c++ - 为什么这个 C++ 代码(来自 C++ Primer 5th 3.4.2)不能正常工作？

Question

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

using namespace std;

int main(){

    vector<string> text = {"a", "v"};
    vector<string>::iterator beg = text.begin();
    vector<string>::iterator end = text.end();
    vector<string>::iterator mid = text.begin() + (end - beg) / 2;

    while (mid != end && *mid != "l") {
        if ("l" < *mid) {
            end = mid;
        } else {
            beg = mid;
        }
        mid = beg + (end - beg) / 2;
    }

    if (*mid == "l")
        cout << "Have Found!" << endl;
    else
        cout << "Not Found!" << endl;
    return 0;
}

我不知道为什么会死循环？请帮助我!我在 VisulStudio 2013 和 Code::Block 中构建代码，但结果都是不好的!

Answer 1

好吧，我正在抓紧解决方案，因为我认为您的算法出错了，这使得 mid 指向未识别的位置。我发现您使用的是 C++11。不过，这个解决方案可以正常工作。

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

using namespace std;

void populateVector(vector<string> &text)
{
    for(int i=97;i<=118;i++)
    {
        text.push_back(string(1,(char)i));
    }
}

void printVector(vector<string> &text)
{
    for(int i=0;i<text.size();i++)
    {
        cout<<text[i]<<"\n";
    }
}

bool found(vector<string> &text,string &toBeFound)
{
    vector<string>::iterator beg = text.begin();
    vector<string>::iterator end = text.end();
    vector<string>::iterator mid ;

    while(beg<=end)
    {

        mid = beg + (end - beg) / 2;
        //This piece of code had to be added 
        if(mid==end)
        {
         break;
        }
        //This piece of code had to be added 
        //@DEBUG: cout<<*mid<<"\n";
        if(*mid==toBeFound)
        {
            return true;
        }
        if(*mid<toBeFound)
        {
            beg=mid+1;
        }
        else
        {
            end=mid-1;
        }
    }
    return false;

}

int main()
{

    vector<string> text;
    string toBeFound = "l";
    populateVector(text);
    printVector(text);

    if(found(text,toBeFound))
    {
        cout<<"Found\n";
    }
    else
    {
        cout<<"Not Found\n";
    }



    return 0;
}

程序是自清的。如果您有任何问题，请告诉我。我想这应该可行。

希望对你有帮助