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Visual Studio 2013 Release Build Not Working

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The most annoying one is that you can't easily see into MFC because you don't have the symbols included for the MFC DLL. share|improve this answer answered Oct 9 '08 at 10:14 Vhaerun 4,090103137 add a comment| up vote 0 down vote You might run your software with Global Flags enabled (Look in Debugging share|improve this answer edited Nov 19 '09 at 13:12 answered Nov 19 '09 at 10:03 Georg Schölly 83.3k31165228 2 Never assume it is the compiler's fault. It would be really nice if I could get Windows to print out a stack trace, or something other than simply terminating the program as if it had exited cleanly. Check This Out

Sapan Sarvaiya15-Apr-05 8:50 Sapan Sarvaiya15-Apr-05 8:50 I have a struct and a enum in my code. Driving through Croatia: can someone tell me where I took this photo? I'm developing a computer game and I *was* geting an Access Violation in release mode. I know that debug contains debugging symbols and release does not have...

Visual Studio 2013 Release Build Not Working

I don't know how that works with MS technologies. Sign In·ViewThread·Permalink Problems with DEBUG preprocessor definition pjvv121-Sep-07 6:30 pjvv121-Sep-07 6:30 Hi all, I realized that my app doesn't use the DEBUG define I've made at the project options. I'm working on VS2010, c++, directX, Windows 7. Yes No Additional feedback? 1500 characters remaining Submit Skip this Thank you!

  • to use RTL in same mode as debug also ...
  • visual-studio-2010 debugging memory share|improve this question asked Mar 20 '11 at 12:50 SeniorLee 4251922 You can debug application built in release mode just as well.
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thx steve Sign In·ViewThread·Permalink Re: crashes with global optimization on jerikat30-Dec-03 8:30 jerikat30-Dec-03 8:30 To suppress a single warning--for example, 4100: #pragma warning (disable: 4100) To suppress multiple warnings: Example: The async and await features you are using in your code are C# 5+ only. Window.Resources) the control works fine. Visual Studio Release Vs Debug Look around your Visual Studio project options to see how to generate pdb files for your release build, etc.

More magic values. –RustyX Apr 4 at 9:56 add a comment| up vote 16 down vote A common pitfall is using an expression with side effect inside an ASSERT. Some time ago there were problems with some compiler optimizations breaking code, but I have not read complaints lately. I would check your code for uninitialized variables. read this article share|improve this answer answered Nov 19 '09 at 9:51 Matthew Scharley 62.2k39151197 add a comment| up vote 1 down vote You'd need to give a lot more information, but yes, it's

You're trying to figure out what settings/properties are set differently based on Debug/Release configurations. Visual Studio Release Mode Are You Listening to Your Compiler? Pro When Logging Isn't Enough: A Modern Approach to Monitoring Performance in Production Pro The Importance of Monitoring Containers Surviving the Release Version 10 Ways to Boost COBOL Application Development Show It is possible to debug in Release mode.

C# Release Build

Is the Compiler Generating Bad Code? You might want to also consider the following: You may want disable your optimization settings when debugging your release version (though this isn't absolutely necessary). Visual Studio 2013 Release Build Not Working Sign In·ViewThread·Permalink Great niry1-Dec-01 19:40 niry1-Dec-01 19:40 Sign In·ViewThread·Permalink Good Job Gil Rivlis2-Aug-01 14:51 Gil Rivlis2-Aug-01 14:51 Good Article, thanks! Visual Studio Debug Release Build The problem is that when I run in release mode I get your classic "Illegal Operation" error.

Read on. « The Doh! http://wiiemulator.net/visual-studio/f10-not-working-in-visual-studio-2013.html Simply by turning the Warning Level to 4, I detected the problem. Browse other questions tagged visual-studio c#-4.0 tfs visual-studio-2013 or ask your own question. Not the answer you're looking for? Difference Between Debug And Release Mode In Visual Studio

Thus some local variables where merely mapping to CPU registers (for optimization) which were shared with the aforementioned function leading to serious memory corruption. share|improve this answer answered Nov 19 '09 at 12:25 Alex Budovski 11.3k43650 add a comment| up vote 1 down vote There are compiler optimizations that can break valid code because they share|improve this answer answered May 28 '14 at 9:28 Mohamad mehdi Kharatizadeh 12915 add a comment| up vote 0 down vote As my experience, that are most being memory corruption issues. http://wiiemulator.net/visual-studio/visual-studio-build-stuck.html In particular, the ASSERT macro evaluates to nothing in a release build, so none of the code found in ASSERTs will be executed.

In Visual Studio, debug builds explicitely initialize allocated memory to given values, see e.g. Debug Vs Release Build So if you have the same problem as me, open the .csproj file and delete this node, this should solve it. Your error is therefore almost certainly that you are using something like you are using a pointer before it has been properly initialized and you're getting away with it in the

I don't use metrowerks compilers anymore. –deft_code Nov 19 '09 at 15:52 I haven't yet seen such a case myself. :) –Georg Schölly Nov 19 '09 at 21:56

Does boiling tap water make it potable? So, no, the debug version is not OK just because you can run it without getting an error. Making Debug Mode more like Release Mode If your Release Mode problem is caused by code being unintentionally removed. If the PDB's were put in the same output folder as your EXE or DLL's they will probably be picked up automatically.

It seems that Visual Studio is not linking Themes/Generic.xaml to the control. Are human fetal cells used to produce Pepsi? Would you want to play games released as DEBUG builds? navigate here But then I realized why: in Release mode, _CrtCheckMemory is #defined's as ((int)1). –Brian Morearty Feb 16 '11 at 19:28 add a comment| up vote 7 down vote Even though you

if (p == 0) { // do stuff } In debug mode the code in the if is not executed but in release mode p contains an undefined value, which is