将imageNamed转换为imageWithContentsOfFile：

Question

我的图像代码是

-(IBAction)start:(id)sender
{
    animation.animationImages = [NSArray arrayWithObjects:
                                 [UIImage imageNamed:@"Paddle 1.png"],
                                 [UIImage imageNamed:@"Paddle 2.png"],
                                 [UIImage imageNamed:@"Paddle 3.png"],
                                 [UIImage imageNamed:@"Paddle 4.png"],
nil];
    [animation setAnimationRepeatCount:0];
    animation.animationDuration = 2.5;
    [animation startAnimating];
}

这是缓存太多的内存，我在前面的一个问题中被告知要将我的代码交换为

[UIImage imageWithContentsOfFile: GetImgWithoutCaching(@"Paddle 1.jpg")]

和

UIImage* GetImgWithoutCaching(NSString* imgName)
{
    NSString *imagePath = [[NSBundle mainBundle] pathForResource:imgName ofType:nil];
    return [UIImage imageWithContentsOfFile:imagePath];
}

编写代码的正确方法是什么？我是否将该代码按原样放置在我的.m中？

Answer 1

首先，您应该检查是否使用视网膜图片：

BOOL isHighResolution = NO;
if ([[UIScreen mainScreen] respondsToSelector:@selector(scale)]) {
    if ([UIScreen mainScreen].scale > 1) {
        isHighResolution = YES;
    }
}

如果您使用的是视网膜图片，请在图像名称中添加@2x，如下所示：

NSString *noExtFileName = [name stringByDeletingPathExtension];
if (isHighResolution) {
    if (![noExtFileName hasSuffix:@"@2x"]) {
       noExtFileName = [noExtFileName stringByAppendingString:@"@2x"];
    }
}

 //if image only "png" type
return [[NSBundle mainBundle] pathForResource:noExtFileName ofType:@"png"];

Answer 2

由于GetImgWithoutCaching函数返回需要的UIImage：

-(IBAction)start:(id)sender
{
    animation.animationImages = [NSArray arrayWithObjects:
                                 GetImgWithoutCaching(@"Paddle 1.jpg"),
                                 GetImgWithoutCaching(@"Paddle 2.jpg"),
                                 GetImgWithoutCaching(@"Paddle 3.jpg"),
                                 GetImgWithoutCaching(@"Paddle 4.jpg"),
nil];
    [animation setAnimationRepeatCount:0];
    animation.animationDuration = 2.5;
    [animation startAnimating];
}