user: 请从给定的汇编指令推导出C源代码：

```asm
<parseLine>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x498,%rsp
  mov    %rdi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x488(%rsp)
  xor    %eax,%eax
  call   <getc@plt>
  mov    %eax,%ebx
  mov    $0x0,%r14d
  movabs $0x2000000300000400,%r15
  mov    D19(%rip),%rax
  cmp    (%rax),%bl
  jne    L34
L8:
  mov    0x488(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x498,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L45:
  mov    $0x1,%ebp
  jmp    L177
L10:
  movslq %r13d,%rax
  movb   $0x21,0x400(%rsp,%rax,1)
  add    $0x1,%r13d
  movslq %r13d,%r13
  movb   $0x0,0x400(%rsp,%r13,1)
  lea    0x400(%rsp),%rbx
  mov    %rbx,%rdi
  call   <strdup@plt>
  mov    %rax,%rdx
  movslq %r14d,%rax
  mov    %rdx,(%rsp,%rax,8)
  add    $0x1,%r14d
  mov    $0x80,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    %r12,%rdi
  call   <getc@plt>
  mov    %eax,%ebx
  jmp    L50
L152:
  add    $0x1,%rbp
L177:
  mov    %bl,0x3ff(%rsp,%rbp,1)
  mov    %ebp,%r13d
  mov    %r12,%rdi
  call   <getc@plt>
  mov    %eax,%ebx
  cmp    $0x3d,%al
  ja     L152
  add    $0x1,%rbp
  bt     %rbx,%r15
  jae    L177
L67:
  cmp    $0x21,%bl
  je     L10
  movslq %r13d,%r13
  movb   $0x0,0x400(%rsp,%r13,1)
  lea    0x400(%rsp),%rbp
  mov    %rbp,%rdi
  call   <strdup@plt>
  mov    %rax,%rdx
  movslq %r14d,%rax
  mov    %rdx,(%rsp,%rax,8)
  add    $0x1,%r14d
  mov    $0x80,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
L50:
  cmp    $0x20,%bl
  je     L171
  cmp    $0x3d,%bl
  jne    L165
L171:
  mov    %r12,%rdi
  call   <getc@plt>
  mov    %eax,%ebx
  cmp    $0x20,%al
  je     L171
  cmp    $0x3d,%al
  je     L171
L165:
  cmp    $0xa,%bl
  jne    L9
  movslq %r14d,%r14
  movq   $0x0,(%rsp,%r14,8)
  mov    %rsp,%rdi
  call   <sysCall@plt>
  mov    D0(%rip),%rax
  cmp    %r12,(%rax)
  je     L8
L179:
  mov    %r12,%rdi
  call   <getc@plt>
  mov    %eax,%ebx
  cmp    $0xa,%al
  je     L179
  cmp    $0x20,%al
  je     L179
  mov    $0x0,%r14d
L9:
  mov    D19(%rip),%rax
  cmp    %bl,(%rax)
  je     L8
L34:
  cmp    $0x3d,%bl
  ja     L45
  mov    $0x1,%ebp
  mov    $0x0,%r13d
  bt     %rbx,%r15
  jae    L177
  jmp    L67
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void parseLine(FILE *stream) {
  char *args[128];
  char buffer[128];
  int argc = 0;
  int i = 0;
  char c = getc(stream);
  while (c != EOF) {
    while (c != ' ' && c != '\n' && c != '=' && c != '!') {
      buffer[i] = c;
      i++;
      c = getc(stream);
    }
    if (c == '!') {
      buffer[i] = c;
      i++;
    }
    buffer[i] = '\0';
    args[argc] = strdup(buffer);
    argc++;
    memset(&buffer[0], 0, sizeof(buffer));
    i = 0;
    if (c == '!')
      c = getc(stream);
    while (c == ' ' || c == '=')
      c = getc(stream);
    if (c == '\n') {
      args[argc] = NULL;
      sysCall(args);
      argc = 0;
      int j = 0;
      while (j <= (argc - 1)) {
        free(args[j]);
        args[j] = NULL;
        j++;
      }
    }
    if (c == '\n' && stream == stdin)
      return;
    while (c == '\n' || c == ' ')
      c = getc(stream);
    if (c == EOF && stream != stdin)
      return;
  }
}
```