How does FORCEINLINE_DEBUGGABLE work?

Development Programming & Scripting C++
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1

According to the source code(UE4.26.2)

// Engine/Source/Runtime/Core/Public/Misc/CoreMiscDefines.h
#if UE_BUILD_DEBUG
	#define FORCEINLINE_DEBUGGABLE FORCEINLINE_DEBUGGABLE_ACTUAL
#else
	#define FORCEINLINE_DEBUGGABLE FORCEINLINE
#endif

// Engine/Source/Runtime/Core/Public/HAL/Platform.h
#ifndef FORCEINLINE_DEBUGGABLE_ACTUAL
	#define FORCEINLINE_DEBUGGABLE_ACTUAL inline
#endif

// Engine/Source/Runtime/Core/Public/Windows/WIndowsPlatform.h
#define FORCEINLINE __forceinline									/* Force code to be inline */

So I don’t understand how the macro FORCEINLINE_DEBUGGABLE can make a function inlined but also debuggable.

1 Like
2

0 information about this.

I think in debug builds its inline but in release__forceinline. idk how that aids debugability.

3

inline is a soft suggestion while __forceinline is a stronger request but the compiler may still ignore it if inlining is impossible.

So in debug build the code is less likely to be inlined (because it use a soft suggestion), which might aids the debugability.

4

I used to think that inlined code is not debugable (this is why I asked the question), however it seems modern debuggers can still show some inlined code if debug symbols are enabled.

I think I can close this question.

5

@NameOccupied You should update the solution before closing, pardon if you were doing it already

6

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