英文:
How to bypass Go int64 value limits while shifting integers?
问题
我正在尝试使用Go语言获取KiB、MiB、...、ZiB、Yib的值，它们分别对应KibiByte、MebiByte、...、ZebiByte、YobiByte。
我的Golang代码如下：
package main
import (
"fmt"
)
func main() {
s := []string{"KiB", "MiB", "GiB", "TiB", "PiB", "EiB", "ZiB", "YiB"}
for k,v := range(s) {
fmt.Printf("%s: %v\n", v, 1 << uint64(10 * (k+1)))
}
}
但是，ZiB和YiB的值超出了Go的uint64范围，这就是为什么我得到以下输出的原因：
KiB: 1024
MiB: 1048576
GiB: 1073741824
TiB: 1099511627776 // 超过了1 << 32
PiB: 1125899906842624
EiB: 1152921504606846976
ZiB: 0 // 超过了1 << 64
YiB: 0 // 超过了1 << 64
另外，使用相同的位移逻辑在Python3中，可以得到以下正确的输出：
a = ["KiB", "MiB", "GiB", "TiB", "PiB", "EiB", "ZiB", "YiB"]
for k,v in enumerate(a):
print("{}: {}".format(v, 1 << (10 *(k+1))))
输出结果如下：
KiB: 1024
MiB: 1048576
GiB: 1073741824
TiB: 1099511627776
PiB: 1125899906842624
EiB: 1152921504606846976
ZiB: 1180591620717411303424
YiB: 1208925819614629174706176
所以，我应该如何绕过Go的uint64限制，使用位移整数来获取正确的值，就像在Python中使用位移整数一样。
谢谢。
英文:
I'm trying with Go to get values of KiB, MiB, ..., ZiB, Yib which are respectively KibiByte, MebiByte, ..., ZebiByte, YobiByte.
My code in Golang is:
package main
import (
"fmt"
)
func main() {
s := []string{"KiB", "MiB", "GiB", "TiB", "PiB", "EiB", "ZiB", "YiB"}
for k,v := range(s) {
fmt.Printf("%s: %v\n", v, 1 << uint64(10 * (k+1)))
}
}
But, the values of ZiB and YiB overflows Go uint64 and this why I'm having this output:
KiB: 1024
MiB: 1048576
GiB: 1073741824
TiB: 1099511627776 // exceeds 1 << 32
PiB: 1125899906842624
EiB: 1152921504606846976
ZiB: 0 // exceeds 1 << 64
YiB: 0 // exceeds 1 << 64
Otherwise, with the same shifting logic in Python3 within this code:
a = ["KiB", "MiB", "GiB", "TiB", "PiB", "EiB", "ZiB", "YiB"]
for k,v in enumerate(a):
print("{}: {}".format(v, 1 << (10 *(k+1))))
The output is correct, like the output below:
KiB: 1024
MiB: 1048576
GiB: 1073741824
TiB: 1099511627776
PiB: 1125899906842624
EiB: 1152921504606846976
ZiB: 1180591620717411303424
YiB: 1208925819614629174706176
So, how can I bypass Go uint64 limits and get the correct values using shifting integers like what I can get from shifting integers using Python.
Thanks.
答案1
得分: 9
你不能使用原始的uint64来处理需要超过64位的数字。Python具有任意精度整数，要在Go中实现相同的功能，你需要使用math/big包。
s := []string{"KiB", "MiB", "GiB", "TiB", "PiB", "EiB", "ZiB", "YiB"}
one := big.NewInt(1)
for k, v := range s {
fmt.Printf("%s: %v\n", v, new(big.Int).Lsh(one, uint(10*(k+1))))
}
链接：https://play.golang.org/p/i5v5P5QgQb
英文:
You can't work with numbers that require more than 64bits with a primitive uint64. Python has arbitrary precision integers, and to get the same in Go you need to use the math/big package.
s := []string{"KiB", "MiB", "GiB", "TiB", "PiB", "EiB", "ZiB", "YiB"}
one := big.NewInt(1)
for k, v := range s {
fmt.Printf("%s: %v\n", v, new(big.Int).Lsh(one, uint(10*(k+1))))
}
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