如何在c ++中使用带有用户输入的枚举

Question

Im making a simple rock paper scissors game and I need to use the enumeration data structure. My problem is that i cannot compile the following code because of an invalid conversion from int (userInput) to Throws (userThrow).

enum Throws {R, P, S};
int userInput;
cout << "What is your throw : ";
cin >> userInput;
Throws userThrow = userInput;

Help?!

Answer 1

You can do it like this:

int userInput;
std::cin >> userInput;
Throws userThrow = static_cast<Throws>(userInput);

Answer 2

R, P and S are technically now identifiers for numbers (0,1 and 2, respectively). Your program now does not know that that 0, 1 and 2 once mapped to letters or strings.

Instead, you must take the input and manually compare it to "R", "P" and "S" and if it matches one, set the userThrow variable accordingly.

Answer 3

enums in C++ are just integer constants. They are resolved at compile time and turned into numbers.

You have to override the >> operator to provide a correct conversion by looking for the correct enum item. I found this link useful.

Basically you read an int from stdin and use it to build a Throws item by using Throws(val) .

If, instead, you want to input directly the representation of the enum field by placing as input the string then it doesn't exist by itself, you have to do it manually because, as stated at the beginning, enum names just disappear at compile time.

Answer 4

Try this out:

enum Throws {R = 'R', P = 'P', S = 'S'};
char userInput;
cout << "What is your throw : ";
cin >> userInput;
Throws userThrow = (Throws)userInput;

Answer 5

由于编译器将枚举视为整数，因此必须匹配手动设置每个枚举的整数以对应ASCII码，然后将整数输入强制转换为枚举。

Answer 6

You can try this:

int userOption;
std::cin >> userOption;

If you do not want to assign user input data, just you want to check then use below code

Throws userThrow = static_cast<Throws>(userOption);

if you want to assign userinput in your Enum then use following code

Throws R = static_cast<Throws>(userOption);

here you choose R or P or S based on need.