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](https://en.cppreference.com/w/cpp/language/constant_expression) can be assigned to a `constexpr` variable. — NathanOliver, Aug 19 '20 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, Aug 19 '20 at 14:05
You can't , please look in cppreference (https://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, Aug 19 '20 at 14:18
AdabH, I just did. Wow.. things have changed a lot since 2017... — Sj L, Aug 19 '20 at 14:34
Can you please tell us what problem you are trying to solve by converting a dynamic value to constexpr? — Kit., Aug 19 '20 at 14:34
I am intending to use a switch case and create a a constexpr for each case similar to : https://stackoverflow.com/questions/16388510/evaluate-a-string-with-a-switch-in-c/16388594 except in my case, I want to use integers instead of strings for each case. — Sj L, Aug 19 '20 at 14:39
I don't see anything similar there... no `constexpr` for sure. Why do you need `constexpr` for your switch? — Kit., Aug 19 '20 at 15:01
@Kit. I am referring to: constexpr unsigned int str2int(const char* str, int h = 0) { return !str[h] ? 5381 : (str2int(str, h+1) * 33) ^ str[h]; } switch (str2int(s)) { case str2int("Value1"): break; case str2int("Value2"): break; } — Sj L, Aug 19 '20 at 16:08
[Works for me](https://godbolt.org/z/nq1aar). What's your problem, again? — Kit., Aug 19 '20 at 16:47
@Kit. ,I was able to modify it such that constexpr unsigned int str2int(const char* str, int h = 0) { return !str[h] ? 5381 : (str2int(str, h+1) * 33) ^ str[h]; } can take an array or vector of strings and used the str2int to set the case. This worked, but not when I switched integers. — Sj L, Aug 19 '20 at 17:11
[Still works](https://godbolt.org/z/bEbjYK). Can you update your question with the example of code that works and the example of code that doesn't? Otherwise it's completely unknown what you are trying to achieve, as I hope you already understand that you cannot use case labels with values that aren't known to the compiler at compile time. — Kit., Aug 19 '20 at 17:54
Let us [continue this discussion in chat](https://chat.stackoverflow.com/rooms/220106/discussion-between-sj-l-and-kit). — Sj L, Aug 19 '20 at 18:36

max66 · Accepted Answer · 2020-08-19T14:30:24.357

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 '20 at 14:30

answered Aug 19 '20 at 14:24

max66

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@idclev463035818 - Why not? Done. – max66 Aug 19 '20 at 14:31
Okay, that looks quite promising. Is there a way to generalize to a[i], or does the index have to be well defined? – Sj L Aug 19 '20 at 14:32
@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. – max66 Aug 19 '20 at 14:51
I see, do you have a suggestion on how to approach this if I want iterate this in a for loop? – Sj L Aug 19 '20 at 14:53
1

@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). – max66 Aug 19 '20 at 14:59

Rohan Bari · Answer 2 · 2020-08-19T14:31:11.340

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 '20 at 14:31

answered Aug 19 '20 at 14:08

Rohan Bari

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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. – einpoklum Aug 19 '20 at 14:20

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

2 Answers2