So I am new to Go and fairly inexperienced with programming in general so I hope I don't get downvoted again for asking stupid questions.
I am working my way through the project euler problems and at problem 25 "1000-digit Fibonacci number" I encountered what seems to be strange behavior. The following is the code I wrote that resulted in this behavior.
package main
import (
"fmt"
"math/big"
)
func main() {
index := 2
l := new(big.Int)
pl := big.NewInt(1)
i := big.NewInt(1)
for {
l = i
i.Add(i, pl)
pl = l
index++
if len(i.String()) == 1000 {
break
}
}
fmt.Println(i, "\nindex: ", index)
}
Naturally this did not generate the correct answer so in the process of determining why I discovered that I had inadvertently discovered a neat way to generate powers of 2. I made the following changes and this did generate the correct result.
package main
import (
"fmt"
"math/big"
)
func main() {
index := 2
l := new(big.Int)
pl := big.NewInt(1)
i := big.NewInt(1)
for {
l.Set(i)
i.Add(i, pl)
pl.Set(l)
index++
if len(i.String()) == 1000 {
break
}
}
fmt.Println(i, "\nindex: ", index)
}
My question is what is happening in the first example that causes each big Int variable to be set to the value of i and why this did not generate an error if this was not the correct way to assign a big Int var value? Is i = l, etc a legitimate big Int operation that is simply incorrect for this situation?
The lines
l = i
and
pl = l
aren't doing what you think they are.
l, pl, and i are pointers, and assigning them to each other copies the pointer value, not the big.Int value.
After executing l = i, l is now the same pointer value as i, pointing to the same big.Int. When you use l.Set(i), it sets l's big.Int value to i's big.Int value, but l and i still point to two separate values.
Related
I am taking the "Tour of Go", and had a question regarding the Exercise: Slices example. Currently I can create the picture by iterating over each index using the the [] operator, just like you could in C.
func Pic(dx, dy int) [][]uint8 {
pic := make([][]uint8, dy)
for i := range pic {
pic[i] = make([]uint8, dx)
for j := range pic[i] {
pic[i][j] = uint8(1)
}
}
return pic
}
However, when I try to do something like below, I get an panic: runtime error: index out of range error. I tried adding print statements and calling Pic(3, 3), which printed out a 3x3 array just fine.
func Pic(dx, dy int) [][]uint8 {
pic := make([][]uint8, dy)
for _, y := range pic {
y = make([]uint8, dx)
for _, x := range y {
x = uint8(1)
_ = x // x has to be used
//fmt.Print("1")
}
//fmt.Print("\n")
}
return pic
}
Any thoughts on what I am doing wrong?
The main problem is your attempt to do assignment. Check my example using your code; https://play.golang.org/p/lwoe79jQ70
What you actually get out of the latter implementation is a 3x0 array, all of the inner arrays are empty. The reason for this is because you're using the range variable for assignment which doesn't work. If the current index is 0, y != pic[0], pic[0] is assigned to y however, y is temporary storage, it typically is the same address and is over written on each iteration. So after the latter example executes, all your x direction arrays are empty, indexing into one causes a panic.
Basically you should just be using your first implementation because it works fine and is the way you would typically do this. But the take away is, when you do a, b := range Something b != Something[a], it is it's on instance, it goes out of scope at the bottom of the loop and assigning to it will not cause a state change to the collection Something, instead you must assign to Something[a] if you want to modify Something[a].
range copies the values from the slice you're iterating over.
See: http://golang.org/ref/spec#RangeClause
To clarify what happens see this simple code example and its output:
package main
import "fmt"
func main() {
s := "hi"
//s[0] = 'H' // cannot assign to s[0]
for _, v := range s {
fmt.Printf("%T, %[1]v, %X\n", v, &v)
v = 'H' // has no effect: this is local var not ref
}
fmt.Println(s)
}
The output is:
int32, 104, C0820042D4
int32, 105, C0820042D4
hi
As you see the address of variable v is not changing (C0820042D4) and v is local variable and range copies value to it, so changing v has no effect.
Here v is rune (int32 alias), A rune is an integer value identifying a Unicode code point, and you cannot assign to s[0] and this won’t compile: s[0] = 'H'
so v = 'H' has no effect on s, it is just local variable.
I have a large dataset where I needed to do some string manipulation (I know strings are immutable). The Replace() function in the strings package does exactly what I need, except I need it to search in reverse.
Say I have this string: AA-BB-CC-DD-EE
Run this script:
package main
import (
"fmt"
"strings"
)
func main() {
fmt.Println(strings.Replace("AA-BB-CC-DD-EE", "-", "", 1))
}
It outputs: AABB-CC-DD-EE
What I need is: AA-BBCCDDEE, where the first instance of the search key is found, and the rest discarded.
Splitting the string, inserting the dash, and joining it back together works. But, I'm thinking there is a more performant way to achieve this.
String slices!
in := "AA-BB-CC-DD-EE"
afterDash := strings.Index(in, "-") + 1
fmt.Println(in[:afterDash] + strings.Replace(in[afterDash:], "-", "", -1))
(might require some tweaking to get the behavior you want in the case that the input has no dashes).
This can be another solution
package main
import (
"strings"
"fmt"
)
func Reverse(s string) string {
n := len(s)
runes := make([]rune, n)
for _, rune := range s {
n--
runes[n] = rune
}
return string(runes[n:])
}
func main() {
S := "AA-BB-CC-DD-EE"
S = Reverse(strings.Replace(Reverse(S), "-", "", strings.Count(S, "-")-1))
fmt.Println(S)
}
Another solution:
package main
import (
"fmt"
"strings"
)
func main() {
S := strings.Replace("AA-BB-CC-DD-EE", "-", "*", 1)
S = strings.Replace(S, "-", "", -1)
fmt.Println(strings.Replace( S, "*", "-", 1))
}
I think you want to use strings.Map rather than rigging things with compositions of functions. It's basically meant for this scenario: character replacement with more complex requirements than Replace and cousins can handle. The definition:
Map returns a copy of the string s with all its characters modified according to the mapping function. If mapping returns a negative value, the character is dropped from the string with no replacement.
Your mapping function can be built with a fairly simple closure:
func makeReplaceFn(toReplace rune, skipCount int) func(rune) rune {
count := 0
return func(r rune) rune {
if r == toReplace && count < skipCount {
count++
} else if r == toReplace && count >= skipCount {
return -1
}
return r
}
}
From there, it's a very straightforward program:
strings.Map(makeReplaceFn('-', 1), "AA-BB-CC-DD-EE")
Playground, this produces the desired output:
AA-BBCCDDEE
Program exited.
I'm not sure whether this is faster or slower than other solutions without benchmarking, because on one hand it has to call a function for each rune in the string, while on the other hand it doesn't have to convert (and thus copy) between a []byte/[]rune and string between each function call (though the subslicing answer by hobbs is probably overall the best).
In addition, the method can be easily adapted to other scenarios (e.g. retaining every other dash), with the caveat that strings.Map can only do rune to rune mapping, and not rune to string mapping like strings.Replace does.
This was a fun question to answer. While the solutions offered work neatly, splitting and replacing, to say nothing of calling Replace 3 times doesn't seem likely to be performant.
The answer? Don't reinvent the wheel, the go standard library has already almost solved this problem with Replace(), let's tweak it. I stumbled a bit over how the API of our new function should work, finally settling on leaving the signature unchanged, but deciding on minimal change from strings.Replace:
func ReplaceAfter(s,old,new string,skip int) string
The variable skip replaces n to clarify what it does since the caller will specify how many instances of old to skip replacing. skip==0 is defined as replacing every instance and skip==-1 is defined as replacing no instances.
From here there were really only a few bits of the function that needed changing.
func ReplaceAfter(s, old, new string, skip int) string {
if old == new || skip == -1 { // changed
return s // avoid allocation
}
// Compute number of replacements.
m := strings.Count(s, old)
if m == 0 || m < skip { // changed
return s // avoid allocation
} // changed (removed else if)
// Apply replacements to buffer.
n := m - skip // changed, n means the same thing but is calculated
t := make([]byte, len(s)+n*(len(new)-len(old))) // longer buffer
w := 0
start := 0
for i := 0; i < m; i++ {
j := start
if len(old) == 0 {
if i > 0 {
_, wid := utf8.DecodeRuneInString(s[start:])
j += wid
}
} else {
j += strings.Index(s[start:], old)
}
if i >= skip { // changed, replace
w += copy(t[w:], s[start:j])
w += copy(t[w:], new)
} else { // changed, skip ahead
w += copy(t[w:], s[start:j+len(old)])
}
start = j + len(old)
}
w += copy(t[w:], s[start:])
return string(t[0:w])
}
Here's a playground link with a working demo. If you're interested, I also copied and adapted the relevant Test functions from go/src/strings/, to make sure that the function as written behaved itself predictably.
Is there a way to loop over intervals between two big int values, x and y, in Go?
for i: = x; i < y; i++ {
// do something
}
Working with big numbers can be kind of clunky because you need to create a big.Int for constants. Other than that, it is a straight forward replacement of each segment of the for statement to one made to deal with big ints.
http://play.golang.org/p/pLSd8yf9Lz
package main
import (
"fmt"
"math/big"
)
var one = big.NewInt(1)
func main() {
start := big.NewInt(1)
end := big.NewInt(5)
// i must be a new int so that it does not overwrite start
for i := new(big.Int).Set(start); i.Cmp(end) < 0; i.Add(i, one) {
fmt.Println(i)
}
}
I imported the math library in my program, and I was trying to find the minimum of three numbers in the following way:
v1[j+1] = math.Min(v1[j]+1, math.Min(v0[j+1]+1, v0[j]+cost))
where v1 is declared as:
t := "stackoverflow"
v1 := make([]int, len(t)+1)
However, when I run my program I get the following error:
./levenshtein_distance.go:36: cannot use int(v0[j + 1] + 1) (type int) as type float64 in argument to math.Min
I thought it was weird because I have another program where I write
fmt.Println(math.Min(2,3))
and that program outputs 2 without complaining.
so I ended up casting the values as float64, so that math.Min could work:
v1[j+1] = math.Min(float64(v1[j]+1), math.Min(float64(v0[j+1]+1), float64(v0[j]+cost)))
With this approach, I got the following error:
./levenshtein_distance.go:36: cannot use math.Min(int(v1[j] + 1), math.Min(int(v0[j + 1] + 1), int(v0[j] + cost))) (type float64) as type int in assignment
so to get rid of the problem, I just casted the result back to int
I thought this was extremely inefficient and hard to read:
v1[j+1] = int(math.Min(float64(v1[j]+1), math.Min(float64(v0[j+1]+1), float64(v0[j]+cost))))
I also wrote a small minInt function, but I think this should be unnecessary because the other programs that make use of math.Min work just fine when taking integers, so I concluded this has to be a problem of my program and not the library per se.
Is there anything that I'm doing terrible wrong?
Here's a program that you can use to reproduce the issues above, line 36 specifically:
package main
import (
"math"
)
func main() {
LevenshteinDistance("stackoverflow", "stackexchange")
}
func LevenshteinDistance(s string, t string) int {
if s == t {
return 0
}
if len(s) == 0 {
return len(t)
}
if len(t) == 0 {
return len(s)
}
v0 := make([]int, len(t)+1)
v1 := make([]int, len(t)+1)
for i := 0; i < len(v0); i++ {
v0[i] = i
}
for i := 0; i < len(s); i++ {
v1[0] = i + 1
for j := 0; j < len(t); j++ {
cost := 0
if s[i] != t[j] {
cost = 1
}
v1[j+1] = int(math.Min(float64(v1[j]+1), math.Min(float64(v0[j+1]+1), float64(v0[j]+cost))))
}
for j := 0; j < len(v0); j++ {
v0[j] = v1[j]
}
}
return v1[len(t)]
}
Until Go 1.18 a one-off function was the standard way; for example, the stdlib's sort.go does it near the top of the file:
func min(a, b int) int {
if a < b {
return a
}
return b
}
You might still want or need to use this approach so your code works on Go versions below 1.18!
Starting with Go 1.18, you can write a generic min function which is just as efficient at run time as the hand-coded single-type version, but works with any type with < and > operators:
func min[T constraints.Ordered](a, b T) T {
if a < b {
return a
}
return b
}
func main() {
fmt.Println(min(1, 2))
fmt.Println(min(1.5, 2.5))
fmt.Println(min("Hello", "世界"))
}
There's been discussion of updating the stdlib to add generic versions of existing functions, but if that happens it won't be until a later version.
math.Min(2, 3) happened to work because numeric constants in Go are untyped. Beware of treating float64s as a universal number type in general, though, since integers above 2^53 will get rounded if converted to float64.
There is no built-in min or max function for integers, but it’s simple to write your own. Thanks to support for variadic functions we can even compare more integers with just one call:
func MinOf(vars ...int) int {
min := vars[0]
for _, i := range vars {
if min > i {
min = i
}
}
return min
}
Usage:
MinOf(3, 9, 6, 2)
Similarly here is the max function:
func MaxOf(vars ...int) int {
max := vars[0]
for _, i := range vars {
if max < i {
max = i
}
}
return max
}
For example,
package main
import "fmt"
func min(x, y int) int {
if x < y {
return x
}
return y
}
func main() {
t := "stackoverflow"
v0 := make([]int, len(t)+1)
v1 := make([]int, len(t)+1)
cost := 1
j := 0
v1[j+1] = min(v1[j]+1, min(v0[j+1]+1, v0[j]+cost))
fmt.Println(v1[j+1])
}
Output:
1
Though the question is quite old, maybe my package imath can be helpful for someone who does not like reinventing a bicycle. There are few functions, finding minimal of two integers: ix.Min (for int), i8.Min (for int8), ux.Min (for uint) and so on. The package can be obtained with go get, imported in your project by URL and functions referred as typeabbreviation.FuncName, for example:
package main
import (
"fmt"
"<Full URL>/go-imath/ix"
)
func main() {
a, b := 45, -42
fmt.Println(ix.Min(a, b)) // Output: -42
}
As the accepted answer states, with the introduction of generics in go 1.18 it's now possible to write a generic function that provides min/max for different numeric types (there is not one built into the language). And with variadic arguments we can support comparing 2 elements or a longer list of elements.
func Min[T constraints.Ordered](args ...T) T {
min := args[0]
for _, x := range args {
if x < min {
min = x
}
}
return min
}
func Max[T constraints.Ordered](args ...T) T {
max := args[0]
for _, x := range args {
if x > max {
max = x
}
}
return max
}
example calls:
Max(1, 2) // 2
Max(4, 5, 3, 1, 2) // 5
Could use https://github.com/pkg/math:
import (
"fmt"
"github.com/pkg/math"
)
func main() {
a, b := 45, -42
fmt.Println(math.Min(a, b)) // Output: -42
}
Since the issue has already been resolved, I would like to add a few words. Always remember that the math package in Golang operates on float64. You can use type conversion to cast int into a float64. Keep in mind to account for type ranges. For example, you cannot fit a float64 into an int16 if the number exceeds the limit for int16 which is 32767. Last but not least, if you convert a float into an int in Golang, the decimal points get truncated without any rounding.
If you want the minimum of a set of N integers you can use (assuming N > 0):
import "sort"
func min(set []int) int {
sort.Slice(set, func(i, j int) bool {
return set[i] < set[j]
})
return set[0]
}
Where the second argument to min function is your less function, that is, the function that decides when an element i of the passed slice is less than an element j
Check it out here in Go Playground: https://go.dev/play/p/lyQYlkwKrsA
I'm newbie in Golan, this should be an easy question for experienced golang devs. I try to do the same test from Spotify to see how fast we can go in Golang :)
The usual bit-twiddling C solutions translate immediately to Go.
package main
import "fmt"
func BitReverse32(x uint32) uint32 {
x = (x&0x55555555)<<1 | (x&0xAAAAAAAA)>>1
x = (x&0x33333333)<<2 | (x&0xCCCCCCCC)>>2
x = (x&0x0F0F0F0F)<<4 | (x&0xF0F0F0F0)>>4
x = (x&0x00FF00FF)<<8 | (x&0xFF00FF00)>>8
return (x&0x0000FFFF)<<16 | (x&0xFFFF0000)>>16
}
func main() {
cases := []uint32{0x1, 0x100, 0x1000, 0x1000000, 0x10000000, 0x80000000, 0x89abcdef}
for _, c := range cases {
fmt.Printf("%08x -> %08x\n", c, BitReverse32(c))
}
}
Note: since 2013, you now have a dedicate math/bits package with Go 1.9 (August 2017).
And it does come with a collection of Reverse() and ReverseBytes() functions: no need to implement one anymore.
Plus, on most architectures, functions in this package are additionally recognized by the compiler and treated as intrinsics for additional performance.
The most straight-forward solution would be converting the bits into a number with strconv and then reversing the number by shifting the bits. I'm not sure how fast it would be, but it should work.
package main
import "fmt"
import "strconv"
func main() {
bits := "10100001"
bits_number := 8
number, _ := strconv.ParseUint(bits, 2, bits_number)
r_number := number - number // reserve type
for i := 0; i < bits_number; i++ {
r_number <<= 1
r_number |= number & 1
number >>= 1
}
fmt.Printf("%s [%d]\n", strconv.FormatUint(r_number, 2), r_number)
}
http://play.golang.org/p/YLS5wkY-iv