内存中的long long类型表示形式

问题描述 投票：2回答：2

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

c++ long-integer
2个回答
1
投票

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.
}
0
投票

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），向左移动给出的都是整数。

使用中

 0xFFull << offset

解决了问题（因为后缀ull告诉它应被视为unsigned long long。]

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