pecl / solr似乎无法正常安装(pecl/solr cant seem to install normaly)

 你在写main()还是_start ？ 
 
 如果你正在编写main ，它是一个正常的函数，其args在rdi ， rsi ，遵循正常的调用约定。 有关x86-64 ABI的链接，请参阅x86标记wiki 。 
 
 如果您正在编写_start ，那么数据就在堆栈中，如ABI的进程启动部分所述： [rsp] = argc ，以及指针数组， char *arg[]从rsp+8 。 它是堆栈上的实际数组，而不是像main获取的数组指针 。 
 
 除非你初始化它，否则rbp没有意义。 它拥有呼叫者留在其中的任何内容。 
 
 
 你的代码片段也很愚蠢：你永远不会初始化rbp。 你应该假设它在进程输入时持有垃圾。 只有rsp才有用。 
 
 lea只是一个使用有效地址语法/编码的移位和添加指令。 mov是加载/存储的助记符。 
 
    ;; your code with comments, also assuming that RBP was initialized
    bits 64
    lea r8, [rbp-0x10]      ; r8 = rbp-0x10
    mov r9, [r8]            ; should have just done mov r9, [rbp-0x10]
    mov r10, [r9+0x10]
    jmp r10                 ; jump to argv[2]???
 
 你把机器代码字节放在argv[2]吗？ 跳转到字符串通常不是很有用。 
 
 当然，由于rbp没有初始化，它实际上并没有访问argv[2] 。 
 
 
 工作实例 
 
 如果你想看看发生了什么，在调试器中单步执行此操作。 
 
; get argc and argv from the stack, for x86-64 SysV ABI
global _start
_start:
    mov   ecx,  [rsp]             ;   load argc (assuming it's smaller than 2^32)

    cmp   ecx, 3
    jb  .argc_below_3
                                  ;   argv[0] is at rsp+8
    mov   rsi,  [rsp+8 +  8*2]    ;   argv[2]  (the 3rd element)
    movzx eax,  byte [rsi]        ;   first char of argv[2]

    ; if you stop here in a debugger, you can see the character from the second arg.

    ; fall through and exit
.argc_below_3:
    xor edi, edi
    mov eax, 231                  ;  exit_group(0)
    syscall

Are you writing main() or _start?
 
If you're writing main, it's a normal function with its args in rdi, rsi, following the normal calling convention. See x86 tag wiki for links to the x86-64 ABI.
 
If you're writing _start, then data is on the stack, as documented in process startup section of the ABI: [rsp] = argc, and above that an array of pointers, char *arg[] starting at rsp+8. It's an actual array right there on the stack, not a pointer to an array like main gets.
 
rbp is meaningless unless you initialize it. It has whatever the caller left in it.
 
 
Your code fragment is silly, too: you never initialize rbp. You should assume it holds garbage on process entry. Only rsp is guaranteed to be useful.
 
lea is just a shift & add instruction that uses effective-address syntax / encoding. mov is the mnemonic for load / store.
 
    ;; your code with comments, also assuming that RBP was initialized
    bits 64
    lea r8, [rbp-0x10]      ; r8 = rbp-0x10
    mov r9, [r8]            ; should have just done mov r9, [rbp-0x10]
    mov r10, [r9+0x10]
    jmp r10                 ; jump to argv[2]???
 
Did you put machine code bytes in argv[2]? Jumping to a string is not normally useful.
 
Of course, since rbp isn't initialized, it's not actually accessing argv[2].
 
 
Working example
 
single-step this in a debugger if you want to see what's going on.
 
; get argc and argv from the stack, for x86-64 SysV ABI
global _start
_start:
    mov   ecx,  [rsp]             ;   load argc (assuming it's smaller than 2^32)

    cmp   ecx, 3
    jb  .argc_below_3
                                  ;   argv[0] is at rsp+8
    mov   rsi,  [rsp+8 +  8*2]    ;   argv[2]  (the 3rd element)
    movzx eax,  byte [rsi]        ;   first char of argv[2]

    ; if you stop here in a debugger, you can see the character from the second arg.

    ; fall through and exit
.argc_below_3:
    xor edi, edi
    mov eax, 231                  ;  exit_group(0)
    syscall