I wrote a function to convert a byte slice to an integer.

The function I created is actually a loop-based implemtation of what Rob Pike published here:

http://commandcenter.blogspot.com/2012/04/byte-order-fallacy.html

Here is Rob's code:

i = (data[0]<<0) | (data[1]<<8) | (data[2]<<16) | (data[3]<<24);

My first implementation (toInt2 in the playground) doesn't work as I expected because it appears to initialize the int value as a uint. This seems really strange but it must be platform specific because the go playground reports a different result than my machine (a mac).

Can anyone explain why these functions behave differently on my mac?

Here's the link to the playground with the code: http://play.golang.org/p/FObvS3W4UD

Here's the code from the playground (for convenience):

/*

Output on my machine:

    amd64 darwin go1.3 input: [255 255 255 255]
    -1
    4294967295

Output on the go playground:

    amd64p32 nacl go1.3 input: [255 255 255 255]
    -1
    -1

*/

package main

import (
    "fmt"
    "runtime"
)

func main() {
    input := []byte{255, 255, 255, 255}
    fmt.Println(runtime.GOARCH, runtime.GOOS, runtime.Version(), "input:", input)
    fmt.Println(toInt(input))
    fmt.Println(toInt2(input))
}

func toInt(bytes []byte) int {
    var value int32 = 0 // initialized with int32

    for i, b := range bytes {
        value |= int32(b) << uint(i*8)
    }
    return int(value) // converted to int
}

func toInt2(bytes []byte) int {
    var value int = 0 // initialized with plain old int

    for i, b := range bytes {
        value |= int(b) << uint(i*8)
    }
    return value
}

vector<int> vi(10); 报错

zhch152

2个回复