我正在尝试从InputStream中调整一张图片的大小,所以我使用了Strange out of memory issue while loading an image to a Bitmap object中的代码,但我不知道为什么这段代码总是返回没有图像的Drawable。
这个代码可以正常工作:
private Drawable decodeFile(InputStream f){
try {
InputStream in2 = new BufferedInputStream(f);
BitmapFactory.Options o2 = new BitmapFactory.Options();
o2.inSampleSize=2;
return new BitmapDrawable(BitmapFactory.decodeStream(in2, null, o2));
} catch (Exception e) {
return null;
}
}
这个不起作用:
private Drawable decodeFile(InputStream f){
try {
InputStream in1 = new BufferedInputStream(f);
InputStream in2 = new BufferedInputStream(f);
//Decode image size
BitmapFactory.Options o = new BitmapFactory.Options();
o.inJustDecodeBounds = true;
BitmapFactory.decodeStream(in1,null,o);
//The new size we want to scale to
final int IMAGE_MAX_SIZE=90;
//Find the correct scale value. It should be the power of 2.
int scale = 2;
if (o.outHeight > IMAGE_MAX_SIZE || o.outWidth > IMAGE_MAX_SIZE) {
scale = (int)Math.pow(2, (int) Math.round(Math.log(IMAGE_MAX_SIZE /
(double) Math.max(o.outHeight, o.outWidth)) / Math.log(0.5)));
}
BitmapFactory.Options o2 = new BitmapFactory.Options();
o2.inJustDecodeBounds = false;
o2.inSampleSize=scale;
return new BitmapDrawable(BitmapFactory.decodeStream(in2, null, o2));
} catch (Exception e) {
return null;
}
}
为什么一个选项会影响另一个选项?如果我使用两个不同的InputStream和Options,这是如何可能的?