constexpr function contains a const - will it ever be evaluated at compile time?
Please can someone clarify (I'm using Visual Studio 15.9.2):
In the following code given that Pi_cnst is evaluated at run time (because defining Pi in this way requires a run time calculation), will RAD2DEG_cnst ever be evaluated at compile time or is the use of "constexpr" always reverting to "const" ?
edit - add: If it is always reverting to const then should I expect a warning, or is it bad in some other way i.e. it seems odd to be able to so easily declare a constexpr for the body to be accepted but ALWAYS be such that it is never actually a constexpr. What have I missed?
constexpr double Pi_error = std::acos(-1.0); //error function call must have a constant value in a constant expression (does not compile)
const double Pi_cnst = std::acos(-1.0); //ok evaluated at run time
constexpr double Pi_expr = 3.1415926; //ok valid constexpr
constexpr double RAD2DEG_cnst(double rad) { return rad * 180.0 / Pi_cnst; } //no error or warning BUT is this ever evaluated at compile time?
constexpr double RAD2DEG_expr(double rad) { return rad * 180.0 / Pi_expr; } //ok can be evaluated at compile time
const double result1 = RAD2DEG_cnst(0.1); //evaluated at run time?
const double result2 = RAD2DEG_expr(0.2); //evaluated at compile time?
double myVariable = 0.3;
const double result3 = RAD2DEG_expr(myVariable); //ok - but can only be evaluated at run time
const double myOtherVariable = 0.4;
const double result4 = RAD2DEG_expr(myOtherVariable); //ok - evaluated at compile time because parameter is CONST?
I found also constexpr function and its parameter and constexpr vs const vs constexpr const .
c++ constexpr
asked Nov 21 '18 at 15:15
Ben SeniorBen Senior
85
|
show 1 more comment
Please can someone clarify (I'm using Visual Studio 15.9.2):
In the following code given that Pi_cnst is evaluated at run time (because defining Pi in this way requires a run time calculation), will RAD2DEG_cnst ever be evaluated at compile time or is the use of "constexpr" always reverting to "const" ?
edit - add: If it is always reverting to const then should I expect a warning, or is it bad in some other way i.e. it seems odd to be able to so easily declare a constexpr for the body to be accepted but ALWAYS be such that it is never actually a constexpr. What have I missed?
constexpr double Pi_error = std::acos(-1.0); //error function call must have a constant value in a constant expression (does not compile)
const double Pi_cnst = std::acos(-1.0); //ok evaluated at run time
constexpr double Pi_expr = 3.1415926; //ok valid constexpr
constexpr double RAD2DEG_cnst(double rad) { return rad * 180.0 / Pi_cnst; } //no error or warning BUT is this ever evaluated at compile time?
constexpr double RAD2DEG_expr(double rad) { return rad * 180.0 / Pi_expr; } //ok can be evaluated at compile time
const double result1 = RAD2DEG_cnst(0.1); //evaluated at run time?
const double result2 = RAD2DEG_expr(0.2); //evaluated at compile time?
double myVariable = 0.3;
const double result3 = RAD2DEG_expr(myVariable); //ok - but can only be evaluated at run time
const double myOtherVariable = 0.4;
const double result4 = RAD2DEG_expr(myOtherVariable); //ok - evaluated at compile time because parameter is CONST?
I found also constexpr function and its parameter and constexpr vs const vs constexpr const .
c++ constexpr
asked Nov 21 '18 at 15:15
Ben SeniorBen Senior
85
What's the question? You saw thatacosdidn't work within a constexpr.
– Matthieu Brucher
Nov 21 '18 at 15:17
@MatthieuBrucher:Pi_cnstcannot beconstexprbecause ofacosbutRAD2DEG_cnstisconstexpralthough it relies onPi_cnst. I believe the question is: how can it be declared asconstexprwhile performing unconstexprable operations, and what does it mean in practice about compile-time evaluation?
– papagaga
Nov 21 '18 at 15:21
Oh, OK. I think constexpr doesn't always been a build time...
– Matthieu Brucher
Nov 21 '18 at 15:23
I get an error forRAD2DEG_cnstwith gcc9 though: "In function 'constexpr double RAD2DEG_cnst(double)': error: the value of 'Pi_cnst' is not usable in a constant expression"
– papagaga
Nov 21 '18 at 15:23
1
Possible duplicate of When does a constexpr function get evaluated at compile time?
– Matthieu Brucher
Nov 21 '18 at 15:23
|
show 1 more comment
Please can someone clarify (I'm using Visual Studio 15.9.2):
In the following code given that Pi_cnst is evaluated at run time (because defining Pi in this way requires a run time calculation), will RAD2DEG_cnst ever be evaluated at compile time or is the use of "constexpr" always reverting to "const" ?
edit - add: If it is always reverting to const then should I expect a warning, or is it bad in some other way i.e. it seems odd to be able to so easily declare a constexpr for the body to be accepted but ALWAYS be such that it is never actually a constexpr. What have I missed?
constexpr double Pi_error = std::acos(-1.0); //error function call must have a constant value in a constant expression (does not compile)
const double Pi_cnst = std::acos(-1.0); //ok evaluated at run time
constexpr double Pi_expr = 3.1415926; //ok valid constexpr
constexpr double RAD2DEG_cnst(double rad) { return rad * 180.0 / Pi_cnst; } //no error or warning BUT is this ever evaluated at compile time?
constexpr double RAD2DEG_expr(double rad) { return rad * 180.0 / Pi_expr; } //ok can be evaluated at compile time
const double result1 = RAD2DEG_cnst(0.1); //evaluated at run time?
const double result2 = RAD2DEG_expr(0.2); //evaluated at compile time?
double myVariable = 0.3;
const double result3 = RAD2DEG_expr(myVariable); //ok - but can only be evaluated at run time
const double myOtherVariable = 0.4;
const double result4 = RAD2DEG_expr(myOtherVariable); //ok - evaluated at compile time because parameter is CONST?
I found also constexpr function and its parameter and constexpr vs const vs constexpr const .
c++ constexpr
asked Nov 21 '18 at 15:15
Ben SeniorBen Senior
85
Please can someone clarify (I'm using Visual Studio 15.9.2):
In the following code given that Pi_cnst is evaluated at run time (because defining Pi in this way requires a run time calculation), will RAD2DEG_cnst ever be evaluated at compile time or is the use of "constexpr" always reverting to "const" ?
edit - add: If it is always reverting to const then should I expect a warning, or is it bad in some other way i.e. it seems odd to be able to so easily declare a constexpr for the body to be accepted but ALWAYS be such that it is never actually a constexpr. What have I missed?
constexpr double Pi_error = std::acos(-1.0); //error function call must have a constant value in a constant expression (does not compile)
const double Pi_cnst = std::acos(-1.0); //ok evaluated at run time
constexpr double Pi_expr = 3.1415926; //ok valid constexpr
constexpr double RAD2DEG_cnst(double rad) { return rad * 180.0 / Pi_cnst; } //no error or warning BUT is this ever evaluated at compile time?
constexpr double RAD2DEG_expr(double rad) { return rad * 180.0 / Pi_expr; } //ok can be evaluated at compile time
const double result1 = RAD2DEG_cnst(0.1); //evaluated at run time?
const double result2 = RAD2DEG_expr(0.2); //evaluated at compile time?
double myVariable = 0.3;
const double result3 = RAD2DEG_expr(myVariable); //ok - but can only be evaluated at run time
const double myOtherVariable = 0.4;
const double result4 = RAD2DEG_expr(myOtherVariable); //ok - evaluated at compile time because parameter is CONST?
I found also constexpr function and its parameter and constexpr vs const vs constexpr const .
c++ constexpr
c++ constexpr
asked Nov 21 '18 at 15:15
Ben SeniorBen Senior
85
asked Nov 21 '18 at 15:15
Ben SeniorBen Senior
85
edited Nov 21 '18 at 15:21
Ben Senior
asked Nov 21 '18 at 15:15
Ben SeniorBen Senior
85
asked Nov 21 '18 at 15:15
Ben SeniorBen Senior
85
asked Nov 21 '18 at 15:15
Ben SeniorBen Senior
85
85
What's the question? You saw thatacosdidn't work within a constexpr.
– Matthieu Brucher
Nov 21 '18 at 15:17
@MatthieuBrucher:Pi_cnstcannot beconstexprbecause ofacosbutRAD2DEG_cnstisconstexpralthough it relies onPi_cnst. I believe the question is: how can it be declared asconstexprwhile performing unconstexprable operations, and what does it mean in practice about compile-time evaluation?
– papagaga
Nov 21 '18 at 15:21
Oh, OK. I think constexpr doesn't always been a build time...
– Matthieu Brucher
Nov 21 '18 at 15:23
I get an error forRAD2DEG_cnstwith gcc9 though: "In function 'constexpr double RAD2DEG_cnst(double)': error: the value of 'Pi_cnst' is not usable in a constant expression"
– papagaga
Nov 21 '18 at 15:23
1
Possible duplicate of When does a constexpr function get evaluated at compile time?
– Matthieu Brucher
Nov 21 '18 at 15:23
|
show 1 more comment
What's the question? You saw thatacosdidn't work within a constexpr.
– Matthieu Brucher
Nov 21 '18 at 15:17
@MatthieuBrucher:Pi_cnstcannot beconstexprbecause ofacosbutRAD2DEG_cnstisconstexpralthough it relies onPi_cnst. I believe the question is: how can it be declared asconstexprwhile performing unconstexprable operations, and what does it mean in practice about compile-time evaluation?
– papagaga
Nov 21 '18 at 15:21
Oh, OK. I think constexpr doesn't always been a build time...
– Matthieu Brucher
Nov 21 '18 at 15:23
I get an error forRAD2DEG_cnstwith gcc9 though: "In function 'constexpr double RAD2DEG_cnst(double)': error: the value of 'Pi_cnst' is not usable in a constant expression"
– papagaga
Nov 21 '18 at 15:23
1
Possible duplicate of When does a constexpr function get evaluated at compile time?
– Matthieu Brucher
Nov 21 '18 at 15:23
What's the question? You saw that
acosdidn't work within a constexpr.– Matthieu Brucher
Nov 21 '18 at 15:17
What's the question? You saw that
acosdidn't work within a constexpr.– Matthieu Brucher
Nov 21 '18 at 15:17
@MatthieuBrucher:
Pi_cnst cannot be constexpr because of acos but RAD2DEG_cnst is constexpr although it relies on Pi_cnst. I believe the question is: how can it be declared as constexpr while performing unconstexprable operations, and what does it mean in practice about compile-time evaluation?– papagaga
Nov 21 '18 at 15:21
@MatthieuBrucher:
Pi_cnst cannot be constexpr because of acos but RAD2DEG_cnst is constexpr although it relies on Pi_cnst. I believe the question is: how can it be declared as constexpr while performing unconstexprable operations, and what does it mean in practice about compile-time evaluation?– papagaga
Nov 21 '18 at 15:21
Oh, OK. I think constexpr doesn't always been a build time...
– Matthieu Brucher
Nov 21 '18 at 15:23
Oh, OK. I think constexpr doesn't always been a build time...
– Matthieu Brucher
Nov 21 '18 at 15:23
I get an error for
RAD2DEG_cnst with gcc9 though: "In function 'constexpr double RAD2DEG_cnst(double)': error: the value of 'Pi_cnst' is not usable in a constant expression"– papagaga
Nov 21 '18 at 15:23
I get an error for
RAD2DEG_cnst with gcc9 though: "In function 'constexpr double RAD2DEG_cnst(double)': error: the value of 'Pi_cnst' is not usable in a constant expression"– papagaga
Nov 21 '18 at 15:23
1
1
Possible duplicate of When does a constexpr function get evaluated at compile time?
– Matthieu Brucher
Nov 21 '18 at 15:23
Possible duplicate of When does a constexpr function get evaluated at compile time?
– Matthieu Brucher
Nov 21 '18 at 15:23
|
show 1 more comment
1 Answer
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Your code is ill-formed, no diagnostic required. In
constexpr double RAD2DEG_cnst(double rad) { return rad * 180.0 / Pi_cnst; }
There is no input which will ever make the function a core constant expression which is specified by [dcl.constexpr]/5
For a constexpr function or constexpr constructor that is neither defaulted nor a template, if no argument values exist such that an invocation of the function or constructor could be an evaluated subexpression of a core constant expression, or, for a constructor, a constant initializer for some object ([basic.start.static]), the program is ill-formed, no diagnostic required.
answered Nov 21 '18 at 15:23
NathanOliverNathanOliver
95.6k16135208
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1 Answer
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Your code is ill-formed, no diagnostic required. In
constexpr double RAD2DEG_cnst(double rad) { return rad * 180.0 / Pi_cnst; }
There is no input which will ever make the function a core constant expression which is specified by [dcl.constexpr]/5
For a constexpr function or constexpr constructor that is neither defaulted nor a template, if no argument values exist such that an invocation of the function or constructor could be an evaluated subexpression of a core constant expression, or, for a constructor, a constant initializer for some object ([basic.start.static]), the program is ill-formed, no diagnostic required.
answered Nov 21 '18 at 15:23
NathanOliverNathanOliver
95.6k16135208
add a comment |
Your code is ill-formed, no diagnostic required. In
constexpr double RAD2DEG_cnst(double rad) { return rad * 180.0 / Pi_cnst; }
There is no input which will ever make the function a core constant expression which is specified by [dcl.constexpr]/5
For a constexpr function or constexpr constructor that is neither defaulted nor a template, if no argument values exist such that an invocation of the function or constructor could be an evaluated subexpression of a core constant expression, or, for a constructor, a constant initializer for some object ([basic.start.static]), the program is ill-formed, no diagnostic required.
answered Nov 21 '18 at 15:23
NathanOliverNathanOliver
95.6k16135208
add a comment |
Your code is ill-formed, no diagnostic required. In
constexpr double RAD2DEG_cnst(double rad) { return rad * 180.0 / Pi_cnst; }
There is no input which will ever make the function a core constant expression which is specified by [dcl.constexpr]/5
For a constexpr function or constexpr constructor that is neither defaulted nor a template, if no argument values exist such that an invocation of the function or constructor could be an evaluated subexpression of a core constant expression, or, for a constructor, a constant initializer for some object ([basic.start.static]), the program is ill-formed, no diagnostic required.
answered Nov 21 '18 at 15:23
NathanOliverNathanOliver
95.6k16135208
Your code is ill-formed, no diagnostic required. In
constexpr double RAD2DEG_cnst(double rad) { return rad * 180.0 / Pi_cnst; }
There is no input which will ever make the function a core constant expression which is specified by [dcl.constexpr]/5
For a constexpr function or constexpr constructor that is neither defaulted nor a template, if no argument values exist such that an invocation of the function or constructor could be an evaluated subexpression of a core constant expression, or, for a constructor, a constant initializer for some object ([basic.start.static]), the program is ill-formed, no diagnostic required.
answered Nov 21 '18 at 15:23
NathanOliverNathanOliver
95.6k16135208
answered Nov 21 '18 at 15:23
NathanOliverNathanOliver
95.6k16135208
answered Nov 21 '18 at 15:23
NathanOliverNathanOliver
95.6k16135208
answered Nov 21 '18 at 15:23
NathanOliverNathanOliver
95.6k16135208
95.6k16135208
add a comment |
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What's the question? You saw that
acosdidn't work within a constexpr.– Matthieu Brucher
Nov 21 '18 at 15:17
@MatthieuBrucher:
Pi_cnstcannot beconstexprbecause ofacosbutRAD2DEG_cnstisconstexpralthough it relies onPi_cnst. I believe the question is: how can it be declared asconstexprwhile performing unconstexprable operations, and what does it mean in practice about compile-time evaluation?– papagaga
Nov 21 '18 at 15:21
Oh, OK. I think constexpr doesn't always been a build time...
– Matthieu Brucher
Nov 21 '18 at 15:23
I get an error for
RAD2DEG_cnstwith gcc9 though: "In function 'constexpr double RAD2DEG_cnst(double)': error: the value of 'Pi_cnst' is not usable in a constant expression"– papagaga
Nov 21 '18 at 15:23
1
Possible duplicate of When does a constexpr function get evaluated at compile time?
– Matthieu Brucher
Nov 21 '18 at 15:23