我目前正在使用以下代码在我的程序中右修剪所有std::字符串:
std::string s;
s.erase(s.find_last_not_of(" \n\r\t")+1);
它工作得很好,但我想知道是否有一些最终情况下它可能会失败?
当然,有优雅的选择和左修剪解决方案的答案是受欢迎的。
我目前正在使用以下代码在我的程序中右修剪所有std::字符串:
std::string s;
s.erase(s.find_last_not_of(" \n\r\t")+1);
它工作得很好,但我想知道是否有一些最终情况下它可能会失败?
当然,有优雅的选择和左修剪解决方案的答案是受欢迎的。
当前回答
穷人的绳子装饰(仅限空格):
std::string trimSpaces(const std::string& str)
{
int start, len;
for (start = 0; start < str.size() && str[start] == ' '; start++);
for (len = str.size() - start; len > 0 && str[start + len - 1] == ' '; len--);
return str.substr(start, len);
}
其他回答
有点晚了,不过没关系。c++ 11在这里,我们有lambda和auto变量。所以我的版本,也处理全空白和空字符串,是:
#include <cctype>
#include <string>
#include <algorithm>
inline std::string trim(const std::string &s)
{
auto wsfront=std::find_if_not(s.begin(),s.end(),[](int c){return std::isspace(c);});
auto wsback=std::find_if_not(s.rbegin(),s.rend(),[](int c){return std::isspace(c);}).base();
return (wsback<=wsfront ? std::string() : std::string(wsfront,wsback));
}
我们可以从wsfront创建一个反向迭代器,并在第二个find_if_not中使用它作为终止条件,但这只在全空白字符串的情况下有用,gcc 4.8至少不足以用auto推断反向迭代器(std::string::const_reverse_iterator)的类型。我不知道构造反向迭代器有多贵,这里是YMMV。修改后,代码如下所示:
inline std::string trim(const std::string &s)
{
auto wsfront=std::find_if_not(s.begin(),s.end(),[](int c){return std::isspace(c);});
return std::string(wsfront,std::find_if_not(s.rbegin(),std::string::const_reverse_iterator(wsfront),[](int c){return std::isspace(c);}).base());
}
这个好吗?(因为这篇文章完全需要另一个答案:)
string trimBegin(string str)
{
string whites = "\t\r\n ";
int i = 0;
while (whites.find(str[i++]) != whites::npos);
str.erase(0, i);
return str;
}
类似的情况下,trimEnd,只是反转极化,指数。
我不确定您的环境是否相同,但在我的环境中,空字符串情况将导致程序中止。我要么用if(!s.empty())包装这个擦除调用,要么像前面提到的那样使用Boost。
对噪音做出我的解决方案。Trim默认创建一个新字符串并返回修改后的字符串,而trim_in_place则修改传递给它的字符串。trim函数支持c++11 move语义。
#include <string>
// modifies input string, returns input
std::string& trim_left_in_place(std::string& str) {
size_t i = 0;
while(i < str.size() && isspace(str[i])) { ++i; };
return str.erase(0, i);
}
std::string& trim_right_in_place(std::string& str) {
size_t i = str.size();
while(i > 0 && isspace(str[i - 1])) { --i; };
return str.erase(i, str.size());
}
std::string& trim_in_place(std::string& str) {
return trim_left_in_place(trim_right_in_place(str));
}
// returns newly created strings
std::string trim_right(std::string str) {
return trim_right_in_place(str);
}
std::string trim_left(std::string str) {
return trim_left_in_place(str);
}
std::string trim(std::string str) {
return trim_left_in_place(trim_right_in_place(str));
}
#include <cassert>
int main() {
std::string s1(" \t\r\n ");
std::string s2(" \r\nc");
std::string s3("c \t");
std::string s4(" \rc ");
assert(trim(s1) == "");
assert(trim(s2) == "c");
assert(trim(s3) == "c");
assert(trim(s4) == "c");
assert(s1 == " \t\r\n ");
assert(s2 == " \r\nc");
assert(s3 == "c \t");
assert(s4 == " \rc ");
assert(trim_in_place(s1) == "");
assert(trim_in_place(s2) == "c");
assert(trim_in_place(s3) == "c");
assert(trim_in_place(s4) == "c");
assert(s1 == "");
assert(s2 == "c");
assert(s3 == "c");
assert(s4 == "c");
}
For what it's worth, here is a trim implementation with an eye towards performance. It's much quicker than many other trim routines I've seen around. Instead of using iterators and std::finds, it uses raw c strings and indices. It optimizes the following special cases: size 0 string (do nothing), string with no whitespace to trim (do nothing), string with only trailing whitespace to trim (just resize the string), string that's entirely whitespace (just clear the string). And finally, in the worst case (string with leading whitespace), it does its best to perform an efficient copy construction, performing only 1 copy and then moving that copy in place of the original string.
void TrimString(std::string & str)
{
if(str.empty())
return;
const auto pStr = str.c_str();
size_t front = 0;
while(front < str.length() && std::isspace(int(pStr[front]))) {++front;}
size_t back = str.length();
while(back > front && std::isspace(int(pStr[back-1]))) {--back;}
if(0 == front)
{
if(back < str.length())
{
str.resize(back - front);
}
}
else if(back <= front)
{
str.clear();
}
else
{
str = std::move(std::string(str.begin()+front, str.begin()+back));
}
}