有没有什么特别的方法可以在 C++ 中使用 getline 作为字符串数据类型来读取 540 个字符的字符串？

Question

我必须阅读这个字符串： efBZw WH EDC EVOh qjfzJ oXkXL QUHXWMl kXRSIyGHb TxQkBWhPI yPTJEZ KyYfFQR ZQvYBPUmZ bkuKbRBWW mjJ WRgIBFNM ojGP XbkJNUhXF SeZVBZ SooFP XXE VIebym gOXP pDXRQ

给定字符串没有输出。

    #include <iostream>
    #include <string>
    #include <ctype.h>
    #include <deque>

    using namespace std;

    int isPiSong(string str,deque<int> numbers){
        int noOfwords = 0, noOfletters = 0, ctr = 0;
        int flag = 1;
        for(std::string::iterator it = str.begin(); it != str.end(); ++it) {

            if(*it != ' ' && flag){
                if(isalpha(*it))
                noOfletters++;
            }
            else{
                if(noOfletters != numbers[ctr++])
                    flag = 0;
                noOfletters = 0;
            }

        }
        return flag;
    }

    int main()
    {
        deque<int> numbers;
              // 31415926535897932384626433833
        numbers.push_back(3);
        numbers.push_back(1);
        numbers.push_back(4);
        numbers.push_back(1);
        numbers.push_back(5);
        numbers.push_back(9);
        numbers.push_back(2);
        numbers.push_back(6);
        numbers.push_back(5);
        numbers.push_back(3);
        numbers.push_back(5);
        numbers.push_back(8);
        numbers.push_back(9);
        numbers.push_back(7);
        numbers.push_back(9);
        numbers.push_back(3);
        numbers.push_back(2);
        numbers.push_back(3);
        numbers.push_back(8);
        numbers.push_back(4);
        numbers.push_back(6);
        numbers.push_back(2);
        numbers.push_back(6);
        numbers.push_back(4);
        numbers.push_back(3);
        numbers.push_back(3);
        numbers.push_back(8);
        numbers.push_back(3);
        numbers.push_back(3);

        int T, flag;
        cin>>T;

        string* pisong = new string[T];

        for(int i = 0; i < T; i++){
           getline(cin>>ws, pisong[i]); //cin >> ws gets rid of leading whitespace 
                                       //first so that getline won't think that it's already                         //reached the end of the line
                                      //It doesn't seem to take the above input string
        }

        for(int i = 0; i < T; i++){
            flag = isPiSong(pisong[i],numbers);
            if(flag){
                cout<<"It's a pi song."<<endl;
            }else{
                cout<<"It's not a pi song."<<endl;
            }
        }
        delete [] pisong;
        return 0;
    }

Answer 1

你想安装 valgrind 并使用它。它立即揭示了这个问题：

==6962== Conditional jump or move depends on uninitialised value(s)
==6962==    at 0x40117D: isPiSong(std::string, std::deque<int, std::allocator<int> >) (stackoverflow-31686882.cpp:19)
==6962==    by 0x401670: main (stackoverflow-31686882.cpp:77)
==6962== 
==6962== Use of uninitialised value of size 8
==6962==    at 0x401173: isPiSong(std::string, std::deque<int, std::allocator<int> >) (stackoverflow-31686882.cpp:19)
==6962==    by 0x401670: main (stackoverflow-31686882.cpp:77)
==6962== 
==6962== Invalid read of size 4
==6962==    at 0x401173: isPiSong(std::string, std::deque<int, std::allocator<int> >) (stackoverflow-31686882.cpp:19)
==6962==    by 0x401670: main (stackoverflow-31686882.cpp:77)
==6962==  Address 0x0 is not stack'd, malloc'd or (recently) free'd
==6962== 
==6962== 
==6962== Process terminating with default action of signal 11 (SIGSEGV)
==6962==  Access not within mapped region at address 0x0
==6962==    at 0x401173: isPiSong(std::string, std::deque<int, std::allocator<int> >) (stackoverflow-31686882.cpp:19)
==6962==    by 0x401670: main (stackoverflow-31686882.cpp:77)

第 19 行是这一行：

if(noOfletters != numbers[ctr++])

ctr 每增加一个空格。一旦数字双端队列用尽（即 ctr > 长度），您将在双端队列中获得非法访问。所以，要么你必须在某个时候将 ctr 重置为零，要么你必须检查 ctr 是否不会太大。

另外还有很大的改进空间。

一旦标志设置为零，它就不能再改变，所以在某些时候，你会返回 false。那么，为什么不立即返回呢？所以不要使用 flag = 0，而是返回 0。

您首先计算字母的数量，然后当您到达下一个空格时，您检查它。如果你给这个字符串：“abc a abcd xxxxxxxxxxxxxxxx”怎么办？等等，让我测试一下：

It's a pi song. [abc a abcd xxxxxxxxxxxxxxxx]

仅当我在最后一个 x 后添加一个空格时，程序才能正确检测到错误：

It's not a pi song. [abc a abcd xxxxxxxxxxxxxxxx ]

解决方案：一旦检测到第一个 isalpha 字符，设置一些 n = numbers[ctr++] - 1 （检查后是否还有数字）。然后对后面的每个 isalpha 执行 n-- 并检查它之前是否为零（即“if (n--) {...}”。如果是 -> 返回 0。如果不是，继续直到空格。空格然后 test 将检查 n 是否达到零（这意味着正好 numbers[ctr++] isalphas 已被跳过）。如果不是 -> 返回 0。

最后你必须决定要做什么，如果数字双端队列用尽但输入跟随。