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In the following x86 assembly code:

dd 0x1BADB002
dd 0x00
dd - (0x1BADB002+0x00)

The values don't seem to be assigned to any variables. So what does this snippet of code do? I've heard something about it being stored in memory, but where exactly?

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dd is a "pseudo-instruction" that assembles 4-byte constants into the output, the same way that add eax,eax assembles 0x01 0xc0 into the output.

The NASM manual section 3.2 Pseudo-Instructions describes db/dw/dd and so on.

In this case, as @MichaelPetch points out, those specific constants are used to assemble a multiboot header into the output file. https://www.gnu.org/software/grub/manual/multiboot/multiboot.html#OS-image-format

How does this assembly bootloader code work?

Related:

How are dw and dd different from db directives for strings?

What is the use of .byte assembler directive in gnu assembly?

x86 assembly - Which variable size to use (db, dw, dd)

于 2017-10-01T19:56:27.563 回答