Template argument deduction is a compiler attempt to deduce template arguments when some are omitted.
Questions tagged [template-argument-deduction]
692 questions
9
votes
3 answers
Get function arguments type as tuple
Problem
Given any function (or callable) type Function, how can I get its all arguments types as a tuple type ?
For example, I need a trait function_traits::arguments, where:
int f();
typename function_traits::arguments //…
Zang MingJie
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9
votes
2 answers
Template Argument Deduction Failure and Function Parameters/Arguments Mismatch
Consider the following program:
template struct A { using X = typename T::X; };
template ::X* = nullptr> void f(T, int);
void f(...);
template void g(T, int, typename A::X* = nullptr); // #
void…
xskxzr
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9
votes
2 answers
Using std::function with templates
So in it's most distilled form I have something like this going on,
template
bool f(const T &a, const T &b, std::function func)
{
return func(a,b);
}
template
bool g(const T &a, const T &b)
{
…
Arii
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9
votes
1 answer
Why does std::is_same give a different result for the two types?
In the code below, why do the two ways of invoking fun: fun(num) and fun(num), give a different result when compiling?
#include
using namespace std;
template
Ryan
- 158
- 5
9
votes
1 answer
Why can't the compiler deduce auto template parameter unless I add const?
I recently had problem with code like this:
constexpr auto lambda = []{};
template
struct Lambda {};
template
void test(Lambda) {}
int main() {
test(Lambda{});
}
Both clang and GCC tells that it can't deduce…
Guillaume Racicot
- 39,621
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9
votes
1 answer
Partial ordering of forwarding reference and normal reference with deduction guides
gcc 8.0.0 and clang 5.0.0 disagree on the behavior of this program:
#include
template
struct A {
A(const T&) { std::cout << __PRETTY_FUNCTION__ << '\n'; }
A(T&&) { std::cout << __PRETTY_FUNCTION__ << '\n';…
Barry
- 286,269
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- 977
9
votes
1 answer
Deduction guide and variadic templates
Consider the following code:
#include
#include
template
struct custom_wrapper
{
template
custom_wrapper(Arg arg): data(arg) {}
T data;
};
template
custom_wrapper(Arg arg) ->…
Vincent
- 57,703
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9
votes
2 answers
Deduction guides and variadic class templates with variadic template constructors - mismatched argument pack lengths
Consider the following class definition and deduction guide:
template
struct foo : Ts...
{
template
foo(Us&&... us) : Ts{us}... { }
};
template
foo(Us&&... us) -> foo;
If I try to…
Vittorio Romeo
- 90,666
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9
votes
2 answers
Function with a size_t Template Parameter
I'm trying to understand the template function. The ultimate goal is to pass an entire array to a function. There seem to be many different ways to implement this but they all use the template function. Here's one of the simpler examples I've…
Ben Marconi
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9
votes
5 answers
Template Argument Type Deduction Fails with C++11
I'm trying to understand how to use C++(11) .
Here's my trivial test program
#include
template
inline U add(typename std::enable_if::value,U>::type a,
typename…
Nordlöw
- 11,838
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- 99
9
votes
2 answers
The difference between int a[5] and int (&a)[5] in template parameter deduction
This question is about functions that take arrays of statically known size.
Take for example the following minimal program:
#include
template
void arrfun_a(int a[N])
{
for(size_t i = 0; i < N; ++i)
std::cout <<…
danielschemmel
- 10,885
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8
votes
3 answers
Getting the right value_type
In my class I have a member:
std::vector memory_;
Now I'd like to have a fnc returning what's in the memory's first element but I do not want to specify std::string as a return type in case later I decide to use different type for this…
smallB
- 16,662
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8
votes
2 answers
Deducing knowledge of original types, whilst simultaneously forwarding
Summary: I want to end up with a function that deduces the exact types it was called with and takes (e.g.) a tuple that forwards them (the types of which will be different from the exact types the function was called with).
I'm stuck trying to…
Flexo
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8
votes
1 answer
Is std::decay_t decay_copy(T&&) equivalent to auto decay_copy(auto&&)?
http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2011/n3255.html defines decay_copy as follows:
template
std::decay_t decay_copy(T&& v)
{
return std::forward(v);
}
I just wonder:
Is it equivalent to the following simpler…
xmllmx
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8
votes
6 answers
Default argument for a functor in a templated parameter
I would like to have a default lambda for a functor argument in my function.
I am aware it is possible using a struct and operator() like this:
struct AddOne {
int operator()(int a) {
return a+1;
}
};
template
Salgar
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