Try this on your computer, not in the Arduino but it's the same language:
typedef unsigned char byte;
void sendMsg(byte msgString)
// print the address of the things, is it a copy? no, it's just a pointer!
printf("Address of the array in sendMsg: %x\n", msgString);
// first test: print the address of things
printf("Address of the array in main: %x\n", things);
// second test
The program will print the same address!
Address of the array in main: bffff70c
Address of the array in sendMsg: bffff70c
In reality, when you declare void sendMsg(byte msgString)
, msgString is a pointer to the array sent as a parameter of the function, nothing is copied, but all the values are accessible in your function because it's the original array address that is being given.
Be very careful though: the C and the C++ do not keep the "length" of the array (given as a parameter, or inside the main function), you have
to know it, otherwise you'll get a buffer overflow somewhere.
- there is no copy (or it has to be explicite with functions like memcpy)
- you have to be careful not to access values outside the array (keep the length, and the largest index is (length-1))
- this was not a stupid question, it's actually a problem for a lot of programmers