英文:
How can I get "index out of range" when I am not indexing
问题
长期从事编程工作,但对Go语言完全不熟悉。
这是我的代码。这是我第一次尝试生成器。我试图生成一个LFSR序列。基本上,每次循环都向右移动一位。如果刚刚移出一个1
位,则与tap
值进行异或运算。
package main
import (
"fmt"
"math/big"
)
// lfsr返回一个LFSR生成器。
func lfsr(tap, start big.Int) func() big.Int {
// 跟踪下一个值。
next := &start
// 生成器。
return func() big.Int {
// 记住当前位置。
ret := *next
// 计算下一个值。
fmt.Println("next", next.String(), "bit(0)", next.Bit(0))
// 当前值是否为奇数?
odd := next.Bit(0)
// 右移一位。
next = next.Rsh(next, 1)
// 如果是奇数-进行异或运算!
if odd != 0 {
// 异或运算!
next = next.Xor(next, &tap)
fmt.Printf("Tap!", next.String())
}
// 返回之前的值。
return ret
}
}
func main() {
ten := new(big.Int)
ten.SetString("10", 32)
f := lfsr(*ten, *ten)
for i := 0; i < 10; i++ {
n := f()
fmt.Println("lfsr ", n.String())
}
}
我得到的输出是:
next 32 bit(0) 0
lfsr 16
next 16 bit(0) 0
lfsr 8
next 8 bit(0) 0
lfsr 4
next 4 bit(0) 0
lfsr 2
next 2 bit(0) 0
lfsr 1
next 1 bit(0) 1
Tap! 0
panic: runtime error: index out of range
我做错了什么,为什么看起来是对的?
Play - 有趣的是 - 它输出:
...
next 1 bit(0) 1
Tap!%!(EXTRA string=0)panic: runtime error: index out of range
goroutine 1 [running]:
math/big.nat.string(0xc010045150, 0x1, 0x5, 0x12b23d0, 0xa, ...)
go/src/pkg/math/big/nat.go:819 +0x67f
math/big.nat.decimalString(0xc010045150, 0x1, 0x5, 0x1, 0x1, ...)
go/src/pkg/math/big/nat.go:731 +0x8f
math/big.(*Int).String(0x7f851e0eff40, 0xc010045150, 0x1)
go/src/pkg/math/big/int.go:331 +0xfe
main.main()
/tmpfs/gosandbox-94dce1ec_430947f1_6360662e_01c3d6ad_a7071d20/prog.go:41 +0x120
这有点不同,并且暗示fmt.Println("lfsr ", n.String())
是出错的,但我还没有找出原因。
补充
经过尝试(将ten.SetString("10", 32)
更改为ten.SetString("10", 10)
),我现在得到:
lfsr 5
next 5 bit(0) 1
Tap!%!(EXTRA string=0)panic: runtime error: index out of range
现在要去睡觉了,希望有人能帮忙解决。
英文:
Long time programmer - total newbie in go.
Here's the code. It is my first attempt at a generator. I am trying to generate an lfsr sequence. Essentially, every time around you shift right one. If you just shifted out a 1
bit, xor with the tap
value.
package main
import (
"fmt"
"math/big"
)
// lfsr returns an lfsr generator.
func lfsr(tap, start big.Int) func() big.Int {
// Keep track of next.
next := &start
// The generator.
return func() big.Int {
// Remember where we are.
ret := *next
// Work out next.
fmt.Println("next", next.String(), "bit(0)", next.Bit(0))
// Is it currently odd?
odd := next.Bit(0)
// Shift right one.
next = next.Rsh(next, 1)
// If odd - tap!
if odd != 0 {
// Tap!
next = next.Xor(next, &tap)
fmt.Printf("Tap!", next.String())
}
// Return where we were.
return ret
}
}
func main() {
ten := new(big.Int)
ten.SetString("10", 32)
f := lfsr(*ten, *ten)
for i := 0; i < 10; i++ {
n := f()
fmt.Println("lfsr ", n.String())
}
}
The printout I am getting is:
next 32 bit(0) 0
lfsr 16
next 16 bit(0) 0
lfsr 8
next 8 bit(0) 0
lfsr 4
next 4 bit(0) 0
lfsr 2
next 2 bit(0) 0
lfsr 1
next 1 bit(0) 1
Tap! 0
panic: runtime error: index out of range
What am I doing wrong - and why does it look right?
Play - Interestingly - it outputs:
...
next 1 bit(0) 1
Tap!%!(EXTRA string=0)panic: runtime error: index out of range
goroutine 1 [running]:
math/big.nat.string(0xc010045150, 0x1, 0x5, 0x12b23d0, 0xa, ...)
go/src/pkg/math/big/nat.go:819 +0x67f
math/big.nat.decimalString(0xc010045150, 0x1, 0x5, 0x1, 0x1, ...)
go/src/pkg/math/big/nat.go:731 +0x8f
math/big.(*Int).String(0x7f851e0eff40, 0xc010045150, 0x1)
go/src/pkg/math/big/int.go:331 +0xfe
main.main()
/tmpfs/gosandbox-94dce1ec_430947f1_6360662e_01c3d6ad_a7071d20/prog.go:41 +0x120
which is a) slightly different and b) suggesting that the fmt.Println("lfsr ", n.String())
is what is failing but I am no nearer working out why.
Added
After experimentation (changing ten.SetString("10", 32)
to ten.SetString("10", 10)
) I now get:
lfsr 5
next 5 bit(0) 1
Tap!%!(EXTRA string=0)panic: runtime error: index out of range
Going to sleep now - hope someone can help.
答案1
得分: 1
似乎是由于big.Int
上的String()
实现引起的。您可以在n
上省略String()
调用,让Println
自己决定如何打印参数:
func main() {
ten := new(big.Int)
ten.SetString("10", 32)
f := lfsr(*ten, *ten)
for i := 0; i < 10; i++ {
n := f()
fmt.Println("lfsr ", n)
}
}
英文:
It seems to be caused by the String() implementation on big.Int. You can leave out the String() call on n
and let Println figure out how to print the argument itself:
func main() {
ten := new(big.Int)
ten.SetString("10", 32)
f := lfsr(*ten, *ten)
for i := 0; i < 10; i++ {
n := f()
fmt.Println("lfsr ", n)
}
}
答案2
得分: 1
复制big.Int
是不安全的。特别是将*big.Int
解引用并赋值给一个值是行不通的。底层数组会被别名引用而不是被复制,因为数组可能以破坏不变性的方式发生变化。
ret := *next
这一行是你的错误来源。
http://play.golang.org/p/W_qOCDsO2A
i := big.NewInt(1)
j := *i
fmt.Println("i:", *i, "j:", j) // i: {false [1]} j: {false [1]}
i = i.Xor(i, i)
fmt.Println("i:", *i, "j:", j) // i: {false []} j: {false [0]}
//j的状态已经改变,它现在违反了不变性
j.String() //boom
复制big.Int
的方法是ret := *new(big.Int).Set(next)
。我认为应该有一个func Copy(n *big.Int) *big.Int
来完成这个操作,因为语法有点丑陋。此外,我建议完全避免将big.Int
用作值,并将lfsr
更改为返回指针。
工作示例:http://play.golang.org/p/WEFRnGlU1H。
英文:
It is not safe to copy big.Int
. In particular dereferencing a *big.Int
and assigning it to a value does not work. The underlying array is aliased instead of copied. as the array can mutate in ways that break the invariants.
The line ret := *next
is the source of your bug.
http://play.golang.org/p/W_qOCDsO2A
i := big.NewInt(1)
j := *i
fmt.Println("i:", *i, "j:", j) // i: {false [1]} j: {false [1]}
i = i.Xor(i, i)
fmt.Println("i:", *i, "j:", j) // i: {false []} j: {false [0]}
//j's state has changed, it now violates the invariants
j.String() //boom
The way to copy a big.Int
is ret := *new(big.Int).Set(next)
. I think there should be func Copy(n *big.Int) *big.Int
that does this, as the syntax is a bit ugly. Also I would avoid using big.Int
as a value at all and change lfsr
to return a pointer.
Working example http://play.golang.org/p/WEFRnGlU1H.
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