转换字节数组双 - ç [英] converting byte array to double - c

问题描述

我想从16个元素的字节数组的数字（双）值，如下所示：

I'm trying to get the numerical (double) value from a byte array of 16 elements, as follows:

unsigned char input[16];
double output;
...
double a  = input[0];
distance = a;
for (i=1;i<16;i++){
    a = input[i] << 8*i;
    output += a;
}

，但它不工作。
似乎包含左移位的结果的临时变量只能存储32位，因为之后的8位的4移位运算溢出

but it does not work. It seems that the temporary variable that contains the result of the left-shift can store only 32 bits, because after 4 shift operations of 8 bits it overflows.

我知道我可以使用类似

a = input[i] * pow(2,8*i);

但是，出于好奇，我想知道是否有使用移位运算符对这个问题的任何解决方案...

but, for curiosity, I was wondering if there's any solution to this problem using the shift operator...

推荐答案

编辑：这不是没有类似 __ int128工作（见注释）

this won't work (see comment) without something like __int128.

a = input[i] << 8*i;

这位前pression 输入[I] 提升为 INT （ 6.3.1.1 ），这是你的机器上32位。为了克服这个问题，左侧操作数必须是64位，如在

The expression input[i] is promoted to int (6.3.1.1) , which is 32bit on your machine. To overcome this issue, the lefthand operand has to be 64bit, like in

a = (1L * input[i]) << 8*i;

或

a = (long long unsigned) input[i] << 8*i;

和记住字节顺序

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