当你没有进行索引操作时，如何出现“索引超出范围”的错误？

huangapple
go评论174阅读模式
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。

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"math/big"
  6. )
  8. // lfsr返回一个LFSR生成器。
  9. func lfsr(tap, start big.Int) func() big.Int {
  10. 	// 跟踪下一个值。
  11. 	next := &start
  12. 	// 生成器。
  13. 	return func() big.Int {
  14. 		// 记住当前位置。
  15. 		ret := *next
  16. 		// 计算下一个值。
  17. 		fmt.Println("next", next.String(), "bit(0)", next.Bit(0))
  18. 		// 当前值是否为奇数？
  19. 		odd := next.Bit(0)
  20. 		// 右移一位。
  21. 		next = next.Rsh(next, 1)
  23. 		// 如果是奇数-进行异或运算！
  24. 		if odd != 0 {
  25. 			// 异或运算！
  26. 			next = next.Xor(next, &tap)
  27. 			fmt.Printf("Tap!", next.String())
  28. 		}
  30. 		// 返回之前的值。
  31. 		return ret
  32. 	}
  33. }
  35. func main() {
  36. 	ten := new(big.Int)
  37. 	ten.SetString("10", 32)
  38. 	f := lfsr(*ten, *ten)
  39. 	for i := 0; i < 10; i++ {
  40. 		n := f()
  41. 		fmt.Println("lfsr ", n.String())
  42. 	}
  43. }

我得到的输出是：

  1. next 32 bit(0) 0
  2. lfsr  16
  3. next 16 bit(0) 0
  4. lfsr  8
  5. next 8 bit(0) 0
  6. lfsr  4
  7. next 4 bit(0) 0
  8. lfsr  2
  9. next 2 bit(0) 0
  10. lfsr  1
  11. next 1 bit(0) 1
  12. Tap! 0
  13. 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：

  1. ...
  2. next 1 bit(0) 1
  3. Tap!%!(EXTRA string=0)panic: runtime error: index out of range
  5. goroutine 1 [running]:
  6. math/big.nat.string(0xc010045150, 0x1, 0x5, 0x12b23d0, 0xa, ...)
  7. 	go/src/pkg/math/big/nat.go:819 +0x67f
  8. math/big.nat.decimalString(0xc010045150, 0x1, 0x5, 0x1, 0x1, ...)
  9. 	go/src/pkg/math/big/nat.go:731 +0x8f
  10. math/big.(*Int).String(0x7f851e0eff40, 0xc010045150, 0x1)
  11. 	go/src/pkg/math/big/int.go:331 +0xfe
  12. main.main()
  13. 	/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：

  1. lfsr  5
  2. next 5 bit(0) 1
  3. 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.

  1. package main
  3. import (
  4. 	&quot;fmt&quot;
  5. 	&quot;math/big&quot;
  6. )
  8. // lfsr returns an lfsr generator.
  9. func lfsr(tap, start big.Int) func() big.Int {
  10. 	// Keep track of next.
  11. 	next := &amp;start
  12. 	// The generator.
  13. 	return func() big.Int {
  14. 		// Remember where we are.
  15. 		ret := *next
  16. 		// Work out next.
  17. 		fmt.Println(&quot;next&quot;, next.String(), &quot;bit(0)&quot;, next.Bit(0))
  18. 		// Is it currently odd?
  19. 		odd := next.Bit(0)
  20. 		// Shift right one.
  21. 		next = next.Rsh(next, 1)
  23. 		// If odd - tap!
  24. 		if odd != 0 {
  25. 			// Tap!
  26. 			next = next.Xor(next, &amp;tap)
  27. 			fmt.Printf(&quot;Tap!&quot;, next.String())
  28. 		}
  30. 		// Return where we were.
  31. 		return ret
  32. 	}
  33. }
  35. func main() {
  36. 	ten := new(big.Int)
  37. 	ten.SetString(&quot;10&quot;, 32)
  38. 	f := lfsr(*ten, *ten)
  39. 	for i := 0; i &lt; 10; i++ {
  40. 		n := f()
  41. 		fmt.Println(&quot;lfsr &quot;, n.String())
  42. 	}
  43. }

The printout I am getting is:

  1. next 32 bit(0) 0
  2. lfsr  16
  3. next 16 bit(0) 0
  4. lfsr  8
  5. next 8 bit(0) 0
  6. lfsr  4
  7. next 4 bit(0) 0
  8. lfsr  2
  9. next 2 bit(0) 0
  10. lfsr  1
  11. next 1 bit(0) 1
  12. Tap! 0
  13. panic: runtime error: index out of range

What am I doing wrong - and why does it look right?

Play - Interestingly - it outputs:

  1. ...
  2. next 1 bit(0) 1
  3. Tap!%!(EXTRA string=0)panic: runtime error: index out of range
  5. goroutine 1 [running]:
  6. math/big.nat.string(0xc010045150, 0x1, 0x5, 0x12b23d0, 0xa, ...)
  7. 	go/src/pkg/math/big/nat.go:819 +0x67f
  8. math/big.nat.decimalString(0xc010045150, 0x1, 0x5, 0x1, 0x1, ...)
  9. 	go/src/pkg/math/big/nat.go:731 +0x8f
  10. math/big.(*Int).String(0x7f851e0eff40, 0xc010045150, 0x1)
  11. 	go/src/pkg/math/big/int.go:331 +0xfe
  12. main.main()
  13. 	/tmpfs/gosandbox-94dce1ec_430947f1_6360662e_01c3d6ad_a7071d20/prog.go:41 +0x120

which is a) slightly different and b) suggesting that the fmt.Println(&quot;lfsr &quot;, n.String()) is what is failing but I am no nearer working out why.

Added

After experimentation (changing ten.SetString(&quot;10&quot;, 32) to ten.SetString(&quot;10&quot;, 10)) I now get:

  1. lfsr  5
  2. next 5 bit(0) 1
  3. 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：

  1. func main() {
  2.     ten := new(big.Int)
  3.     ten.SetString("10", 32)
  4.     f := lfsr(*ten, *ten)
  5.     for i := 0; i < 10; i++ {
  6.         n := f()
  7.         fmt.Println("lfsr ", n)
  8.     }
  9. }
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:

  1. func main() {
  2.     ten := new(big.Int)
  3.     ten.SetString(&quot;10&quot;, 32)
  4.     f := lfsr(*ten, *ten)
  5.     for i := 0; i &lt; 10; i++ {
  6.         n := f()
  7.         fmt.Println(&quot;lfsr &quot;, n)
  8.     }
  9. }

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

  1. i := big.NewInt(1)
  2. j := *i
  3. fmt.Println("i:", *i, "j:", j) // i: {false [1]} j: {false [1]}
  4. i = i.Xor(i, i)
  5. fmt.Println("i:", *i, "j:", j) // i: {false []} j: {false [0]}
  6. //j的状态已经改变，它现在违反了不变性
  7. 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

  1. i := big.NewInt(1)
  2. j := *i
  3. fmt.Println(&quot;i:&quot;, *i, &quot;j:&quot;, j) // i: {false [1]} j: {false [1]}
  4. i = i.Xor(i, i)
  5. fmt.Println(&quot;i:&quot;, *i, &quot;j:&quot;, j) // i: {false []} j: {false [0]}
  6. //j&#39;s state has changed, it now violates the invariants
  7. 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.

huangapple
  • 本文由 发表于 2013年9月27日 07:06:11
  • 转载请务必保留本文链接：https://go.coder-hub.com/19040138.html
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