector.
    //reduce the size of 'tokens' by one through deleting the first element 
}
//use 'line' string one by one in the subsequent code

假设代币＆＃39;包含两个元素，例如

猫，狗，公牛

在第一次迭代中，我想要提取“猫”。从向量中移出并将其大小减小1。它现在将包含：

狗，公牛

现在，我想使用这个提取的元素＆＃39; cat＆＃39; （存储在字符串＆＃39; line＆＃39;中）在我的后续代码中。下次我想选择“狗”＃39;并使向量仅包含

牛

。等等。提前感谢您的想法！

Answer 1

当在容器上进行迭代以清空它时，只要容器不为空，就更容易进行迭代。

#include <iostream>
#include <string>
#include <vector>

int main()
{
    std::vector<std::string> tokens = {"cat", "dog", "bull"};
    while(tokens.empty() == false)
    {
        // Extract the data from first element in the list
        auto line = std::move(tokens.front());

        // Pop the moved element from the vector
        tokens.erase(tokens.begin());

        // Do work with 'line'
        std::cout << line << '\n';
    }

    return 0;
}

也就是说，从向量的前面移除元素是非常低效的，因为所有后续元素都必须向上移动。请考虑从后面删除元素。

Answer 2

tokens.erase(tokens.begin());

就是这样。

Answer 3

std :: vector提供了你在另一端看起来想要的操作......

std::vector<std::string>  tokens;

// the last token can be accessed: 
std::string tok = tokens.back();

// removing the last tok
assert(tokens.size());  // pop_back on an empty vector is UB
tokens.pop_back();      // removes element from vector - constant time!

// and if you get confused on how to load tokens in reverse order
// you can load as planned, then reverse the elements to do your work
std::reverse(std::begin(tokens), std::end(tokens))

试一试。我发现在矢量背面易于适应。

Answer 4

惯用法是：

std::vector<std::string> tokens;
// populate the vector 'tokens'
// ...

for (const auto& token : tokens) {
    // use 'token' string one by one in the subsequent code, as:
    std::cout << token << std::endl;
}
tokens.clear(); // If needed.

c ++ - 如何从向量<string>中提取第一个元素，并在循环中将向量大小减少1

4 个答案: