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Why does generic method with constraint of T: class result in boxing? [duplicate]

Why a generic method which constrains T to class would have boxing instructions in the generates MSIL code?

I was quite surprised by this since surely since T is being constrained to a reference type the generated code should not need to perform any boxing.

Here is the c# code:

protected void SetRefProperty<T>(ref T propertyBackingField, T newValue) where T : class
{
    bool isDifferent = false;

    // for reference types, we use a simple reference equality check to determine
    // whether the values are 'equal'.  We do not use an equality comparer as these are often
    // unreliable indicators of equality, AND because value equivalence does NOT indicate
    // that we should share a reference type since it may be a mutable.

    if (propertyBackingField != newValue)
    {
        isDifferent = true;
    }
}

Here is the generated IL:

.method family hidebysig instance void SetRefProperty<class T>(!!T& propertyBackingField, !!T newValue) cil managed
{
    .maxstack 2
    .locals init (
        [0] bool isDifferent,
        [1] bool CS$4$0000)
    L_0000: nop 
    L_0001: ldc.i4.0 
    L_0002: stloc.0 
    L_0003: ldarg.1 
    L_0004: ldobj !!T
    L_0009: box !!T
    L_000e: ldarg.2 
    L_000f: box !!T
    L_0014: ceq 
    L_0016: stloc.1 
    L_0017: ldloc.1 
    L_0018: brtrue.s L_001e
    L_001a: nop 
    L_001b: ldc.i4.1 
    L_001c: stloc.0 
    L_001d: nop 
    L_001e: ret 
}

Notice the box !!T instructions.

Why this is being generated?

How to avoid this?

like image 837
Phil Avatar asked Sep 09 '09 15:09

Phil


1 Answers

You don't have to worry about any performance-degradations from the box instruction because if its argument is a reference type, the box instruction does nothing. Though it's still strange that the box instruction has even been created (maybe lazyiness/easier design at code generation?).

like image 134
Thomas Danecker Avatar answered Oct 13 '22 08:10

Thomas Danecker