Convert image to RGBA for libnx

[Cid2mizard]

Hello,

I want to convert a jpeg to rgba for libnx but i have a bug...

I'm on Windows 10 64bits, i have test with ImageMagick "magick convert image.jpg image.rgba", next step rename image.rgba to image.bin, but the display is bad...
I've test with imagemagick Q16-x64-dll & Q8-x64-dll but same result...

What is the good way for convert image for libnx ?

Code: Select all

#include <string.h>
#include <stdio.h>
#include <switch.h>

#include "image_bin.h"//Your own raw RGB888 1280x720 image at "data/image.bin" is required.

//See also libnx gfx.h.

int main(int argc, char **argv)
{
	u32* framebuf;
	u32  cnt=0;
	u8*  imageptr = (u8*)image_bin;

	//Enable max-1080p support. Remove for 720p-only resolution.
	//gfxInitResolutionDefault();

	gfxInitDefault();

	//Set current resolution automatically depending on current/changed OperationMode. Only use this when using gfxInitResolution*().
	//gfxConfigureAutoResolutionDefault(true);

	while(appletMainLoop())
	{
		//Scan all the inputs. This should be done once for each frame
		hidScanInput();

		//hidKeysDown returns information about which buttons have been just pressed (and they weren't in the previous frame)
		u32 kDown = hidKeysDown(CONTROLLER_P1_AUTO);

		if (kDown & KEY_PLUS) break; // break in order to return to hbmenu

		u32 width, height;
		u32 pos;
		framebuf = (u32*) gfxGetFramebuffer((u32*)&width, (u32*)&height);

		if(cnt==60)
		{
			cnt=0;
		}
		else
		{
			cnt++;
		}

		//Each pixel is 4-bytes due to RGBA8888.
		u32 x, y;
		for (y=0; y<height; y++)//Access the buffer linearly.
		{
			for (x=0; x<width; x++)
			{
				pos = y * width + x;
				framebuf[pos] = RGBA8_MAXALPHA(imageptr[pos*3+0], imageptr[pos*3+1], imageptr[pos*3+2]);
			}
		}

		gfxFlushBuffers();
		gfxSwapBuffers();
		gfxWaitForVsync();
	}

	gfxExit();
	return 0;
}

The code is same of example without #ifdef DISPLAY_IMAGE and Fade Grey.

Thank you.