ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<build_rotmatrix>:
  endbr64
  movsd  0x18(%rsi),%xmm0
  movsd  (%rsi),%xmm3
  movsd  0x8(%rsi),%xmm2
  movsd  0x10(%rsi),%xmm1
  mulsd  %xmm3,%xmm3
  mulsd  %xmm0,%xmm0
  mulsd  %xmm2,%xmm2
  mulsd  %xmm1,%xmm1
  addsd  %xmm3,%xmm0
  subsd  %xmm2,%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdi)
  movsd  (%rsi),%xmm0
  movsd  0x10(%rsi),%xmm1
  mulsd  0x8(%rsi),%xmm0
  mulsd  0x18(%rsi),%xmm1
  subsd  %xmm1,%xmm0
  addsd  %xmm0,%xmm0
  movsd  %xmm0,0x8(%rdi)
  movsd  0x10(%rsi),%xmm0
  movsd  0x8(%rsi),%xmm1
  mulsd  (%rsi),%xmm0
  mulsd  0x18(%rsi),%xmm1
  movq   $0x0,0x18(%rdi)
  addsd  %xmm1,%xmm0
  addsd  %xmm0,%xmm0
  movsd  %xmm0,0x10(%rdi)
  movsd  (%rsi),%xmm0
  movsd  0x10(%rsi),%xmm1
  mulsd  0x8(%rsi),%xmm0
  mulsd  0x18(%rsi),%xmm1
  addsd  %xmm1,%xmm0
  addsd  %xmm0,%xmm0
  movsd  %xmm0,0x20(%rdi)
  movsd  0x18(%rsi),%xmm0
  movsd  (%rsi),%xmm3
  movsd  0x8(%rsi),%xmm2
  movsd  0x10(%rsi),%xmm1
  mulsd  %xmm3,%xmm3
  mulsd  %xmm0,%xmm0
  mulsd  %xmm2,%xmm2
  mulsd  %xmm1,%xmm1
  subsd  %xmm3,%xmm0
  addsd  %xmm2,%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,0x28(%rdi)
  movsd  0x8(%rsi),%xmm0
  movsd  (%rsi),%xmm1
  mulsd  0x10(%rsi),%xmm0
  mulsd  0x18(%rsi),%xmm1
  movq   $0x0,0x38(%rdi)
  subsd  %xmm1,%xmm0
  addsd  %xmm0,%xmm0
  movsd  %xmm0,0x30(%rdi)
  movsd  0x10(%rsi),%xmm0
  movsd  0x8(%rsi),%xmm1
  mulsd  (%rsi),%xmm0
  mov    D4(%rip),%rax
  mulsd  0x18(%rsi),%xmm1
  subsd  %xmm1,%xmm0
  addsd  %xmm0,%xmm0
  movsd  %xmm0,0x40(%rdi)
  movsd  0x8(%rsi),%xmm0
  movsd  (%rsi),%xmm1
  mulsd  0x10(%rsi),%xmm0
  mulsd  0x18(%rsi),%xmm1
  addsd  %xmm1,%xmm0
  addsd  %xmm0,%xmm0
  movsd  %xmm0,0x48(%rdi)
  movsd  0x18(%rsi),%xmm0
  movsd  (%rsi),%xmm3
  movsd  0x8(%rsi),%xmm2
  movsd  0x10(%rsi),%xmm1
  mov    %rax,0x78(%rdi)
  mulsd  %xmm0,%xmm0
  movq   $0x0,0x58(%rdi)
  mulsd  %xmm3,%xmm3
  movq   $0x0,0x60(%rdi)
  mulsd  %xmm2,%xmm2
  movq   $0x0,0x68(%rdi)
  mulsd  %xmm1,%xmm1
  movq   $0x0,0x70(%rdi)
  subsd  %xmm3,%xmm0
  subsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,0x50(%rdi)
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D4
data_type: f64

tool: D4: 1

assistant: ```c
void build_rotmatrix(double m[4][4], double quat[4]) {
  m[0][0] = ((quat[3]) * (quat[3])) + ((quat[0]) * (quat[0])) - ((quat[1]) * (quat[1])) - ((quat[2]) * (quat[2]));
  m[0][1] = 2.0 * (quat[0] * quat[1] - quat[2] * quat[3]);
  m[0][2] = 2.0 * (quat[2] * quat[0] + quat[1] * quat[3]);
  m[0][3] = 0.0;
  m[1][0] = 2.0 * (quat[0] * quat[1] + quat[2] * quat[3]);
  m[1][1] = ((quat[3]) * (quat[3])) - ((quat[0]) * (quat[0])) + ((quat[1]) * (quat[1])) - ((quat[2]) * (quat[2]));
  m[1][2] = 2.0 * (quat[1] * quat[2] - quat[0] * quat[3]);
  m[1][3] = 0.0;
  m[2][0] = 2.0 * (quat[2] * quat[0] - quat[1] * quat[3]);
  m[2][1] = 2.0 * (quat[1] * quat[2] + quat[0] * quat[3]);
  m[2][2] = ((quat[3]) * (quat[3])) - ((quat[0]) * (quat[0])) - ((quat[1]) * (quat[1])) + ((quat[2]) * (quat[2]));
  m[2][3] = 0.0;
  m[3][0] = 0.0;
  m[3][1] = 0.0;
  m[3][2] = 0.0;
  m[3][3] = 1.0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rank>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  movsbl -0x4(%rbp),%eax
  sub    $0x31,%eax
  cmp    $0x20,%eax
  ja     L4
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D7(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D7(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x0,%eax
  jmp    L23
  mov    $0x1,%eax
  jmp    L23
  mov    $0x2,%eax
  jmp    L23
  mov    $0x3,%eax
  jmp    L23
  mov    $0x4,%eax
  jmp    L23
  mov    $0x5,%eax
  jmp    L23
  mov    $0x6,%eax
  jmp    L23
  mov    $0x7,%eax
  jmp    L23
  mov    $0x8,%eax
  jmp    L23
  mov    $0x9,%eax
  jmp    L23
  mov    $0xa,%eax
  jmp    L23
  mov    $0xb,%eax
  jmp    L23
  mov    $0xc,%eax
  jmp    L23
L4:
  mov    $0xffffffff,%eax
L23:
  pop    %rbp
  ret
```

assistant: ```c
int rank(c)
char c;
{
  switch (c) {
  case 'A':
    return 0;
    break;
  case '1':
    return 1;
    break;
  case '2':
    return 2;
    break;
  case '3':
    return 3;
    break;
  case '4':
    return 4;
    break;
  case '5':
    return 5;
    break;
  case '6':
    return 6;
    break;
  case '7':
    return 7;
    break;
  case '8':
    return 8;
    break;
  case '9':
    return 9;
    break;
  case 'J':
    return 10;
    break;
  case 'Q':
    return 11;
    break;
  case 'K':
    return 12;
    break;
  default:
    return -1;
    break;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<rank>:
  endbr64
  sub    $0x31,%edi
  cmp    $0x20,%dil
  ja     L6
  movzbl %dil,%edi
  lea    D9(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x1,%eax
  ret
  mov    $0x2,%eax
  ret
  mov    $0x3,%eax
  ret
  mov    $0x4,%eax
  ret
  mov    $0x5,%eax
  ret
  mov    $0x6,%eax
  ret
  mov    $0x7,%eax
  ret
  mov    $0x8,%eax
  ret
  mov    $0x9,%eax
  ret
  mov    $0xa,%eax
  ret
  mov    $0xb,%eax
  ret
  mov    $0xc,%eax
  ret
L6:
  mov    $0xffffffff,%eax
  ret
  mov    $0x0,%eax
  ret
```

assistant: ```c
int rank(c)
char c;
{
  switch (c) {
  case 'A':
    return 0;
    break;
  case '1':
    return 1;
    break;
  case '2':
    return 2;
    break;
  case '3':
    return 3;
    break;
  case '4':
    return 4;
    break;
  case '5':
    return 5;
    break;
  case '6':
    return 6;
    break;
  case '7':
    return 7;
    break;
  case '8':
    return 8;
    break;
  case '9':
    return 9;
    break;
  case 'J':
    return 10;
    break;
  case 'Q':
    return 11;
    break;
  case 'K':
    return 12;
    break;
  default:
    return -1;
    break;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  unsigned char x;
  x;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
  unsigned char x;
  x;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  unsigned char x;
  x;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  unsigned char x;
  x;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<isAuthenticatedAsRoot>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <getuid@plt>
  test   %rax,%rax
  jne    L19
  mov    $0x1,%eax
  jmp    L1
L19:
  mov    $0x0,%eax
L1:
  pop    %rbp
  ret
```

assistant: ```c
int isAuthenticatedAsRoot(void) {
  if (getuid() == 0) {
    return 1;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<isAuthenticatedAsRoot>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <getuid@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int isAuthenticatedAsRoot(void) {
  if (getuid() == 0) {
    return 1;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<isAuthenticatedAsRoot>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <getuid@plt>
  test   %rax,%rax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int isAuthenticatedAsRoot(void) {
  if (getuid() == 0) {
    return 1;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<isAuthenticatedAsRoot>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <getuid@plt>
  test   %rax,%rax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int isAuthenticatedAsRoot(void) {
  if (getuid() == 0) {
    return 1;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<get_note_date>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movq   $0x0,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L38
  mov    $0x0,%eax
  jmp    L6
L38:
  mov    -0x10(%rbp),%rax
  lea    D11(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x8(%rbp)
  lea    D11(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x8(%rbp)
  lea    D11(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L12
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  jmp    L6
L12:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L48
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  jmp    L6
L48:
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x18(%rbp),%rax
L6:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D11
type_guessed: string

tool: D11: 	

assistant: ```c
char *get_note_date(char *line) {
  char *date = NULL;
  char *tmpline = strdup(line);
  char *datetoken = NULL;
  if (tmpline == NULL)
    return NULL;
  datetoken = strtok(tmpline, "\t");
  datetoken = strtok(NULL, "\t");
  datetoken = strtok(NULL, "\t");
  if (datetoken == NULL) {
    free(tmpline);
    return NULL;
  }
  date = malloc((strlen(datetoken) + 1) * sizeof(char));
  if (date == NULL) {
    free(tmpline);
    return NULL;
  }
  strcpy(date, datetoken);
  free(tmpline);
  return date;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_note_date>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  call   <strdup@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L6
  lea    D1(%rip),%rbp
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rbp,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rbp,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L16
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L20
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rdi
  call   <free@plt>
L24:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  mov    %rbx,%rdi
  call   <free@plt>
  mov    %rbp,%r12
  jmp    L24
L20:
  mov    %rbx,%rdi
  call   <free@plt>
  jmp    L24
L6:
  mov    %rax,%r12
  jmp    L24
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
type_infered: string

tool: D1: 	

assistant: ```c
char *get_note_date(char *line) {
  char *date = NULL;
  char *tmpline = strdup(line);
  char *datetoken = NULL;
  if (tmpline == NULL)
    return NULL;
  datetoken = strtok(tmpline, "\t");
  datetoken = strtok(NULL, "\t");
  datetoken = strtok(NULL, "\t");
  if (datetoken == NULL) {
    free(tmpline);
    return NULL;
  }
  date = malloc((strlen(datetoken) + 1) * sizeof(char));
  if (date == NULL) {
    free(tmpline);
    return NULL;
  }
  strcpy(date, datetoken);
  free(tmpline);
  return date;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_note_date>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  call   <strdup@plt>
  test   %rax,%rax
  je     L33
  lea    D9(%rip),%r12
  mov    %rax,%rdi
  mov    %rax,%rbp
  mov    %r12,%rsi
  call   <strtok@plt>
  mov    %r12,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %r12,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L16
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L7
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L7:
  mov    %rbp,%rdi
  call   <free@plt>
L27:
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L16:
  mov    %rbp,%rdi
  xor    %r13d,%r13d
  call   <free@plt>
  jmp    L27
L33:
  xor    %r13d,%r13d
  jmp    L27
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
type_guessed: string

tool: D9: 	

assistant: ```c
char *get_note_date(char *line) {
  char *date = NULL;
  char *tmpline = strdup(line);
  char *datetoken = NULL;
  if (tmpline == NULL)
    return NULL;
  datetoken = strtok(tmpline, "\t");
  datetoken = strtok(NULL, "\t");
  datetoken = strtok(NULL, "\t");
  if (datetoken == NULL) {
    free(tmpline);
    return NULL;
  }
  date = malloc((strlen(datetoken) + 1) * sizeof(char));
  if (date == NULL) {
    free(tmpline);
    return NULL;
  }
  strcpy(date, datetoken);
  free(tmpline);
  return date;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_note_date>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  call   <strdup@plt>
  test   %rax,%rax
  je     L8
  lea    D5(%rip),%r12
  mov    %rax,%rdi
  mov    %rax,%rbp
  mov    %r12,%rsi
  call   <strtok@plt>
  mov    %r12,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %r12,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L32
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L41
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L41:
  mov    %rbp,%rdi
  call   <free@plt>
L28:
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L32:
  mov    %rbp,%rdi
  xor    %r13d,%r13d
  call   <free@plt>
  jmp    L28
L8:
  xor    %r13d,%r13d
  jmp    L28
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
datatype: string

tool: D5: 	

assistant: ```c
char *get_note_date(char *line) {
  char *date = NULL;
  char *tmpline = strdup(line);
  char *datetoken = NULL;
  if (tmpline == NULL)
    return NULL;
  datetoken = strtok(tmpline, "\t");
  datetoken = strtok(NULL, "\t");
  datetoken = strtok(NULL, "\t");
  if (datetoken == NULL) {
    free(tmpline);
    return NULL;
  }
  date = malloc((strlen(datetoken) + 1) * sizeof(char));
  if (date == NULL) {
    free(tmpline);
    return NULL;
  }
  strcpy(date, datetoken);
  free(tmpline);
  return date;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<_main>:
  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    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <_computeFib@plt>
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D13
type_guessed: string
name: read_data
arguments: label: D8
type_guessed: string

tool: D13: %ld

tool: D8: %ld


assistant: ```c
long _main() {
  long EV_input;
  scanf("%ld", &EV_input);
  printf("%ld\n", (long)_computeFib(EV_input));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    D1(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rdi
  call   <_computeFib@plt>
  mov    %rax,%rsi
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D1
type: string
name: parse_data
arguments: data_label: D17
type: string

tool: D1: %ld

tool: D17: %ld


assistant: ```c
long _main() {
  long EV_input;
  scanf("%ld", &EV_input);
  printf("%ld\n", (long)_computeFib(EV_input));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_main>:
  endbr64
  sub    $0x18,%rsp
  lea    D16(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%rdi
  call   <_computeFib@plt>
  lea    D2(%rip),%rdi
  mov    %rax,%rsi
  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: retrieve_data
arguments: label: D16
type_infered: string
name: retrieve_data
arguments: label: D2
type_infered: string

tool: D16: %ld

tool: D2: %ld


assistant: ```c
long _main() {
  long EV_input;
  scanf("%ld", &EV_input);
  printf("%ld\n", (long)_computeFib(EV_input));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_main>:
  endbr64
  sub    $0x18,%rsp
  lea    D14(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%rdi
  call   <_computeFib@plt>
  lea    D11(%rip),%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D14
datatype: string
name: unpack_data
arguments: tag: D11
datatype: string

tool: D14: %ld

tool: D11: %ld


assistant: ```c
long _main() {
  long EV_input;
  scanf("%ld", &EV_input);
  printf("%ld\n", (long)_computeFib(EV_input));
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<p3_add_seq>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  lea    -0x10(%rax),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x6,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x70,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  lea    0x10(%rdx),%eax
  pop    %rbp
  ret
```

assistant: ```c
int p3_add_seq(seq, add)
int seq;
int add;
{ return (seq - 0x10 + add) % (0x7f - 0x10 + 1) + 0x10; }
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<p3_add_seq>:
  endbr64
  lea    -0x10(%rdi,%rsi,1),%eax
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x6,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x70,%edx,%edx
  sub    %edx,%eax
  add    $0x10,%eax
  ret
```

assistant: ```c
int p3_add_seq(seq, add)
int seq;
int add;
{ return (seq - 0x10 + add) % (0x7f - 0x10 + 1) + 0x10; }
```
user: What is the c source code of the assembly code below:

```asm
<p3_add_seq>:
  endbr64
  lea    -0x10(%rdi,%rsi,1),%eax
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0xffffffff92492493,%rdx,%rdx
  sar    $0x1f,%ecx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x6,%edx
  sub    %ecx,%edx
  imul   $0x70,%edx,%edx
  sub    %edx,%eax
  add    $0x10,%eax
  ret
```

assistant: ```c
int p3_add_seq(seq, add)
int seq;
int add;
{ return (seq - 0x10 + add) % (0x7f - 0x10 + 1) + 0x10; }
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<p3_add_seq>:
  endbr64
  lea    -0x10(%rdi,%rsi,1),%eax
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0xffffffff92492493,%rdx,%rdx
  sar    $0x1f,%ecx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x6,%edx
  sub    %ecx,%edx
  imul   $0x70,%edx,%edx
  sub    %edx,%eax
  add    $0x10,%eax
  ret
```

assistant: ```c
int p3_add_seq(seq, add)
int seq;
int add;
{ return (seq - 0x10 + add) % (0x7f - 0x10 + 1) + 0x10; }
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<linkat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %r8d,-0x2c(%rbp)
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L7
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L7
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <link@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L22
L7:
  mov    D2(%rip),%rax
  mov    (%rax),%edx
  mov    D6(%rip),%rax
  mov    %edx,(%rax)
  movl   $0xffffffff,-0x4(%rbp)
L22:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int linkat(int olddirfd, const char *oldpath, int newdirfd, const char *newpath, int flags) {
  int result;
  if ((oldpath[0] == '/') && (newpath[0] == '/')) {
    result = link(oldpath, newpath);
  } else {
    errno = EINVAL;
    result = -1;
  }
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<linkat>:
  endbr64
  mov    %rsi,%rdi
  cmpb   $0x2f,(%rsi)
  jne    L13
  mov    %rcx,%rsi
  cmpb   $0x2f,(%rcx)
  jne    L13
  sub    $0x8,%rsp
  call   <link@plt>
  add    $0x8,%rsp
  ret
L13:
  mov    D11(%rip),%rax
  mov    (%rax),%edx
  mov    D8(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int linkat(int olddirfd, const char *oldpath, int newdirfd, const char *newpath, int flags) {
  int result;
  if ((oldpath[0] == '/') && (newpath[0] == '/')) {
    result = link(oldpath, newpath);
  } else {
    errno = EINVAL;
    result = -1;
  }
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<linkat>:
  endbr64
  mov    %rsi,%rdi
  mov    %rcx,%rsi
  cmpb   $0x2f,(%rdi)
  jne    L12
  cmpb   $0x2f,(%rcx)
  jne    L12
  jmp    1050 <link@plt>
  nopl   0x0(%rax)
L12:
  mov    D9(%rip),%rax
  mov    (%rax),%edx
  mov    D18(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int linkat(int olddirfd, const char *oldpath, int newdirfd, const char *newpath, int flags) {
  int result;
  if ((oldpath[0] == '/') && (newpath[0] == '/')) {
    result = link(oldpath, newpath);
  } else {
    errno = EINVAL;
    result = -1;
  }
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<linkat>:
  endbr64
  mov    %rsi,%rdi
  mov    %rcx,%rsi
  cmpb   $0x2f,(%rdi)
  jne    L4
  cmpb   $0x2f,(%rcx)
  jne    L4
  jmp    1050 <link@plt>
  nopl   0x0(%rax)
L4:
  mov    D14(%rip),%rax
  mov    (%rax),%edx
  mov    D2(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int linkat(int olddirfd, const char *oldpath, int newdirfd, const char *newpath, int flags) {
  int result;
  if ((oldpath[0] == '/') && (newpath[0] == '/')) {
    result = link(oldpath, newpath);
  } else {
    errno = EINVAL;
    result = -1;
  }
  return result;
}
```
user: What is the c source code of the assembly code below:

```asm
<sx_prefix_destroy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L6
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L6:
  nop
  leave
  ret
```

assistant: ```c
void sx_prefix_destroy(struct sx_prefix *p) {
  if (p)
    free(p);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<sx_prefix_destroy>:
  endbr64
  test   %rdi,%rdi
  je     L5
  sub    $0x8,%rsp
  call   <free@plt>
  add    $0x8,%rsp
  ret
L5:
  ret
```

assistant: ```c
void sx_prefix_destroy(struct sx_prefix *p) {
  if (p)
    free(p);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sx_prefix_destroy>:
  endbr64
  test   %rdi,%rdi
  je     L5
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L5:
  ret
```

assistant: ```c
void sx_prefix_destroy(struct sx_prefix *p) {
  if (p)
    free(p);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<sx_prefix_destroy>:
  endbr64
  test   %rdi,%rdi
  je     L6
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L6:
  ret
```

assistant: ```c
void sx_prefix_destroy(struct sx_prefix *p) {
  if (p)
    free(p);
}
```