C ++使用标准算法与字符串，count_if与isdigit，函数转换 [英] C++ using standard algorithms with strings, count_if with isdigit, function cast

问题描述

我想以最短代码的方式计算字符串中的所有数字。我试过这样：

I want to count all numbers in string in shortest code way. I tried that way:

#include <string>
#include <algorithm>

unsigned countNumbers(const std::string s) {
    return count_if(s.begin(), s.end(), isdigit);
}

错误讯息是：

a.cc: In function ‘unsigned int countNumbers(std::string)’:
a.cc:5:45: error: no matching function for call to ‘count_if(std::basic_string<char>::const_iterator, std::basic_string<char>::const_iterator, <unresolved overloaded function type>)’
a.cc:5:45: note: candidate is:
/usr/include/c++/4.6/bits/stl_algo.h:4607:5: note: template<class _IIter, class _Predicate> typename std::iterator_traits<_InputIterator>::difference_type std::count_if(_IIter, _IIter, _Predicate)

我知道count_if（）想函数像：
bool（* f）（char）;作为第三个参数，所以我试图转换函数：

I know that count_if() wants function like: bool (*f)(char); as a third argument, so I tried to cast the function:

unsigned countNumbers(const std::string s) {
    return count_if(s.begin(), s.end(), reinterpret_cast<bool (*)( char )>(isdigit));
}

错误讯息是：

a.cc: In function ‘unsigned int countNumbers(std::string)’:
a.cc:5:80: error: overloaded function with no contextual type information

我也尝试了更长的版本，这给出了相同的编译错误：

I tried also a bit longer version, which gives the same compilation error:

unsigned countNumbers(const std::string s) {
    typedef bool ( * f_ptr )( char );
    f_ptr ptr = reinterpret_cast<f_ptr>(isdigit);
    return count_if(s.begin(), s.end(), ptr);
}

我想避免的解决方案是创建一个函数，适配器：

The solution that I want to avoid is to create a function which would be an adapter:

#include <string>
#include <algorithm>

bool is_digit(char c) {
    return isdigit(c);
}

unsigned countNumbers(const std::string s) {
    return count_if(s.begin(), s.end(), is_digit);
}

< （int）in std :: algorithm's function which bool（* f）（int）without creating adapter-function and without using lambda expression？

有更多的问题，当我知道如何解决这个问题，将解决例如：

I have more issues which would be solved when I get know how to solve the problem, e.g.:

• 检查字符串是否可打印：find_if_not


• 检查字符串是否包含，。！？...：find_if（s.begin（），s.end（） ），ispunct）
  和更多...

我只想知道如何在标准C ++感谢std :: algorithms
我在互联网上搜索很长时间，我发现类似问题，但我没有找到解决方案

I just want to know how to have much more string possibilities in standard C++ thanks to std::algorithms I was searching at the Internet long time, I found similar problem, but I found no solution

推荐答案

您可以使用静态转换来解析函数。或者，如果这是你想要做的事，你可以使用模板来解决它：

You can resolve the function by using a static cast. Alternatively if this is something you want to do a lot you can use a template to resolve it:

#include <string>
#include <cctype>
#include <algorithm>

unsigned count(const std::string& s) {
  return std::count_if(s.begin(), s.end(), static_cast<int(*)(int)>(std::isdigit));
}

template <int(*Pred)(int)> 
unsigned foo(const std::string& s) {
  return std::count_if(s.begin(), s.end(), Pred);
}

int main() {
  count("");
  foo<std::isdigit>("");
  foo<std::isprint>("");
}

static_cast 通常的方式解决模糊 - 它总是做你期望的，可以成为一个更大的表达式的一部分。

static_cast is the "usual" way of resolving ambiguous - it always does what you expect and can be part of a larger expression.

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