[英]User input of integer followed by garbage
In my simple Fraction
class, I have the following method to get user input for the numerator
, which works fine for checking garbage input like garbage
, but will not recognize user input which starts with an integer, and is followed by garbage, 1 garbage
or 1garbage
. 在我简单的
Fraction
类中,我具有以下方法来获取numerator
用户输入,该方法可以很好地用于检查诸如garbage
类的garbage
输入,但无法识别以整数开头,后跟垃圾, 1 garbage
或1garbage
。
void Fraction::inputNumerator()
{
int inputNumerator;
// loop forever until the code hits a BREAK
while (true) {
std::cout << "Enter the numerator: ";
// attempt to get the int value from standard input
std::cin >> inputNumerator;
// check to see if the input stream read the input as a number
if (std::cin.good()) {
numerator = inputNumerator;
break;
} else {
// the input couldn't successfully be turned into a number, so the
// characters that were in the buffer that couldn't convert are
// still sitting there unprocessed. We can read them as a string
// and look for the "quit"
// clear the error status of the standard input so we can read
std::cin.clear();
std::string str;
std::cin >> str;
// Break out of the loop if we see the string 'quit'
if (str == "quit") {
std::cout << "Goodbye!" << std::endl;
exit(EXIT_SUCCESS);
}
// some other non-number string. give error followed by newline
std::cout << "Invalid input (type 'quit' to exit)" << std::endl;
}
}
}
I saw a few posts on using the getline
method for this, but they didn't compile when I tried them, and I'm having trouble finding the original post, sorry. 我看到了一些关于使用
getline
方法的帖子,但是当我尝试使用它们时它们并没有编译,抱歉,我在查找原始帖子时遇到了麻烦。
Better check as follows: 更好的检查如下:
// attempt to get the int value from standard input
if(std::cin >> inputNumerator)
{
numerator = inputNumerator;
break;
} else { // ...
Or yes: Follow recommendations to parse a complete input line combining std::getline()
and std::istringstream
appropriately. 或是:按照建议来解析完整的输入行,以适当地组合
std::getline()
和std::istringstream
。
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