ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<test_onepath>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D10(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  call   <read_quaddobl_target_system@plt>
  mov    $0x0,%eax
  call   <read_quaddobl_start_system@plt>
  mov    $0x0,%eax
  call   <copy_quaddobl_start_solutions_to_container@plt>
  lea    0x4(%rsp),%rdi
  call   <solcon_number_of_quaddobl_solutions@plt>
  mov    0x4(%rsp),%esi
  lea    D20(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  pxor   %xmm1,%xmm1
  movsd  D1(%rip),%xmm0
  mov    $0x1,%edi
  call   <ade_onepath_qd@plt>
  lea    D19(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  call   <solcon_write_quaddobl_solutions@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D10
type_guessed: string
name: parse_data
arguments: label: D20
type_guessed: string
name: parse_data
arguments: label: D1
type_guessed: f64
name: parse_data
arguments: label: D19
type_guessed: string

tool: D10: 
Tracking one solution path ...

tool: D20: The solution container has size %d.


tool: D1: 1

tool: D19: The solutions after Newton's method :

assistant: ```c
void test_onepath(void) {
  int fail, len;
  printf("\nTracking one solution path ...\n");
  fail = read_quaddobl_target_system();
  fail = read_quaddobl_start_system();
  fail = copy_quaddobl_start_solutions_to_container();
  fail = solcon_number_of_quaddobl_solutions(&len);
  printf("The solution container has size %d.\n", len);
  fail = ade_onepath_qd(1, 1.0, 0.0);
  printf("The solutions after Newton's method :\n");
  fail = solcon_write_quaddobl_solutions();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<test_onepath>:
  endbr64
  sub    $0x18,%rsp
  lea    D27(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  xor    %eax,%eax
  call   <read_quaddobl_target_system@plt>
  xor    %eax,%eax
  call   <read_quaddobl_start_system@plt>
  xor    %eax,%eax
  call   <copy_quaddobl_start_solutions_to_container@plt>
  lea    0x4(%rsp),%rdi
  call   <solcon_number_of_quaddobl_solutions@plt>
  mov    0x4(%rsp),%esi
  lea    D21(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  pxor   %xmm1,%xmm1
  mov    $0x1,%edi
  movsd  D32(%rip),%xmm0
  call   <ade_onepath_qd@plt>
  lea    D24(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  call   <solcon_write_quaddobl_solutions@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D27
type: string
name: parse_data
arguments: tag: D21
type: string
name: parse_data
arguments: tag: D32
type: f64
name: parse_data
arguments: tag: D24
type: string

tool: D27: 
Tracking one solution path ...

tool: D21: The solution container has size %d.


tool: D32: 1

tool: D24: The solutions after Newton's method :

assistant: ```c
void test_onepath(void) {
  int fail, len;
  printf("\nTracking one solution path ...\n");
  fail = read_quaddobl_target_system();
  fail = read_quaddobl_start_system();
  fail = copy_quaddobl_start_solutions_to_container();
  fail = solcon_number_of_quaddobl_solutions(&len);
  printf("The solution container has size %d.\n", len);
  fail = ade_onepath_qd(1, 1.0, 0.0);
  printf("The solutions after Newton's method :\n");
  fail = solcon_write_quaddobl_solutions();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test_onepath>:
  endbr64
  sub    $0x18,%rsp
  lea    D25(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  xor    %eax,%eax
  call   <read_quaddobl_target_system@plt>
  xor    %eax,%eax
  call   <read_quaddobl_start_system@plt>
  xor    %eax,%eax
  call   <copy_quaddobl_start_solutions_to_container@plt>
  lea    0x4(%rsp),%rdi
  call   <solcon_number_of_quaddobl_solutions@plt>
  mov    0x4(%rsp),%esi
  lea    D23(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  pxor   %xmm1,%xmm1
  mov    $0x1,%edi
  movsd  D1(%rip),%xmm0
  call   <ade_onepath_qd@plt>
  lea    D20(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  call   <solcon_write_quaddobl_solutions@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D25
datatype: string
name: unpack_data
arguments: data_tag: D23
datatype: string
name: unpack_data
arguments: data_tag: D1
datatype: f64
name: unpack_data
arguments: data_tag: D20
datatype: string

tool: D25: 
Tracking one solution path ...

tool: D23: The solution container has size %d.


tool: D1: 1

tool: D20: The solutions after Newton's method :

assistant: ```c
void test_onepath(void) {
  int fail, len;
  printf("\nTracking one solution path ...\n");
  fail = read_quaddobl_target_system();
  fail = read_quaddobl_start_system();
  fail = copy_quaddobl_start_solutions_to_container();
  fail = solcon_number_of_quaddobl_solutions(&len);
  printf("The solution container has size %d.\n", len);
  fail = ade_onepath_qd(1, 1.0, 0.0);
  printf("The solutions after Newton's method :\n");
  fail = solcon_write_quaddobl_solutions();
}
```
user: What is the c source code of the assembly code below:

```asm
<getbitu>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L1
L18:
  mov    -0x8(%rbp),%eax
  lea    (%rax,%rax,1),%esi
  mov    -0x4(%rbp),%eax
  lea    0x7(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x3,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  shr    $0x1d,%ecx
  add    %ecx,%eax
  and    $0x7,%eax
  sub    %ecx,%eax
  mov    %eax,%ecx
  mov    $0x7,%eax
  sub    %ecx,%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  and    $0x1,%eax
  add    %esi,%eax
  mov    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L1:
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x4(%rbp)
  jl     L18
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int getbitu(const unsigned char *buff, int pos, int len) {
  unsigned int bits = 0;
  int i;
  for (i = pos; i < pos + len; i++) {
    bits = (bits << 1) + ((buff[i / 8] >> (7 - i % 8)) & 1u);
  }
  return bits;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<getbitu>:
  endbr64
  mov    %rdi,%r8
  mov    %esi,%eax
  lea    (%rsi,%rdx,1),%r9d
  cmp    %r9d,%esi
  jge    L5
  mov    $0x0,%edi
  mov    $0x7,%r10d
L27:
  lea    0x7(%rax),%edx
  test   %eax,%eax
  cmovns %eax,%edx
  sar    $0x3,%edx
  movslq %edx,%rdx
  movzbl (%r8,%rdx,1),%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  shr    $0x1d,%ecx
  lea    (%rax,%rcx,1),%esi
  and    $0x7,%esi
  sub    %ecx,%esi
  mov    %r10d,%ecx
  sub    %esi,%ecx
  sar    %cl,%edx
  and    $0x1,%edx
  lea    (%rdx,%rdi,2),%edi
  add    $0x1,%eax
  cmp    %r9d,%eax
  jne    L27
L2:
  mov    %edi,%eax
  ret
L5:
  mov    $0x0,%edi
  jmp    L2
```

assistant: ```c
unsigned int getbitu(const unsigned char *buff, int pos, int len) {
  unsigned int bits = 0;
  int i;
  for (i = pos; i < pos + len; i++) {
    bits = (bits << 1) + ((buff[i / 8] >> (7 - i % 8)) & 1u);
  }
  return bits;
}
```
user: What is the c source code of the assembly code below?

```asm
<getbitu>:
  endbr64
  lea    (%rsi,%rdx,1),%r9d
  mov    %esi,%eax
  xor    %r8d,%r8d
  cmp    %r9d,%esi
  jge    L16
  mov    $0x7,%r10d
  nopl   0x0(%rax,%rax,1)
L12:
  test   %eax,%eax
  lea    0x7(%rax),%edx
  mov    %eax,%ecx
  cmovns %eax,%edx
  sar    $0x1f,%ecx
  shr    $0x1d,%ecx
  lea    (%rax,%rcx,1),%esi
  sar    $0x3,%edx
  add    $0x1,%eax
  movslq %edx,%rdx
  and    $0x7,%esi
  sub    %ecx,%esi
  movzbl (%rdi,%rdx,1),%edx
  mov    %r10d,%ecx
  sub    %esi,%ecx
  sar    %cl,%edx
  and    $0x1,%edx
  lea    (%rdx,%r8,2),%r8d
  cmp    %r9d,%eax
  jne    L12
L16:
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned int getbitu(const unsigned char *buff, int pos, int len) {
  unsigned int bits = 0;
  int i;
  for (i = pos; i < pos + len; i++) {
    bits = (bits << 1) + ((buff[i / 8] >> (7 - i % 8)) & 1u);
  }
  return bits;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getbitu>:
  endbr64
  lea    (%rsi,%rdx,1),%r9d
  mov    %esi,%eax
  xor    %r8d,%r8d
  cmp    %r9d,%esi
  jge    L18
  mov    $0x7,%r10d
  nopl   0x0(%rax,%rax,1)
L1:
  test   %eax,%eax
  lea    0x7(%rax),%edx
  mov    %eax,%ecx
  cmovns %eax,%edx
  sar    $0x1f,%ecx
  shr    $0x1d,%ecx
  lea    (%rax,%rcx,1),%esi
  sar    $0x3,%edx
  add    $0x1,%eax
  movslq %edx,%rdx
  and    $0x7,%esi
  sub    %ecx,%esi
  movzbl (%rdi,%rdx,1),%edx
  mov    %r10d,%ecx
  sub    %esi,%ecx
  sar    %cl,%edx
  and    $0x1,%edx
  lea    (%rdx,%r8,2),%r8d
  cmp    %r9d,%eax
  jne    L1
L18:
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned int getbitu(const unsigned char *buff, int pos, int len) {
  unsigned int bits = 0;
  int i;
  for (i = pos; i < pos + len; i++) {
    bits = (bits << 1) + ((buff[i / 8] >> (7 - i % 8)) & 1u);
  }
  return bits;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<safeStrnlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L35
  cmpl   $0x0,-0x1c(%rbp)
  jg     L12
L35:
  mov    $0xffffffff,%eax
  jmp    L74
L12:
  movl   $0x0,-0x4(%rbp)
  jmp    L71
L22:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L81
  addl   $0x1,-0x4(%rbp)
L71:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L22
  jmp    L75
L81:
  nop
L75:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jne    L39
  mov    $0xffffffff,%eax
  jmp    L74
L39:
  mov    -0x4(%rbp),%eax
L74:
  pop    %rbp
  ret
```

assistant: ```c
int safeStrnlen(const char *buf, int buf_size) {
  int i = 0;
  if (buf == NULL || buf_size <= 0)
    return -1;
  for (i = 0; i < buf_size; i++) {
    if (buf[i] == '\0')
      break;
  }
  if (i == buf_size)
    return -1;
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<safeStrnlen>:
  endbr64
  test   %rdi,%rdi
  je     L46
  test   %esi,%esi
  jle    L46
  mov    $0x0,%eax
L30:
  cmpb   $0x0,(%rdi)
  je     L52
  add    $0x1,%eax
  add    $0x1,%rdi
  cmp    %eax,%esi
  jg     L30
L52:
  cmp    %eax,%esi
  je     L4
L50:
  ret
L4:
  mov    $0xffffffff,%eax
  ret
L46:
  mov    $0xffffffff,%eax
  jmp    L50
```

assistant: ```c
int safeStrnlen(const char *buf, int buf_size) {
  int i = 0;
  if (buf == NULL || buf_size <= 0)
    return -1;
  for (i = 0; i < buf_size; i++) {
    if (buf[i] == '\0')
      break;
  }
  if (i == buf_size)
    return -1;
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<safeStrnlen>:
  endbr64
  test   %rdi,%rdi
  je     L18
  test   %esi,%esi
  jle    L18
  xor    %eax,%eax
  nop
L47:
  cmpb   $0x0,(%rdi)
  je     L9
  add    $0x1,%eax
  add    $0x1,%rdi
  cmp    %eax,%esi
  jg     L47
L9:
  cmp    %eax,%esi
  je     L18
  ret
L18:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int safeStrnlen(const char *buf, int buf_size) {
  int i = 0;
  if (buf == NULL || buf_size <= 0)
    return -1;
  for (i = 0; i < buf_size; i++) {
    if (buf[i] == '\0')
      break;
  }
  if (i == buf_size)
    return -1;
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<safeStrnlen>:
  endbr64
  test   %rdi,%rdi
  je     L3
  test   %esi,%esi
  jle    L3
  xor    %eax,%eax
  nop
L2:
  cmpb   $0x0,(%rdi)
  je     L29
  add    $0x1,%eax
  add    $0x1,%rdi
  cmp    %eax,%esi
  jg     L2
L29:
  cmp    %eax,%esi
  je     L3
  ret
L3:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int safeStrnlen(const char *buf, int buf_size) {
  int i = 0;
  if (buf == NULL || buf_size <= 0)
    return -1;
  for (i = 0; i < buf_size; i++) {
    if (buf[i] == '\0')
      break;
  }
  if (i == buf_size)
    return -1;
  return i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cur_ms_ts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    -0x10(%rbp),%eax
  imul   $0x3e8,%eax,%ecx
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%esi
  mov    %edx,%eax
  sub    %esi,%eax
  add    %ecx,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
uint32_t cur_ms_ts(void) {
  struct timeval tv;
  gettimeofday(&tv, 0);
  return tv.tv_sec * 1000 + tv.tv_usec / 1000;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cur_ms_ts>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  imul   $0x3e8,(%rsp),%eax
  mov    0x4(%rsp),%ecx
  movslq %ecx,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x26,%rdx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  add    %edx,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint32_t cur_ms_ts(void) {
  struct timeval tv;
  gettimeofday(&tv, 0);
  return tv.tv_sec * 1000 + tv.tv_usec / 1000;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<cur_ms_ts>:
  endbr64
  sub    $0x18,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  movslq 0x4(%rsp),%rdx
  imul   $0x3e8,(%rsp),%eax
  mov    %rdx,%rcx
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x26,%rdx
  sub    %ecx,%edx
  add    %edx,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint32_t cur_ms_ts(void) {
  struct timeval tv;
  gettimeofday(&tv, 0);
  return tv.tv_sec * 1000 + tv.tv_usec / 1000;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<cur_ms_ts>:
  endbr64
  sub    $0x18,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  movslq 0x4(%rsp),%rdx
  imul   $0x3e8,(%rsp),%eax
  mov    %rdx,%rcx
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x26,%rdx
  sub    %ecx,%edx
  add    %edx,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint32_t cur_ms_ts(void) {
  struct timeval tv;
  gettimeofday(&tv, 0);
  return tv.tv_sec * 1000 + tv.tv_usec / 1000;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fmaxl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x8,%rsp
  push   0x18(%rbp)
  push   0x10(%rbp)
  call   <__isnanl@plt>
  add    $0x10,%rsp
  test   %rax,%rax
  je     L57
  fldt   0x20(%rbp)
  jmp    L10
L57:
  push   0x28(%rbp)
  push   0x20(%rbp)
  call   <__isnanl@plt>
  add    $0x10,%rsp
  test   %rax,%rax
  je     L11
  fldt   0x10(%rbp)
  jmp    L10
L11:
  push   0x18(%rbp)
  push   0x10(%rbp)
  call   <__builtin_signbit@plt>
  add    $0x10,%rsp
  mov    %rax,%rbx
  push   0x28(%rbp)
  push   0x20(%rbp)
  call   <__builtin_signbit@plt>
  add    $0x10,%rsp
  cmp    %rax,%rbx
  je     L21
  push   0x18(%rbp)
  push   0x10(%rbp)
  call   <__builtin_signbit@plt>
  add    $0x10,%rsp
  test   %rax,%rax
  je     L23
  fldt   0x20(%rbp)
  jmp    L10
L23:
  fldt   0x10(%rbp)
  jmp    L10
L21:
  fldt   0x20(%rbp)
  fldt   0x10(%rbp)
  fcomip %st(1),%st
  fstp   %st(0)
  jbe    L67
  fldt   0x10(%rbp)
  jmp    L10
L67:
  fldt   0x20(%rbp)
L10:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
long double fmaxl(long double x, long double y) {
  if (((sizeof(x) == sizeof(float)) ? __isnanf(x) : (sizeof(x) == sizeof(double)) ? __isnan(x) : __isnanl(x)))
    return (y);
  if (((sizeof(y) == sizeof(float)) ? __isnanf(y) : (sizeof(y) == sizeof(double)) ? __isnan(y) : __isnanl(y)))
    return (x);
  if (__builtin_signbit(x) != __builtin_signbit(y))
    if (__builtin_signbit(x))
      return (y);
    else
      return (x);
  return (x > y ? x : y);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<fmaxl>:
  endbr64
  push   %rbx
  push   0x18(%rsp)
  push   0x18(%rsp)
  call   <__isnanl@plt>
  add    $0x10,%rsp
  test   %rax,%rax
  je     L4
L48:
  fldt   0x20(%rsp)
  pop    %rbx
  ret
L4:
  push   0x28(%rsp)
  push   0x28(%rsp)
  call   <__isnanl@plt>
  add    $0x10,%rsp
  test   %rax,%rax
  je     L10
  fldt   0x10(%rsp)
  fstpt  0x20(%rsp)
  jmp    L48
L10:
  push   0x18(%rsp)
  push   0x18(%rsp)
  call   <__builtin_signbit@plt>
  mov    %rax,%rbx
  push   0x38(%rsp)
  push   0x38(%rsp)
  call   <__builtin_signbit@plt>
  add    $0x20,%rsp
  cmp    %rax,%rbx
  jne    L24
  fldt   0x20(%rsp)
  fldt   0x10(%rsp)
  fucomi %st(1),%st
  fcmovbe %st(1),%st
  fstp   %st(1)
  fstpt  0x20(%rsp)
  jmp    L48
L24:
  push   0x18(%rsp)
  push   0x18(%rsp)
  call   <__builtin_signbit@plt>
  add    $0x10,%rsp
  test   %rax,%rax
  fldt   0x20(%rsp)
  fldt   0x10(%rsp)
  fcmovne %st(1),%st
  fstp   %st(1)
  fstpt  0x20(%rsp)
  jmp    L48
```

assistant: ```c
long double fmaxl(long double x, long double y) {
  if (((sizeof(x) == sizeof(float)) ? __isnanf(x) : (sizeof(x) == sizeof(double)) ? __isnan(x) : __isnanl(x)))
    return (y);
  if (((sizeof(y) == sizeof(float)) ? __isnanf(y) : (sizeof(y) == sizeof(double)) ? __isnan(y) : __isnanl(y)))
    return (x);
  if (__builtin_signbit(x) != __builtin_signbit(y))
    if (__builtin_signbit(x))
      return (y);
    else
      return (x);
  return (x > y ? x : y);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<fmaxl>:
  endbr64
  push   %rbx
  push   0x18(%rsp)
  push   0x18(%rsp)
  call   <__isnanl@plt>
  pop    %r8
  pop    %r9
  test   %rax,%rax
  je     L29
L41:
  fldt   0x20(%rsp)
  pop    %rbx
  ret
  nopl   0x0(%rax)
L29:
  push   0x28(%rsp)
  push   0x28(%rsp)
  call   <__isnanl@plt>
  pop    %rsi
  pop    %rdi
  test   %rax,%rax
  jne    L36
  push   0x18(%rsp)
  push   0x18(%rsp)
  call   <__builtin_signbit@plt>
  push   0x38(%rsp)
  push   0x38(%rsp)
  mov    %rax,%rbx
  call   <__builtin_signbit@plt>
  add    $0x20,%rsp
  cmp    %rax,%rbx
  jne    L33
  fldt   0x20(%rsp)
  fldt   0x10(%rsp)
  fcomip %st(1),%st
  fstp   %st(0)
  jbe    L41
L36:
  fldt   0x10(%rsp)
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L33:
  push   0x18(%rsp)
  push   0x18(%rsp)
  call   <__builtin_signbit@plt>
  pop    %rdx
  pop    %rcx
  test   %rax,%rax
  jne    L41
  jmp    L36
```

assistant: ```c
long double fmaxl(long double x, long double y) {
  if (((sizeof(x) == sizeof(float)) ? __isnanf(x) : (sizeof(x) == sizeof(double)) ? __isnan(x) : __isnanl(x)))
    return (y);
  if (((sizeof(y) == sizeof(float)) ? __isnanf(y) : (sizeof(y) == sizeof(double)) ? __isnan(y) : __isnanl(y)))
    return (x);
  if (__builtin_signbit(x) != __builtin_signbit(y))
    if (__builtin_signbit(x))
      return (y);
    else
      return (x);
  return (x > y ? x : y);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fmaxl>:
  endbr64
  push   %rbx
  push   0x18(%rsp)
  push   0x18(%rsp)
  call   <__isnanl@plt>
  pop    %r8
  pop    %r9
  test   %rax,%rax
  je     L56
L5:
  fldt   0x20(%rsp)
  pop    %rbx
  ret
  nopl   0x0(%rax)
L56:
  push   0x28(%rsp)
  push   0x28(%rsp)
  call   <__isnanl@plt>
  pop    %rsi
  pop    %rdi
  test   %rax,%rax
  jne    L53
  push   0x18(%rsp)
  push   0x18(%rsp)
  call   <__builtin_signbit@plt>
  push   0x38(%rsp)
  push   0x38(%rsp)
  mov    %rax,%rbx
  call   <__builtin_signbit@plt>
  add    $0x20,%rsp
  cmp    %rax,%rbx
  jne    L15
  fldt   0x20(%rsp)
  fldt   0x10(%rsp)
  fcomip %st(1),%st
  fstp   %st(0)
  jbe    L5
L53:
  fldt   0x10(%rsp)
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L15:
  push   0x18(%rsp)
  push   0x18(%rsp)
  call   <__builtin_signbit@plt>
  pop    %rdx
  pop    %rcx
  test   %rax,%rax
  jne    L5
  jmp    L53
```

assistant: ```c
long double fmaxl(long double x, long double y) {
  if (((sizeof(x) == sizeof(float)) ? __isnanf(x) : (sizeof(x) == sizeof(double)) ? __isnan(x) : __isnanl(x)))
    return (y);
  if (((sizeof(y) == sizeof(float)) ? __isnanf(y) : (sizeof(y) == sizeof(double)) ? __isnan(y) : __isnanl(y)))
    return (x);
  if (__builtin_signbit(x) != __builtin_signbit(y))
    if (__builtin_signbit(x))
      return (y);
    else
      return (x);
  return (x > y ? x : y);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<printf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L3
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L3:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb8(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  mov    -0xb8(%rbp),%edx
  mov    -0xc8(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <vfprintf@plt>
  mov    %eax,-0xb4(%rbp)
  mov    -0xb8(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xb4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int printf(const char *fmt, ...) {
  int len;
  va_list ap;
  va_start(ap, fmt);
  len = vfprintf(stdout, fmt, ap);
  va_end(ap);
  return (len);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<printf>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    D9(%rip),%rax
  mov    (%rax),%edi
  call   <vfprintf@plt>
  mov    %eax,%ebx
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int printf(const char *fmt, ...) {
  int len;
  va_list ap;
  va_start(ap, fmt);
  len = vfprintf(stdout, fmt, ap);
  va_end(ap);
  return (len);
}
```
user: Translate the below assembly code into C programming language:

```asm
<printf>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  mov    %rdi,%rsi
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r13d,%edi
  call   <va_start@plt>
  mov    D0(%rip),%rax
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    (%rax),%edi
  call   <vfprintf@plt>
  mov    %r13d,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int printf(const char *fmt, ...) {
  int len;
  va_list ap;
  va_start(ap, fmt);
  len = vfprintf(stdout, fmt, ap);
  va_end(ap);
  return (len);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<printf>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  mov    %rdi,%rsi
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r13d,%edi
  call   <va_start@plt>
  mov    D1(%rip),%rax
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    (%rax),%edi
  call   <vfprintf@plt>
  mov    %r13d,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int printf(const char *fmt, ...) {
  int len;
  va_list ap;
  va_start(ap, fmt);
  len = vfprintf(stdout, fmt, ap);
  va_end(ap);
  return (len);
}
```