Question

我遇到一个小问题，当数组board1[i]或board2[i]仅包含0时，似乎While循环不会停止。

编写while ((board1[i] == 0) || (board2[i] == 0))是正确的，因为我想要的是当某些电路板只包含0时，我希望循环停止。

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

我想要的是当某些主板只包含0时我想要的循环太停。

我想你写的与你想要的完全相反：

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

如果board1 [i]等于零，则上述将运行;如果board2 [i]等于零，则运行。如果你想停止，如果两者中的任何一个为零，你应该写

while (!((board1[i] == 0) || (board2[i] == 0)))
注意！（不）在开头。同样你也可以写

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

Answer 2

只写而不是

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

如果你想要两个电路板需要同时为0时写

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

否则如果其中一个板需要为0才能退出

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

Answer 3

你所说的是，如果任何一块电路板都保持零。你想说的是，直到一块板子没有，继续前进。

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

此段错误。你从索引零开始并检查EITHER board1或board2在该索引中是否有0，然后你做while循环的东西UNTIL board1 [i]！= 0＆amp;＆amp; board2 [i]！= 0.这意味着一旦找到零指数，你就永远不会停止循环，这意味着游戏不会结束。

你需要做的是分别浏览每个数组，看看是否符合条件然后决定做什么。亲身？我在两个阵列中寻找一个。两个数组中都不是全零。这样你就不会寻找停止的理由。让你的程序变得懒惰：停止，除非有理由继续。

以下行中的某些内容，以查看是否全部为零：

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
    }
}

基本上，你首先要说的是每次迭代＆＃34;直到我给出一个不这么认为的理由，董事会都是零＆＃34;。然后寻求证明董事会AREN＆＃39;全部为零。然后，如果其中任何一个都是零，那么就不要做任何事情并停止游戏。

C ++＆＃34;或＃34; while循环中的语句混合起来

3 个答案: