正確的循環邏輯？

我的hang子手程序有問題。我需要讓玩家對每個猜測都顯示棋盤，如果答案正確，棋盤不會改變。如果答案不正確，該板將改變。我試圖做循環，以便當用戶做出6次錯誤的猜測時，最終棋盤將顯示，意味着玩家輸掉了比賽。這是到目前爲止我的代碼...正確的循環邏輯？

#include <iostream> 
#include <fstream> 
#include <string> 
#include <cctype> 
#include <algorithm> 
using namespace std; 

const string boardOne = " ------|\n |  |\n  |\n  |\n  |\n -------\n\n"; 
const string boardTwo = " ------|\n |  |\n 0  |\n  |\n  |\n -------\n\n"; 
const string boardThree = " ------|\n |  |\n 0  |\n |  |\n  |\n -------\n\n"; 
const string boardFour = " ------|\n |  |\n 0  |\n-|  |\n  |\n -------\n\n"; 
const string boardFive = " ------|\n |  |\n 0  |\n-|- |\n  |\n -------\n\n"; 
const string boardSix = " ------|\n |  |\n 0  |\n-|- |\n \\ |\n -------\n\n"; 
const string boardSeven = " ------|\n |  |\n 0  |\n-|- |\n/ \\ |\n -------\n\n"; 

int main() 
{ 
// Declarations 
string fileWord; 
ifstream inFile; 
int wrongGuess = 0; 
char letterGuess = 0; 



// while loop to get the last word from file 
inFile.open("hangman.dat"); 
while (!inFile.eof()) 
{ 
    cout << fileWord << endl; 
    inFile >> fileWord; 
} 

inFile.close(); 

//capitalize the word to use in the game of hangman 
std::transform(fileWord.begin(), fileWord.end(), fileWord.begin(), toupper); 
cout << "\nWord to Guess: " << fileWord << endl; 

cout << "\n" << boardOne << endl; 

while (wrongGuess != 6) 
{ 
    cout << "\nEnter a letter to guess: "; 
    cin >> letterGuess; 
    letterGuess = toupper(letterGuess); 
    cout << "You guessed the letter: " << letterGuess << endl; 
    bool found = false; 

    for (int i = 0; i < fileWord.length(); i++) 
    { 
     if (fileWord[i] == letterGuess) 
     { 
      cout << "\n" << letterGuess << " is in the letter to guess." << endl; 
      found = true; 
     } 

    } 
    // if not found - increment wrong guesses 
    if (!found) 
    { 
     wrongGuess++; 
     cout << "\n" << letterGuess << " is not in the word to guess." << endl; 
     //print the board that corresponds to the wrongGuess 

     if (wrongGuess = 0) 
      cout << boardOne << endl; 
     if (wrongGuess = 1) 
      cout << boardTwo << endl; 
     if (wrongGuess = 2) 
      cout << boardThree << endl; 
     if (wrongGuess = 3) 
      cout << boardFour << endl; 
     if (wrongGuess = 4) 
      cout << boardFive << endl; 
     if (wrongGuess = 5) 
      cout << boardSix << endl; 
     if (wrongGuess = 6) 
      cout << boardSeven << endl; 
    } 
} 


cout << "\n\n"; 
system("pause"); 
return 0; 
}

來源

2015-11-19 rfallon

你做得='和'在==不等於'的'分配的如果聲明。例如'if（wrongGuess = 0）'應該是'if（wrongGuess == 0）'。 – Arite

會有更好的方式來編碼，而不是所有這些if語句？ – rfallon

您可改爲爲您的電路板提供一組「字符串」。例如： 'const string board [] = {「Board 1 Content」，「Board 2 Content」，\t「Board 3 Content」，「Board 4 Content」};'。然後打印相關的板： 'cout << board [wrongGuess] << endl;'。這個假設'wrongGuess'不是越界。 – Arite

if (wrongGuess = 0) 
     cout << boardOne << endl; 
    if (wrongGuess = 1) 
     cout << boardTwo << endl; 
    if (wrongGuess = 2) 
     cout << boardThree << endl; 
    if (wrongGuess = 3) 
     cout << boardFour << endl; 
    if (wrongGuess = 4) 
     cout << boardFive << endl; 
    if (wrongGuess = 5) 
     cout << boardSix << endl; 
    if (wrongGuess = 6) 
     cout << boardSeven << endl;

所有這些都是錯誤的，你在你的情況做分配。分配

wrongGuess = 3

的值是3。因此，他們都將計算爲true除了

wrongGuess = 0

來源

2015-11-19 23:40:54 jensa

正確的循環邏輯？

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