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比如go的一个函数myFunc,在入rule映射为MyFunc:
func myFunc()(int, string){ }
在rule中应用:
rule "ruleId" begin a, str := MyFunc() Print(str) end
实际在执行rule的过程中,只能拿到a的值,str的值是拿不到的吗,后续能支持吗
The text was updated successfully, but these errors were encountered:
啊,这个不支持返回多个值?
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感觉不是这个意思+_+,是在rule自己的函数拿不到多个返回
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比如go的一个函数myFunc,在入rule映射为MyFunc:
在rule中应用:
实际在执行rule的过程中,只能拿到a的值,str的值是拿不到的吗,后续能支持吗
The text was updated successfully, but these errors were encountered: