You can use this method.
public static File getRobotCacheFile(Context context) throws IOException {
File cacheFile = new File(context.getCacheDir(), "robot.png");
try {
InputStream inputStream = context.getAssets().open("robot.png");
try {
FileOutputStream outputStream = new FileOutputStream(cacheFile);
try {
byte[] buf = new byte[1024];
int len;
while ((len = inputStream.read(buf)) > 0) {
outputStream.write(buf, 0, len);
}
} finally {
outputStream.close();
}
} finally {
inputStream.close();
}
} catch (IOException e) {
throw new IOException("Could not open robot png", e);
}
return cacheFile;
}
You should never use InputStream.available() in such cases. It returns only bytes that are buffered. Method with .available() will never work with bigger files and will not work on some devices at all.
In Kotlin (;D):
@Throws(IOException::class)
fun getRobotCacheFile(context: Context): File = File(context.cacheDir, "robot.png")
.also {
it.outputStream().use { cache -> context.assets.open("robot.png").use { it.copyTo(cache) } }
}