ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<cec_ops_request_short_audio_descriptor>:
  endbr64
  mov    %rdi,%r8
  mov    %rsi,%r9
  mov    %rdx,%r10
  mov    (%rdi),%eax
  sub    $0x2,%eax
  cmp    $0x4,%eax
  jg     L13
  mov    %eax,(%rsi)
  test   %eax,%eax
  jne    L16
  ret
L13:
  movl   $0x4,(%rsi)
L16:
  mov    $0x0,%eax
L24:
  lea    0x2(%rax),%edi
  mov    %eax,%esi
  mov    0x8(%r8),%rdx
  mov    (%rdx,%rdi,4),%edx
  sar    $0x6,%edx
  mov    %edx,(%r10,%rsi,4)
  mov    0x8(%r8),%rdx
  mov    (%rdx,%rdi,4),%edx
  and    $0x3f,%edx
  mov    %edx,(%rcx,%rsi,4)
  add    $0x1,%eax
  cmp    %eax,(%r9)
  ja     L24
  ret
```

assistant: ```c
void cec_ops_request_short_audio_descriptor(const struct cec_msg *msg, __u8 *num_descriptors, __u8 *audio_format_id, __u8 *audio_format_code) {
  unsigned int i;
  *num_descriptors = msg->len - 2;
  if (*num_descriptors > 4)
    *num_descriptors = 4;
  for (i = 0; i < *num_descriptors; i++) {
    audio_format_id[i] = msg->msg[2 + i] >> 6;
    audio_format_code[i] = msg->msg[2 + i] & 0x3f;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cec_ops_request_short_audio_descriptor>:
  endbr64
  mov    (%rdi),%eax
  mov    %rsi,%r9
  sub    $0x2,%eax
  cmp    $0x4,%eax
  jg     L36
  mov    %eax,(%rsi)
  test   %eax,%eax
  je     L11
L6:
  mov    0x8(%rdi),%r10
  xor    %eax,%eax
  nopl   (%rax)
L18:
  lea    0x2(%rax),%esi
  mov    %eax,%edi
  add    $0x1,%eax
  lea    (%r10,%rsi,4),%r8
  mov    (%r8),%esi
  sar    $0x6,%esi
  mov    %esi,(%rdx,%rdi,4)
  mov    (%r8),%esi
  and    $0x3f,%esi
  mov    %esi,(%rcx,%rdi,4)
  cmp    %eax,(%r9)
  ja     L18
  ret
  nopl   0x0(%rax)
L11:
  ret
  nopl   0x0(%rax)
L36:
  movl   $0x4,(%rsi)
  jmp    L6
```

assistant: ```c
void cec_ops_request_short_audio_descriptor(const struct cec_msg *msg, __u8 *num_descriptors, __u8 *audio_format_id, __u8 *audio_format_code) {
  unsigned int i;
  *num_descriptors = msg->len - 2;
  if (*num_descriptors > 4)
    *num_descriptors = 4;
  for (i = 0; i < *num_descriptors; i++) {
    audio_format_id[i] = msg->msg[2 + i] >> 6;
    audio_format_code[i] = msg->msg[2 + i] & 0x3f;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<cec_ops_request_short_audio_descriptor>:
  endbr64
  mov    (%rdi),%eax
  mov    %rsi,%r9
  sub    $0x2,%eax
  cmp    $0x4,%eax
  jg     L13
  mov    %eax,(%rsi)
  test   %eax,%eax
  je     L36
L27:
  mov    0x8(%rdi),%r10
  xor    %eax,%eax
  nopl   (%rax)
L28:
  lea    0x2(%rax),%esi
  mov    %eax,%edi
  add    $0x1,%eax
  lea    (%r10,%rsi,4),%r8
  mov    (%r8),%esi
  sar    $0x6,%esi
  mov    %esi,(%rdx,%rdi,4)
  mov    (%r8),%esi
  and    $0x3f,%esi
  mov    %esi,(%rcx,%rdi,4)
  cmp    %eax,(%r9)
  ja     L28
  ret
  nopl   0x0(%rax)
L36:
  ret
  nopl   0x0(%rax)
L13:
  movl   $0x4,(%rsi)
  jmp    L27
```

assistant: ```c
void cec_ops_request_short_audio_descriptor(const struct cec_msg *msg, __u8 *num_descriptors, __u8 *audio_format_id, __u8 *audio_format_code) {
  unsigned int i;
  *num_descriptors = msg->len - 2;
  if (*num_descriptors > 4)
    *num_descriptors = 4;
  for (i = 0; i < *num_descriptors; i++) {
    audio_format_id[i] = msg->msg[2 + i] >> 6;
    audio_format_code[i] = msg->msg[2 + i] & 0x3f;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<compare_vst_NSS_n>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x34(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L36:
  cmp    %rdx,%rsp
  je     L46
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L36
L46:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L33
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L33:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x20(%rbp)
  movl   $0x0,-0x2c(%rbp)
  jmp    L57
L21:
  mov    $0xff,%esi
  mov    $0x0,%edi
  call   <Frama_C_interval@plt>
  mov    -0x20(%rbp),%rcx
  mov    -0x34(%rbp),%edx
  movslq %edx,%rdx
  mov    %al,(%rcx,%rdx,1)
  addl   $0x1,-0x2c(%rbp)
L57:
  mov    -0x2c(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L21
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <compare_vst_NSS@plt>
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int compare_vst_NSS_n(int n) {
  unsigned char input[n];
  int i;
  for (i = 0; i < n; i++)
    input[n] = Frama_C_interval(0, 255);
  return compare_vst_NSS(input);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<compare_vst_NSS_n>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  movslq %edi,%rax
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L44:
  cmp    %rdx,%rsp
  je     L25
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L44
L25:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L14
  orq    $0x0,-0x8(%rsp,%rax,1)
L14:
  mov    %rsp,%r13
  test   %r12d,%r12d
  jle    L1
  mov    $0x0,%ebx
  movslq %r12d,%r14
L4:
  mov    $0xff,%esi
  mov    $0x0,%edi
  call   <Frama_C_interval@plt>
  mov    %al,0x0(%r13,%r14,1)
  add    $0x1,%ebx
  cmp    %ebx,%r12d
  jne    L4
L1:
  mov    %r13,%rdi
  call   <compare_vst_NSS@plt>
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L41
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L41:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int compare_vst_NSS_n(int n) {
  unsigned char input[n];
  int i;
  for (i = 0; i < n; i++)
    input[n] = Frama_C_interval(0, 255);
  return compare_vst_NSS(input);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<compare_vst_NSS_n>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  movslq %edi,%r13
  push   %r12
  mov    %r13,%r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  lea    0xf(%r13),%rax
  mov    %rsp,%rcx
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L40
L22:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L22
L40:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L16
L12:
  mov    %rsp,%r14
  test   %r12d,%r12d
  jle    L1
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L30:
  mov    $0xff,%esi
  xor    %edi,%edi
  add    $0x1,%ebx
  call   <Frama_C_interval@plt>
  mov    %al,(%r14,%r13,1)
  cmp    %ebx,%r12d
  jne    L30
L1:
  mov    %r14,%rdi
  call   <compare_vst_NSS@plt>
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L27
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nopl   0x0(%rax)
L16:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L12
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int compare_vst_NSS_n(int n) {
  unsigned char input[n];
  int i;
  for (i = 0; i < n; i++)
    input[n] = Frama_C_interval(0, 255);
  return compare_vst_NSS(input);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<compare_vst_NSS_n>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  movslq %edi,%r13
  push   %r12
  mov    %r13,%r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  lea    0xf(%r13),%rax
  mov    %rsp,%rcx
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L53
L63:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L63
L53:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L22
L37:
  mov    %rsp,%r14
  test   %r12d,%r12d
  jle    L14
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L41:
  mov    $0xff,%esi
  xor    %edi,%edi
  add    $0x1,%ebx
  call   <Frama_C_interval@plt>
  mov    %al,(%r14,%r13,1)
  cmp    %ebx,%r12d
  jne    L41
L14:
  mov    %r14,%rdi
  call   <compare_vst_NSS@plt>
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L67
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nopl   0x0(%rax)
L22:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L37
L67:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int compare_vst_NSS_n(int n) {
  unsigned char input[n];
  int i;
  for (i = 0; i < n; i++)
    input[n] = Frama_C_interval(0, 255);
  return compare_vst_NSS(input);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<divide>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L0
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <merge@plt>
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <merge@plt>
  mov    -0x20(%rbp),%ecx
  mov    -0x4(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <merge@plt>
L0:
  nop
  leave
  ret
```

assistant: ```c
void divide(int arr[], int left, int right) {
  int mid;
  if (left < right) {
    mid = (left + right);
    merge(arr, left, mid);
    merge(arr, mid + 1, right);
    merge(arr, left, mid, right);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<divide>:
  endbr64
  cmp    %edx,%esi
  jl     L8
  ret
L8:
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%ebx
  mov    %edx,%ebp
  lea    (%rsi,%rdx,1),%r13d
  mov    %r13d,%edx
  mov    $0x0,%eax
  call   <merge@plt>
  lea    0x1(%r13),%esi
  mov    %ebp,%edx
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <merge@plt>
  mov    %ebp,%ecx
  mov    %r13d,%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <merge@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void divide(int arr[], int left, int right) {
  int mid;
  if (left < right) {
    mid = (left + right);
    merge(arr, left, mid);
    merge(arr, mid + 1, right);
    merge(arr, left, mid, right);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<divide>:
  endbr64
  cmp    %edx,%esi
  jl     L5
  ret
  nopl   0x0(%rax)
L5:
  push   %r14
  lea    (%rsi,%rdx,1),%r14d
  xor    %eax,%eax
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %edx,%r12d
  mov    %r14d,%edx
  push   %rbp
  mov    %esi,%ebp
  sub    $0x8,%rsp
  call   <merge@plt>
  lea    0x1(%r14),%esi
  mov    %r12d,%edx
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <merge@plt>
  add    $0x8,%rsp
  mov    %r12d,%ecx
  mov    %r14d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1050 <merge@plt>
```

assistant: ```c
void divide(int arr[], int left, int right) {
  int mid;
  if (left < right) {
    mid = (left + right);
    merge(arr, left, mid);
    merge(arr, mid + 1, right);
    merge(arr, left, mid, right);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<divide>:
  endbr64
  cmp    %edx,%esi
  jl     L2
  ret
  nopl   0x0(%rax)
L2:
  push   %r14
  lea    (%rsi,%rdx,1),%r14d
  xor    %eax,%eax
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %edx,%r12d
  mov    %r14d,%edx
  push   %rbp
  mov    %esi,%ebp
  sub    $0x8,%rsp
  call   <merge@plt>
  lea    0x1(%r14),%esi
  mov    %r12d,%edx
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <merge@plt>
  add    $0x8,%rsp
  mov    %r12d,%ecx
  mov    %r14d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1050 <merge@plt>
```

assistant: ```c
void divide(int arr[], int left, int right) {
  int mid;
  if (left < right) {
    mid = (left + right);
    merge(arr, left, mid);
    merge(arr, mid + 1, right);
    merge(arr, left, mid, right);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<timeFactor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  movq   %xmm0,%rcx
  movapd %xmm1,%xmm0
  mov    $0x0,%eax
  mov    $0x0,%edx
  mov    %rcx,%rax
  movq   %xmm0,%rdx
  mov    %rax,-0x40(%rbp)
  mov    %rdx,-0x38(%rbp)
  movsd  %xmm2,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movsd  -0x40(%rbp),%xmm0
  movsd  %xmm0,-0x30(%rbp)
  movsd  -0x38(%rbp),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x48(%rbp),%xmm1
  movsd  -0x48(%rbp),%xmm0
  movsd  D9(%rip),%xmm3
  movapd %xmm0,%xmm2
  subsd  %xmm3,%xmm2
  movsd  -0x40(%rbp),%xmm0
  mulsd  %xmm0,%xmm2
  movsd  D3(%rip),%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x28(%rbp)
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <normalize@plt>
  mov    -0x30(%rbp),%rax
  mov    -0x28(%rbp),%rdx
  mov    %rax,%rsi
  mov    %rdx,%rdi
  mov    %rax,%rcx
  mov    %rdx,%rbx
  mov    %rsi,%rax
  movq   %rbx,%xmm1
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  movq   %rax,%xmm0
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D9
type: f64
name: parse_data
arguments: label: D3
type: byte[8]

tool: D9: 1

tool: {"D3": [0x00,0x00,0x00,0x00,0x65,0xcd,0xcd,0x41]}

assistant: ```c
struct timespec timeFactor(struct timespec a, double f) {
  struct timespec r;
  r.tv_sec = a.tv_sec;
  r.tv_nsec = f * a.tv_nsec + (f - 1.) * a.tv_sec * 1000000000;
  normalize(&r);
  return r;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<timeFactor>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movsd  %xmm0,(%rsp)
  movapd %xmm2,%xmm3
  subsd  D1(%rip),%xmm3
  mulsd  %xmm3,%xmm0
  mulsd  D3(%rip),%xmm0
  mulsd  %xmm1,%xmm2
  addsd  %xmm2,%xmm0
  movsd  %xmm0,0x8(%rsp)
  mov    %rsp,%rdi
  call   <normalize@plt>
  movsd  (%rsp),%xmm0
  movsd  0x8(%rsp),%xmm1
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D1
type_guessed: f64
name: read_data
arguments: tag: D3
type_guessed: byte[8]

tool: D1: 1

tool: {"D3": [0x00,0x00,0x00,0x00,0x65,0xcd,0xcd,0x41]}

assistant: ```c
struct timespec timeFactor(struct timespec a, double f) {
  struct timespec r;
  r.tv_sec = a.tv_sec;
  r.tv_nsec = f * a.tv_nsec + (f - 1.) * a.tv_sec * 1000000000;
  normalize(&r);
  return r;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<timeFactor>:
  endbr64
  sub    $0x28,%rsp
  movapd %xmm2,%xmm3
  subsd  D14(%rip),%xmm3
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movsd  %xmm0,(%rsp)
  mulsd  %xmm1,%xmm2
  mov    %rsp,%rdi
  mulsd  %xmm3,%xmm0
  mulsd  D18(%rip),%xmm0
  addsd  %xmm2,%xmm0
  movsd  %xmm0,0x8(%rsp)
  call   <normalize@plt>
  movsd  (%rsp),%xmm0
  movsd  0x8(%rsp),%xmm1
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x28,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D14
type_infered: f64
name: parse_data
arguments: data_label: D18
type_infered: byte[8]

tool: D14: 1

tool: {"D18": [0x00,0x00,0x00,0x00,0x65,0xcd,0xcd,0x41]}

assistant: ```c
struct timespec timeFactor(struct timespec a, double f) {
  struct timespec r;
  r.tv_sec = a.tv_sec;
  r.tv_nsec = f * a.tv_nsec + (f - 1.) * a.tv_sec * 1000000000;
  normalize(&r);
  return r;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<timeFactor>:
  endbr64
  sub    $0x28,%rsp
  movapd %xmm2,%xmm3
  subsd  D0(%rip),%xmm3
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mulsd  %xmm1,%xmm2
  mov    %rsp,%rdi
  mulsd  %xmm0,%xmm3
  mulsd  D11(%rip),%xmm3
  addsd  %xmm2,%xmm3
  unpcklpd %xmm3,%xmm0
  movaps %xmm0,(%rsp)
  call   <normalize@plt>
  movsd  (%rsp),%xmm0
  movsd  0x8(%rsp),%xmm1
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x28,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D0
type: f64
name: extract_symbolic_data
arguments: tag: D11
type: byte[8]

tool: D0: 1

tool: {"D11": [0x00,0x00,0x00,0x00,0x65,0xcd,0xcd,0x41]}

assistant: ```c
struct timespec timeFactor(struct timespec a, double f) {
  struct timespec r;
  r.tv_sec = a.tv_sec;
  r.tv_nsec = f * a.tv_nsec + (f - 1.) * a.tv_sec * 1000000000;
  normalize(&r);
  return r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<nt_hash_password>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <nt_unicode@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L12
  lea    -0x15(%rbp),%rax
  mov    %rax,%rdi
  call   <MD4_Init@plt>
  mov    -0x14(%rbp),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x10(%rbp),%rcx
  lea    -0x15(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <MD4_Update@plt>
  lea    -0x15(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <MD4_Final@plt>
  lea    -0x15(%rbp),%rax
  mov    $0x1,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x14(%rbp),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  jmp    L4
L12:
  nop
L4:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

assistant: ```c
void nt_hash_password(unsigned char *hash, const char *pass) {
  int len;
  unsigned char *nt_pw;
  MD4_CTX context;
  len = strlen(pass);
  if ((nt_pw = nt_unicode(pass, len)) == NULL)
    return;
  MD4_Init(&context);
  MD4_Update(&context, nt_pw, 2 * len);
  MD4_Final(hash, &context);
  memset(&context, 0, sizeof context);
  memset(nt_pw, 0, 2 * len);
  free(nt_pw);
}
```
user: What is the c source code of the assembly code below:

```asm
<nt_hash_password>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %eax,%esi
  mov    %rbp,%rdi
  call   <nt_unicode@plt>
  test   %rax,%rax
  je     L12
  mov    %rax,%rbp
  lea    0x7(%rsp),%r13
  mov    %r13,%rdi
  call   <MD4_Init@plt>
  add    %ebx,%ebx
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <MD4_Update@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <MD4_Final@plt>
  mov    $0x1,%edx
  mov    $0x0,%esi
  mov    %r13,%rdi
  call   <memset@plt>
  mov    %ebx,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    %rbp,%rdi
  call   <free@plt>
L12:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void nt_hash_password(unsigned char *hash, const char *pass) {
  int len;
  unsigned char *nt_pw;
  MD4_CTX context;
  len = strlen(pass);
  if ((nt_pw = nt_unicode(pass, len)) == NULL)
    return;
  MD4_Init(&context);
  MD4_Update(&context, nt_pw, 2 * len);
  MD4_Final(hash, &context);
  memset(&context, 0, sizeof context);
  memset(nt_pw, 0, 2 * len);
  free(nt_pw);
}
```
user: What is the c source code of the assembly code below:

```asm
<nt_hash_password>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  mov    %rsi,%rdi
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  mov    %eax,%r12d
  call   <nt_unicode@plt>
  test   %rax,%rax
  je     L14
  lea    0x7(%rsp),%r14
  mov    %rax,%rbp
  add    %r12d,%r12d
  mov    %r14,%rdi
  call   <MD4_Init@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <MD4_Update@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <MD4_Final@plt>
  mov    $0x1,%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  call   <memset@plt>
  mov    %rbp,%rdi
  mov    %r12d,%edx
  xor    %esi,%esi
  call   <memset@plt>
  mov    %rbp,%rdi
  call   <free@plt>
L14:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void nt_hash_password(unsigned char *hash, const char *pass) {
  int len;
  unsigned char *nt_pw;
  MD4_CTX context;
  len = strlen(pass);
  if ((nt_pw = nt_unicode(pass, len)) == NULL)
    return;
  MD4_Init(&context);
  MD4_Update(&context, nt_pw, 2 * len);
  MD4_Final(hash, &context);
  memset(&context, 0, sizeof context);
  memset(nt_pw, 0, 2 * len);
  free(nt_pw);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<nt_hash_password>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  mov    %rsi,%rdi
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  mov    %eax,%r12d
  call   <nt_unicode@plt>
  test   %rax,%rax
  je     L13
  lea    0x7(%rsp),%r14
  mov    %rax,%rbp
  add    %r12d,%r12d
  mov    %r14,%rdi
  call   <MD4_Init@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <MD4_Update@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <MD4_Final@plt>
  mov    $0x1,%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  call   <memset@plt>
  mov    %rbp,%rdi
  mov    %r12d,%edx
  xor    %esi,%esi
  call   <memset@plt>
  mov    %rbp,%rdi
  call   <free@plt>
L13:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void nt_hash_password(unsigned char *hash, const char *pass) {
  int len;
  unsigned char *nt_pw;
  MD4_CTX context;
  len = strlen(pass);
  if ((nt_pw = nt_unicode(pass, len)) == NULL)
    return;
  MD4_Init(&context);
  MD4_Update(&context, nt_pw, 2 * len);
  MD4_Final(hash, &context);
  memset(&context, 0, sizeof context);
  memset(nt_pw, 0, 2 * len);
  free(nt_pw);
}
```