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Old 01-13-2024, 04:48 PM   #1
Kite
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Default C++ question - overlapping arrays?

I need to do something rather unorthodox in C++. Can't find the answer anywhere online.

I need to define an array that overlaps another array, so both arrays read from and write to the same memory locations. Related, I also need to define an array of chars that overlaps an array of strings.

Something like this:
Code:
byte originalArray[256];
byte myArray = originalArray + 128; // an array of bytes, it starts halfway in

if (myArray[0] == myArray[1]) {     // read from originalArray[128] and [129]
  myArray[2] = 3;                   // write to originalArray[130]
}
Do I need to define the length of myArray? If so, how?

2nd example, for strings:
Code:
char originalArray [10][32];        // 10 null-terminated strings, length <= 31
char myArray = originalArray + 128; // an array of chars, not null-terminated

if (myArray[0] == ' ') {            // read/write 1st char of 5th string
  myArray[0] = '!';
}
I know this is not good programming practice. But I absolutely positively have to do it this way! (Long story.)
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Old 01-13-2024, 07:37 PM   #2
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Code:
char * myArray = (char *)originalArray + 128; // an array of chars, not null-terminated
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Old 01-14-2024, 12:22 AM   #3
Kite
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Thanks, will test this out soon!
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Old 01-18-2024, 04:23 AM   #4
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Works great!

Next question, what if I want to access myArray[i][j]? Something like this:

Code:
byte originalArray[200];
byte * myArray[0] = (byte *)originalArray + 100; 
byte * myArray[1] = (byte *)originalArray + 108; 

if (myArray[0][2] == myArray[1][2]) {  // read from originalArray[102] and [110]
  myArray[0][3] = 4;                   // write to originalArray[103]
}
or maybe this:

Code:
byte originalArray[2][200];
byte * myArray[0] = (byte *)originalArray + 100;  

if (myArray[0][2] == myArray[1][2]) {  // read from originalArray[0][102] and [1][102]
  myArray[0][3] = 4;                   // write to originalArray[0][103]
}
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Last edited by Kite; 01-18-2024 at 05:21 AM.
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Old 01-18-2024, 05:45 PM   #5
Kite
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Also, is it possible for a data structure to overlap an array of bytes?

Code:
byte originalArray[200];

struct note {
  byte pitch;       // midi note number
  byte duration;    // measured in 16th notes
};
note * Notes[61] = (byte *)originalArray;

n = 10;
if (Notes[n].pitch > 60) {   // read from originalArray[20]
  Notes[n].duration = 8;     // write to originalArray[21]
}
And is it possible to mix data types?
Code:
byte originalArray[200];
signed char * myArray1 = (byte *)originalArray;
    boolean * myArray2 = (byte *)originalArray + 20;
        int * myArray3 = (byte *)originalArray + 60;
In my use case, I would never ever read from or write to originalArray directly. I would only read/write myArray. So I wouldn't be attempting to interpret a byte as a boolean, or four bytes as an int.

And finally, what about just accessing one element of originalArray as a simple variable, via an alias? Something like:

Code:
byte originalArray[200];
byte * myVar1 = originalArray + 10;
byte * myVar2 = originalArray + 11;

if (myVar1 == 0) {       // read from originalArray [10]
  myVar2 = 3;            // write to originalArray [11]
}
Perhaps there is some general method of making a variable or an array or a data structure (or an array of structures, a structure of arrays, etc. etc.) overlap an array of bytes?
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Old 01-18-2024, 06:21 PM   #6
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Quote:
And is it possible to mix data types?
Code:
byte originalArray[200];
signed char * myArray1 = (byte *)originalArray;
    boolean * myArray2 = (byte *)originalArray + 20;
        int * myArray3 = (byte *)originalArray + 60;
Yes but you need to properly cast like this example.
Rule is that the right side (cast) must match the left side type.

Code:
char            originalArray[200];
signed char   * myArray1 = (signed char *)    originalArray;
bool          * myArray2 = (bool *)           originalArray + 20;
int           * myArray3 = (int  *)           originalArray + 60;
unsigned char * myArray4 = (unsigned char *)  originalArray + 80;
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Old 01-18-2024, 07:01 PM   #7
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Quote:
Code:
byte originalArray[2][200];
byte * myArray[0] = (byte *)originalArray + 100;  

if (myArray[0][2] == myArray[1][2]) {  // read from originalArray[0][102] and [1][102]
  myArray[0][3] = 4;                   // write to originalArray[0][103]
}
That won't work. It won't even compile without error.
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Last edited by jacksoonbrowne; 01-18-2024 at 07:43 PM.
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Old 01-18-2024, 07:25 PM   #8
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I am not sure how familiar you are with c++.

Pointers to objects are only easy to use once you have a good grasp of C++ pointers, casting them, and then using them to access data.

Perhaps it would be better to pose 1 question at a time.
That would make it easier for me to provide a working example and an explanation of it.

I am here to help you as needed,
Cheers,
Roy
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Old 01-18-2024, 08:34 PM   #9
Kite
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Thanks again for your help, Roy! I'm not very familiar with C++, but I am very familiar with jesusonic. I find I can mostly teach myself C++ from online tutorials. That doesn't work when I'm doing something this unorthodox, of course.

So, as you say, one question at a time. Can I set up a 2D array that overlaps a 1D array? Obviously this code wouldn't work:

Code:
byte originalArray[200];
byte * myArray = (byte *)originalArray + 100; 

if (myArray[0][2] == myArray[1][2]) {  // read from originalArray[102] and [110]
  myArray[0][3] = 4;                   // write to originalArray[103]
}
Because the compiler doesn't know how long the rows of myArray are. Supposing I want rows of 8 bytes, how would I declare myArray?
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Old 01-18-2024, 09:05 PM   #10
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Quote:
Originally Posted by Kite View Post
Thanks again for your help, Roy! I'm not very familiar with C++, but I am very familiar with jesusonic. I find I can mostly teach myself C++ from online tutorials. That doesn't work when I'm doing something this unorthodox, of course.

So, as you say, one question at a time. Can I set up a 2D array that overlaps a 1D array? Obviously this code wouldn't work:

Code:
byte originalArray[200];
byte * myArray = (byte *)originalArray + 100; 

if (myArray[0][2] == myArray[1][2]) {  // read from originalArray[102] and [110]
  myArray[0][3] = 4;                   // write to originalArray[103]
}
Because the compiler doesn't know how long the rows of myArray are. Supposing I want rows of 8 bytes, how would I declare myArray?


I did this a few years ago in my production code.
Need to refresh my memory and look at my old code.

Standby ...
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Old 01-18-2024, 09:40 PM   #11
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Ok,

Here's an example of using a 2d array pointing into a 1d array

Code:
byte   originalArray[200];
byte  (*myArray)[2][100]   = (byte(*)[2][100]) (originalArray + 16);

if ((*myArray)[0][2] == (*myArray)[1][2]) {  // read from originalArray[102] and [110]
	(*myArray)[0][3] = 4;                   // write to originalArray[103]
}
Beware that if you use indices are outside the boundaries of either the 1d or 2d arrays you will get a memory violation.

For example: if you reference the following (based on the above example) you will get a memory violation:
Code:
byte x = (*myArray)[2][101]
I
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Old 01-19-2024, 05:03 PM   #12
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Great! Next question, how would I make a data structure overlap an array of bytes?

Code:
byte originalArray[400];

struct note {
  byte pitch;       // midi note number
  int duration;     // measured in milliseconds
};
note myNote;

(somehow make myNote start on originalArray[10])

if (myNote.pitch > 60) {   // read from originalArray[10]
  myNote.duration = 800;   // write to originalArray[11-44]
}
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Old 01-21-2024, 03:15 AM   #13
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Quote:
Originally Posted by Kite View Post
Great! Next question, how would I make a data structure overlap an array of bytes?
Well, you could do something like:

Code:
byte originalArray[400];

struct note {
  byte pitch;       // midi note number
  int duration;     // measured in milliseconds
};

note* myNote = (note*)&originalArray[10];

if (myNote->pitch > 60) {   // read from originalArray[10]
  myNote->duration = 800;   // write to originalArray[14]
}
However, depending on your CPU architecture you might not want to do that, unless you are sure that originalArray is properly aligned. Standard C/C++ rules say that the int type should be 4-byte aligned, and for some architectures (like ARM) this actually is important, but for others (e.g. x86) it doesn't really matter.
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Old 01-21-2024, 10:40 PM   #14
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The code runs on an arduino.
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Old 01-22-2024, 12:28 AM   #15
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I don't know what the alignment is on an Arduino. A quick Google search says that the int type could be either 16 or 32-bit (i.e. 2 or 4 bytes), further complicating things. Maybe it would be better to allocate your array as type note, and then cast it to byte when needed i.e.:

Code:
struct note {
  byte pitch;       // midi note number
  int duration;     // measured in milliseconds
};

note originalArray[400 / sizeof(note)]; // 100 or 50 elements, depending on sizeof(int)
byte* byteArray = (byte*)&originalArray[0]; // 400 elements

note* myNote = &note[3]; // Same address as byteArray[12] or [24]

if (myNote->pitch > 60) {
  myNote->duration = 800;
}
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Old 01-22-2024, 03:15 AM   #16
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Quote:
Originally Posted by Kite View Post
The code runs on an arduino.
I am very familiar with Arduino

Which board are you using?
Uno, Mega, ESP32 ... etc?
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Old 01-22-2024, 04:53 PM   #17
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Arduino Due ARM chip.
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Old 01-22-2024, 10:29 PM   #18
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Quote:
Originally Posted by Kite View Post
Arduino Due ARM chip.
"int" on DUE is 32 bit.

@Tale provided a viable option
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Old 01-23-2024, 07:00 AM   #19
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Making sure I understand, I think this code will work?

Code:
maxStringLen = 30;           // including the null at the end
char originalArray[10][maxStringLen];  

struct note {
  byte pitch;                // midi note number
  byte duration;             // in 16th notes
  signed char tuning;        // range is ±50 cents
  boolean hasStem[2];        // [0] = up-stem, [1] = down-stem
};

// overlap starting on the 20th char of the 3rd message
note* myNote = (note*) &originalArray + 2*maxStringLen + 20;

if (myNote->pitch > 60) {    // read from originalArray[80]
  myNote->duration = 8;      // write to originalArray[81]
}

BTW, sorry for the newbie question, but what does "->" mean? I can't manage to google it haha. Does myNote->pitch mean the same as myNote.pitch?
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Old 01-23-2024, 08:54 AM   #20
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No, for starters both originalArray[80] and myNote are out of bounds and originalArray[81] is 30 bytes after [80]. originalArray[80] is 2400 bytes after the start of originalArray and myNote is 400 bytes after the start. (note*)&originalArray+n adds n*sizeof(note) bytes.

What do you need this for exactly? What are the real input and expected output? This sounds like a XY problem.

Tale's suggestion of having an array of notes for working with then casting it for viewing is way better. But is seems like you need each note to be null-terminated?


-> is the "member of pointer" operator. myNote->pitch is equivalent to (*myNote).pitch and myNote.pitch is invalid. (It can also be overloaded to implement it on non-pointer objects.)

Last edited by cfillion; 01-23-2024 at 01:33 PM.
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