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When passing an array to a function, you usually pass the length of the array along with it. You can compute the size of the array with sizeof(myArray)/sizeof(myArray[0]), which works just fine.

However, if you try to do this within the function that you pass the array to, this doesn't return the length of the array - eg. if it is an int array, it returns 1, if it is a double array, it returns 0.

Why is that so and is there any way to compute the size of an array that was passed as a parameter?

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  • If you don't use it often (every different type/size makes new instantiation od f function), then you can use templates like: template <typename T, size_t N> void f(T(& myArray)[N])
    – KIIV
    Commented Nov 12, 2019 at 11:44

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You can't. You are passing a pointer to the array, and sizeof will give you the size of that pointer (2 on an 8-bit Arduino).

Instead the normal way is to pass the size of the array as a parameter to your function.

myFunction(myArray, sizeof(myArray) / sizeof(myArray[0]));
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