Question

I m试图在将星阵列转换成uint64_t时标出一个灯泡(unsign long)。

我知道你可以这样做,但这不是我的问题。我希望以多种重复的方式这样做。我想知道,为什么在有些情形下,重复做法没有按计划进行。

该法典的结果是正确的:

#include <stdio.h>

int main()
{
    unsigned char byte[8];
    unsigned long long x;

    byte[0] = 0x15;
    byte[1] = 0x15; 
    byte[2] = 0x17;
    byte[3] = 0x18; 
    byte[4] = 0x19;
    byte[5] = 0x20;
    byte[6] = 0x21;
    byte[7] = 0x12;
            
    x = (byte[0]*0x100000000000000) +
        (byte[1]*0x1000000000000) + 
        (byte[2]*0x10000000000) + 
        (byte[3]*0x100000000);

    x = x +
        (byte[4]<<24) + 
        (byte[5]<<16) + 
        (byte[6]<<8) + 
        byte[7];

    printf("%llx
", x);
}

结果是15171819202112,这是正确的。

但就本案而言,我用这些 by子代替:

byte[0] = 0x15;
byte[1] = 0x92;  // changed 
byte[2] = 0x53;  // changed
byte[3] = 0x22;  // changed
byte[4] = 0xec;  // changed
byte[5] = 0x33;  // changed
byte[6] = 0x99;  // changed
byte[7] = 0x12;

......结果是错误的。

I get:
15925321ec339912.

But, it should be:
15925322ec339912.

Why? What is the bug?

Answer 1

在这一次低压中:

byte[4]<<24

页: 1 这一数值首先见promoted至int。详情见。关于算术转换的标准 :

The following may be used in an expression wherever an int or unsigned int may be used:

An object or expression with an integer type (other than int or unsigned int) whose integer conversion rank is less than or equal to the rank of int and unsigned int.

A bit-field of type _Bool, int, signed int, or unsigned int.

If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions. All other types are unchanged by the integer promotions.

因此,<代码>逐[4]的升值有<代码>int的型号,已签署,(很可能)32比值。然后,将<条码>int值留给24位。鉴于原始价值为0xec,因此,价值1的数值被移至所产生的<代码>int数值的标尺。

在上,将1个字移入了标栏。 http://www. open-std.org/jtc1/sc22/wg14/www/docs/n1570.pdf” rel=“nofollow noreferer” 关于双向转移操作者的标准 :

The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. If E1 has an unsigned type, the value of the result is E1 × 2^E2, reduced modulo one more than the maximum value representable in the result type. If E1 has a signed type and nonnegative value, and E1 × 2^E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.

您可以首先将数值投到<代码>,长期寄出<> 代码>,以便转换有效。

x=x+((unsigned long long)byte[4]<<24)+(byte[5]<<16)+(byte[6]<<8)+byte[7];

友情链接