fun f6() int {
var i int = 0
defer func(a int) {
fmt.Println("i f6:", a)
i = i+5
}(i)
i = 1
iPP
return i
}
according to the rewriting rule, the return statement is separated into two sentences., return xxx will be rewritten as:
return value = xxx
call defer function
empty return
iTun0
return value r = i (i equals 0)
iTun1
iPP (i equals 2)
return r (I think it should be 0)
Why is the result 2?
I don't understand go, but take a look at the defer function
. Isn't that how it works
fun f6() int {
var i int = 0
i = 1
iPP
// return xxx
// defer RET
r = i
func(a int) {
fmt.Println("i f6:", a)
i = i+5
}(i)
return
}so
return I is a single instruction multiple operation is
in the return I code is changed to r = I; return,
but the defer function will be between assignment and return, and your function does not affect the value of the external scope I
so the return must be 2, not 0.
certainly does not change I to r in advance, which is illogical
I am also a beginner of go. This is how I understand
.`
func f6() int { // fun ==> func
var i int = 0
defer func(a int) { //
fmt.Println("i f6:", a) // a = i = 0 print 0
i = i+5
}(i)
i = 1
iPP // i = 2
return i // return 2
}
`
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