Chosing suffix (l-b-w) for mov instruction
assembly, att, x86
Solution
In `movb $-17,(%esp)` the destination is not the register `%esp` but the memory location whose address is in `%esp`. Because of the `b` in `movb`, a single byte will be stored at that memory location. The value stored there will be -17 (which is equivalent to the unsigned byte 0xef).
`movw $-17,(%esp)` and `movl $-17,(%esp)` would also be legal instructions and they'd do different things, storing the 2 or 4 byte values 0xffef or 0xffffffef at memory locations `%esp` through `%esp+1` or `%esp+3`.
This instruction needs the `b` or `w` or `l` to disambiguate the meaning, unlike your other examples, because neither `$-17` nor `(%esp)` is a fixed-size entity. If you try `mov $-17,(%esp)` the assembler will complain.
UPDATE: I just noticed question #5, `push $0xFF` which also seems like it could be ambiguous (`pushl $0xFF` and `pushw $0xFF` are both legal), but there is a special rule for `push` that assumes `l` whenever there is an ambiguity. 16-bit pushes are very rare (the sysv ABI keeps everything aligned on the stack in multiples of 4 bytes so you always push 32 bits for a function argument, even if it's a `short` or `char`)
Problem
I am new to assembly.I am reading computers system programmer's perspective. I don't understand how I choose suffix for `mov` instruction. I know each register and bit count. Suffix usage is determined by bit count (32 bit `l`, 16 bit `w`, 8 bit `b`). Few example is not valid for prior sentence. For example `%esp` is 32-bit register but for 4. step suffix `b` is used instead of `l`. Please give an explanation for using suffix. questions : answer : `l-w-b-b-l-w-l` Source: Computer Systems: A Programmer's Perspective (CSAPP) by Bryant, O'Hallaron