Ale*_*eev 5 c++ validation c++11
我正在编写程序的一部分,它解析并验证程序控制台参数中的一些用户输入.我选择使用stringstream用于此目的,但遇到无符号类型读取的问题.
下一个模板用于从给定字符串中读取请求的类型:
#include <iostream>
#include <sstream>
#include <string>
using std::string;
using std::stringstream;
using std::cout;
using std::endl;
template<typename ValueType>
ValueType read_value(string s)
{
stringstream ss(s);
ValueType res;
ss >> res;
if (ss.fail() or not ss.eof())
throw string("Bad argument: ") + s;
return res;
}
// +template specializations for strings, etc.
int main(void)
{
cout << read_value<unsigned int>("-10") << endl;
}
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如果类型是无符号的,输入字符串包含负数,我希望看到异常抛出(由引起ss.fail() = true
).但是stringstream会生成转换为无符号类型的值(书面示例中为4294967286).
如何修复此样本以实现所需的行为(最好不回退到c函数)?我知道它可以通过简单的第一个符号检查完成,但我可以放置前导空格.我可以编写自己的解析器,但不相信问题是如此不可预测,标准库无法解决它.
对于无符号类型,隐藏在stringstream运算符深处的函数是strtoull和strtoul.它们以描述的方式工作,但提到的功能是低级的.为什么stringstream不提供一些验证级别?(我只是希望我错了,但确实需要一些动作来实现这一点).
num_get
支持显式检查签名的方面。'-'
对于无符号类型,拒绝任何以 a 开头(空格之后)的非零数字,并使用默认的 C 语言环境num_get
进行实际转换。
#include <locale>
#include <istream>
#include <ios>
#include <algorithm>
template <class charT, class InputIterator = std::istreambuf_iterator<charT> >
class num_get_strictsignedness : public std::num_get <charT, InputIterator>
{
public:
typedef charT char_type;
typedef InputIterator iter_type;
explicit num_get_strictsignedness(std::size_t refs = 0)
: std::num_get<charT, InputIterator>(refs)
{}
~num_get_strictsignedness()
{}
private:
#define DEFINE_DO_GET(TYPE) \
virtual iter_type do_get(iter_type in, iter_type end, \
std::ios_base& str, std::ios_base::iostate& err, \
TYPE& val) const override \
{ return do_get_templ(in, end, str, err, val); } // MACRO END
DEFINE_DO_GET(unsigned short)
DEFINE_DO_GET(unsigned int)
DEFINE_DO_GET(unsigned long)
DEFINE_DO_GET(unsigned long long)
// not sure if a static locale::id is required..
template <class T>
iter_type do_get_templ(iter_type in, iter_type end, std::ios_base& str,
std::ios_base::iostate& err, T& val) const
{
using namespace std;
if(in == end)
{
err |= ios_base::eofbit;
return in;
}
// leading white spaces have already been discarded by the
// formatted input function (via sentry's constructor)
// (assuming that) the sign, if present, has to be the first character
// for the formatting required by the locale used for conversion
// use the "C" locale; could use any locale, e.g. as a data member
// note: the signedness check isn't actually required
// (because we only overload the unsigned versions)
bool do_check = false;
if(std::is_unsigned<T>{} && *in == '-')
{
++in; // not required
do_check = true;
}
in = use_facet< num_get<charT, InputIterator> >(locale::classic())
.get(in, end, str, err, val);
if(do_check && 0 != val)
{
err |= ios_base::failbit;
val = 0;
}
return in;
}
};
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使用示例:
#include <sstream>
#include <iostream>
int main()
{
std::locale loc( std::locale::classic(),
new num_get_strictsignedness<char>() );
std::stringstream ss("-10");
ss.imbue(loc);
unsigned int ui = 42;
ss >> ui;
std::cout << "ui = "<<ui << std::endl;
if(ss)
{
std::cout << "extraction succeeded" << std::endl;
}else
{
std::cout << "extraction failed" << std::endl;
}
}
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笔记:
1
函数中初始化引用计数器char
),您需要添加自己的方面(可以是模板的不同实例)wchar_t
charXY_t
num_get_strictsignedness
"-0"
被接受