golang - string permutation - slice bounds out of range - string

Here is the Go Playground code for this problem.
I am trying to write a golang string permutation using recursion.
The permutation function takes two arguments, prefix which is an empty string ("") and str which is "abc". The code is below
func main() {
str := "abc"
perm("", str)
}
func perm(prefix string, str string) {
n := len(str)
fmt.Println(n)
if n == 0 {
fmt.Println(prefix)
} else {
for i := 0; i < n; n++ {
perm(prefix+str[i:i+1], str[0:i]+str[(i+1):n])
}
}
}
When I run this code, I am shown 3,2,1,0 for the value of n as expected.
I successfully get "abc" but then I get a "panic: runtime error: slice bounds out of range" error.
It never shows the second round of 3,2,1,0 so it doesn't even get to the b combinations. I feel like if an error would occur is when it gets to the c portion of the string but since it doesn't even get to the b portion, I am not sure what is wrong.

Simple:
for i := 0; i < n; n++ {
^^^
?
Replace n by i, which is supposed to go from 0 to n-1.
Resulting playground.

Related

Can I make a prefilled string in golang with make or new?

I am trying to optimize my stringpad library in Go. So far the only way I have found to fill a string (actually bytes.Buffer) with a known character value (ex. 0 or " ") is with a for loop.
the snippet of code is:
// PadLeft pads string on left side with p, c times
func PadLeft(s string, p string, c int) string {
var t bytes.Buffer
if c <= 0 {
return s
}
if len(p) < 1 {
return s
}
for i := 0; i < c; i++ {
t.WriteString(p)
}
t.WriteString(s)
return t.String()
}
The larger the string pad I believe there is more memory copies of the t buffer. Is there a more elegant way to make a known size buffer with a known value on initialization?
You can only use make() and new() to allocate buffers (byte slices or arrays) that are zeroed. You may use composite literals to obtain slices or arrays that initially contain non-zero values, but you can't describe the initial values dynamically (indices must be constants).
Take inspiration from the similar but very efficient strings.Repeat() function. It repeats the given string with given count:
func Repeat(s string, count int) string {
// Since we cannot return an error on overflow,
// we should panic if the repeat will generate
// an overflow.
// See Issue golang.org/issue/16237
if count < 0 {
panic("strings: negative Repeat count")
} else if count > 0 && len(s)*count/count != len(s) {
panic("strings: Repeat count causes overflow")
}
b := make([]byte, len(s)*count)
bp := copy(b, s)
for bp < len(b) {
copy(b[bp:], b[:bp])
bp *= 2
}
return string(b)
}
strings.Repeat() does a single allocation to obtain a working buffer (which will be a byte slice []byte), and uses the builtin copy() function to copy the repeatable string. One thing noteworthy is that it uses the working copy and attempts to copy the whole of it incrementally, meaning e.g. if the string has already been copied 4 times, copying this buffer will make it 8 times, etc. This will minimize the calls to copy(). Also the solution takes advantage of that copy() can copy bytes from a string without having to convert it to a byte slice.
What we want is something similar, but we want the result to be prepended to a string.
We can account for that, simply allocating a buffer that is used inside Repeat() plus the length of the string we're left-padding.
The result (without checking the count param):
func PadLeft(s, p string, count int) string {
ret := make([]byte, len(p)*count+len(s))
b := ret[:len(p)*count]
bp := copy(b, p)
for bp < len(b) {
copy(b[bp:], b[:bp])
bp *= 2
}
copy(ret[len(b):], s)
return string(ret)
}
Testing it:
fmt.Println(PadLeft("aa", "x", 1))
fmt.Println(PadLeft("aa", "x", 2))
fmt.Println(PadLeft("abc", "xy", 3))
Output (try it on the Go Playground):
xaa
xxaa
xyxyxyabc
See similar / related question: Is there analog of memset in go?

Bitmasking conversion of CPU ids with Go

I have a mask that contains a binary counting of cpu_ids (0xA00000800000 for 3 CPUs) which I want to convert into a string of comma separated cpu_ids: "0,2,24".
I did the following Go implementation (I am a Go starter). Is it the best way to do it? Especially the handling of byte buffers seems to be inefficient!
package main
import (
"fmt"
"os"
"os/exec"
)
func main(){
cpuMap := "0xA00000800000"
cpuIds = getCpuIds(cpuMap)
fmt.Println(cpuIds)
}
func getCpuIds(cpuMap string) string {
// getting the cpu ids
cpu_ids_i, _ := strconv.ParseInt(cpuMap, 0, 64) // int from string
cpu_ids_b := strconv.FormatInt(cpu_ids_i, 2) // binary as string
var buff bytes.Buffer
for i, runeValue := range cpu_ids_b {
// take care! go returns code points and not the string
if runeValue == '1' {
//fmt.Println(bitString, i)
buff.WriteString(fmt.Sprintf("%d", i))
}
if (i+1 < len(cpu_ids_b)) && (runeValue == '1') {
//fmt.Println(bitString)
buff.WriteString(string(","))
}
}
cpuIds := buff.String()
// remove last comma
cpuIds = cpuIds[:len(cpuIds)-1]
//fmt.Println(cpuIds)
return cpuIds
}
Returns:
"0,2,24"
What you're doing is essentially outputting the indices of the "1"'s in the binary representation from left-to-right, and starting index counting from the left (unusal).
You can achieve the same using bitmasks and bitwise operators, without converting it to a binary string. And I would return a slice of indices instead of its formatted string, easier to work with.
To test if the lowest (rightmost) bit is 1, you can do it like x&0x01 == 1, and to shift a whole number bitwise to the right: x >>= 1. After a shift, the rightmost bit "disappears", and the previously 2nd bit becomes the 1st, so you can test again with the same logic. You may loop until the number is greater than 0 (which means it sill has 1-bits).
See this question for more examples of bitwise operations: Difference between some operators "|", "^", "&", "&^". Golang
Of course if we test the rightmost bit and shift right, we get the bits (indices) in reverse order (compared to what you want), and the indices are counted from right, so we have to correct this before returning the result.
So the solution looks like this:
func getCpuIds(cpuMap string) (r []int) {
ci, err := strconv.ParseInt(cpuMap, 0, 64)
if err != nil {
panic(err)
}
count := 0
for ; ci > 0; count, ci = count+1, ci>>1 {
if ci&0x01 == 1 {
r = append(r, count)
}
}
// Indices are from the right, correct it:
for i, v := range r {
r[i] = count - v - 1
}
// Result is in reverse order:
for i, j := 0, len(r)-1; i < j; i, j = i+1, j-1 {
r[i], r[j] = r[j], r[i]
}
return
}
Output (try it on the Go Playground):
[0 2 24]
If for some reason you need the result as a comma separated string, this is how you can obtain that:
buf := &bytes.Buffer{}
for i, v := range cpuIds {
if i > 0 {
buf.WriteString(",")
}
buf.WriteString(strconv.Itoa(v))
}
cpuIdsStr := buf.String()
fmt.Println(cpuIdsStr)
Output (try it on the Go Playground):
0,2,24

Go: convert rune (string) to string representation of the binary

This is just in case someone else is learning Golang and is wondering how to convert from a string to a string representation in binary.
Long story short, I have been looking at the standard library without being able to find the right call. So I started with something similar to the following:
func RuneToBinary(r rune) string {
var buf bytes.Buffer
b := []int64{128, 64, 32, 16, 8, 4, 2, 1}
v := int64(r)
for i := 0; i < len(b); i++ {
t := v-b[i]
if t >= 0 {
fmt.Fprintf(&buf, "1")
v = t
} else {
fmt.Fprintf(&buf, "0")
}
}
return buf.String()
}
This is all well and dandy, but after a couple of days looking around I found that I should have been using the fmt package instead and just format the rune with %b%:
var r rune
fmt.Printf("input: %b ", r)
Is there a better way to do this?
Thanks
Standard library support
fmt.Printf("%b", r) - this solution is already very compact and easy to write and understand. If you need the result as a string, you can use the analog Sprintf() function:
s := fmt.Sprintf("%b", r)
You can also use the strconv.FormatInt() function which takes a number of type int64 (so you first have to convert your rune) and a base where you can pass 2 to get the result in binary representation:
s := strconv.FormatInt(int64(r), 2)
Note that in Go rune is just an alias for int32, the 2 types are one and the same (just you may refer to it by 2 names).
Doing it manually ("Simple but Naive"):
If you'd want to do it "manually", there is a much simpler solution than your original. You can test the lowest bit with r & 0x01 == 0 and shift all bits with r >>= 1. Just "loop" over all bits and append either "1" or "0" depending on the bit:
Note this is just for demonstration, it is nowhere near optimal regarding performance (generates "redundant" strings):
func RuneToBin(r rune) (s string) {
if r == 0 {
return "0"
}
for digits := []string{"0", "1"}; r > 0; r >>= 1 {
s = digits[r&1] + s
}
return
}
Note: negative numbers are not handled by the function. If you also want to handle negative numbers, you can first check it and proceed with the positive value of it and start the return value with a minus '-' sign. This also applies the other manual solution below.
Manual Performance-wise solution:
For a fast solution we shouldn't append strings. Since strings in Go are just byte slices encoded using UTF-8, appending a digit is just appending the byte value of the rune '0' or '1' which is just one byte (not multi). So we can allocate a big enough buffer/array (rune is 32 bits so max 32 binary digits), and fill it backwards so we won't even have to reverse it at the end. And return the used part of the array converted to string at the end. Note that I don't even call the built-in append function to append the binary digits, I just set the respective element of the array in which I build the result:
func RuneToBinFast(r rune) string {
if r == 0 {
return "0"
}
b, i := [32]byte{}, 31
for ; r > 0; r, i = r>>1, i-1 {
if r&1 == 0 {
b[i] = '0'
} else {
b[i] = '1'
}
}
return string(b[i+1:])
}

How can I assign a new char into a string in Go?

I'm trying to alter an existing string in Go but I keep getting this error "cannot assign to new_str[i]"
package main
import "fmt"
func ToUpper(str string) string {
new_str := str
for i:=0; i<len(str); i++{
if str[i]>='a' && str[i]<='z'{
chr:=uint8(rune(str[i])-'a'+'A')
new_str[i]=chr
}
}
return new_str
}
func main() {
fmt.Println(ToUpper("cdsrgGDH7865fxgh"))
}
This is my code, I wish to change lowercase to uppercase but I cant alter the string. Why? How can I alter it?
P.S I wish to use ONLY the fmt package!
Thanks in advance.
You can't... they are immutable. From the Golang Language Specification:
Strings are immutable: once created, it is impossible to change the contents of a string.
You can however, cast it to a []byte slice and alter that:
func ToUpper(str string) string {
new_str := []byte(str)
for i := 0; i < len(str); i++ {
if str[i] >= 'a' && str[i] <= 'z' {
chr := uint8(rune(str[i]) - 'a' + 'A')
new_str[i] = chr
}
}
return string(new_str)
}
Working sample: http://play.golang.org/p/uZ_Gui7cYl
Use range and avoid unnecessary conversions and allocations. Strings are immutable. For example,
package main
import "fmt"
func ToUpper(s string) string {
var b []byte
for i, c := range s {
if c >= 'a' && c <= 'z' {
if b == nil {
b = []byte(s)
}
b[i] = byte('A' + rune(c) - 'a')
}
}
if b == nil {
return s
}
return string(b)
}
func main() {
fmt.Println(ToUpper("cdsrgGDH7865fxgh"))
}
Output:
CDSRGGDH7865FXGH
In Go strings are immutable. Here is one very bad way of doing what you want (playground)
package main
import "fmt"
func ToUpper(str string) string {
new_str := ""
for i := 0; i < len(str); i++ {
chr := str[i]
if chr >= 'a' && chr <= 'z' {
chr = chr - 'a' + 'A'
}
new_str += string(chr)
}
return new_str
}
func main() {
fmt.Println(ToUpper("cdsrgGDH7865fxgh"))
}
This is bad because
you are treating your string as characters - what if it is UTF-8? Using range str is the way to go
appending to strings is slow - lots of allocations - a bytes.Buffer would be a good idea
there is a very good library routine to do this already strings.ToUpper
It is worth exploring the line new_str += string(chr) a bit more. Strings are immutable, so what this does is make a new string with the chr on the end, it doesn't extend the old string. This is wildly inefficient for long strings as the allocated memory will tend to the square of the string length.
Next time just use strings.ToUpper!

Find index of a substring in a string, with start index specified

I know there is strings.Index and strings.LastIndex, but they just find the first and last. Is there any function I can use, where I can specify the start index? Something like the last line in my example.
Example:
s := "go gopher, go"
fmt.Println(strings.Index(s, "go")) // Position 0
fmt.Println(strings.LastIndex(s, "go")) // Postion 11
fmt.Println(strings.Index(s, "go", 1)) // Position 3 - Start looking for "go" begining at index 1
It's an annoying oversight, you have to create your own function.
Something like:
func indexAt(s, sep string, n int) int {
idx := strings.Index(s[n:], sep)
if idx > -1 {
idx += n
}
return idx
}
No, but it might be simpler to apply strings.Index on a slice of the string
strings.Index(s[1:], "go")+1
strings.Index(s[n:], "go")+n
See example (for the case where the string isn't found, see OneOfOne's answer), but, as commented by Dewy Broto, one can simply test it with a 'if' statement including a simple statement:
(also called 'if' with an initialization statement)
if i := strings.Index(s[n:], sep) + n; i >= n {
...
}

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