memcpy - taking the address of temporary array error

Question

I am working on Arduino and trying to change the elements of an array. Before setup, I initialized the array like this:

bool updateArea[5] = { false };

And then I wanted to change the array like this:

updateArea[0] => false,
updateArea[1] => true,
updateArea[2] => false,
updateArea[3] => false,
updateArea[4] => true

by using:

memcpy(&updateArea[0], (bool []) {false, true, false, false, true}, 5);

However, I get the "taking address of temporary array" error.

I also tried to initialize the array in setup and loop functions but get the same error.

Does this answer your question? C++ error: "taking address of temporary array" — Juraj
– Juraj, Commented Sep 8, 2021 at 9:44
what is your compiler? "compound literals" are not temporaries but rather unnamed local/global variables. Is it a compiler bug? — tstanisl
– tstanisl, Commented Sep 8, 2021 at 10:09

Daniel Kleinstein · Accepted Answer · 2021-09-08 11:03:44Z

6

This sort of syntax is valid in C, but not in C++ - which is the language underlying the Arduino IDE.

But you have a few easy solutions:

Since you're willing to write out the array anyways, why not just:
```
bool updateArea[5] = {false, true, false, false, true};
```

You can declare the array as a non-temporary array and then pass it to memcpy:

static const bool newArray[5] = {false, true, false, false, true};
memcpy(updateArea, newArray, sizeof(updateArea));

If you can assume that sizeof(bool) == 1, then you can use this hacky solution:
```
memcpy(updateArea, "\x00\x01\x00\x00\x01", sizeof(updateArea));
```
which will copy the bytes directly.

edited Sep 8, 2021 at 11:03

answered Sep 8, 2021 at 9:45

Daniel Kleinstein

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1 Comment

Stack Exchange Broke The Law Over a year ago

probably better to declare newArray as static const so the code generation is (presumably) more efficient. Otherwise the compiler may initialize this temporary variable every time the function is called and then copy it

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