搜索链接到 cout 的变量的名称！像 a:cout c++

Question

当我在互联网上通过源代码搜索新信息时，我看到有人用 goto 给一个变量，这个变量被链接到一个 cout 语句，与 std::cout.

相同

他写了a:cout。

你能帮我找到这个函数的名字吗？

void Main_Menu() {
    int i;
    cout << "\n\n\n\n\n\n\n\n\n\n\n\n\t\t\t\t\t\t\t\t  HOSPITAL MANAGEMENT SYSTEM \n\n";
    cout << "\n\n\t\t\t\t\t\tPlease,  Choose from the following Options: \n\n";
    cout << "\t\t\t\t\t\t _________________________________________________________________ \n";
    cout << "\t\t\t\t\t\t|                                                                |\n";
    cout << "\t\t\t\t\t\t|             1  >> Add New Patient Record                        |\n";
    cout << "\t\t\t\t\t\t|             2  >> Add Diagnosis Information                     |\n";
    cout << "\t\t\t\t\t\t|             3  >> Full History of the Patient                   |\n";
    cout << "\t\t\t\t\t\t|             4  >> Information About the Hospital                |\n";
    cout << "\t\t\t\t\t\t|             5  >> Exit the Program                              |\n";
    cout << "\t\t\t\t\t\t|_________________________________________________________________|\n\n";
a:  cout << "\t\t\t\t\t\tEnter your choice: "; 
cin >> i;
if (i > 5 || i < 1) { 
    cout << "\n\n\t\t\t\t\t\tInvalid Choice\n"; 
    cout << "\t\t\t\t\t\tTry again...........\n\n"; 
    goto a; 
} //if inputed choice is other than given choice

Answer 1

没有链接到 cout 的变量 a。 a: 是一个标签，您可以使用 goto a; 跳转到该标签。代码也可以这样写

a:
  cout << "\t\t\t\t\t\tEnter your choice: ";
  cin >> i;
  if (i > 5 || i < 1) {
    cout << "\n\n\t\t\t\t\t\tInvalid Choice\n";
    cout << "\t\t\t\t\t\tTry again...........\n\n";
    goto a;
  } //if inputed choice is other than given choice

有趣的旁注：为什么

void f()
{
  http://whosebug.com
  https://whosebug.com
}

C 有效吗？因为 http 和 https 被视为标签并且 // 开始评论。

搜索链接到 cout 的变量的名称！像 a:cout c++

searching for the name of a variable linked to a cout! like a:cout c++

c++

windows

cout

goto