如何修剪 std::string? [英] How to trim a std::string?

问题描述

我目前正在使用以下代码对程序中的所有 std::strings 进行右修剪:

I'm currently using the following code to right-trim all the std::strings in my programs:

std::string s;
s.erase(s.find_last_not_of(" \n\r\t")+1);

它工作正常，但我想知道是否有一些最终情况可能会失败?

It works fine, but I wonder if there are some end-cases where it might fail?

当然，欢迎提供优雅的替代方案和左装饰解决方案的答案.

Of course, answers with elegant alternatives and also left-trim solution are welcome.

推荐答案

EDIT 从 c++17 开始，标准库的某些部分被删除.幸运的是，从 c++11 开始，我们有了 lambdas，这是一个很好的解决方案.

EDIT Since c++17, some parts of the standard library were removed. Fortunately, starting with c++11, we have lambdas which are a superior solution.

#include <algorithm> 
#include <cctype>
#include <locale>

// trim from start (in place)
static inline void ltrim(std::string &s) {
    s.erase(s.begin(), std::find_if(s.begin(), s.end(), [](unsigned char ch) {
        return !std::isspace(ch);
    }));
}

// trim from end (in place)
static inline void rtrim(std::string &s) {
    s.erase(std::find_if(s.rbegin(), s.rend(), [](unsigned char ch) {
        return !std::isspace(ch);
    }).base(), s.end());
}

// trim from both ends (in place)
static inline void trim(std::string &s) {
    ltrim(s);
    rtrim(s);
}

// trim from start (copying)
static inline std::string ltrim_copy(std::string s) {
    ltrim(s);
    return s;
}

// trim from end (copying)
static inline std::string rtrim_copy(std::string s) {
    rtrim(s);
    return s;
}

// trim from both ends (copying)
static inline std::string trim_copy(std::string s) {
    trim(s);
    return s;
}

感谢 https://stackoverflow.com/a/44973498/524503 提出现代解决方案.

Thanks to https://stackoverflow.com/a/44973498/524503 for bringing up the modern solution.

我倾向于使用这 3 种中的一种来满足我的修剪需求:

I tend to use one of these 3 for my trimming needs:

#include <algorithm> 
#include <functional> 
#include <cctype>
#include <locale>

// trim from start
static inline std::string &ltrim(std::string &s) {
    s.erase(s.begin(), std::find_if(s.begin(), s.end(),
            std::not1(std::ptr_fun<int, int>(std::isspace))));
    return s;
}

// trim from end
static inline std::string &rtrim(std::string &s) {
    s.erase(std::find_if(s.rbegin(), s.rend(),
            std::not1(std::ptr_fun<int, int>(std::isspace))).base(), s.end());
    return s;
}

// trim from both ends
static inline std::string &trim(std::string &s) {
    return ltrim(rtrim(s));
}

它们是不言自明的，而且效果很好.

They are fairly self-explanatory and work very well.

EDIT:顺便说一句，我有 std::ptr_fun 来帮助消除 std::isspace 的歧义，因为实际上还有第二个定义它支持语言环境.这本来可以是一个演员阵容，但我更喜欢这个.

EDIT: BTW, I have std::ptr_fun in there to help disambiguate std::isspace because there is actually a second definition which supports locales. This could have been a cast just the same, but I tend to like this better.

EDIT:解决一些关于通过引用接受参数、修改和返回参数的评论.我同意.我可能更喜欢的一个实现是两组函数，一组用于就地，一组用于制作副本.一组更好的例子是:

EDIT: To address some comments about accepting a parameter by reference, modifying and returning it. I Agree. An implementation that I would likely prefer would be two sets of functions, one for in place and one which makes a copy. A better set of examples would be:

#include <algorithm> 
#include <functional> 
#include <cctype>
#include <locale>

// trim from start (in place)
static inline void ltrim(std::string &s) {
    s.erase(s.begin(), std::find_if(s.begin(), s.end(),
            std::not1(std::ptr_fun<int, int>(std::isspace))));
}

// trim from end (in place)
static inline void rtrim(std::string &s) {
    s.erase(std::find_if(s.rbegin(), s.rend(),
            std::not1(std::ptr_fun<int, int>(std::isspace))).base(), s.end());
}

// trim from both ends (in place)
static inline void trim(std::string &s) {
    ltrim(s);
    rtrim(s);
}

// trim from start (copying)
static inline std::string ltrim_copy(std::string s) {
    ltrim(s);
    return s;
}

// trim from end (copying)
static inline std::string rtrim_copy(std::string s) {
    rtrim(s);
    return s;
}

// trim from both ends (copying)
static inline std::string trim_copy(std::string s) {
    trim(s);
    return s;
}

我保留上面的原始答案，但为了上下文和保持高票答案仍然可用.

I am keeping the original answer above though for context and in the interest of keeping the high voted answer still available.

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