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 个字符的字符串吗？

Question

我必须读入这个字符串： efBZw WH EDC EVOh qjfzJ oXkXL QUHXWMl kXRSIyGHb TxQkBWhPI yPTJEZ KyYfFQR ZQvYBPUmZ bkuKbRBWW mjJ WRgIBFNM ojGP XbkJNUhXF SeZVBZ SooFP XXE Viebym gOXP pDXRQ=[=116=]

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

您想安装并使用 valgrind。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> >) (Whosebug-31686882.cpp:19)
==6962==    by 0x401670: main (Whosebug-31686882.cpp:77)
==6962== 
==6962== Use of uninitialised value of size 8
==6962==    at 0x401173: isPiSong(std::string, std::deque<int, std::allocator<int> >) (Whosebug-31686882.cpp:19)
==6962==    by 0x401670: main (Whosebug-31686882.cpp:77)
==6962== 
==6962== Invalid read of size 4
==6962==    at 0x401173: isPiSong(std::string, std::deque<int, std::allocator<int> >) (Whosebug-31686882.cpp:19)
==6962==    by 0x401670: main (Whosebug-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> >) (Whosebug-31686882.cpp:19)
==6962==    by 0x401670: main (Whosebug-31686882.cpp:77)

第 19 行是这个：

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

ctr 每增加一个 space。一旦数字双端队列耗尽（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 是否太大。

此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 return false。那么，为什么不直接 return 呢？所以不是 flag = 0，而是 return 0.

你先数字母的个数，然后当你到达下一个space时，你检查它。如果你给这个字符串："abc a abcd xxxxxxxxxxxxxxxx" 会怎样？等等，让我测试一下：

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

只有我在最后一个x后面加一个space，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，还有数字）。然后对于每个后续的 isalpha 做 n-- 并检查它之前是否为零（即 "if (n--) {...}"。如果它是 -> return 0。如果不是，继续直到 space。space 测试然后将检查 n 是否已达到零（这意味着，恰好 numbers[ctr++] isalphas 已被跳过）。如果没有 -> return 0.

最后你必须决定做什么，如果数字双端队列已用完但输入如下。