ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<debug_loop>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  call   <stack_so_deep@plt>
  jmp    L4
```

assistant: ```c
int debug_loop() {
  while (1) {
    int stop_point_1 = 100;
    stack_so_deep();
    int stop_point_2 = 200;
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<__kern_shell_time>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rsi
  lea    -0x10(%rbp),%rcx
  lea    -0x14(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rsi,%r8
  lea    D43(%rip),%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <ksscanf@plt>
  test   %rax,%rax
  je     L0
  mov    -0xc(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <kprintf@plt>
  mov    -0xc(%rbp),%eax
  movsbl %al,%edx
  mov    -0x10(%rbp),%eax
  movsbl %al,%ecx
  mov    -0x14(%rbp),%eax
  movsbl %al,%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <__kern_rtc_tset@plt>
  jmp    L2
L0:
  mov    -0x28(%rbp),%rax
  lea    D44(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <ksscanf@plt>
  test   %rax,%rax
  je     L31
  lea    -0x15(%rbp),%rdx
  lea    -0x16(%rbp),%rcx
  lea    -0x17(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <__kern_rtc_tget@plt>
  movzbl -0x15(%rbp),%eax
  movsbl %al,%ecx
  movzbl -0x16(%rbp),%eax
  movsbl %al,%edx
  movzbl -0x17(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <kprintf@plt>
  jmp    L2
L31:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <kprintf@plt>
L2:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D43
type: string
name: unpack_data
arguments: data_tag: D8
type: string
name: unpack_data
arguments: data_tag: D44
type: string
name: unpack_data
arguments: data_tag: D11
type: string
name: unpack_data
arguments: data_tag: D20
type: string

tool: D43: time %d:%d:%d

tool: D8: Setting clock to %02d:%02d:%02d


tool: D44: time

tool: D11: Current time: %02d:%02d:%02d


tool: D20: %s: command syntax error


assistant: ```c
void __kern_shell_time(char *cstr) {
  char h, m, s;
  int ih, im, is;
  if (ksscanf(cstr, "time %d:%d:%d", &ih, &im, &is)) {
    kprintf("Setting clock to %02d:%02d:%02d\n", ih, im, is);
    __kern_rtc_tset((char)ih, (char)im, (char)is);
  } else if (ksscanf(cstr, "time")) {
    __kern_rtc_tget(&h, &m, &s);
    kprintf("Current time: %02d:%02d:%02d\n", h, m, s);
  } else {
    kprintf("%s: command syntax error\n", cstr);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<__kern_shell_time>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x14(%rsp),%r8
  lea    D3(%rip),%rsi
  call   <ksscanf@plt>
  test   %rax,%rax
  je     L20
  mov    0x14(%rsp),%ecx
  mov    0x10(%rsp),%edx
  mov    0xc(%rsp),%esi
  lea    D30(%rip),%rdi
  mov    $0x0,%eax
  call   <kprintf@plt>
  movsbl 0x14(%rsp),%edx
  movsbl 0x10(%rsp),%esi
  movsbl 0xc(%rsp),%edi
  call   <__kern_rtc_tset@plt>
L29:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x20,%rsp
  pop    %rbx
  ret
L20:
  lea    D9(%rip),%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <ksscanf@plt>
  test   %rax,%rax
  je     L48
  lea    0xb(%rsp),%rdx
  lea    0xa(%rsp),%rsi
  lea    0x9(%rsp),%rdi
  call   <__kern_rtc_tget@plt>
  movsbl 0xb(%rsp),%ecx
  movsbl 0xa(%rsp),%edx
  movsbl 0x9(%rsp),%esi
  lea    D24(%rip),%rdi
  mov    $0x0,%eax
  call   <kprintf@plt>
  jmp    L29
L48:
  mov    %rbx,%rsi
  lea    D26(%rip),%rdi
  mov    $0x0,%eax
  call   <kprintf@plt>
  jmp    L29
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D3
data_type: string
name: extract_symbolic_data
arguments: tag: D30
data_type: string
name: extract_symbolic_data
arguments: tag: D9
data_type: string
name: extract_symbolic_data
arguments: tag: D24
data_type: string
name: extract_symbolic_data
arguments: tag: D26
data_type: string

tool: D3: time %d:%d:%d

tool: D30: Setting clock to %02d:%02d:%02d


tool: D9: time

tool: D24: Current time: %02d:%02d:%02d


tool: D26: %s: command syntax error


assistant: ```c
void __kern_shell_time(char *cstr) {
  char h, m, s;
  int ih, im, is;
  if (ksscanf(cstr, "time %d:%d:%d", &ih, &im, &is)) {
    kprintf("Setting clock to %02d:%02d:%02d\n", ih, im, is);
    __kern_rtc_tset((char)ih, (char)im, (char)is);
  } else if (ksscanf(cstr, "time")) {
    __kern_rtc_tget(&h, &m, &s);
    kprintf("Current time: %02d:%02d:%02d\n", h, m, s);
  } else {
    kprintf("%s: command syntax error\n", cstr);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<__kern_shell_time>:
  endbr64
  push   %rbp
  lea    D47(%rip),%rsi
  mov    %rdi,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x14(%rsp),%r8
  call   <ksscanf@plt>
  test   %rax,%rax
  je     L31
  mov    0x10(%rsp),%edx
  mov    0xc(%rsp),%esi
  lea    D8(%rip),%rdi
  xor    %eax,%eax
  mov    0x14(%rsp),%ecx
  call   <kprintf@plt>
  movsbl 0x14(%rsp),%edx
  movsbl 0x10(%rsp),%esi
  movsbl 0xc(%rsp),%edi
  call   <__kern_rtc_tset@plt>
L34:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L44
  add    $0x20,%rsp
  pop    %rbp
  ret
L31:
  xor    %eax,%eax
  lea    D33(%rip),%rsi
  mov    %rbp,%rdi
  call   <ksscanf@plt>
  test   %rax,%rax
  je     L10
  lea    0xb(%rsp),%rdx
  lea    0xa(%rsp),%rsi
  lea    0x9(%rsp),%rdi
  call   <__kern_rtc_tget@plt>
  movsbl 0xb(%rsp),%ecx
  xor    %eax,%eax
  movsbl 0xa(%rsp),%edx
  movsbl 0x9(%rsp),%esi
  lea    D49(%rip),%rdi
  call   <kprintf@plt>
  jmp    L34
  nopl   0x0(%rax)
L10:
  mov    %rbp,%rsi
  lea    D17(%rip),%rdi
  xor    %eax,%eax
  call   <kprintf@plt>
  jmp    L34
L44:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D47
type: string
name: disassemble_data
arguments: data_label: D8
type: string
name: disassemble_data
arguments: data_label: D33
type: string
name: disassemble_data
arguments: data_label: D49
type: string
name: disassemble_data
arguments: data_label: D17
type: string

tool: D47: time %d:%d:%d

tool: D8: Setting clock to %02d:%02d:%02d


tool: D33: time

tool: D49: Current time: %02d:%02d:%02d


tool: D17: %s: command syntax error


assistant: ```c
void __kern_shell_time(char *cstr) {
  char h, m, s;
  int ih, im, is;
  if (ksscanf(cstr, "time %d:%d:%d", &ih, &im, &is)) {
    kprintf("Setting clock to %02d:%02d:%02d\n", ih, im, is);
    __kern_rtc_tset((char)ih, (char)im, (char)is);
  } else if (ksscanf(cstr, "time")) {
    __kern_rtc_tget(&h, &m, &s);
    kprintf("Current time: %02d:%02d:%02d\n", h, m, s);
  } else {
    kprintf("%s: command syntax error\n", cstr);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<__kern_shell_time>:
  endbr64
  push   %rbp
  lea    D8(%rip),%rsi
  mov    %rdi,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x14(%rsp),%r8
  call   <ksscanf@plt>
  test   %rax,%rax
  je     L43
  mov    0x10(%rsp),%edx
  mov    0xc(%rsp),%esi
  lea    D23(%rip),%rdi
  xor    %eax,%eax
  mov    0x14(%rsp),%ecx
  call   <kprintf@plt>
  movsbl 0x14(%rsp),%edx
  movsbl 0x10(%rsp),%esi
  movsbl 0xc(%rsp),%edi
  call   <__kern_rtc_tset@plt>
L36:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x20,%rsp
  pop    %rbp
  ret
L43:
  xor    %eax,%eax
  lea    D31(%rip),%rsi
  mov    %rbp,%rdi
  call   <ksscanf@plt>
  test   %rax,%rax
  je     L22
  lea    0xb(%rsp),%rdx
  lea    0xa(%rsp),%rsi
  lea    0x9(%rsp),%rdi
  call   <__kern_rtc_tget@plt>
  movsbl 0xb(%rsp),%ecx
  xor    %eax,%eax
  movsbl 0xa(%rsp),%edx
  movsbl 0x9(%rsp),%esi
  lea    D12(%rip),%rdi
  call   <kprintf@plt>
  jmp    L36
  nopl   0x0(%rax)
L22:
  mov    %rbp,%rsi
  lea    D16(%rip),%rdi
  xor    %eax,%eax
  call   <kprintf@plt>
  jmp    L36
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D8
type: string
name: unpack_data
arguments: data_tag: D23
type: string
name: unpack_data
arguments: data_tag: D31
type: string
name: unpack_data
arguments: data_tag: D12
type: string
name: unpack_data
arguments: data_tag: D16
type: string

tool: D8: time %d:%d:%d

tool: D23: Setting clock to %02d:%02d:%02d


tool: D31: time

tool: D12: Current time: %02d:%02d:%02d


tool: D16: %s: command syntax error


assistant: ```c
void __kern_shell_time(char *cstr) {
  char h, m, s;
  int ih, im, is;
  if (ksscanf(cstr, "time %d:%d:%d", &ih, &im, &is)) {
    kprintf("Setting clock to %02d:%02d:%02d\n", ih, im, is);
    __kern_rtc_tset((char)ih, (char)im, (char)is);
  } else if (ksscanf(cstr, "time")) {
    __kern_rtc_tget(&h, &m, &s);
    kprintf("Current time: %02d:%02d:%02d\n", h, m, s);
  } else {
    kprintf("%s: command syntax error\n", cstr);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movsd  -0x18(%rbp),%xmm1
  movsd  D18(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x18(%rbp),%xmm0
  addsd  -0x10(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D23
type_guessed: string
name: read_data
arguments: data_tag: D18
type_guessed: f64
name: read_data
arguments: data_tag: D13
type_guessed: string

tool: D23: %lf

tool: D18: 0.25

tool: D13: %.2lf


assistant: ```c
int main() {
  double salary = 0.0;
  double plus = 0.0;
  scanf("%lf", &salary);
  plus = salary * 0.25;
  printf("%.2lf\n", plus + salary);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsp,%rsi
  lea    D8(%rip),%rdi
  call   <scanf@plt>
  movsd  (%rsp),%xmm1
  movapd %xmm1,%xmm0
  mulsd  D1(%rip),%xmm0
  addsd  %xmm1,%xmm0
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D8
data_type: string
name: disassemble_data
arguments: tag: D1
data_type: f64
name: disassemble_data
arguments: tag: D9
data_type: string

tool: D8: %lf

tool: D1: 0.25

tool: D9: %.2lf


assistant: ```c
int main() {
  double salary = 0.0;
  double plus = 0.0;
  scanf("%lf", &salary);
  plus = salary * 0.25;
  printf("%.2lf\n", plus + salary);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D29(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movq   $0x0,(%rsp)
  call   <scanf@plt>
  movsd  (%rsp),%xmm1
  movsd  D7(%rip),%xmm0
  lea    D13(%rip),%rdi
  mulsd  %xmm1,%xmm0
  addsd  %xmm1,%xmm0
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D29
datatype: string
name: parse_data
arguments: data_tag: D7
datatype: f64
name: parse_data
arguments: data_tag: D13
datatype: string

tool: D29: %lf

tool: D7: 0.25

tool: D13: %.2lf


assistant: ```c
int main() {
  double salary = 0.0;
  double plus = 0.0;
  scanf("%lf", &salary);
  plus = salary * 0.25;
  printf("%.2lf\n", plus + salary);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D28(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movq   $0x0,(%rsp)
  call   <scanf@plt>
  movsd  (%rsp),%xmm1
  movsd  D22(%rip),%xmm0
  lea    D14(%rip),%rdi
  mulsd  %xmm1,%xmm0
  addsd  %xmm1,%xmm0
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D28
type_infered: string
name: read_data
arguments: tag: D22
type_infered: f64
name: read_data
arguments: tag: D14
type_infered: string

tool: D28: %lf

tool: D22: 0.25

tool: D14: %.2lf


assistant: ```c
int main() {
  double salary = 0.0;
  double plus = 0.0;
  scanf("%lf", &salary);
  plus = salary * 0.25;
  printf("%.2lf\n", plus + salary);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<GetLine>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L30
L35:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L30:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jge    L14
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x8(%rbp)
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  je     L14
  cmpl   $0xa,-0x8(%rbp)
  jne    L35
L14:
  cmpl   $0xa,-0x8(%rbp)
  jne    L11
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L11:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int GetLine(char s[], int limit) {
  int c, i;
  for (i = 0; i < limit - 1 && (c = getchar()) != EOF && c != '\n'; i++) {
    s[i] = c;
  }
  if (c == '\n') {
    s[i] = c;
    i++;
  }
  s[i] = '\0';
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<GetLine>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  cmp    $0x1,%esi
  jle    L53
  mov    %rdi,%rbp
  lea    -0x1(%rsi),%r12d
  mov    $0x0,%ebx
L40:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D9(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L55
  cmp    $0xa,%eax
  je     L55
  mov    %al,0x0(%rbp)
  add    $0x1,%ebx
  add    $0x1,%rbp
  cmp    %r12d,%ebx
  jne    L40
  mov    %r12d,%ebx
  jmp    L21
L53:
  mov    $0x0,%ebx
L55:
  cmp    $0xa,%eax
  je     L15
L21:
  movslq %ebx,%rax
  movb   $0x0,0x0(%r13,%rax,1)
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  movslq %ebx,%rax
  movb   $0xa,0x0(%r13,%rax,1)
  add    $0x1,%ebx
  jmp    L21
```

assistant: ```c
int GetLine(char s[], int limit) {
  int c, i;
  for (i = 0; i < limit - 1 && (c = getchar()) != EOF && c != '\n'; i++) {
    s[i] = c;
  }
  if (c == '\n') {
    s[i] = c;
    i++;
  }
  s[i] = '\0';
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<GetLine>:
  endbr64
  push   %r14
  mov    %rdi,%rcx
  xor    %r14d,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  test   %ebp,%ebp
  jle    L12
  mov    D9(%rip),%r12
  mov    %rdi,%rbx
  xor    %r14d,%r14d
  jmp    L27
  nopl   0x0(%rax,%rax,1)
L46:
  lea    0x1(%r14),%edx
  cmp    $0xa,%eax
  je     L50
  mov    %al,(%rcx)
  lea    0x1(%rcx),%rbx
  cmp    %edx,%ebp
  je     L26
  mov    %edx,%r14d
L27:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rbx,%rcx
  cmp    %eax,(%r12)
  jne    L46
  cmp    $0xa,%eax
  je     L42
L12:
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L42:
  lea    0x1(%r14),%edx
L50:
  movslq %edx,%rcx
  movb   $0xa,(%rbx)
  mov    %rcx,%r14
  add    %r13,%rcx
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L26:
  movslq %ebp,%rcx
  mov    %ebp,%r14d
  add    %r13,%rcx
  mov    %r14d,%eax
  movb   $0x0,(%rcx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int GetLine(char s[], int limit) {
  int c, i;
  for (i = 0; i < limit - 1 && (c = getchar()) != EOF && c != '\n'; i++) {
    s[i] = c;
  }
  if (c == '\n') {
    s[i] = c;
    i++;
  }
  s[i] = '\0';
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<GetLine>:
  endbr64
  push   %r14
  mov    %rdi,%rcx
  xor    %r14d,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  test   %ebp,%ebp
  jle    L7
  mov    D1(%rip),%r12
  mov    %rdi,%rbx
  xor    %r14d,%r14d
  jmp    L21
  nopl   0x0(%rax,%rax,1)
L2:
  lea    0x1(%r14),%edx
  cmp    $0xa,%eax
  je     L43
  mov    %al,(%rcx)
  lea    0x1(%rcx),%rbx
  cmp    %edx,%ebp
  je     L12
  mov    %edx,%r14d
L21:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rbx,%rcx
  cmp    %eax,(%r12)
  jne    L2
  cmp    $0xa,%eax
  je     L20
L7:
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L20:
  lea    0x1(%r14),%edx
L43:
  movslq %edx,%rcx
  movb   $0xa,(%rbx)
  mov    %rcx,%r14
  add    %r13,%rcx
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L12:
  movslq %ebp,%rcx
  mov    %ebp,%r14d
  add    %r13,%rcx
  mov    %r14d,%eax
  movb   $0x0,(%rcx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int GetLine(char s[], int limit) {
  int c, i;
  for (i = 0; i < limit - 1 && (c = getchar()) != EOF && c != '\n'; i++) {
    s[i] = c;
  }
  if (c == '\n') {
    s[i] = c;
    i++;
  }
  s[i] = '\0';
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<parseTransformArgs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  movl   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L132
L172:
  addq   $0x1,-0x8(%rbp)
L132:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L108
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x28,%al
  jne    L172
L108:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L169
  mov    $0x1,%eax
  jmp    L47
L169:
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L72
L8:
  addq   $0x1,-0x10(%rbp)
L72:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L138
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x29,%al
  jne    L8
L138:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L162
  mov    $0x1,%eax
  jmp    L47
L14:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isnum@plt>
  test   %rax,%rax
  je     L136
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x24(%rbp)
  jg     L11
  mov    $0x0,%eax
  jmp    L47
L11:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <atof@plt>
  mov    %rax,%rsi
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%ecx
  mov    -0x30(%rbp),%rdx
  mov    %ecx,(%rdx)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  cvtsi2ss %rsi,%xmm0
  movss  %xmm0,(%rax)
  jmp    L32
L1:
  addq   $0x1,-0x8(%rbp)
L32:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jae    L162
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isnum@plt>
  test   %rax,%rax
  jne    L1
  jmp    L162
L136:
  addq   $0x1,-0x8(%rbp)
L162:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jb     L14
  mov    -0x10(%rbp),%rax
  sub    -0x18(%rbp),%rax
L47:
  leave
  ret
```

assistant: ```c
int parseTransformArgs(const char *str, float *args, int maxNa, int *na) {
  const char *end;
  const char *ptr;
  *na = 0;
  ptr = str;
  while (*ptr && *ptr != '(')
    ++ptr;
  if (*ptr == 0)
    return 1;
  end = ptr;
  while (*end && *end != ')')
    ++end;
  if (*end == 0)
    return 1;
  while (ptr < end) {
    if (isnum(*ptr)) {
      if (*na >= maxNa)
        return 0;
      args[(*na)++] = (float)atof(ptr);
      while (ptr < end && isnum(*ptr))
        ++ptr;
    } else {
      ++ptr;
    }
  }
  return (int)(end - str);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<parseTransformArgs>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r15
  mov    %edx,%r14d
  mov    %rcx,%r12
  movl   $0x0,(%rcx)
  movzbl (%rdi),%eax
  test   %al,%al
  je     L51
  cmp    $0x28,%al
  je     L51
  mov    %rdi,%rbx
L90:
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  test   %al,%al
  je     L130
  cmp    $0x28,%al
  jne    L90
L130:
  mov    $0x1,%edx
  test   %al,%al
  je     L33
  movzbl (%rbx),%eax
  test   %al,%al
  je     L197
  cmp    $0x29,%al
  je     L197
  mov    %rbx,%rbp
L85:
  add    $0x1,%rbp
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L31
  cmp    $0x29,%al
  jne    L85
L31:
  mov    $0x1,%edx
  test   %al,%al
  je     L33
  cmp    %rbx,%rbp
  ja     L210
  jmp    L229
L51:
  mov    %r13,%rbx
  jmp    L130
L34:
  cmp    %r14d,(%r12)
  jge    L42
  mov    %rbx,%rdi
  call   <atof@plt>
  mov    %rax,%rdx
  mov    (%r12),%eax
  lea    0x1(%rax),%ecx
  mov    %ecx,(%r12)
  cltq
  pxor   %xmm0,%xmm0
  cvtsi2ss %rdx,%xmm0
  movss  %xmm0,(%r15,%rax,4)
  cmp    %rbx,%rbp
  jbe    L229
L209:
  movsbl (%rbx),%edi
  call   <isnum@plt>
  test   %rax,%rax
  je     L39
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L209
L229:
  mov    %ebp,%edx
  sub    %r13d,%edx
L33:
  mov    %edx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L39:
  cmp    %rbp,%rbx
  jae    L229
L210:
  movsbl (%rbx),%edi
  call   <isnum@plt>
  test   %rax,%rax
  jne    L34
  add    $0x1,%rbx
  jmp    L39
L42:
  mov    $0x0,%edx
  jmp    L33
L197:
  mov    $0x1,%edx
  mov    %rbx,%rbp
  test   %al,%al
  je     L33
  jmp    L229
```

assistant: ```c
int parseTransformArgs(const char *str, float *args, int maxNa, int *na) {
  const char *end;
  const char *ptr;
  *na = 0;
  ptr = str;
  while (*ptr && *ptr != '(')
    ++ptr;
  if (*ptr == 0)
    return 1;
  end = ptr;
  while (*end && *end != ')')
    ++end;
  if (*end == 0)
    return 1;
  while (ptr < end) {
    if (isnum(*ptr)) {
      if (*na >= maxNa)
        return 0;
      args[(*na)++] = (float)atof(ptr);
      while (ptr < end && isnum(*ptr))
        ++ptr;
    } else {
      ++ptr;
    }
  }
  return (int)(end - str);
}
```