While循環中的C ++“或”語句將事情混在一起

Question

我面臨一個小問題，當數組board1[i]或board2[i]僅包含0時，While循環不會停止。

所以寫while ((board1[i] == 0) || (board2[i] == 0))是正確的，因為我想要的是某些板僅包含0時我希望循環也停止。

void ShootAtShip(int board1[], int board2[], string names[], int cap){

const int hit = 0;
int shot = 0;

for (int i = 0; i < cap; i++){

    while ((board1[i] == 0) || (board2[i] == 0)){ //detects if any board has all their ships shot down



        cout << names[1] << " set a position to shoot." << endl;
        cin >> shot;

        while ((shot >= cap) || (shot < 0)) {       //detects if the number is allowed

            cout << "That number is not allowed, "<<  names[1] << " set a position to shoot." << endl;
            cin >> shot;

        }


        if (board1[shot] != 0){

            board1[shot] = 0;
            cout << "Hit!" << endl;
        }

        else{

            cout << "You missed." << endl;
        }

        shot = 0;


        cout << names[0] << " set a position to shoot." << endl;
        cin >> shot;

        while ((shot >= cap) || (shot < 0)) {       //detects if the number is allowed

            cout << "That number is not allowed, " << names[0] << " set a position to shoot." << endl;
            cin >> shot;

        }

        if (board2[shot] != 0){

            board2[shot] = 0;
            cout << "Hit!" << endl;
        }

        else{

            cout << "You missed." << endl;
        }



    }

Answer 1

我想要的是某些板僅包含0時，我希望循環也停止。

我想您寫的恰恰與您想要的相反：

while ((board1[i] == 0) || (board2[i] == 0))

如果board1 [i]等於零，或者如果board2 [i]等於零 ， 則將執行上述操作。 如果您想在兩者之一為零時停止它，則應該寫

while (!((board1[i] == 0) || (board2[i] == 0)))

注意！ （不是）開頭。 同樣，你也可以寫

while (board1[i] != 0 && board2[i] != 0)

Answer 2

只是寫而不是

while ((board1[i] == 0) || (board2[i] == 0))

如果您希望兩個板同時需要為0，請寫

while (board1[i] != 0 && board2[i] != 0)

否則，如果其中一個董事會需要為0退出，則

while (board1[i] != 0 || board2[i] != 0)

Answer 3

您要說的是，如果任一板的零位繼續前進。 您要說的是，直到一個董事會沒有董事會為止。

for (int i = 0; i < cap; i++){

while ((board1[i] == 0) || (board2[i] == 0)){ //ACTUALLY DETECTS IF EITHER 
//BOARD HAS A ZERO AT INDEX i 

這部分是錯誤的。 您從索引零開始，檢查是否EITHER board1或board2在該索引中為0，然后執行while循環填充UNTIL board1 [i]！= 0 && board2 [i]！=0。這意味着您一旦找到零索引就永遠不會停止循環，這意味着游戲不會結束。

您需要做的是改為分別遍歷每個數組，看看是否滿足條件，然后決定要做什么。 親身？ 我會在兩個陣列中尋找一個。 在這兩個數組中並非全為零。 這樣，您就無需尋找停止的理由。 讓您的程序變得懶惰：除非有理由繼續，否則請停止。

遵循以下內容，看其中一個是否全為零：

bool boardOneAllZeroes = false;
bool boardTwoAllZeroes = false;
//So if board one is all zeroes OR board two is all zeroes, stop looping.
while(!boardOneAllZeroes && !boardTwoAllZeroes)
{
    boardOneAllZeroes = true;
    boardTwoAllZeroes = true;
    //The above two lines basically say "This loop isn't going to keep 
    //going unless you give me a good reason to later on.
    //Next we go through each index in both arrays looking for a one.
    for(int i = 0; i < cap; i++)
    {
        //If we find a one in board one, then board one is not all zeroes. 
        //Set it back to false
        if(board1[i] == 1)
        {
             boardOneAllZeroes = false;
        }
        //Same thing with board two.
        if(board2[i] == 1)
        {
             boardTwoAllZeroes = false;
        }
    }
    //Check to make sure both still have a one, because we don't need to keep going 
    //if both are all zeros.
    if(!boardOneAllZeroes && !boardTwoAllZeroes)
    {
        do game things
    }
}

從本質上講，您首先在每次迭代中都說“直到我有理由認為否則，木板都是零”。 然后設法證明木板不是全零。 然后，如果其中一個全為零，則不執行任何操作並停止游戲。