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我对这个 C++ 世界非常陌生，并尝试为数字密码编写输入验证函数。这是我到目前为止得到的：

#include <iostream>
#include <limits>
using namespace std;

void isNumeric(int &iN)
{
    while (1) {
        cin >> iN;

        if (cin.fail()) {
            cin.clear();
            cin.ignore(numeric_limits<streamsize>::max(), '\n');
            cout << "Only 'numeric' value(s) are allowed: ";
            continue;
        }

        // alpha-numeric entry also not allowed 
        cin.ignore(numeric_limits<streamsize>::max(), '\n');
        if (cin.gcount() >  1) continue;

        // check against the -ve value
        if (iN <= 0 ) continue;

    break;
    }
}

int main()
{
    int x;

    cout << "Enter your number: ";
    isNumeric(x);
    cout << "You've entered: " << x << endl;

    return 0;
}

对于不正确的值，它工作得很好，但在有效输入时不会跳出循环。知道我在这里缺少什么吗？干杯！！

来自 James Kanze 脚本的错误：

test.cpp: In function ‘bool parseNumber(const string&, int&)’:
test.cpp:11:20: error: no match for ‘operator>>’ in ‘text >> results’
test.cpp:11:20: note: candidates are:
/usr/include/c++/4.6/bits/basic_string.tcc:998:5: note: template<class _CharT, class _Traits, class _Alloc> std::basic_istream<_CharT, _Traits>& std::operator>>(std::basic_istream<_CharT, _Traits>&, std::basic_string<_CharT, _Traits, _Alloc>&)
/usr/include/c++/4.6/bits/istream.tcc:957:5: note: template<class _CharT2, class _Traits2> std::basic_istream<_CharT, _Traits>& std::operator>>(std::basic_istream<_CharT, _Traits>&, _CharT2*)
/usr/include/c++/4.6/bits/istream.tcc:925:5: note: template<class _CharT, class _Traits> std::basic_istream<_CharT, _Traits>& std::operator>>(std::basic_istream<_CharT, _Traits>&, _CharT&)
/usr/include/c++/4.6/istream:709:5: note: template<class _Traits> std::basic_istream<char, _Traits>& std::operator>>(std::basic_istream<char, _Traits>&, unsigned char&)
/usr/include/c++/4.6/istream:714:5: note: template<class _Traits> std::basic_istream<char, _Traits>& std::operator>>(std::basic_istream<char, _Traits>&, signed char&)
/usr/include/c++/4.6/istream:756:5: note: template<class _Traits> std::basic_istream<char, _Traits>& std::operator>>(std::basic_istream<char, _Traits>&, unsigned char*)
/usr/include/c++/4.6/istream:761:5: note: template<class _Traits> std::basic_istream<char, _Traits>& std::operator>>(std::basic_istream<char, _Traits>&, signed char*)
test.cpp:11:42: error: ‘const string’ has no member named ‘peek’
test.cpp:11:52: error: ‘EOF’ was not declared in this scope

新代码：使用getline()和验证作为字符串 感谢大家（尤其是 James Kanze）帮助我。这东西在这里工作得差不多了。

void isNumeric( int &iN )
{
    string sN; 

    while (1) {
        getline(cin, sN);

        bool valNum = true;
        for ( unsigned iDx=0; iDx < sN.length(); iDx++ )
            if ( !isdigit(sN[iDx]) ) { 
                valNum = false;
                break;
            }   

        if ( !valNum ) { 
            cout << "Wrong entry; Try again: ";
            continue;
        }   

        stringstream sStream (sN );
        sStream >> iN; 

        if ( iN<=0 ) {
            cout << "Cannot be 0; Try again: "; 
            continue;
        }     
    break;   
    }   
}

那里还有改进的余地吗？干杯！！

4

2 回答 2

2

如果转换失败，则流本身的计算结果为 false，因此您可以执行以下操作：

int get_int() {
    int i;
    std::cout << "Please enter a number: " << std::endl;
    while(!(std::cin >> i)) {
        std::cin.clear();   //clear flags
        //discard bad input
        std::cin.ignore(std::numeric_limits<std::streamsize>::max()); 
        std::cout << "Incorrect, must be numeric: " << std::endl;
    }
    return i;
 }
于 2012-04-16T17:36:55.080 回答
2

这看起来像面向行的输入。在这种情况下，通常的解决方案是使用getline： 

bool parseNumber( std::string const& text, int& results )
{
    std::istringstream parser( text );
    return parser >> results >> std::ws && parser.peek() == EOF;
}

int getNumber()
{
    int results;
    std::string line;
    while ( ! std::getline( std::cin, line ) || ! parseNumber( line, results ) ) 
    {
        std::cin.clear();
        std::cout << "Only 'numeric' value(s) allowed:";
    }
    return results;
}
于 2012-04-16T18:08:17.390 回答