C++: cast array/vector integer element to constexpr

Question

I couldn't really find a solution online anywhere for this. I am a bit of a dinosaur from before C++11 and I couldn't figure out typecasting a constexpr.

Does anyone know how to convert a C-style array and/or a std::vector (integer) element into a constexpr?

That is to say, let's say

int a[]={1,2};
vector<int> v={1,2};

how would I convert a[1] and v[1] into constexpr?

constexpr int b=a[1];

for example, a compiler would complain in a for loop.

error: the value of ‘a’ is not usable in a constant expression
     constexpr int b=a[i];

I am pretty much out of ideas for the time being. Thanks

You can't. Only constant expressions can be assigned to a constexpr variable. — NathanOliver
– NathanOliver, Commented Aug 19, 2020 at 14:02
you'd need a timemachine to turn a value that is only known at runtime into one that is already known at compile-time — 463035818_is_not_an_ai
– 463035818_is_not_an_ai, Commented Aug 19, 2020 at 14:05
You can't , please look in cppreference (en.cppreference.com/w/cpp/language/constexpr ) "A constexpr variable must satisfy the following requirements: its type must be a LiteralType. it must be immediately initialized the full-expression of its initialization, including all implicit conversions, constructors calls, etc, must be a constant expression " — AdabH
– AdabH, Commented Aug 19, 2020 at 14:18
AdabH, I just did. Wow.. things have changed a lot since 2017... — Sj L
– Sj L, Commented Aug 19, 2020 at 14:34
Can you please tell us what problem you are trying to solve by converting a dynamic value to constexpr? — Kit.
– Kit., Commented Aug 19, 2020 at 14:34

max66 · Accepted Answer · 2020-08-19 14:30:24Z

3

Suggestion: if you can use at least C++14, instead of std::vector, if you can, use a std::array

Declaring it constexpr

#include <array>

int main () 
 {
   constexpr std::array<int, 2u> a {1, 2};

   constexpr auto b = a[1];
 }

std::array is a type compatible with constexpr and so it's operator[]() (const version), or also it's at() method (const version).

C++14 is required because in C++11 std::array::operator[]() const and std::array::at() const aren't constexpr methods so can't be used in a constant expression.

Unfortunately, std::vector require memory allocation so isn't compatible (before C++20) with constexpr.

For the C-style array case, you have only to declare it constexpr

int main () 
 {
// VVVVVVVVV   
   constexpr int a[] = {1,2};

   constexpr auto b = a[1];
 }

edited Aug 19, 2020 at 14:30

answered Aug 19, 2020 at 14:24

max66

67k11 gold badges76 silver badges121 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

max66 Over a year ago

@idclev463035818 - Why not? Done.

Sj L Over a year ago

Okay, that looks quite promising. Is there a way to generalize to a[i], or does the index have to be well defined?

max66 Over a year ago

@SjL - Obviously the index has to be a constant expression too; so constexpr auto index = 0u; constexpr auto b = a[index]; works, auto index = 0u; constexpr auto b = a[index]; doesn't.

Sj L Over a year ago

I see, do you have a suggestion on how to approach this if I want iterate this in a for loop?

max66 Over a year ago

@SjL - If you just know the use of variadic templates... I usually pass through an helper function that accept a std::index_sequence (that contain indexes as template values, so a sequence of constant expressions) and calling it using std::make_index_sequence (that generate the template sequence of indexes).

Rohan Bari · Accepted Answer · 2020-08-19 14:31:11Z

0

The constexpr must have a constant expression, i.e. they must be known at compile time. You can't perform an assignment to b with a, it's an error in this case.

In the given code snippet, the a isn't defined as constexpr, thus, it doesn't explicitly tells the compiler that the expression is a constant.

edited Aug 19, 2020 at 14:31

answered Aug 19, 2020 at 14:08

Rohan Bari

7,7393 gold badges18 silver badges42 bronze badges

1 Comment

einpoklum Over a year ago

But a is "known at compile time", or rather we know what it holds, at compile time. Please explain to OP why this is not enough.

Collectives™ on Stack Overflow

C++: cast array/vector integer element to constexpr

2 Answers 2

5 Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

5 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related