如何將單詞排序爲C++中的結構數組？

Question

我對C++非常陌生，所以對於我可能會或可能不會明白的單詞很容易... 我們給出了一個字符串形式的段落，我們應該從字符串中獲取每個單詞並填充一個數組，顯示數組。我們今天瞭解了結構，所以這就是爲什麼我們應該輸出docWordCount（以及爲什麼可能出現錯誤的原因......）

我想要做的是沿着字符串移動，直到找到空間，當我這樣做時，我使用.substr命令將單詞複製到數組中。我最初嘗試使用static_cast來查找是否存在空間，我不確定問題是不是這樣工作，或者我做了錯誤的事情（可能是後者）。每當我沿着字符串移動時，我會將字數增加1，所以它會輸出所有字，而不是輸出字。另外我應該提到，當我編譯代碼時，它輸出文本，然後它給我一個「調試斷言失敗！[...]表達式：字符串下標超出範圍。」在另一個窗口中出錯。

#include <string> 
#include <iostream> 
using namespace std; 

int main() 
{ 

    struct wordCount 
    { 
     string word; 
     int count; 
    }; 

    wordCount docWordCount [500]; 
    for(int i = 0; i < 500; i++) 
    { 
     docWordCount[i].word = ""; 
     docWordCount[i].count = 0; 
    } 

string text ="If there's one good thing that's come out of the European debt crisis\ 
     it's that the U.S. has been able to shield itself from much of \ 
      the mess. The sentiment will continue in 2012 as the U.S. \ 
      economy is expected to grow faster than its European counterparts. \ 
     That's according to the Organization for Economic Cooperation and \ 
     Development which says the U.S. economy will expand at a 2.9% \ 
      annual rate in the first quarter and then a 2.8% rate in the second quarter."; 
cout << text << endl; 

int wordLength = 0; 

for(int i = 0; i < 500; i++) 
    { 
    if (text[i] == ' ') //if there isnt a space 
     wordLength++; 

    if (text[i] == !' ') //if there is a space 
     docWordCount[i].word = text.substr(i - wordLength, i); 
    } 

for (int i = 0; i < 100; i++) 
cout << docWordCount[i].word << endl; 
    return 0; 

} 

它應該是什麼樣子是

If 
Theres 
one 
good 
thing 

等...是什麼，我試圖做的聲音？有沒有更簡單的方法來解決這個問題？

Answer 1

一些錯誤，但其他聲音。

1. 您將該索引與text混淆，並將索引轉換爲單詞數組;爲後者創建一個新變量wordIndex。
2. 檢查空間與否的操作員是錯誤的。 ==表示是，使用!=表示不是。
3. 你應該迭代i爲text.length()
4. 最好將500硬編碼更改爲常量。
5. 危急地，你想text.substr(i - wordLength, wordLength);即第二個參數是長度，而不是子串的結束索引。

Answer 2

您的代碼包含幾個錯誤。例如，此聲明

if (text[i] == !' ') //if there is a space 

無效。它比較字符text[i]與布爾值false（!' '）

請嘗試下面的代碼（未經測試）。也許它含有較少的錯誤比你的代碼。:)

#include <string> 
#include <iostream> 

int main() 
{ 
    struct wordCount 
    { 
     std::string word; 
     int count; 
    }; 

    const size_t N = 500; 
    wordCount docWordCount[N] = {}; 

    std::string text ="If there's one good thing that's come out of the European debt crisis\ 
         it's that the U.S. has been able to shield itself from much of \ 
         the mess. The sentiment will continue in 2012 as the U.S. \ 
         economy is expected to grow faster than its European counterparts. \ 
         That's according to the Organization for Economic Cooperation and \ 
         Development which says the U.S. economy will expand at a 2.9% \ 
         annual rate in the first quarter and then a 2.8% rate in the second quarter."; 

    std::cout << text << std::endl; 

    size_t n = 0; 
    std::string::size_type i = 0; 

    while (n < N && i < text.size()) 
    { 
     while (i < text.size() && text[i] == ' ') ++i; 

     std::string::size_type j = i; 
     while (i < text.size() && text[i] != ' ') ++i; 

     if (j != i) docWordCount[n++].word.assign(text, j, i - j); 
    } 


    for (size_t i = 0; i < n; i++) std::cout << docWordCount[i].word << std::endl; 

    return 0; 
} 

這將是簡單的，如果你會使用類性病的成員函數::字符串作爲例如find

順便說線程的名稱不符合其描述。關於排序沒有提及。