BigInt pointer in Go

Question

I was playing around with various packages and data types and I came across this function(big.NewInt()) from "math/big" package. so this function returns a pointer to the type bigInt but when i print it out i get the actual value not the address where it is stored can anyone help me understand this better?

package main

import "fmt"
import "math/big"
import "reflect"

func main(){
    target := big.NewInt(1)
    a := 5
    s := &a
    fmt.Println(reflect.TypeOf(target))
    fmt.Println(reflect.TypeOf(s))
    fmt.Println(s)
    fmt.Println(target)
}

This is a feature of fmt. Which invokes the appropriate String function and has nothing to do withe pointers or big.Int. — Volker, Jun 04 '20 at 12:36
but y does it automatically print the address in case of a integer pointer and not in case of a big int pointer — Chandru, Jun 04 '20 at 12:40
@Chandru: because an `*int` does not have a `String()` method to make it a `fmt.Stringer`. — JimB, Jun 04 '20 at 12:48

score 1 · Accepted Answer · answered Jun 04 '20 at 12:36

1

fmt.Printf("%p\n", target)      // 0xc00000c080

answered Jun 04 '20 at 12:36

aMike

852
5
14

which value does the target hold?"0xc000000c080" or 1? – Chandru Jun 04 '20 at 12:41
It printed '1' before it the address. And this address was "around" the first address that it printed. My test was just a copy and paste of your example with that one line added. – aMike Jun 04 '20 at 13:26

BigInt pointer in Go

1 Answers1