How many times will len(p) run? Only one and the result will be saved, or will it run for each iteration?
func main() {
p := []int{2, 3, 5, 7, 11, 13}
for i:=0;i<len(p);i++ {}
}
len() on a slice is optimized by the compiler, it's like accessing a local variable. It's not really a function call.
You can verify with:
$ cat x.go
package main
import "fmt"
func main() {
a := []int{1,2, 3}
fmt.Println(len(a))
}
And then look at the compiler output:
$ go tool 6g -S x.go
--- prog list "main" --- 0000 (x.go:3) TEXT main+0(SB),$128-0 0001 (x.go:4) MOVQ $0,autotmp_0002+-80(SP) 0002 (x.go:4) MOVQ $0,autotmp_0002+-72(SP) 0003 (x.go:4) MOVQ $0,autotmp_0002+-64(SP) 0004 (x.go:4) LEAQ autotmp_0002+-80(SP),BX 0005 (x.go:4) MOVQ BX,autotmp_0001+-56(SP) 0006 (x.go:4) MOVQ autotmp_0001+-56(SP),BX 0007 (x.go:4) MOVQ statictmp_0000+0(SB),BP 0008 (x.go:4) MOVQ BP,(BX) 0009 (x.go:4) MOVQ statictmp_0000+8(SB),BP 0010 (x.go:4) MOVQ BP,8(BX) 0011 (x.go:4) MOVQ statictmp_0000+16(SB),BP 0012 (x.go:4) MOVQ BP,16(BX) 0013 (x.go:4) MOVQ autotmp_0001+-56(SP),BX 0014 (x.go:4) MOVQ $3,CX 0015 (x.go:5) LEAQ autotmp_0005+-16(SP),DI 0016 (x.go:5) MOVQ $0,AX 0017 (x.go:5) STOSQ , 0018 (x.go:5) STOSQ , 0019 (x.go:5) LEAQ autotmp_0005+-16(SP),BX 0020 (x.go:5) MOVQ BX,autotmp_0004+-48(SP) 0021 (x.go:5) MOVQ autotmp_0004+-48(SP),BX 0022 (x.go:5) MOVQ $1,SI 0023 (x.go:5) MOVQ $1,DX 0024 (x.go:5) MOVQ BX,autotmp_0003+-40(SP) 0025 (x.go:5) MOVQ autotmp_0003+-40(SP),BX 0026 (x.go:5) MOVQ $type.int+0(SB),AX 0027 (x.go:5) MOVQ AX,(BX) 0028 (x.go:5) MOVQ CX,8(BX) 0029 (x.go:5) MOVQ autotmp_0003+-40(SP),BX 0030 (x.go:5) MOVQ BX,(SP) 0031 (x.go:5) MOVQ SI,8(SP) 0032 (x.go:5) MOVQ DX,16(SP) 0033 (x.go:5) CALL ,fmt.Println+0(SB) 0034 (x.go:6) RET ,
Notice the CALL to fmt.Println, but no call to len.
From the algorithmic point of view condition is evaluated for each step of the loop for. In this case it seems to be six times of the evaluation of the len(p).
Try an experiment! http://play.golang.org/p/Eksb6bQovC
Or read the docs: http://golang.org/ref/spec#For_statements (TL;DR: "The condition is evaluated before each iteration.")
In your example, len is part of the condition so it is evaluated before each iteration. There will be an iteration for the six times the condition evaluates to true, then one more evaluation where it evaluates to false and the for loop exits. So, seven times in all.
A second experiment:
func Test_X(t * testing.T) {
list := []int{1, 2, 3}
fmt.Printf("# b: len(list) %d
", len(list))
for i := 0; i < len(list); i++ {
fmt.Printf("# %d: len(list) %d, list[%d] == %d
", i, len(list), i, list[i])
if i == 2 {
list = append(list, 4)
}
}
fmt.Printf("# a: len(list) %d
", len(list))
}
output:
=== RUN Test_X
# b: len(list) 3
# 0: len(list) 3, list[0] == 1
# 1: len(list) 3, list[1] == 2
# 2: len(list) 3, list[2] == 3
# 3: len(list) 4, list[3] == 4
# a: len(list) 4
--- PASS: Test_X (0.05 seconds)
It gives a reason for not optimizing away then len() calls.