将uintptr_t转换为id< MTLTexture> [英] Convert uintptr_t to id<MTLTexture>
问题描述
我想创建简单的iOS插件,可以将纹理绘制到统一的Texture2D。我已经通过CreateExternalTexture()和UpdateExternalTexture()完成了它,它工作正常,但我很好奇我是否可以直接从iOS端填充Unity纹理。这是我的iOS插件代码:
I wanted to create simple iOS plugin which can draw the texture to unity Texture2D. I've done it by CreateExternalTexture() and UpdateExternalTexture(), it's working fine, but I'm curious if I can actually fill the Unity texture straight from iOS side. Here's my code of iOS plugin:
//
// testTexturePlugin.m
// Unity-iPhone
//
// Created by user on 18/01/16.
//
//
#import <OpenGLES/ES2/gl.h>
#import <OpenGLES/ES2/glext.h>
#import <UIKit/UIKit.h>
#include "UnityMetalSupport.h"
#include <stdlib.h>
#include <stdint.h>
static UIImage* LoadImage()
{
NSString* imageName = @"logo"; //[NSString stringWithUTF8String: filename];
NSString* imagePath = [[NSBundle mainBundle] pathForResource: imageName ofType: @"png"];
return [UIImage imageWithContentsOfFile: imagePath];
}
// you need to free this pointer
static void* LoadDataFromImage(UIImage* image)
{
CGImageRef imageData = image.CGImage;
unsigned imageW = CGImageGetWidth(imageData);
unsigned imageH = CGImageGetHeight(imageData);
// for the sake of the sample we enforce 128x128 textures
//assert(imageW == 128 && imageH == 128);
void* textureData = ::malloc(imageW * imageH * 4);
::memset(textureData, 0x00, imageW * imageH * 4);
CGContextRef textureContext = CGBitmapContextCreate(textureData, imageW, imageH, 8, imageW * 4, CGImageGetColorSpace(imageData), kCGImageAlphaPremultipliedLast);
CGContextSetBlendMode(textureContext, kCGBlendModeCopy);
CGContextDrawImage(textureContext, CGRectMake(0, 0, imageW, imageH), imageData);
CGContextRelease(textureContext);
return textureData;
}
static void CreateMetalTexture(uintptr_t texRef, void* data, unsigned w, unsigned h)
{
#if defined(__IPHONE_8_0) && !TARGET_IPHONE_SIMULATOR
NSLog(@"texRef iOS = %lu", texRef);
id<MTLTexture> tex = (id<MTLTexture>)(size_t)texRef;
MTLRegion r = MTLRegionMake3D(0, 0, 0, w, h, 1);
[tex replaceRegion: r mipmapLevel: 0 withBytes: data bytesPerRow: w * 4];
#else
#endif
}
extern "C" void FillUnityTexture(uintptr_t texRef)
{
UIImage* image = LoadImage();
void* textureData = LoadDataFromImage(image);
if (UnitySelectedRenderingAPI() == apiMetal)
CreateMetalTexture(texRef, textureData, image.size.width, image.size.height);
::free(textureData);
}
这里是Unity代码:
And here's the Unity code:
using UnityEngine;
using System;
using System.Collections;
using System.Runtime.InteropServices;
public class TextureHandler : MonoBehaviour {
[SerializeField]
private Renderer _mesh;
private Texture2D _meshTexture;
[DllImport("__Internal")]
private static extern void FillUnityTexture(IntPtr texRef);
void Start () {
_meshTexture = new Texture2D(200, 200, TextureFormat.ARGB32, false);
_mesh.material.SetTextureScale ("_MainTex", new Vector2 (-1, -1));
_mesh.material.mainTexture = _meshTexture;
IntPtr texPtr = _meshTexture.GetNativeTexturePtr();
Debug.Log("texPtr Unity = " + texPtr);
FillUnityTexture(texPtr);
}
}
Unity纹理上的指针传递给iOS插件没错,我查了一下。但我在iOS插件的这一行崩溃:
Pointer on the Unity texture is passing to the iOS plugin correctly, I checked. But I have the crash in this line on iOS plugin:
[tex replaceRegion: r mipmapLevel: 0 withBytes: data bytesPerRow: w * 4];
我很确定我有这个问题,因为Unity纹理上的指针转换错误(uintptr_t)到金属纹理(id)。
and I'm pretty sure that I have this problem because of wrong converting of the pointer on Unity texture (uintptr_t) to metal texture (id).
所以我的问题是 - 如何正确地将纹理指针转换为MTLTexture?
So my question is - how can I convert pointer to the texture to MTLTexture properly?
推荐答案
我想你应该读一下ARC。
I guess you should read about ARC.
您可以使用 __ bridge_retained
来转移新创建的 id< MTLTexture> <的所有权/ code>对象
uintptr_t
代码。如果要将 uintptr_t
转换回 id< MTLTexture>
,请使用 __ bridge
如果你不想转回所有权,或者当你这样做时使用 __ bridge_transfer
。
You can use __bridge_retained
to transfer the ownership of newly created id<MTLTexture>
object to uintptr_t
code. When you want to convert the uintptr_t
back to id<MTLTexture>
use __bridge
if you don't want to transfer the ownership back or use __bridge_transfer
when you do.
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