Hi i am new to the golang programming language. I can get the bigint
value from the factoral function but it is not working with the add
function.
i have had the add function accepting bigint but when i try to add a
.Mod and .Div methods it returns 0 fro some reason. the if else
statement in the add function is an old statement i was using when i
had int values coming from the factoral function.
it worked perfectly when it was as int value. When i attempted to
alter the if else statement to accept bigint values i couldn't get it
to work at all.
I have tried the .Mod and .Div methods and they are printing out the
correct values. But when i try to .Add them together it always returns
0. even if the values are "22", "2". I've put the if else statement back to the original int values for now.
if anyone can help me out with this one i would be very greatful.
package main
import (
"fmt"
"math/big"
)
func factoral(n uint64) (r *big.Int) {
one, bn := big.NewInt(1), new(big.Int).SetUint64(n)
r = big.NewInt(1)
if bn.Cmp(one) <= 0 {
return
}
for i := big.NewInt(2); i.Cmp(bn) <= 0; i.Add(i, one) {
r.Mul(r, i)
}
return
}
func add(number *big.Int) *big.Int {
//this the statement that works with normal int values
if number/10 < 10 {
return sum + number/10
} else {
return sum + add(number/10)
}
}
func main() {
fmt.Println(add(factoral(100)))
}
Fixed it
package main
import (
"fmt"
"math/big"
)
func factoral(n uint64) (r *big.Int) {
one, bn := big.NewInt(1), new(big.Int).SetUint64(n)
r = big.NewInt(1)
if bn.Cmp(one) <= 0 {
return
}
for i := big.NewInt(2); i.Cmp(bn) <= 0; i.Add(i, one) {
r.Mul(r, i)
}
return
}
func add(number *big.Int) *big.Int {
ten := big.NewInt(10)
sum := big.NewInt(0)
mod := big.NewInt(0)
for ten.Cmp(number)<0 {
sum.Add(sum, mod.Mod(number,ten))
number.Div(number,ten)
}
sum.Add(sum,number)
return sum
}
func main() {
fmt.Println(add(factoral(100)))
}
Seems your issue was likely with the way the Big Int object whose method you invoke is going to be the one the value is assigned to, and not necessarily one of the arguments.
See: https://golang.org/pkg/math/big/#Int.Div
Related
I am trying to implement fibonacci recursion in golang using n goroutines with communicating via channels.
I am returning an integer from the function, but i am actually just sending the sum of f(n-1) +f(n-2) over channel c but this is not working correctly. It prints the first two values correct, and every value after is just 1.
package main
import "fmt"
// Fibonacci in a recursive version
func fiboR(n int, c chan int ) int {
if(n == 0){
c <- 0
return 0
} else if n == 1 {
c <- 1
return 1
} else{
c <- fiboR(n-1,c) + fiboR(n-2,c)
return fiboR(n-1,c) + fiboR(n-2,c)
}
}
func main() {
for i := 0; i < 10; i++ {
procchan := make(chan int)
go fiboR(i,procchan)
fmt.Println(i,<-procchan )
}
}
Also is it possible to use channels for receiving the two recursive calls?
Your solution will try to output more than the one value you extract from the channel as you increase the value of i.
What your code will try to send to the channel for each i:
0: 0
1: 1
2: 1,0,1
3: 1,0,1,1,2
4: 1,0,1,1,2,1,0,1,3
...
Since you create a new channel for each i and then only extract one value you will always get the first value in the line above.
If you try to run it with these modifications it will output what you wanted (https://play.golang.org/p/_mn3l5x8iZ).
package main
import "fmt"
// Fibonacci in a recursive version
func fiboRchan(n int, c chan int) {
c <- fiboR(n)
}
func fiboR(n int) int {
if n == 0 {
return 0
} else if n == 1 {
return 1
} else {
return fiboR(n-1) + fiboR(n-2)
}
}
func main() {
for i := 0; i < 10; i++ {
procchan := make(chan int)
go fiboRchan(i, procchan)
fmt.Println(i, <-procchan)
}
}
Adding to #nissefors answer, the main process there is most likely a sequential one because in the for loop you would be waiting on the channel to return and then proceed to the next iteration.
A minor modification in the main function could fire all the fibonaccis at once and then in a separate for loop the channels that are corresponding to each go routine can be accessed
Playground URL: https://play.golang.org/p/7e3JnWeSp6
package main
import "fmt"
// Fibonacci in a recursive version
func fiboRchan(n int, c chan int) {
fmt.Println("PROCESSING FOR %d", n)
c <- fiboR(n)
}
func fiboR(n int) int {
if n == 0 {
return 0
} else if n == 1 {
return 1
} else {
return fiboR(n-1) + fiboR(n-2)
}
}
func main() {
var arr[10]chan int
for i := 0; i < 10; i++ {
procchan := make(chan int)
arr[i] = procchan
go fiboRchan(i, procchan)
}
// By now all the go routines are fired
// Now iterate through the channel array and read from the
// respective channel
for i:=0; i< 10; i++ {
fmt.Println(i, <-arr[i])
}
}
package main
import (
"fmt"
)
func main(){
f,val,val1:=fibonacci()
fmt.Println(val,val1)
for i:=0;i<=10;i++ {
fmt.Println(f(i),val,val1)
}
_,val,val1=fibonacci()
fmt.Println(val,val1)
}
func fibonacci()(func(n int)int,int,int){
var val int
var val1 int
f:=func(n int)int{
if n==0||n==1{
val,val1=1,1
}else{
val,val1=val+val1,val
}
return val
}
fmt.Println("fibonacci val =",val,"val1 =",val1)
return f,val,val1
}
Here is my code on sloving fibonacci without using recursion when I
read about lambda function/closure. And the Go Documentary says a
closure will capture some external state. My understanding is the
closure will keep a copy of state of the function which it is
declared. These states are just copies whatever I do on them won't
modify the original, is that so?
from your test code here: https://play.golang.org/p/pajT2bAIe2
your fibonacci function is working out the nth numbers in the sequence provided it's called in an incremental fashion as you are doing. but the values you return from the initial call to fibonacci are not pointers (or references) to those integer values they are just the values of those integers at that time, think of them as being copied out of the function, try using integer pointers instead like this: https://play.golang.org/p/-vLja7Fpsq
package main
import (
"fmt"
)
func main() {
f, val, val1 := fibonacci()
fmt.Println(val, val1)
for i := 0; i <= 10; i++ {
fmt.Println(f(i), *val, *val1) //dereference the pointer to get its value at the current time
}
_, val, val1 = fibonacci()
fmt.Println(*val, *val1)
}
func fibonacci() (func(n int) int, *int, *int) {
var val int
var val1 int
f := func(n int) int {
if n == 0 || n == 1 {
val, val1 = 1, 1
} else {
val, val1 = val+val1, val
}
return val
}
fmt.Println("fibonacci val =", val, "val1 =", val1)
return f, &val, &val1 // return pointers to the closured values instead of just the values
}
Although you have accepted the above answer i'm giving another explanation. The reason why you receive the last value of the loop operation has to do with the Go's lexical scoping. The for loop introduces a new lexical block in which the value is referenced by it's memory address, so by pointer and not by it's value. In order to get the value, you have to de-reference.
Each time the for loop makes an iteration the value processed is pointing to the same memory address. All the function values created by this loop "capture" and share the same variable - and addressable storage location, not it's value at that particular moment. Thus when the last iteration is finished, the variable holds the value from the final step.
A much better approach for these kind of operations would be to use goroutines, because in these cases you are not communicating through sharing the same memory address, but you are sharing the memory through communication.
Here is a more elegant solution using goroutine:
package main
import (
"fmt"
)
func fibonacci(ch chan interface{}, quit chan struct{}) {
x, y := 1, 1
for {
select {
case ch <- x:
x, y = y, x+y
fmt.Println(x , y)
case <-quit:
fmt.Println("Quiting...")
return
}
}
}
func main() {
ch := make(chan interface{})
quit := make(chan struct{})
go func() {
for i := 0; i < 10; i++ {
<-ch
}
quit <- struct{}{}
}()
fibonacci(ch, quit)
}
https://play.golang.org/p/oPQgXWyV9u
I need to convert an int32 to string in Golang. Is it possible to convert int32 to string in Golang without converting to int or int64 first?
Itoa needs an int. FormatInt needs an int64.
One line answer is fmt.Sprint(i).
Anyway there are many conversions, even inside standard library function like fmt.Sprint(i), so you have some options (try The Go Playground):
1- You may write your conversion function (Fastest):
func String(n int32) string {
buf := [11]byte{}
pos := len(buf)
i := int64(n)
signed := i < 0
if signed {
i = -i
}
for {
pos--
buf[pos], i = '0'+byte(i%10), i/10
if i == 0 {
if signed {
pos--
buf[pos] = '-'
}
return string(buf[pos:])
}
}
}
2- You may use fmt.Sprint(i) (Slow)
See inside:
// Sprint formats using the default formats for its operands and returns the resulting string.
// Spaces are added between operands when neither is a string.
func Sprint(a ...interface{}) string {
p := newPrinter()
p.doPrint(a)
s := string(p.buf)
p.free()
return s
}
3- You may use strconv.Itoa(int(i)) (Fast)
See inside:
// Itoa is shorthand for FormatInt(int64(i), 10).
func Itoa(i int) string {
return FormatInt(int64(i), 10)
}
4- You may use strconv.FormatInt(int64(i), 10) (Faster)
See inside:
// FormatInt returns the string representation of i in the given base,
// for 2 <= base <= 36. The result uses the lower-case letters 'a' to 'z'
// for digit values >= 10.
func FormatInt(i int64, base int) string {
_, s := formatBits(nil, uint64(i), base, i < 0, false)
return s
}
Comparison & Benchmark (with 50000000 iterations):
s = String(i) takes: 5.5923198s
s = String2(i) takes: 5.5923199s
s = strconv.FormatInt(int64(i), 10) takes: 5.9133382s
s = strconv.Itoa(int(i)) takes: 5.9763418s
s = fmt.Sprint(i) takes: 13.5697761s
Code:
package main
import (
"fmt"
//"strconv"
"time"
)
func main() {
var s string
i := int32(-2147483648)
t := time.Now()
for j := 0; j < 50000000; j++ {
s = String(i) //5.5923198s
//s = String2(i) //5.5923199s
//s = strconv.FormatInt(int64(i), 10) // 5.9133382s
//s = strconv.Itoa(int(i)) //5.9763418s
//s = fmt.Sprint(i) // 13.5697761s
}
fmt.Println(time.Since(t))
fmt.Println(s)
}
func String(n int32) string {
buf := [11]byte{}
pos := len(buf)
i := int64(n)
signed := i < 0
if signed {
i = -i
}
for {
pos--
buf[pos], i = '0'+byte(i%10), i/10
if i == 0 {
if signed {
pos--
buf[pos] = '-'
}
return string(buf[pos:])
}
}
}
func String2(n int32) string {
buf := [11]byte{}
pos := len(buf)
i, q := int64(n), int64(0)
signed := i < 0
if signed {
i = -i
}
for {
pos--
q = i / 10
buf[pos], i = '0'+byte(i-10*q), q
if i == 0 {
if signed {
pos--
buf[pos] = '-'
}
return string(buf[pos:])
}
}
}
The Sprint function converts a given value to string.
package main
import (
"fmt"
)
func main() {
var sampleInt int32 = 1
sampleString := fmt.Sprint(sampleInt)
fmt.Printf("%+V %+V\n", sampleInt, sampleString)
}
// %!V(int32=+1) %!V(string=1)
See this example.
Use a conversion and strconv.FormatInt to format int32 values as a string. The conversion has zero cost on most platforms.
s := strconv.FormatInt(int64(n), 10)
If you have many calls like this, consider writing a helper function similar to strconv.Itoa:
func formatInt32(n int32) string {
return strconv.FormatInt(int64(n), 10)
}
All of the low-level integer formatting code in the standard library works with int64 values. Any answer to this question using formatting code in the standard library (fmt package included) requires a conversion to int64 somewhere. The only way to avoid the conversion is to write formatting function from scratch, but there's little point in doing that.
func FormatInt32(value int32) string {
return fmt.Sprintf("%d", value)
}
Does this work?
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 am trying to implement this bit of code:
func factorial(x int) (result int) {
if x == 0 {
result = 1;
} else {
result = x * factorial(x - 1);
}
return;
}
as a big.Int so as to make it effective for larger values of x.
The following is returning a value of 0 for fmt.Println(factorial(r))
The factorial of 7 should be 5040?
Any ideas on what I am doing wrong?
package main
import "fmt"
import "math/big"
func main() {
fmt.Println("Hello, playground")
//n := big.NewInt(40)
r := big.NewInt(7)
fmt.Println(factorial(r))
}
func factorial(n *big.Int) (result *big.Int) {
//fmt.Println("n = ", n)
b := big.NewInt(0)
c := big.NewInt(1)
if n.Cmp(b) == -1 {
result = big.NewInt(1)
}
if n.Cmp(b) == 0 {
result = big.NewInt(1)
} else {
// return n * factorial(n - 1);
fmt.Println("n = ", n)
result = n.Mul(n, factorial(n.Sub(n, c)))
}
return result
}
This code on go playground: http://play.golang.org/p/yNlioSdxi4
Go package math.big has func (*Int) MulRange(a, b int64). When called with the first parameter set to 1, it will return b!:
package main
import (
"fmt"
"math/big"
)
func main() {
x := new(big.Int)
x.MulRange(1, 10)
fmt.Println(x)
}
Will produce
3628800
In your int version, every int is distinct. But in your big.Int version, you're actually sharing big.Int values. So when you say
result = n.Mul(n, factorial(n.Sub(n, c)))
The expression n.Sub(n, c) actually stores 0 back into n, so when n.Mul(n, ...) is evaluated, you're basically doing 0 * 1 and you get back 0 as a result.
Remember, the results of big.Int operations don't just return their value, they also store them into the receiver. This is why you see repetition in expressions like n.Mul(n, c), e.g. why it takes n again as the first parameter. Because you could also sayresult.Mul(n, c) and you'd get the same value back, but it would be stored in result instead of n.
Here is your code rewritten to avoid this problem:
func factorial(n *big.Int) (result *big.Int) {
//fmt.Println("n = ", n)
b := big.NewInt(0)
c := big.NewInt(1)
if n.Cmp(b) == -1 {
result = big.NewInt(1)
}
if n.Cmp(b) == 0 {
result = big.NewInt(1)
} else {
// return n * factorial(n - 1);
fmt.Println("n = ", n)
result = new(big.Int)
result.Set(n)
result.Mul(result, factorial(n.Sub(n, c)))
}
return
}
And here is a slightly more cleaned-up/optimized version (I tried to remove extraneous allocations of big.Ints): http://play.golang.org/p/feacvk4P4O
For example,
package main
import (
"fmt"
"math/big"
)
func factorial(x *big.Int) *big.Int {
n := big.NewInt(1)
if x.Cmp(big.NewInt(0)) == 0 {
return n
}
return n.Mul(x, factorial(n.Sub(x, n)))
}
func main() {
r := big.NewInt(7)
fmt.Println(factorial(r))
}
Output:
5040
Non-recursive version:
func FactorialBig(n uint64) (r *big.Int) {
//fmt.Println("n = ", n)
one, bn := big.NewInt(1), new(big.Int).SetUint64(n)
r = big.NewInt(1)
if bn.Cmp(one) <= 0 {
return
}
for i := big.NewInt(2); i.Cmp(bn) <= 0; i.Add(i, one) {
r.Mul(r, i)
}
return
}
playground