How do I ignore extra input integers-CodePudding

I've been building a menu for class, and I can't figure out how to stop calculations and output an error message when more inputs are given than asked for. This is a snippet of what I have(val1 and val2 are unsigned long int and inputError is a boolean):

          case '&':
                std::cout << "Enter values for first and second operand:" << std::endl;
                std::cin >> val1 >> val2;
                inputError = 0;
                checkData1(val1, val2, inputError);
                if (inputError) {
                    continue;
                }
                else {
                    result = val1 & val2;
                    std::cout << "Result: " << result;
                    std::cout << " [Hexadecimal: " << std::setw(bitsize) << std::hex << result << std::dec << "]" << std::endl;
                }
                break;

This is the void function (checkData1):

    void checkData1(unsigned long int val1, unsigned long int val2, bool& inputError) {
    if (std::cin.fail()) {
        inputError = 1;
        std::cout << "Input error, enter choice and values again." << std::endl;
        std::cin.clear();
        while (std::cin.get() != '\n') {
            continue;
        }
    }
    else if ((int)val1 < 0 || (int)val2 < 0) {
        inputError = 1;
        std::cout << "Input error, enter choice and values again." << std::endl;
    }
}

And this is what I get:

Enter one of the following choices:
& - AND
| - OR
^ - XOR
~ - NOT
- - Negate
< - Left shift(logical shift only)
> - Right shift(logical shift only)
# - stop execution

&

Enter values for first and second operand:

1 2 3

Result: 0 [Hexadecimal:        0]
Enter one of the following choices:
& - AND
| - OR
^ - XOR
~ - NOT
- - Negate
< - Left shift(logical shift only)
> - Right shift(logical shift only)
# - stop execution
Invalid choice 3, please try again.
Enter one of the following choices:
& - AND
| - OR
^ - XOR
~ - NOT
- - Negate
< - Left shift(logical shift only)
> - Right shift(logical shift only)
# - stop execution

Instead of doing the calculations with the first two integers and giving an error with the third, I want it to ignore all of it, and put out an error message instead.

CodePudding user response：

You can add this to checkData1. This uses the cin.peek function to check if there's anything other than a new line character in the cin buffer, and if so it sets the error variable and clears out the buffer

void checkData1(unsigned long int val1, unsigned long int val2, bool& inputError) {
    if (std::cin.fail()) {
        inputError = 1;
        std::cout << "Input error, enter choice and values again." << std::endl;
        std::cin.clear();
        while (std::cin.get() != '\n') {
            continue;
        }
    }
    else if ((int)val1 < 0 || (int)val2 < 0) {
        inputError = 1;
        std::cout << "Input error, enter choice and values again." << std::endl;
    }
    if(std::cin.peek() != '\n'){
        inputError = 1;
        std::cout << "Make sure to only put in two numbers" << std::endl;
        while (std::cin.get() != '\n') {
            continue;
        }
    }
    
}

CodePudding user response：

void checkData1(unsigned long int val1, unsigned long int val2, bool& inputError) {
    if (std::cin.fail()) {
        inputError = 1;
        std::cout << "Input error, enter choice and values again." << std::endl;
        std::cin.clear();
        while (std::cin.get() != '\n') {
            continue;
        }
    }
    else if ((int)val1 < 0 || (int)val2 < 0) {
        inputError = 1;
        std::cout << "Input error, enter choice and values again." << std::endl;
    }
    if(std::cin.peek() != '\n'){
        inputError = 1;
        std::cout << "Make sure to only put in two numbers" << std::endl;
        while (std::cin.get() != '\n') {
            continue;
        }
    }
    
}