将dword存储到地址中 [英] Storing dword into address
问题描述
我正在自学汇编语言的某些部分，现在，我专注于在地址中存储数据声明.
I'm in middle of teaching myself certain parts of Assembly Language, and right now, I am focusing on storing data declarations in addresses.
在存储十六进制时，我知道例如要处理字节；
When it comes to storing hex, I know that if I am dealing with bytes, for instance;
1234
我可以这样存储它们:
Address 0000 - 12
Address 0001 - 24
由于双字是32位，我假设每个双字将占用两倍的空间.
Because dwords are 32 bits, I am assuming that each would take up twice as much space.
如果我最后用这个作为dword:
If I end up with this for dword:
54 00 87 D1 49 5A AF 56 32
它们会这样存储吗?
Address 0000 - 54
Address 0002 - 00
Address 0004 - 87
Address 0006 - D1
Address 0008 - 49
Address 000A - 5A
Address 000C - AF
Address 000F - 56
Address 0011 - 32
?
推荐答案
如上所述，您的值超过了dword.
As has been pointed out, your values exceed a dword.
在x86上，字"是16位，因为8086是16位微处理器.在这种情况下，它的意思是两个字节". 双字"是两个字或四个字节，而四字"是四个字或八个字节. x86是小尾数"处理器，因此它开始从寄存器的小端开始写入内存.
On x86, a "word" is 16 bits, because the 8086 is a 16-bit microprocessor. In this context, it means "two bytes". A "double word" is two words, or four bytes, and a "quad word" is four words, or eight bytes. x86 is a "little endian" processor, so it begins writing to memory from the little end of your register.
如果您执行类似操作(使用int语法和gcc样式的十六进制数字):
If you do something like (in intel syntax and gcc style hex numbers):
#Load the lowest 8 bits of the rax register (al) with 0x54
#This is loading a BYTE (1 byte)
mov al,0x54
#Load the lowest 16 bits of the rbx register (bx) with 0x5400
#This is loading a WORD (2 bytes)
mov bx,0x5400
#Load the lowest 32 bits of the rcx register (ecx) with 0x540087D1
#This is loading a DWORD (4 bytes)
mov ecx,0x540087D1
#Load the entire 64 bit register rdx with 0x540087D1495AAF56
#This is loading a QWORD (8 bytes)
mov rdx,0x540087D1495AAF56
然后，如果将它们移动到寄存器rsi中的地址，则会得到:
Then if you were to move these to an address held in register rsi, you would get:
#Put the value of al (0x54) into address at [rsi+0]
mov [rsi],al
#Put the value of bx (0x5400) starting at the address at rsi+0,
# such that [rsi+0] will be 0x00 and [rsi+1] will be 0x54
mov [rsi],bx
#Put the value of ecx (0x540087D1) starting at the address of rsi+0,
# such that [rsi+0] will be 0xD1, [rsi+1] will be 0x87,
# [rsi+3] will be 0x00, and [rsi+4] will be 0x54
mov [rsi],ecx
#Put the value of rdx (0x540087D1495AAF56) starting at the address of rsi+0,
#such that [rsi++0] will be 0x56, [rsi+1] will be 0xAF,
# [rsi+2] will be 0x5A, [rsi+3] will be 0x49,
# [rsi+4] will be 0xD1, [rsi+5] will be 0x87,
# [rsi+6] will be 0x00, and [rsi+7] will be 0x54
mov [rsi],rdx
您的示例值只有9个字节，不适合任何寄存器，也不是机器类型.
Your example value, with 9 bytes, doesn't fit into any of the registers, and isn't a machine type.
因此，您得到的双字公羊看起来像:
So your resulting ram for a double word would look like:
0x540087D1
(小尾数，例如x86):
第一个地址-0xD1
第二个地址-0x87
第三个地址-0x00
第四个地址-0x54
0x540087D1
(little endian, such as x86):
First address- 0xD1
Second address- 0x87
Third address- 0x00
Fourth address- 0x54
(大字节序，例如SPARC):
第一个地址-0x54
第二个地址-0x00
第三个地址-0x87
第四个地址-0xD1
(big endian, such as SPARC):
First address- 0x54
Second address- 0x00
Third address- 0x87
Fourth address- 0xD1
我还将补充一点，在以后的装配问题中，您应该始终讨论有问题的体系结构-几乎没有通用装配问题.
I will also add that in future assembly questions, you should always discuss the architecture in question- there's almost no generic assembly questions.
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