Is a constexpr size needed for initialising an array with an explicit size?

Question

C++ Primer (5th edition) states that when an array is initialised with a size, the size provided has to be a constexpr, like this:

int arr[10];

// or

constexpr size_t sz = 10;
int arr[sz];

However, this code compiles just fine for me:

size_t sz = 10;
int arr[sz];

So, is the constexpr really needed?

`size_t sz = 10; int arr[sz];` is illegal. If you are using gcc, compile using `-pedantic-errors` to stop it from compiling. You **always** need a compile time constant for array sizes in C++. — NathanOliver, Apr 01 '20 at 12:43
Note that `const size_t sz = 10;` is also valid, since `constexpr` is relatively new to the C++ language. — ChrisMM, Apr 01 '20 at 12:45

score 2 · Accepted Answer · answered Apr 01 '20 at 12:47

Compiling (g++ -pedantic --std=c++11 -o t t.cpp) in pedantic mode gives:

t.cpp:5:8: warning: variable length arrays are a C99 feature [-Wvla-extension]
int arr[sz];
       ^
1 warning generated.

You may even reject the code with -pedantic-errors.

[C++11: 8.3.4/1]: In a declaration T D where D has the form

D1 [ constant-expressionopt ] attribute-specifier-seqopt [..]

Thus any constant is acceptable, constexpr or const variables.

1 Answers1