How to assert that a constexpr if else clause never happen?
I want to raise a compile time error when non of the constexpr if conditions is true eg:
if constexpr(condition1){
...
} else if constexpr (condition2) {
....
} else if constexpr (condition3) {
....
} else {
// I want the else clause never taken. But I heard the code below is not allowed
static_assert(false);
}
// I'd rather not repeat the conditions again like this:
static_assert(condition1 || condition2 || condition3);
c++ c++17
asked Dec 27 '18 at 12:57
John Smith
656114
add a comment |
I want to raise a compile time error when non of the constexpr if conditions is true eg:
if constexpr(condition1){
...
} else if constexpr (condition2) {
....
} else if constexpr (condition3) {
....
} else {
// I want the else clause never taken. But I heard the code below is not allowed
static_assert(false);
}
// I'd rather not repeat the conditions again like this:
static_assert(condition1 || condition2 || condition3);
c++ c++17
asked Dec 27 '18 at 12:57
John Smith
656114
add a comment |
I want to raise a compile time error when non of the constexpr if conditions is true eg:
if constexpr(condition1){
...
} else if constexpr (condition2) {
....
} else if constexpr (condition3) {
....
} else {
// I want the else clause never taken. But I heard the code below is not allowed
static_assert(false);
}
// I'd rather not repeat the conditions again like this:
static_assert(condition1 || condition2 || condition3);
c++ c++17
asked Dec 27 '18 at 12:57
John Smith
656114
I want to raise a compile time error when non of the constexpr if conditions is true eg:
if constexpr(condition1){
...
} else if constexpr (condition2) {
....
} else if constexpr (condition3) {
....
} else {
// I want the else clause never taken. But I heard the code below is not allowed
static_assert(false);
}
// I'd rather not repeat the conditions again like this:
static_assert(condition1 || condition2 || condition3);
c++ c++17
c++ c++17
asked Dec 27 '18 at 12:57
John Smith
656114
asked Dec 27 '18 at 12:57
John Smith
656114
asked Dec 27 '18 at 12:57
John Smith
656114
asked Dec 27 '18 at 12:57
John Smith
656114
asked Dec 27 '18 at 12:57
John Smith
656114
656114
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
You have to make the discarded statement dependent of the template parameters
template <class...> constexpr std::false_type always_false{};
if constexpr(condition1){
...
} else if constexpr (condition2) {
....
} else if constexpr (condition3) {
....
} else {
static_assert(always_false<T>);
}
This is so because
[temp.res]/8 - The program is ill-formed, no diagnostic required, if
no valid specialization can be generated for a template or a substatement of a
constexpr ifstatement within a template and the template is not instantiated, or ...
answered Dec 27 '18 at 13:02
Jans
8,09522535
add a comment |
Here's a workaround from cppreference.com, i.e. use a type-dependent expression instead.
Note: the discarded statement can't be ill-formed for every possible specialization:
The common workaround for such a catch-all statement is a type-dependent expression that is always false:
e.g.
template<class T> struct dependent_false : std::false_type {};
then
static_assert(dependent_false<T>::value);
answered Dec 27 '18 at 13:02
songyuanyao
89.9k11170234
2
Ironic really since that's exactly what this gives us! Silly language.
– Lightness Races in Orbit
Dec 27 '18 at 16:42
add a comment |
taking a slightly different tack...
#include <ciso646>
template<auto x> void something();
template<class...Conditions>
constexpr int which(Conditions... cond)
{
int sel = 0;
bool found = false;
auto elect = [&found, &sel](auto cond)
{
if (not found)
{
if (cond)
{
found = true;
}
else
{
++sel;
}
}
};
(elect(cond), ...);
if (not found) throw "you have a logic error";
return sel;
}
template<bool condition1, bool condition2, bool condition3>
void foo()
{
auto constexpr sel = which(condition1, condition2, condition3);
switch(sel)
{
case 0:
something<1>();
break;
case 1:
something<2>();
break;
case 2:
something<3>();
break;
}
}
int main()
{
foo<false, true, false>();
// foo<false, false, false>(); // fails to compile
}
As I understand it, which is evaluated in constexpr context, which means that it is legal unless the program has to follow a code path that is illegal in a constexpr context.
For all expected cases, the throw path is not taken, so the function is legal. When illegal inputs are provided, we go down the ill-formed path, which causes a compiler error.
I'd be interested to know whether this solution is strictly correct from a language-lawyer perspective.
It works on gcc, clang and MSVC.
...or for fans of obfuscated code...
template<class...Conditions>
constexpr int which(Conditions... cond)
{
auto sel = 0;
((cond or (++sel, false)) or ...) or (throw "program is ill-formed", false);
return sel;
}
answered Dec 27 '18 at 17:53
Richard Hodges
55.4k657100
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You have to make the discarded statement dependent of the template parameters
template <class...> constexpr std::false_type always_false{};
if constexpr(condition1){
...
} else if constexpr (condition2) {
....
} else if constexpr (condition3) {
....
} else {
static_assert(always_false<T>);
}
This is so because
[temp.res]/8 - The program is ill-formed, no diagnostic required, if
no valid specialization can be generated for a template or a substatement of a
constexpr ifstatement within a template and the template is not instantiated, or ...
answered Dec 27 '18 at 13:02
Jans
8,09522535
add a comment |
You have to make the discarded statement dependent of the template parameters
template <class...> constexpr std::false_type always_false{};
if constexpr(condition1){
...
} else if constexpr (condition2) {
....
} else if constexpr (condition3) {
....
} else {
static_assert(always_false<T>);
}
This is so because
[temp.res]/8 - The program is ill-formed, no diagnostic required, if
no valid specialization can be generated for a template or a substatement of a
constexpr ifstatement within a template and the template is not instantiated, or ...
answered Dec 27 '18 at 13:02
Jans
8,09522535
add a comment |
You have to make the discarded statement dependent of the template parameters
template <class...> constexpr std::false_type always_false{};
if constexpr(condition1){
...
} else if constexpr (condition2) {
....
} else if constexpr (condition3) {
....
} else {
static_assert(always_false<T>);
}
This is so because
[temp.res]/8 - The program is ill-formed, no diagnostic required, if
no valid specialization can be generated for a template or a substatement of a
constexpr ifstatement within a template and the template is not instantiated, or ...
answered Dec 27 '18 at 13:02
Jans
8,09522535
You have to make the discarded statement dependent of the template parameters
template <class...> constexpr std::false_type always_false{};
if constexpr(condition1){
...
} else if constexpr (condition2) {
....
} else if constexpr (condition3) {
....
} else {
static_assert(always_false<T>);
}
This is so because
[temp.res]/8 - The program is ill-formed, no diagnostic required, if
no valid specialization can be generated for a template or a substatement of a
constexpr ifstatement within a template and the template is not instantiated, or ...
answered Dec 27 '18 at 13:02
Jans
8,09522535
edited Dec 27 '18 at 13:09
answered Dec 27 '18 at 13:02
Jans
8,09522535
answered Dec 27 '18 at 13:02
Jans
8,09522535
answered Dec 27 '18 at 13:02
Jans
8,09522535
8,09522535
add a comment |
add a comment |
Here's a workaround from cppreference.com, i.e. use a type-dependent expression instead.
Note: the discarded statement can't be ill-formed for every possible specialization:
The common workaround for such a catch-all statement is a type-dependent expression that is always false:
e.g.
template<class T> struct dependent_false : std::false_type {};
then
static_assert(dependent_false<T>::value);
answered Dec 27 '18 at 13:02
songyuanyao
89.9k11170234
2
Ironic really since that's exactly what this gives us! Silly language.
– Lightness Races in Orbit
Dec 27 '18 at 16:42
add a comment |
Here's a workaround from cppreference.com, i.e. use a type-dependent expression instead.
Note: the discarded statement can't be ill-formed for every possible specialization:
The common workaround for such a catch-all statement is a type-dependent expression that is always false:
e.g.
template<class T> struct dependent_false : std::false_type {};
then
static_assert(dependent_false<T>::value);
answered Dec 27 '18 at 13:02
songyuanyao
89.9k11170234
2
Ironic really since that's exactly what this gives us! Silly language.
– Lightness Races in Orbit
Dec 27 '18 at 16:42
add a comment |
Here's a workaround from cppreference.com, i.e. use a type-dependent expression instead.
Note: the discarded statement can't be ill-formed for every possible specialization:
The common workaround for such a catch-all statement is a type-dependent expression that is always false:
e.g.
template<class T> struct dependent_false : std::false_type {};
then
static_assert(dependent_false<T>::value);
answered Dec 27 '18 at 13:02
songyuanyao
89.9k11170234
Here's a workaround from cppreference.com, i.e. use a type-dependent expression instead.
Note: the discarded statement can't be ill-formed for every possible specialization:
The common workaround for such a catch-all statement is a type-dependent expression that is always false:
e.g.
template<class T> struct dependent_false : std::false_type {};
then
static_assert(dependent_false<T>::value);
answered Dec 27 '18 at 13:02
songyuanyao
89.9k11170234
answered Dec 27 '18 at 13:02
songyuanyao
89.9k11170234
answered Dec 27 '18 at 13:02
songyuanyao
89.9k11170234
answered Dec 27 '18 at 13:02
songyuanyao
89.9k11170234
89.9k11170234
2
Ironic really since that's exactly what this gives us! Silly language.
– Lightness Races in Orbit
Dec 27 '18 at 16:42
add a comment |
2
Ironic really since that's exactly what this gives us! Silly language.
– Lightness Races in Orbit
Dec 27 '18 at 16:42
2
2
Ironic really since that's exactly what this gives us! Silly language.
– Lightness Races in Orbit
Dec 27 '18 at 16:42
Ironic really since that's exactly what this gives us! Silly language.
– Lightness Races in Orbit
Dec 27 '18 at 16:42
add a comment |
taking a slightly different tack...
#include <ciso646>
template<auto x> void something();
template<class...Conditions>
constexpr int which(Conditions... cond)
{
int sel = 0;
bool found = false;
auto elect = [&found, &sel](auto cond)
{
if (not found)
{
if (cond)
{
found = true;
}
else
{
++sel;
}
}
};
(elect(cond), ...);
if (not found) throw "you have a logic error";
return sel;
}
template<bool condition1, bool condition2, bool condition3>
void foo()
{
auto constexpr sel = which(condition1, condition2, condition3);
switch(sel)
{
case 0:
something<1>();
break;
case 1:
something<2>();
break;
case 2:
something<3>();
break;
}
}
int main()
{
foo<false, true, false>();
// foo<false, false, false>(); // fails to compile
}
As I understand it, which is evaluated in constexpr context, which means that it is legal unless the program has to follow a code path that is illegal in a constexpr context.
For all expected cases, the throw path is not taken, so the function is legal. When illegal inputs are provided, we go down the ill-formed path, which causes a compiler error.
I'd be interested to know whether this solution is strictly correct from a language-lawyer perspective.
It works on gcc, clang and MSVC.
...or for fans of obfuscated code...
template<class...Conditions>
constexpr int which(Conditions... cond)
{
auto sel = 0;
((cond or (++sel, false)) or ...) or (throw "program is ill-formed", false);
return sel;
}
answered Dec 27 '18 at 17:53
Richard Hodges
55.4k657100
add a comment |
taking a slightly different tack...
#include <ciso646>
template<auto x> void something();
template<class...Conditions>
constexpr int which(Conditions... cond)
{
int sel = 0;
bool found = false;
auto elect = [&found, &sel](auto cond)
{
if (not found)
{
if (cond)
{
found = true;
}
else
{
++sel;
}
}
};
(elect(cond), ...);
if (not found) throw "you have a logic error";
return sel;
}
template<bool condition1, bool condition2, bool condition3>
void foo()
{
auto constexpr sel = which(condition1, condition2, condition3);
switch(sel)
{
case 0:
something<1>();
break;
case 1:
something<2>();
break;
case 2:
something<3>();
break;
}
}
int main()
{
foo<false, true, false>();
// foo<false, false, false>(); // fails to compile
}
As I understand it, which is evaluated in constexpr context, which means that it is legal unless the program has to follow a code path that is illegal in a constexpr context.
For all expected cases, the throw path is not taken, so the function is legal. When illegal inputs are provided, we go down the ill-formed path, which causes a compiler error.
I'd be interested to know whether this solution is strictly correct from a language-lawyer perspective.
It works on gcc, clang and MSVC.
...or for fans of obfuscated code...
template<class...Conditions>
constexpr int which(Conditions... cond)
{
auto sel = 0;
((cond or (++sel, false)) or ...) or (throw "program is ill-formed", false);
return sel;
}
answered Dec 27 '18 at 17:53
Richard Hodges
55.4k657100
add a comment |
taking a slightly different tack...
#include <ciso646>
template<auto x> void something();
template<class...Conditions>
constexpr int which(Conditions... cond)
{
int sel = 0;
bool found = false;
auto elect = [&found, &sel](auto cond)
{
if (not found)
{
if (cond)
{
found = true;
}
else
{
++sel;
}
}
};
(elect(cond), ...);
if (not found) throw "you have a logic error";
return sel;
}
template<bool condition1, bool condition2, bool condition3>
void foo()
{
auto constexpr sel = which(condition1, condition2, condition3);
switch(sel)
{
case 0:
something<1>();
break;
case 1:
something<2>();
break;
case 2:
something<3>();
break;
}
}
int main()
{
foo<false, true, false>();
// foo<false, false, false>(); // fails to compile
}
As I understand it, which is evaluated in constexpr context, which means that it is legal unless the program has to follow a code path that is illegal in a constexpr context.
For all expected cases, the throw path is not taken, so the function is legal. When illegal inputs are provided, we go down the ill-formed path, which causes a compiler error.
I'd be interested to know whether this solution is strictly correct from a language-lawyer perspective.
It works on gcc, clang and MSVC.
...or for fans of obfuscated code...
template<class...Conditions>
constexpr int which(Conditions... cond)
{
auto sel = 0;
((cond or (++sel, false)) or ...) or (throw "program is ill-formed", false);
return sel;
}
answered Dec 27 '18 at 17:53
Richard Hodges
55.4k657100
taking a slightly different tack...
#include <ciso646>
template<auto x> void something();
template<class...Conditions>
constexpr int which(Conditions... cond)
{
int sel = 0;
bool found = false;
auto elect = [&found, &sel](auto cond)
{
if (not found)
{
if (cond)
{
found = true;
}
else
{
++sel;
}
}
};
(elect(cond), ...);
if (not found) throw "you have a logic error";
return sel;
}
template<bool condition1, bool condition2, bool condition3>
void foo()
{
auto constexpr sel = which(condition1, condition2, condition3);
switch(sel)
{
case 0:
something<1>();
break;
case 1:
something<2>();
break;
case 2:
something<3>();
break;
}
}
int main()
{
foo<false, true, false>();
// foo<false, false, false>(); // fails to compile
}
As I understand it, which is evaluated in constexpr context, which means that it is legal unless the program has to follow a code path that is illegal in a constexpr context.
For all expected cases, the throw path is not taken, so the function is legal. When illegal inputs are provided, we go down the ill-formed path, which causes a compiler error.
I'd be interested to know whether this solution is strictly correct from a language-lawyer perspective.
It works on gcc, clang and MSVC.
...or for fans of obfuscated code...
template<class...Conditions>
constexpr int which(Conditions... cond)
{
auto sel = 0;
((cond or (++sel, false)) or ...) or (throw "program is ill-formed", false);
return sel;
}
answered Dec 27 '18 at 17:53
Richard Hodges
55.4k657100
edited Dec 27 '18 at 18:36
answered Dec 27 '18 at 17:53
Richard Hodges
55.4k657100
answered Dec 27 '18 at 17:53
Richard Hodges
55.4k657100
answered Dec 27 '18 at 17:53
Richard Hodges
55.4k657100
55.4k657100
add a comment |
add a comment |
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