c++ - 打印星号中上下移动的字母

Question

首先，我是一个初学者，试图弄清楚如何使用星号打印上下移动的字母“OK”。目标是从偏移位置开始并让它们递增和递减 x 次。 (我还没有参与这部分)。但首先，我一直在思考为什么“K”没有按照我想要的方式打印出来。谁能提供帮助？

我尝试设置一个 for 循环来处理每个字符的打印。我试图制作计数器，以便在打印字符时它们看起来是偏移的，因为我知道如果它们都从第 0 行开始，那么它们将并排打印。以下是我到目前为止编写的一些代码。 'O' 打印但 'K' 打印奇怪。

int main() {
char userInputCharacter;   // User input for some character
int  userInputNumber;      // User input for some numerical value

cout << endl
     << "Choose from among the following options: \n"

     << "   2. Display OK as an animation \n"

     << "Your choice -> ";
cin >> userInputCharacter;
cout << endl;


// Display OK as an animation
if( userInputCharacter == '2') {
    cout << "How many sets of letters do you want to display? -> ";
    cin >> userInputNumber;

    for( int setsOfLetters = 0; setsOfLetters < userInputNumber; setsOfLetters++) {

        // Display some number of blank lines.  This starts as a large number the first time, then
        // gets smaller each subsequent time, moving the ENTIRE set of letters vertically.
        for( int numberOfBlankLines = userInputNumber; numberOfBlankLines > setsOfLetters; numberOfBlankLines--) {
            cout << endl;
        }

        // Display one set of letters, going through and printing one "slice" of each letter at a time.

        int i = 0;
        int j  = 4;
        for( int i=0; i<8; i++) {
            if(      i==0)    cout << "       ";
            else if( i==1) cout    << "       ";
            else if( i==2) cout    << "  **   ";
            else if( i==3) cout    << " *  *  ";
            else if( i==4) cout    << "*    * ";
            else if( i==5) cout    << "*    * ";
            else if( i==6) cout    << " *  *  ";
            else if( i==7) cout    << "  **   ";


            if(      j ==0) cout << "      ";
            else if( j ==1) cout<< "       ";
            else if( j ==2) cout << "*  * ";
            else if( j ==3) cout << "* *  ";
            else if( j ==4) cout << "**   ";
            else if( j ==5) cout << "* *  ";
            else if( j ==6) cout << "*  * ";
            else if( j ==7) cout << "       ";

            cout << endl;


        }//end for( int i...)

        // Clear the screen after the letters are displayed.
        this_thread::sleep_for(chrono::milliseconds( 185));    // Sleep for 185 milliseconds
        system( "clear");   // Clear the screen.  Comment out this line if you don't want them erased.

    }//end for( int setsOfLetters...

}//end else if( userInputCharacter == '2' ...

return 0;
}//end main()

如果我们可以建议使用 for 循环和 if 和 else 语句来解决这个问题，那将对我有好处。我还没有涉及函数和字符串，也没有涉及数组，我不想跳到前面而变得更加困惑。提前谢谢大家。

Answer 1

正如我在对您之前帖子的评论中所述，您可能想尝试在没有 for 的情况下打印循环:

std::cout << "  **   *  * \n";
std::cout << " *  *  * *  \n";
std::cout << "*    * **   \n";
std::cout << "*    * * *  \n";
std::cout << " *  *  *  * \n";
std::cout << "  **   *   *\n";

可能，对您来说更简单的解决方案是将文本声明为 C 样式字符串:

static const char ok_lines[] =
{
    "  **   *  * \n"
    " *  *  * *  \n"
    "*    * **   \n"
    "*    * * *  \n"
    " *  *  *  * \n"
    "  **   *   *\n"
};

然后您可以在之前打印行:

std::cout << "\n\n";
std::cout << ok_line;

或打印以下行:

std::cout << ok_line;
std::cout << "\n\n";

您的动画可能看起来像:

static const unsigned int MAX_CYCLES = 10U;
for (unsigned int i = 0; i < MAX_CYCLES; ++i)
{
   Clear_The_Screen();
   if ((i % 2) == 0)
   {
      std::cout << ok_lines;
      std::cout << "\n\n";
   }
   else
   {
      std::cout << "\n\n";
      std::cout << ok_lines;
   }
   Delay();  // You'll need to write or research this one.
}

提醒:简单的设计产生简单的代码。简单的代码更容易调试，注入(inject)的缺陷也更少。

注意:如果你想要更高效的输出，替换std::cout << ok_lines;与 std::cout.write(ok_lines, sizeof(ok_lines) - 1); . -1防止将 nul 终止符写入控制台。