fmt: go1.9 does not properly process big integer

[Codefor]

Please answer these questions before submitting your issue. Thanks!

What version of Go are you using (go version)?

go version go1.9 linux/amd64

Does this issue reproduce with the latest release?

Yes

What operating system and processor architecture are you using (go env)?

GOARCH="amd64"
GOBIN=""
GOEXE=""
GOHOSTARCH="amd64"
GOHOSTOS="linux"
GOOS="linux"
GOPATH="/root/go"
GORACE=""
GOROOT="/usr/local/go"
GOTOOLDIR="/usr/local/go/pkg/tool/linux_amd64"
GCCGO="gccgo"
CC="gcc"
GOGCCFLAGS="-fPIC -m64 -pthread -fmessage-length=0 -fdebug-prefix-map=/tmp/go-build626720835=/tmp/go-build -gno-record-gcc-switches"
CXX="g++"
CGO_ENABLED="1"
CGO_CFLAGS="-g -O2"
CGO_CPPFLAGS=""
CGO_CXXFLAGS="-g -O2"
CGO_FFLAGS="-g -O2"
CGO_LDFLAGS="-g -O2"
PKG_CONFIG="pkg-config"

What did you do?

package main

import (
    "fmt"
)

func main() {
    var (
        a int64 = 83520398946977
        b int64 = 23200110818
        c int64 = 3600
    )

    //shows 3600, expected 3600
    fmt.Println(a / b)

    //shows 83520398944800, expected 83520398946977
    fmt.Println(b * c)

    //shows false, expected true
    fmt.Println(a == b*c)
}

What did you expect to see?

a = b*c

What did you see instead?

a != b*c
c != a/b

[mvdan]

shows 83520398944800, expected 83520398946977

Where did you get that "expected" value? I tried calc as well as Google and they both give 83520398944800.

This value is within the bounds of an int64, and these aren't really big ints, so I would be very surprised if Go was actually wrong here.

[randall77]

Yes, I would agree that your expectations are wrong.

[Codefor]

//shows 3600, expected 3600
fmt.Println(a / b)

Here a/b = 3600, which c is 3600.So shouldn't b *c equals to a?

https://play.golang.org/p/1Y9wPCerkDd

@mvdan @randall77

[Codefor]

sorry，I see.

a % b != 0

[mvdan]

Remember to use https://github.com/golang/go/wiki/Questions for questions - the issue tracker is just for bugs and proposals.