将6个字节的数组复制到long long整数变量 [英] Copy 6 byte array to long long integer variable

问题描述

我从内存中读取了一个6字节的unsigned char数组. 字节序是这里的大字节序(Big Endian). 现在，我想将存储在数组中的值分配给一个整数变量.我认为这必须是long long，因为它必须包含最多6个字节.

I have read from memory a 6 byte unsigned char array. The endianess is Big Endian here. Now I want to assign the value that is stored in the array to an integer variable. I assume this has to be long long since it must contain up to 6 bytes.

目前，我以这种方式进行分配:

At the moment I am assigning it this way:

unsigned char aFoo[6];
long long nBar;
// read values to aFoo[]...
// aFoo[0]: 0x00
// aFoo[1]: 0x00
// aFoo[2]: 0x00
// aFoo[3]: 0x00
// aFoo[4]: 0x26
// aFoo[5]: 0x8e
nBar = (aFoo[0] << 64) + (aFoo[1] << 32) +(aFoo[2] << 24) + (aFoo[3] << 16) + (aFoo[4] << 8) + (aFoo[5]);

memcpy方法会很整洁，但是当我这样做时

A memcpy approach would be neat, but when I do this

memcpy(&nBar, &aFoo, 6);

这6个字节从头开始复制到long long，因此在末尾具有填充零. 有没有比我分配工作更好的方法了?

the 6 bytes are being copied to the long long from the start and thus have padding zeros at the end. Is there a better way than my assignment with the shifting?

推荐答案

您要完成的工作称为反序列化或反编组.

What you want to accomplish is called de-serialisation or de-marshalling.

对于如此宽的值，使用循环是个好主意，除非您确实需要最大值.速度，并且编译器无法向量化循环:

For values that wide, using a loop is a good idea, unless you really need the max. speed and your compiler does not vectorise loops:

uint8_t array[6];
...
uint64_t value = 0;

uint8_t *p = array;
for ( int i = (sizeof(array) - 1) * 8 ; i >= 0 ; i -= 8 )
    value |= (uint64_t)*p++ << i;

//左对齐 值<< = 64-(sizeof(array)* 8);

// left-align value <<= 64 - (sizeof(array) * 8);

请注意使用stdint.h类型和sizeof(uint8_t) cannot differ from 1`.仅保证这些具有预期的位宽.移位值时也请使用无符号整数.右移某些值是实现定义的，而左移会调用未定义的行为.

Note using stdint.h types and sizeof(uint8_t) cannot differ from1`. Only these are guaranteed to have the expected bit-widths. Also use unsigned integers when shifting values. Right shifting certain values is implementation defined, while left shifting invokes undefined behaviour.

如果 f 您需要一个带符号的值，只需

Iff you need a signed value, just

int64_t final_value = (int64_t)value;

移位后.这仍然是实现定义的，但是所有现代实现(可能是较旧的实现)仅复制该值而无需修改.现代的编译器可能会对此进行优化，因此不会有任何损失.

after the shifting. This is still implementation defined, but all modern implementations (and likely the older) just copy the value without modifications. A modern compiler likely will optimize this, so there is no penalty.

当然可以移动声明.我只是将它们放在用于完整性的地方.

The declarations can be moved, of course. I just put them before where they are used for completeness.

这篇关于将6个字节的数组复制到long long整数变量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！