I have following two functions:
void bar(const std::string &s)
{
someCFunctionU(s.c_str());
}
void bar(const std::wstring &s)
{
someCFunctionW(s.c_str());
}
Both of these call some C function which accepts const char * or const wchar_t * and have U or W suffixes respectively. I would like to create a template function to handle both of these cases. I tried following attempt:
template <typename T>
void foo(const std::basic_string<T> &s)
{
if constexpr (std::is_same_v<T, char>)
someCFunctionU(s.c_str());
else
someCFunctionW(s.c_str());
}
But this does not seem to work correctly. If I call:
foo("abc");
this will not compile. Why is that? why a compiler is not able to deduce the proper type T to char? Is it possible to create one function which would handle both std::string and std::wstring?
this will not compile. Why is that? why a compiler is not able to deduce the proper type T to char?
As better explained by others, "abc" is a char[4], so is convertible to a std::basic_string<char> but isn't a std::basic_string<char>, so can't be deduced the T type as char for a template function that accept a std::basic_string<T>.
Is it possible to create one function which would handle both std::string and std::wstring?
Yes, it's possible; but what's wrong with your two-function-in-overloading solution?
Anyway, if you really want a single function and if you accept to write a lot of casuistry, I suppose you can write something as follows
template <typename T>
void foo (T const & s)
{
if constexpr ( std::is_same_v<T, std::string> )
someCFunctionU(s.c_str());
else if constexpr ( std::is_convertible_v<T, char const *>)
someCFunctionU(s);
else if constexpr ( std::is_same_v<T, std::wstring> )
someCFunctionW(s.c_str());
else if constexpr ( std::is_convertible_v<T, wchar_t const *> )
someCFunctionW(s);
// else exception ?
}
or, a little more synthetic but less efficient
template <typename T>
void foo (T const & s)
{
if constexpr ( std::is_convertible_v<T, std::string> )
someCFunctionU(std::string{s}.c_str());
else if constexpr (std::is_convertible_v<T, std::wstring> )
someCFunctionW(std::wstring{s}.c_str());
// else exception ?
}
So you should be able to call foo() with std::string, std::wstring, char *, wchar_t *, char[] or wchar_t[].
The issue here is that in foo("abc");, "abc" is not a std::string or a std::wstring, it is a const char[N]. Since it isn't a std::string or a std::wstring the compiler cannot deduce what T should be and it fails to compile. The easiest solution is to use what you already have. The overloads will be considered and it is a better match to convert "abc" to a std::string so it will call that version of the function.
If you want you could use a std::string_view/std::wstring_view instead of std::string/std::wstring so you don't actually allocate any memory if you pass the function a string literal. That would change the overloads to
void bar(std::string_view s)
{
someCFunctionU(s.data());
}
void bar(std::wstring_view s)
{
someCFunctionW(s.data());
}
Do note that std::basic_string_view can be constructed without having a null terminator so it is possible to pass a std::basic_string_view that won't fulfill the null terminated c-string requirement that your C function has. In that case the code has undefined behavior.
A workaround in C++17 is:
template <typename T>
void foo(const T &s)
{
std::basic_string_view sv{s}; // Class template argument deduction
if constexpr (std::is_same_v<typename decltype(sv)::value_type, char>)
someCFunctionU(sv.data());
else
someCFunctionW(sv.data());
}
And to avoid issue mentioned by Justin about non-null-terminated string
template <typename T> struct is_basic_string_view : std::false_type {};
template <typename T> struct is_basic_string_view<basic_string_view<T>> : std::true_type
{};
template <typename T>
std::enable_if_t<!is_basic_string_view<T>::value> foo(const T &s)
{
std::basic_string_view sv{s}; // Class template argument deduction
if constexpr (std::is_same_v<typename decltype(sv)::value_type, char>)
someCFunctionU(sv.data());
else
someCFunctionW(sv.data());
}
Yes, there exist a type, i.e. std::basic_string<char>, which can be copy initialized from expression "abc". So you can call a function like void foo(std::basic_string<char>) with argument "abc".
And no, you can't call a function template template <class T> void foo(const std::basic_string<T> &s) with argument "abc". Because in order to figure out whether the parameter can be initialized by the argument, the compiler need to determine the template parameter T first. It will try to match const std::basic_string<T> & against const char [4]. And it will fail.
The reason why it will fail is because of the template argument deduction rule. The actual rule is very complicated. But in this case, for std::basic_string<char> to be examined during the deduction, compiler will need to look for a proper "converting constructor", i.e. the constructor which can be called implicitly with argument "abc", and such lookup isn't allowed by the standard during deduction.
Yes, it is possible to handle std::string and std::wstring in one function template:
void foo_impl(const std::string &) {}
void foo_impl(const std::wstring &) {}
template <class T>
auto foo(T &&t) {
return foo_impl(std::forward<T>(t));
}
Related
#include <cstdio>
#include <string>
constexpr char str[] = "/home/qspace/etc/client/mmkvcfgsvr_test_byset_cli.conf";
void test(bool a)
{
printf("b=%d",a);
}
void test(const std::string& s){
printf("s=%s",s.c_str());
}
int main()
{
test(str);
return 0;
}
Like this code, the C++ compiler will convert char* to bool and then call the first function, which is inconsistent with my original intention.
Is there any way to prevent the compiler from performing type conversions that I don't want?
Like "-fno-permissive", but unfortunately, it doesn't work.
How to explicitly call the specified overload function?
Convert the argument at call site: test(std::string(str));
Take expected address of overload function: static_cast<void(*)(const std::string&)>(print)(str);
Is there any way to prevent the compiler from performing type conversions that I don't want?
You might add a catch-all overload as deleted: template <typename T> void test(const T&) = delete;
Alternatively, in C++17, you might do the "dispatching" manually:
template <typename T>
void test(const T& t)
{
static_assert(std::is_constructible_v<std::string, T>
|| std::is_convertible_v<T, bool>);
if constexpr (std::is_constructible_v<std::string, T>) {
const std::string& s = t;
printf("s=%s", s.c_str());
} else if constexpr (std::is_convertible_v<T, bool>) {
printf("b=%d", bool(t));
}
}
You're mixing C and STL types (char array vs std::string). There are two solutions. The immediately obvious solution is to create a temporary std::string object every time you wish to pass a char array into a function expecting std::string.
test(std::string(str));
The other solution, which I prefer, is to avoid C types altogether. To create a string constant, use STL type directly:
const std::string str {"/home/qspace/etc/client/mmkvcfgsvr_test_byset_cli.conf"};
If you wish to retain constexpr see this thread: Is it possible to use std::string in a constexpr?
Let's say I have a function:
#include <optional>
template <typename T>
std::optional<T> foo(T const &input);
It accepts a value, attempts to work with a copy of it and returns said copy on success (std::nullopt on fail).
But the problem is, when a string literal is passed into such function, an error T in optional<T> must meet the Cpp17Destructible requirements occurs.
It's caused by static_assert(is_object_v<_Ty> && is_destructible_v<_Ty> && !is_array_v<_Ty>, ...) defined in <optional>.
The next expression works correctly:
foo((char const*) "bar");
This one fails:
foo("bar");
The question is, how do I force the compiler to implicitly convert char const[] to char const*?
P. S. I know that it could be done by simply overloading the function, by I'm not too keen on code duplication it causes, so I'm curious whether an alternative solution is applicable here.
Edit: Rewrote the answer. With return type deduction, this would be convenient.
template <typename T>
auto foo(T const &input){
auto copy {std::move(input)};
// ...
return std::optional{std::move(copy)};
}
Not really what you asked for, though consider that not much repetition is needed:
template <int s>
std::optional<const char*> foo(const char (&str)[s]) {
return foo(&str[0]);
}
or simpler:
std::optional<char const*> foo(char const *input) {
return foo<char const *>(input);
}
Hmm a strange one in VC2012 I can't seem to work out the syntax for passing a const pointer by const reference into a function of a templated class whose template argument is a non const pointer ie:
template<typename T>
struct Foo
{
void Add( const T& Bar ) { printf(Bar); }
};
void main()
{
Foo<char*> foo;
const char* name = "FooBar";
foo.Add(name); // Causes error
}
So I've simplified my problem here but basically I want the argument to 'Add' to have a const T ie const char*. I've tried:
void Add( const (const T)& Bar );
typedef const T ConstT;
void Add( const (ConstT)& Bar );
void Add( const typename std::add_const<T>::type& Bar );
None of which work. The exact error I'm getting is:
error C2664: 'Foo<T>::Add' : cannot convert parameter 1 from 'const char *' to 'char *const &'
with
[
T=char *
]
Conversion loses qualifiers
which I can see is correct but how do I solve it without const casting 'name' to be non const.
There is a strong difference between a pointer to a constant object (T const*, or const T*) and a constant pointer to a non-constant object (T * const). In your case the signature of the member Add is:
void Foo<char *>::Add(char * const& ); // reference to a constant pointer to a
// non-constant char
I usually recommend that people drop the use of const on the left hand side exactly for this reason, as beginners usually confuse typedefs (or deduced types) with type substitution and when they read:
const T& [T == char*]
They misinterpret
const char*&
If the const is placed in the right place:
T const &
Things are simpler for beginners, as plain mental substitution works:
char * const &
A different problem than what you are asking, but maybe what you think you want, is:
Given a type T have a function that takes a U that is const T if T is not a pointer type, or X const * if T is a pointer to X
template <typename T>
struct add_const_here_or_there {
typedef T const type;
};
template <typename T>
struct add_const_here_or_there<T*> {
typedef T const * type;
};
Then you can use this in your signature:
template <typename T>
void Foo<T>::Add( const typename add_const_here_or_there<T>::type & arg ) {
...
Note that I am adding two const in the signature, so in your case char* will map to char const * const &, as it seems that you want to pass a const& to something and you also want the pointed type to be const.
You might have wondered as of the name for the metafunction: *add_const_here_or_there*, it is like that for a reason: there is no simple way of describing what you are trying to do, which is usually a code smell. But here you have your solution.
It looks like your issue here as that as soon as you have a pointer type mapped to a template type, you can no longer add const-ness to the pointed-to type, only to the pointer itself. What it looks like you're trying to do is automatically add constness to the parameter of your function (so if T is char* the function should accept const char* const& rather than char* const& as you've written). The only way to do that is with another template to add constness to the pointee for pointer types, as follows. I took the liberty of including missing headers and correcting the signature of main:
#include <cstdio>
template<typename T>
struct add_const_to_pointee
{
typedef T type;
};
template <typename T>
struct add_const_to_pointee<T*>
{
typedef const T* type;
};
template<typename T>
struct Foo
{
void Add( typename add_const_to_pointee<T>::type const & Bar ) { printf(Bar); }
};
int main()
{
Foo<char*> foo;
const char* name = "FooBar";
foo.Add(name); // Causes error
}
As mentioned in another another however, this issue just goes away if you use std::string instead of C-style strings.
You need to change the template argument to your Foo object to Foo<const char*>. Because if T=char*, then const T=char*const, not const char*. Trying to coerce it to work is not a good idea and would probably result in undefined behavior.
Use:
Foo<const char*> foo;
const char* name = "FooBar";
foo.Add(name);
And write int main() instead of void main()
If passing const char* instead of char* to Foo is not an option you can finesse the correct type with std::remove_pointer. This will remove the pointer modifier and allow you to provide a more explicit type.
#include <type_traits>
template<typename T>
struct Foo
{
void Add(typename std::remove_pointer<T>::type const*& Bar ) { printf(Bar); }
};
To prevent the pointer value from being modified you can declare the reference as const as well.
void Add(typename std::remove_pointer<T>::type const* const& Bar )
{ Bar = "name"; } // <- fails
If you need to reduce the type from say a pointer to pointer you can use std::decay along with std::remove_pointer
void Add(typename std::remove_pointer<typename std::decay<T>::type>::type const*& Bar)
{
printf(Bar);
}
This really depends on what your requirements for T are. I suggest assuming only the base type (e.g. char) is passed as T and building reference and pointer types from that.
I'm trying to derive a technique for writing string-algorithms that is truly independent of the underlying type of string.
Background: the prototypes for GetIndexOf and FindOneOf are either overloaded or templated variations on:
int GetIndexOf(const char * pszInner, const char * pszString);
const char * FindOneOf(const char * pszString, const char * pszSetOfChars);
This issue comes up in the following template function:
// return index of, or -1, the first occurrence of any given char in target
template <typename T>
inline int FindIndexOfOneOf(const T * str, const T * pszSearchChars)
{
return GetIndexOf(FindOneOf(str, pszSearchChars), str);
}
Objectives:
1. I would like this code to work for CStringT<>, const char *, const wchar_t * (and should be trivial to extend to std::string)
2. I don't want to pass anything by copy (only by const & or const *)
In an attempt to solve these two objectives, I thought I might be able to use a type-selector of sorts to derive the correct interfaces on the fly:
namespace details {
template <typename T>
struct char_type_of
{
// typedef T type; error for invalid types (i.e. anything for which there is not a specialization)
};
template <>
struct char_type_of<const char *>
{
typedef char type;
};
template <>
struct char_type_of<const wchar_t *>
{
typedef wchar_t type;
};
template <>
struct char_type_of<CStringA>
{
typedef CStringA::XCHAR type;
};
template <>
struct char_type_of<CStringW>
{
typedef CStringW::XCHAR type;
};
}
#define CHARTYPEOF(T) typename details::char_type_of<T>::type
Which allows:
template <typename T>
inline int FindIndexOfOneOf(T str, const CHARTYPEOF(T) * pszSearchChars)
{
return GetIndexOf(FindOneOf(str, pszSearchChars), str);
}
This should guarantee that the second argument is passed as const *, and should not determine T (rather only the first argument should determine T).
But the problem with this approach is that T, when str is a CStringT<>, is a copy of the CStringT<> rather than a reference to it: hence we have an unnecessary copy.
Trying to rewrite the above as:
template <typename T>
inline int FindIndexOfOneOf(T & str, const CHARTYPEOF(T) * pszSearchChars)
{
return GetIndexOf(FindOneOf(str, pszSearchChars), str);
}
Makes it impossible for the compiler (VS2008) to generate a correct instance of FindIndexOfOneOf<> for:
FindIndexOfOneOf(_T("abc"), _T("def"));
error C2893: Failed to specialize function template 'int FindIndexOfOneOf(T &,const details::char_type_of<T>::type *)'
With the following template arguments: 'const char [4]'
This is a generic problem I've had with templates since they were introduced (yes, I'm that old): That it's been essentially impossible to construct a way to handle both old C-style arrays and newer class based entities (perhaps best highlighted by const char [4] vs. CString<> &).
The STL/std library "solved" this issue (if one can really call it solving) by instead using pairs of iterators everywhere instead of a reference to the thing itself. I could go this route, except it sucks IMO, and I don't want to have to litter my code with two-arguments everywhere a single argument properly handled should have been.
Basically, I'm interested in an approach - such as using some sort of stringy_traits - that would allow me to write GetIndexOfOneOf<> (and other similar template functions) where the argument is the string (not a pair of (being, end] arguments), and the template that is then generated be correct based on that string-argument-type (either const * or const CString<> &).
So the Question: How might I write FindIndexOfOneOf<> such that its arguments can be any of the following without ever creating a copy of the underlying arguments:
1. FindIndexOfOneOf(_T("abc"), _T("def"));
2. CString str; FindIndexOfOneOf(str, _T("def"));
3. CString str; FindIndexOfOneOf(T("abc"), str);
3. CString str; FindIndexOfOneOf(str, str);
Related threads to this one that have lead me to this point:
A better way to declare a char-type appropriate CString<>
Templated string literals
Try this.
#include <type_traits>
inline int FindIndexOfOneOf(T& str, const typename char_type_of<typename std::decay<T>::type>::type* pszSearchChars)
The problem is that when you make the first argument a reference type T becomes deduced as:
const char []
but you want
const char*
You can use the following to make this conversion.
std::decay<T>::type
The documentation says.
If is_array<U>::value is true, the modified-type type is remove_extent<U>::type *.
You can use Boost's enable_if and type_traits for this:
#include <boost/type_traits.hpp>
#include <boost/utility/enable_if.hpp>
// Just for convenience
using boost::enable_if;
using boost::disable_if;
using boost::is_same;
// Version for C strings takes param #1 by value
template <typename T>
inline typename enable_if<is_same<T, const char*>, int>::type
FindIndexOfOneOf(T str, const CHARTYPEOF(T) * pszSearchChars)
{
return GetIndexOf(FindOneOf(str, pszSearchChars), str);
}
// Version for other types takes param #1 by ref
template <typename T>
inline typename disable_if<is_same<T, const char*>, int>::type
FindIndexOfOneOf(T& str, const CHARTYPEOF(T) * pszSearchChars)
{
return GetIndexOf(FindOneOf(str, pszSearchChars), str);
}
You should probably expand the first case to handle both char and wchar_t strings, which you can do using or_ from Boost's MPL library.
I would also recommend making the version that takes a reference take a const reference instead. This just avoids instantiation of 2 separate versions of the code (as it stands, T will be inferred as a const type for const objects, and a non-const type for non-const objects; changing the parameter type to T const& str means T will always be inferred as a non-const type).
Based on your comments about iterators it seems you've not fully considered options you may have. I can't do anything about personal preference, but then again...IMHO it shouldn't be a formidable obstacle to overcome in order to accept a reasonable solution, which should be weighed and balanced technically.
template < typename Iter >
void my_iter_fun(Iter start, Iter end)
{
...
}
template < typename T >
void my_string_interface(T str)
{
my_iter_fun(str.begin(), str.end());
}
template < typename T >
void my_string_interface(T* chars)
{
my_iter_fun(chars, chars + strlen(chars));
}
Alternative to my previous answer, if you don't want to install tr1.
Add the following template specializations to cover the deduced T type when the first argument is a reference.
template<unsigned int N>
struct char_type_of<const wchar_t[N]>
{
typedef wchar_t type;
};
template<unsigned int N>
struct char_type_of<const char[N]>
{
typedef char type;
};
I have a "set" data type:
template <class V>
struct Set {
void add(const V& value) {}
};
I want to write a top-level function version of Set::add.
template <class V>
void add(const Set<V>& set, const V& value) {}
This doesn't quite work with string literals:
Set<const char*> set;
const char* val = "a";
set.add(val); // ok
set.add("a"); // ok
add(set, val); // ok
add(set, "a"); // ERROR
add<const char*>(set, "a"); // ok
The error message (g++ 4.2.4):
no matching function for call to ‘add(Set<const char*>&, const char [2])’
It looks it has something to do with the fact that "a" has type const char[2] and not const char*. Does anybody know how to get this to work?
The problem is that V gets one type for the left parameter, and another type for the right one. I suspect you also want to be able to say add(setOfLong, 0) - but with that template you couldn't. I recommend to add a separate template parameter to solve this
template <class SetV, class V>
void add(const Set<SetV>& set, const V& value) {}
There's another way to solve this problem (forget where I saw it...).
You can use an "Identity" type wrapper to have the compiler not take a type into account when performing inference.
template <T>
struct Identity {
typedef T type;
};
Then define "add" like this:
template <class V>
void add(const Set<V>& set, const typename Identity<V>::Type& value) {}
This causes 'V' to be deduced solely based on the first argument type. Once that's determined, it goes ahead and uses it for the second argument, which works fine.