如何使用运算符取消谓词函数!在C ++? [英] How to negate a predicate function using operator ! in C++?
问题描述
我要擦除所有不符合条件的元素。例如:删除字符串中不是数字的所有字符。我的解决方案使用boost :: is_digit工作得很好。
I want to erase all the elements that do not satisfy a criterion. For example: delete all the characters in a string that are not digit. My solution using boost::is_digit worked well.
struct my_is_digit {
bool operator()( char c ) const {
return c >= '0' && c <= '9';
}
};
int main() {
string s( "1a2b3c4d" );
s.erase( remove_if( s.begin(), s.end(), !boost::is_digit() ), s.end() );
s.erase( remove_if( s.begin(), s.end(), !my_is_digit() ), s.end() );
cout << s << endl;
return 0;
}
然后我尝试自己的版本,编译器抱怨:
错误C2675:unary'!':'my_is_digit'没有定义此运算符或转换为预定义运算符可接受的类型
Then I tried my own version, the compiler complained :( error C2675: unary '!' : 'my_is_digit' does not define this operator or a conversion to a type acceptable to the predefined operator
我可以使用not1但是我仍然认为操作符!在我的当前上下文中更有意义我如何实现这样的!像boost :: is_digit()?任何想法?
I could use not1() adapter, however I still think the operator ! is more meaningful in my current context. How could I implement such a ! like boost::is_digit() ? Any idea?
更新
按照Charles Bailey的说明,我编译了此代码段,但输出结果为空:
Follow Charles Bailey's instruction, I got this code snippet compiled, however the output is nothing:
struct my_is_digit : std::unary_function<bool, char> {
bool operator()( char c ) const {
return isdigit( c );
}
};
std::unary_negate<my_is_digit> operator !( const my_is_digit& rhs ) {
return std::not1( rhs );
}
int main() {
string s( "1a2b3c4d" );
//s.erase( remove_if( s.begin(), s.end(), !boost::is_digit() ), s.end() );
s.erase( remove_if( s.begin(), s.end(), !my_is_digit() ), s.end() );
cout << s << endl;
return 0;
}
有什么想法吗?
感谢,
Chan
Thanks,
Chan
推荐答案
您应该可以使用 std :: not1
。
std::unary_negate<my_is_digit> operator!( const my_is_digit& x )
{
return std::not1( x );
}
为了这个工作你必须 #include< ; function> ,并从实用程序类
std :: unary_function< code>导出
。这完全是一个typedef助手,并且不会为你的函数添加运行时开销。 my_is_digit
char,bool>
For this to work you have to #include <functional>
and derive your my_is_digit
functor from the utility class std::unary_function< char, bool >
. This is purely a typedef helper and adds no runtime overhead to your functor.
完成工作示例:
#include <string>
#include <algorithm>
#include <functional>
#include <iostream>
#include <ostream>
struct my_is_digit : std::unary_function<char, bool>
{
bool operator()(char c) const
{
return c >= '0' && c <= '9';
}
};
std::unary_negate<my_is_digit> operator!( const my_is_digit& x )
{
return std::not1( x );
}
int main() {
std::string s( "1a2b3c4d" );
s.erase( std::remove_if( s.begin(), s.end(), !my_is_digit() ), s.end() );
std::cout << s << std::endl;
return 0;
}
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