使用GO语言将ASCII数字字符的字节数组转换为int

Question

I see some answers to exactly the same question I have: How to convert Byte array to int in GO programming language?

I wrote below function to convert byte array to int

func convertByteToInt(in []byte) int32 {
    return  (int32(in[0]) << 24 | int32(in[1]) << 16 | int32(in[2]) << 8 | int32(in[3]))
}

Before that, I made sure that byte array has correct(base 256) values. in[0] = 54 (ASCII for 6), in[1] = 54 (ASCII for 6), in[2] = 49 (ASCII for 1), in[3] = 49 (ASCII for 1).

So I am expecting to retrieve integer 6611 value from byte array, but I ended up getting 909521201. I fail to understand what is exactly going on in such a simple conversion. Can anyone flash some light?

THanks

Answer 1

@Volker is right in his comment, you don't have a binary number in your array, you have an ASCII string. Yet, you try to decode it as binary. Note there's no need to validate any input (maybe except the length) if you were dealing with binary number, as all single byte values are valid.

@Ainar-G gave you a way of converting ASCII number into integer.

Compare these two approaches: (http://play.golang.org/p/_wufZ4P_aE)

buf := []byte{54, 54, 49, 49}

x, _ := strconv.Atoi(string(buf))
fmt.Println(x)

This prints 6611; but look at this:

var y int32
_ = binary.Read(bytes.NewReader(buf), binary.BigEndian, &y)
fmt.Println(y)

This prints 909521201, so exactly what you got (and didn't expect). As a side note, you're manually decoding it as BigEndian, so this is not "such a simple conversion" at the end, because there're some more factors to consider.

Your handcrafted conversion from ASCII would look more or less as follows:

var x int32
for _, c := range in {
    x = x*10 + int32(c - '0')
}
return x

But using strconv is the way to go.