我想从8字节类型中提取字节,类似于char func(long long number, size_t offset)
,因此对于偏移量n,我将获得第n
个字节(0 <= n <= 7)。在这样做的时候,我意识到我不知道内存中实际上如何表示8字节变量。希望您能帮助我解决。我首先编写了一个简短的python脚本,以在每个字节中打印由A
s(ascii值为65)组成的数字
sumx = 0
for x in range(8):
sumx += (ord('A')*256**x)
print('x {} sumx {}'.format(x,sumx))
输出为
x 0 sumx 65
x 1 sumx 16705
x 2 sumx 4276545
x 3 sumx 1094795585
x 4 sumx 280267669825
x 5 sumx 71748523475265
x 6 sumx 18367622009667905
x 7 sumx 4702111234474983745
在我看来,每个数字都是一堆A
,后跟0。接下来,我编写了一个简短的C ++代码以提取第n
个字节
#include <iostream>
#include <array>
char func0(long long number, size_t offset)
{
offset <<= 3;
return (number & (0x00000000000000FF << offset)) >> offset;
}
char func1(long long unsigned number, size_t offset)
{
char* ptr = (char*)&number;
return ptr[offset];
}
int main()
{
std::array<long long,8> arr{65,16705,4276545,1094795585,280267669825,71748523475265,18367622009667905,4702111234474983745};
for (int i = 0; i < arr.size(); i++)
for (int j = 0; j < sizeof(long long unsigned); j++)
std::cout << "char " << j << " in number " << i << " (" << arr[i] << ") func0 " << func0(arr[i], j) << " func1 " << func1(arr[i], j) << std::endl;
return 0;
}
这里是程序输出(注意从第5个字节开始的区别)
~ # g++ -std=c++11 prog.cpp -o prog; ./prog
char 0 in number 0 (65) func0 A func1 A
char 1 in number 0 (65) func0 func1
char 2 in number 0 (65) func0 func1
char 3 in number 0 (65) func0 func1
char 4 in number 0 (65) func0 func1
char 5 in number 0 (65) func0 func1
char 6 in number 0 (65) func0 func1
char 7 in number 0 (65) func0 func1
char 0 in number 1 (16705) func0 A func1 A
char 1 in number 1 (16705) func0 A func1 A
char 2 in number 1 (16705) func0 func1
char 3 in number 1 (16705) func0 func1
char 4 in number 1 (16705) func0 func1
char 5 in number 1 (16705) func0 func1
char 6 in number 1 (16705) func0 func1
char 7 in number 1 (16705) func0 func1
char 0 in number 2 (4276545) func0 A func1 A
char 1 in number 2 (4276545) func0 A func1 A
char 2 in number 2 (4276545) func0 A func1 A
char 3 in number 2 (4276545) func0 func1
char 4 in number 2 (4276545) func0 func1
char 5 in number 2 (4276545) func0 func1
char 6 in number 2 (4276545) func0 func1
char 7 in number 2 (4276545) func0 func1
char 0 in number 3 (1094795585) func0 A func1 A
char 1 in number 3 (1094795585) func0 A func1 A
char 2 in number 3 (1094795585) func0 A func1 A
char 3 in number 3 (1094795585) func0 A func1 A
char 4 in number 3 (1094795585) func0 func1
char 5 in number 3 (1094795585) func0 func1
char 6 in number 3 (1094795585) func0 func1
char 7 in number 3 (1094795585) func0 func1
char 0 in number 4 (280267669825) func0 A func1 A
char 1 in number 4 (280267669825) func0 A func1 A
char 2 in number 4 (280267669825) func0 A func1 A
char 3 in number 4 (280267669825) func0 A func1 A
char 4 in number 4 (280267669825) func0 func1 A
char 5 in number 4 (280267669825) func0 func1
char 6 in number 4 (280267669825) func0 func1
char 7 in number 4 (280267669825) func0 func1
char 0 in number 5 (71748523475265) func0 A func1 A
char 1 in number 5 (71748523475265) func0 A func1 A
char 2 in number 5 (71748523475265) func0 A func1 A
char 3 in number 5 (71748523475265) func0 A func1 A
char 4 in number 5 (71748523475265) func0 func1 A
char 5 in number 5 (71748523475265) func0 func1 A
char 6 in number 5 (71748523475265) func0 func1
char 7 in number 5 (71748523475265) func0 func1
char 0 in number 6 (18367622009667905) func0 A func1 A
char 1 in number 6 (18367622009667905) func0 A func1 A
char 2 in number 6 (18367622009667905) func0 A func1 A
char 3 in number 6 (18367622009667905) func0 A func1 A
char 4 in number 6 (18367622009667905) func0 func1 A
char 5 in number 6 (18367622009667905) func0 func1 A
char 6 in number 6 (18367622009667905) func0 func1 A
char 7 in number 6 (18367622009667905) func0 func1
char 0 in number 7 (4702111234474983745) func0 A func1 A
char 1 in number 7 (4702111234474983745) func0 A func1 A
char 2 in number 7 (4702111234474983745) func0 A func1 A
char 3 in number 7 (4702111234474983745) func0 A func1 A
char 4 in number 7 (4702111234474983745) func0 func1 A
char 5 in number 7 (4702111234474983745) func0 func1 A
char 6 in number 7 (4702111234474983745) func0 func1 A
char 7 in number 7 (4702111234474983745) func0 A func1 A
此代码具有2个函数,func1
返回期望值,func0
我以为它应该返回与func1
相同的值,但没有,我不确定为什么。基本上,我理解8字节类型,例如8字节数组,func1
清楚地表明了这种情况。我不确定为什么要使用移位来移至第n
个字节不起作用,并且我不确定我是否完全理解内存中如何安排8个字节的变量
问题在于代码中
0x00000000000000FF << offset
左边的数字0xFF
只是一个整数(无论您放置了多少个零),向左移动给出的整数(实际上是整数的大小...向左移动大于整数的大小不是)便携式代码)。
改为使用:
0xFFull << offset
解决了问题(因为后缀ull
告诉它应被视为unsigned long long
。]
当然,正如另一个答案所述,(number >> (offset * 8)) & 0xFF
更简单且有效。
这是一种非常简单的操作,非常复杂。您甚至不需要考虑字节序问题,因为您不需要仅为了获取字节就访问long long
的内存表示形式。
获得第n个字节仅是掩盖所有其他字节并将该值转换为unsigned char
的问题。像这样:
unsigned char nth_byte(unsigned long long int value, int n)
{
//Assert that n is on the range [0, 8)
value = value >> (8 * n); //Move the desired byte into the first byte.
value = value & 0xFF; //Mask away everything that isn't the first byte.
return unsigned char(value); //Return the first byte.
}
func0中的问题是,您的十六进制文字虽然包含8个字节的数据,但由于未指定精度而被解释为long。使用0xffULL(0xff unsigned long long)代替0x00000000000000ff应该可以为您提供所需的内容。
线索是,它在前32位中运行良好,然后跌落了。不过,我不知所措地解释了第七个A的来源。
分析变量的基础内存表示的正确方法是使用memcpy并将其复制到char
数组(参考:C aliasing rules and memcpy):
#include <cstring>
char get_char(long long num, size_t offs)
{
char array[sizeof(long long)];
memcpy(array, &num, sizeof(long long));
return array[offs];
}
然后为以下示例:
int main()
{
long long var = 0x7766554433221100;
for (size_t idx = 0; idx < sizeof(long long); ++idx)
std::cout << '[' << idx << ']' << '=' << std::hex << static_cast<int>(get_char(var, idx)) << '\n';
}
在小端系统上,我们得到:
[0]=0
[1]=11
[2]=22
[3]=33
[4]=44
[5]=55
[6]=66
[7]=77
在大端系统上,我们得到:
[0]=77
[1]=66
[2]=55
[3]=44
[4]=33
[5]=22
[6]=11
[7]=0