user: 请从给定的汇编指令推导出C源代码：

```asm
<getRegArg>:
  endbr64
  push   %rbx
  lea    D94(%rip),%rsi
  mov    %rdi,%rbx
  call   <strcmp@plt>
  test   %rax,%rax
  je     L95
  lea    D26(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L95
  lea    D2(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L54
  lea    D58(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L54
  lea    D31(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L48
  lea    D11(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L48
  lea    D79(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L86
  lea    D62(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L86
  lea    D39(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L76
  lea    D27(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L76
  lea    D17(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L110
  lea    D70(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L110
  lea    0x1(%rbx),%rdi
  call   <atoi@plt>
  movzbl (%rbx),%ecx
  sub    $0x52,%ecx
  cmp    $0x22,%cl
  ja     L135
  mov    $0x1,%edx
  shl    %cl,%rdx
  movabs $0x100000001,%rcx
  test   %rcx,%rdx
  jne    L138
  movabs $0x400000004,%rcx
  test   %rcx,%rdx
  jne    L88
  movabs $0x200000002,%rcx
  add    $0x8,%eax
  test   %rcx,%rdx
  mov    $0xffffffff,%edx
  cmove  %edx,%eax
L138:
  pop    %rbx
  ret
  nopl   (%rax)
L95:
  mov    $0x1c,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L54:
  mov    $0x1d,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L48:
  mov    $0x1e,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L86:
  mov    $0x1f,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L76:
  mov    $0x20,%eax
  pop    %rbx
  ret
L110:
  mov    $0x21,%eax
  pop    %rbx
  ret
L88:
  add    $0x18,%eax
  pop    %rbx
  ret
L135:
  or     $0xffffffff,%eax
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: label: D94
type_infered: string
name: read_data
arguments: label: D26
type_infered: string
name: read_data
arguments: label: D2
type_infered: string
name: read_data
arguments: label: D58
type_infered: string
name: read_data
arguments: label: D31
type_infered: string
name: read_data
arguments: label: D11
type_infered: string
name: read_data
arguments: label: D79
type_infered: string
name: read_data
arguments: label: D62
type_infered: string
name: read_data
arguments: label: D39
type_infered: string
name: read_data
arguments: label: D27
type_infered: string
name: read_data
arguments: label: D17
type_infered: string
name: read_data
arguments: label: D70
type_infered: string

tool: D94: BP

tool: D26: bp

tool: D2: SP

tool: D58: sp

tool: D31: IP

tool: D11: ip

tool: D79: PTBR

tool: D62: ptbr

tool: D39: PTLR

tool: D27: ptlr

tool: D17: EFR

tool: D70: efr

assistant: ```c
int getRegArg(char *arg) {
  int argvalue;
  if (strcmp(arg, "BP") == 0 || strcmp(arg, "bp") == 0)
    return (28);
  else if (strcmp(arg, "SP") == 0 || strcmp(arg, "sp") == 0)
    return (29);
  else if (strcmp(arg, "IP") == 0 || strcmp(arg, "ip") == 0)
    return (30);
  else if (strcmp(arg, "PTBR") == 0 || strcmp(arg, "ptbr") == 0)
    return (31);
  else if (strcmp(arg, "PTLR") == 0 || strcmp(arg, "ptlr") == 0)
    return (32);
  else if (strcmp(arg, "EFR") == 0 || strcmp(arg, "efr") == 0)
    return (33);
  else
    argvalue = atoi(arg + 1);
  switch (arg[0]) {
  case 'R':
  case 'r':
    if (argvalue >= 0 && argvalue < 8)
      ;
    return (0 + argvalue);
    break;
  case 'S':
  case 's':
    if (argvalue >= 0 && argvalue < 16)
      ;
    return (8 + argvalue);
    break;
  case 'T':
  case 't':
    if (argvalue >= 0 && argvalue < 4)
      ;
    return (24 + argvalue);
    break;
  }
  return -1;
}
```