[android help] For video processing on android , is it better to proceed on client or server side? I need to evaluate performance in terms of frame rate and latency

For video processing on android , is it better to proceed on client or server side? I need to evaluate performance in terms of frame rate and latency

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I have to create an application that takes a video of a QR code stream and then analyzes it in order to extract information from it. should the video processing be done on the android device itself using OpenCv library or it is better to do it on a server (http server for instance). I need exact statistics for performance evaluation in termes of frame rate, CPU consumption and latency asked 3 mins ago

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[android help] Any idea why adb does not recognize my nexus 4 on mac

Any idea why adb does not recognize my nexus 4 on mac

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I can't figure out why my nexus 4 is absent from the list of adb devices. I've tried the following $ adb kill-server; adb devices $ echo 0x18d1 >> ~/.android/adb_usb.ini $ adb kill-server; adb devices Changed usb computer connection type to PTP instead of MTP Restarted the phone etc Adb is on my path. Nexus 4 shows up when I view System Information. I've tried asking on xda devs but no luck. Any suggestions on what I can try? Thanks in advance asked Jul 1 at 5:39

	Its because that you didnt update your SDK you need to update you SDK to the latest one i.e API level 17 your Nexes 4 OS version is 4.2 so you need to update your android sdk to API leve 17 4.2 to detect it. answered Jul 1 at 5:43

	You have to first enable "Developer" mode on the device before trying "adb devices" option. Here is the topic that explains how. How to find and turn on usb debugging mode on Nexus 4. This worked for me. answered 11 hours ago

	Very weird, I switched to a cheap chinese made cable and now adb sees my device. answered 2 mins ago

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[android help] Is there implicit ListFragment margin/padding? How to remove it?

Is there implicit ListFragment margin/padding? How to remove it?

I had this same problem. I'm using ActionBarSherlock and was able to fix this with Styles and Themes.

First, I added this to my themes.xml:

This tells SherlockListFragment to use my custom style for its ListView. Then I added this custom style to my styles.xml:

You could also try one of these settings:

    0
    0 

Pointing the ListView to my custom style fixed the problem for me. Essentially, setting the styles on the ListView Widget should do the trick.

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[android help] logical operation in java

logical operation in java

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	I have been trying to implement the following condition in a more sofisticated manner: if (x > 1) doSomething(); else { doSomething(); } want to code it like: (x > 1) ? doSomething() : doSomething(); Is this not possible at all? asked 5 mins ago

Not possible if doSomething() returns void. Refer the JLS 15.25 The first expression must be of type boolean or Boolean, or a compile-time error occurs. It is a compile-time error for either the second or the third operand expression to be an invocation of a void method. In fact, by the grammar of expression statements (§14.8), it is not permitted for a conditional expression to appear in any context where an invocation of a void method could appear. P.S: Why you want to doSomething() for both true or false? answered 4 mins ago

	Not in full generality; the arguments in the ternary (formally the ternary requires expressions) have to evaluate to the same type. If doSomething() is a void type then certainly not. answered 4 mins ago

	It is dependent on return type.If you define like this void doSomething() then its not possible. answered 1 min ago

	No you can't. In JLS §14.8 ExpressionStatement: StatementExpression ; StatementExpression: Assignment PreIncrementExpression PreDecrementExpression PostIncrementExpression PostDecrementExpression MethodInvocation ClassInstanceCreationExpression Oops, no ConditionalExpression at all. answered 1 min ago

	The ternary is used for inline if conditions. They need a return value, because u could write something like that: System.out.println((x > 1) ? "True" : "False"); with void this would not be possible answered 10 secs ago

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