Initialiser list passed as function parameter for array

Question

How do I make this work:

void foo(uint8_t a[]) { ... }

foo({0x01, 0x02, 0x03});

It gives me an error:

error: cannot conver         


        

Answer 1

This:

void foo(uint8_t a[]) { ... }

is a function that takes a uint8_t*, not an array - arrays are decayed to pointers when used as function arguments. The issue is that an initializer list (like {0x01, 0x02, 0x03}) cannot be converted to a uint8_t*.

If what you want is to pass an arbitrary number of uint8_ts to foo, the simple solution is to use the new std::initializer_list

void foo(std::initializer_list<uint8_t> a) { ... }

foo({0x01, 0x02, 0x03, 0x04, 0x05}); // OK - a has 5 elems in it

Or you could take a variadic pack and construct an array from it internally:

template <typename... Args,
          typename = std::enable_if_t<
              all_true<std::is_convertible<Args, uint8_t>::value...>
              >>
void foo(Args... elems) {
    uint8_t a[] = {elems...};
    // ...
}

That has slightly different usage:

foo({0x01, 0x02, 0x03}); // error
foo(0x01, 0x02, 0x03; // OK - a has 3 elems in it

Answer 2

foo(std::array<uint8_t, 3>{0x01, 0x02, 0x03}.data());

Answer 3

The answers so far haven't addressed the main problem with the question: In the signature

void foo(uint8_t a[])

a is not an array, but a pointer to a uint8_t. This is despite the fact that the declaration of a makes it look like an array. This is even pointed out by the error message:

cannot convert '<brace-enclosed initializer list>' to 'uint8_t* {aka unsigned char*}'

So, in the same way you are not allowed to do this:

uint8_t *a = {0x01, 0x02, 0x03}; // Eek! Error

You can't call foo({0x01, 0x02, 0x03}); With the signature above.

I suggest you take some time to read up on C-style arrays and how they are not first-class citizens in C++.

From the answer you posted to your own question, it seems that you are looking for a function that works for fixed-size arrays. But don't pass it by value! I recommend using the following declaration:

void foo(std::array<uint8_t, 3> const &a);

Answer 4

You cannot. Just construct

uint8_t a[] = {0x01, 0x02, 0x03};

and call foo(a).

Or just use std::array, that is probably better.

Answer 5

This worked for me, I had to change the function signature but it's actually better in my case as it statically checks the array length:

void foo(std::array<uint8_t, 3> a) { /* use a.data() instead of a */ }

foo({0x01, 0x02, 0x03}); // OK

foo({0x01, 0x02}); // Works, at least on GCC 4.9.1. The third value is set to zero.

foo({0x01, 0x02, 0x03, 0x04}); // Compilation error.