I am using std::ptr_fun
as follows:
static inline std::string <rim(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;
}
as presented in this answer .
However this does not compile with C++17 (using Microsoft Visual Studio 2017), with the error:
error C2039: 'ptr_fun': is not a member of 'std'
How can this be fixed?
You use a lambda:
static inline std::string <rim(std::string &s) {
s.erase(s.begin(), std::find_if(s.begin(), s.end(), [](int c) {return !std::isspace(c);}));
return s;
}
The answer you cited is from 2008, well before C++11 and lambdas existed.
Just use a lambda:
[](unsigned char c){ return !std::isspace(c); }
Note that I changed the argument type to unsigned char
, see the notes for std::isspace
for why.
std::ptr_fun
was deprecated in C++11, and will be removed completely in C++17.
According to cppreference , std::ptr_fun
is deprecated since C++11 and discontinued since C++17.
Similarly, std::not1
is deprecated since C++17.
So best don't use either, but a lambda (as explained in other answers).
Alternatively, you might use std::not_fn :
static inline std::string <rim(std::string &s) {
s.erase(s.begin(), std::find_if(s.begin(), s.end(),
std::not_fn(static_cast<int(*)(int)>(std::isspace))));
return s;
}
My answer is similar to the already given answers in this thread. But instead of
int isspace(int c);
function from the standard C
library, I am suggesting to use
bool isspace(char c, const locale& loc);
function instantiation from the standard C++
library ( http://en.cppreference.com/w/cpp/locale/isspace ), which is more type-correct. In this case you don't need to think about char -> unsigned char -> int
conversions and about the current user's locale.
The lambda which searches for non-space will looks like this then:
[](char c) { return !std::isspace(c, std::locale::classic()); }
And the full code of ltrim
function will look like this:
static inline std::string& ltrim(std::string& s) {
auto is_not_space = [](char c) { return !std::isspace(c, std::locale::classic()); };
auto first_non_space = std::find_if(s.begin(), s.end(), is_not_space);
s.erase(s.begin(), first_non_space);
return s;
}
You use Lambda as suggested by Nicol Bolas but you can use auto and type will be deduced there, as follow:-
static inline std::string <rim(std::string &s) {
s.erase(s.begin(), std::find_if(s.begin(), s.end(), [](auto c) {return
!std::isspace(c);}));
return s;
}
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