add explicit default group #3103
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LFSaw
changed the title
| @@ -256,6 +256,7 @@ BusPlug : AbstractFunction { | |||
| ^this | |||
| }; | |||
| if(bus.rate == \control) { "Can't monitor a control rate bus.".warn; monitor.stop; ^this }; | |||
| group = group ?? {this.homeServer.defaultGroup}; | |||
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this can be simplified to ? without braces instead of ??.
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no.
with ??, the latter is only evaluated when group is nil.
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so if i'm understanding correctly, you should update the usage of ? below as well then
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hm... good question. of course, function evaluation costs as well.
TempoClock.default, the value after ? is an instance variable of TempoClock so it only just returns the value. Thus, in the case below, there's not much difference between ? and ??.
Compare:
bench{50000.do{[nil, 1].choose ? TempoClock.default}};
bench{50000.do{[nil, 1].choose ?? {TempoClock.default}}};
In the upper case, defaultGroup is a method of server which requires computation, hence we only want to evaluate it when it is needed.
bench{50000.do{[nil, 1].choose ? s.defaultGroup}};
bench{50000.do{[nil, 1].choose ?? {s.defaultGroup}}};
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ah, you're right! this is good to know. thanks.
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? always evaluates the second operand, whether it will be used or not.
If I recall correctly, ?? inlines the following function if possible, so the cost should be less than pushing a function and calling value. In fact, I just checked: in something ?? { blah }, ?? compiles to a ControlOpcode that skips over the function's byte codes if the receiver is not nil. The cost of the jump should be a lot less.
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Thanks for the clarification, james. So my explanation was slightly off but the effect is the same :) (?)
I'll keep in mind {/**/}.def.dumpByteCode;:
{#[nil, 1].choose ?? {s.defaultGroup}}.def.dumpByteCodes
>>>
BYTECODES: (13)
0 40 PushLiteral instance of Array (0x11fbdf828, size=2, set=2)
1 C1 24 SendSpecialMsg 'choose'
3 8F 17 00 05 ControlOpcode 5 (11)
7 01 14 PushInstVarX 's'
9 B0 TailCallReturnFromFunction
10 A1 00 SendMsg 'defaultGroup'
12 F2 BlockReturn
{#[nil, 1].choose ? s.defaultGroup}.def.dumpByteCodes
>>>
BYTECODES: (10)
0 40 PushLiteral instance of Array (0x11fbdf278, size=2, set=2)
1 C1 24 SendSpecialMsg 'choose'
3 01 14 PushInstVarX 's'
5 A1 00 SendMsg 'defaultGroup'
7 8F 16 ControlOpcode
9 F2 BlockReturn
But I have to say that I cannot read this... what's the ControlOpCode 5 (11) doing? That's the jump you're talking about?
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will wait until @telephon has a chance to look at this before merging |
this fixes #3098