CODEWITHSUNDEEP
c++bitmaparduinoavr
Tomir Rogi Szawok
Tomir Rogi Szawok

Reputation: 609

Simple bitmap rotation for arduino, c++

I'm trying to achieve compromise for my app, but got no luck (or rather knowledge) so far.

I've got bitmap for black-white screen, it looks like this (I use arduino byte style, because it's more readable)

{
    B00111100, B01001000,
    B00100100, B01010000,
    B00111100, B01110000,
    B00100100, B01001000
}

It is array of bytes, each byte representing 8 next horizontal pixels. Problem is that I have to use bitmap, where each byte represents 8 next vertical pixels, so it's like turning it this way

{
    B00000000,
    B00000000,
    B11110000,
    B10100000,
    B11110000,
    B00000000,

    B11110000,
    B00100000,
    B01100000,
    B10010000
}

I tried, but ended up completely without any idea how to do this.

Edit. I could be misunderstod, so I added brackets to code, It's more clear right now.

Upvotes: 1

Views: 792

Answers (1)

Jacques Supcik
Jacques Supcik

Reputation: 659

Here is an example using plain C (gcc):

#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>


typedef uint8_t byte;

void print_bin(byte x) {
    printf("B");
    for (int i = 0; i < 8; i++) {
        printf("%s", (x >> (7-i)) % 2 ? "1" : "0");
    }
    printf("\n");
}

void reverse(byte* in, byte* out, int width, int height) {
    int width_bytes = (width + 7) / 8;
    int height_bytes = (height + 7) / 8;
    // init *out. You can skip the next line if you are sure that *out is clear.
    memset (out, 0, width * height_bytes);
    for (int y = 0; y < height; y++) {
        for (int x = 0; x < width; x++) {
            if (in[(y * width_bytes + x / 8)] & (1 << (7 - x % 8))) {
                out[(x * height_bytes + y / 8)] |= (1 << (7 - y % 8));
            }
        }
    }
}

#define WIDTH  13
#define HEIGHT  4
#define IN_SIZE  (((WIDTH + 7) / 8) * HEIGHT)
#define OUT_SIZE (((HEIGHT + 7) / 8) * WIDTH)

int main() {
    byte in[IN_SIZE] = {
        0b00111100, 0b01001000,
        0b00100100, 0b01010000,
        0b00111100, 0b01110000,
        0b00100100, 0b01001000
    };

    byte* out = calloc(OUT_SIZE, 1);
    reverse (in, out, WIDTH, HEIGHT);
    for (int i = 0; i < OUT_SIZE; i++) {
        print_bin(out[i]);
    }

}

And this is the result:

B00000000
B00000000
B11110000
B10100000
B10100000
B11110000
B00000000
B00000000
B00000000
B11110000
B00100000
B01100000
B10010000

If speed is an issue, you can do the following optimisation:

void reverse(byte* in, byte* out, int width, int height) {
    int width_bytes = (width + 7) / 8;
    int height_bytes = (height + 7) / 8;
    // init *out. You can skip the next line if you are sure that *out is clear.
    memset (out, 0, width * height_bytes);
    for (int y = 0; y < height; y++) {
        for (int x = 0; x < width; x++) {
            int t; // optimisation
            if ((x % 8) == 0) t = in[(y * width_bytes + x / 8)];
            if (t & (1 << (7 - x % 8))) {
                out[(x * height_bytes + y / 8)] |= (1 << (7 - y % 8));
            }
        }
    }
}

Upvotes: 3

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