ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<strtoll>:
  endbr64
  push   %r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  jmp    L54
  nopw   0x0(%rax,%rax,1)
L104:
  add    $0x1,%rbx
L54:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L104
  xor    %r14d,%r14d
  cmpb   $0x2d,(%rbx)
  je     L71
L21:
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strtoull@plt>
  test   %rbp,%rbp
  je     L41
  cmp    %rbx,0x0(%rbp)
  je     L123
L41:
  mov    D34(%rip),%rdx
  mov    (%rdx),%rdx
  cmp    %rax,%rdx
  jae    L44
  movabs $0x8000000000000000,%rcx
  cmp    %rcx,%rax
  jne    L72
  test   %r14d,%r14d
  jne    L93
L72:
  mov    D10(%rip),%rax
  mov    (%rax),%rcx
  mov    D6(%rip),%rax
  mov    %rcx,(%rax)
  mov    %rdx,%rax
  test   %r14d,%r14d
  je     L116
  mov    D42(%rip),%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  mov    (%rax),%rax
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L44:
  mov    %rax,%rdx
  neg    %rdx
  test   %r14d,%r14d
  cmovne %rdx,%rax
L116:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  cs nopw 0x0(%rax,%rax,1)
L123:
  mov    %r12,0x0(%rbp)
  jmp    L41
  cs nopw 0x0(%rax,%rax,1)
L71:
  movsbl 0x1(%rbx),%edi
  call   <isalnum@plt>
  test   %rax,%rax
  je     L21
  add    $0x1,%rbx
  mov    $0xffffffff,%r14d
  jmp    L21
L93:
  mov    D6(%rip),%rax
  movq   $0x0,(%rax)
  mov    %rcx,%rax
  jmp    L116
```

assistant: ```c
long long int strtoll(const char *nptr, char **endptr, int base) {
  int neg = 0;
  unsigned long long int v;
  const char *orig = nptr;
  while (isspace(*nptr)) {
    nptr++;
  }
  if (*nptr == '-' && isalnum(nptr[1])) {
    neg = -1;
    nptr++;
  }
  v = strtoull(nptr, endptr, base);
  if (endptr && *endptr == nptr) {
    *endptr = (char *)orig;
  }
  if (v > LLONG_MAX) {
    if (v == 0x8000000000000000ull && neg) {
      errno = 0;
      return v;
    }
    errno = ERANGE;
    return (neg ? LLONG_MIN : LLONG_MAX);
  }
  return (neg ? -v : v);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<imagine_root>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  movsd  %xmm0,-0x30(%rbp)
  movsd  -0x30(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%edi
  call   <abs@plt>
  movq   %xmm0,%rax
  movsd  D30(%rip),%xmm0
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <pow@plt>
  movq   %xmm0,%rbx
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <abs@plt>
  movapd %xmm0,%xmm1
  addsd  %xmm0,%xmm1
  movq   %rbx,%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x28(%rbp),%xmm0
  movq   D45(%rip),%xmm1
  xorpd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x24(%rbp),%xmm0
  movapd %xmm0,%xmm2
  addsd  %xmm0,%xmm2
  divsd  %xmm2,%xmm1
  movapd %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x20(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  movsd  -0x20(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D30
type_infered: f64
name: unpack_data
arguments: data_label: D45
type_infered: byte[8]
name: unpack_data
arguments: data_label: D38
type_infered: string
name: unpack_data
arguments: data_label: D36
type_infered: string
name: unpack_data
arguments: data_label: D13
type_infered: string

tool: D30: 0.5

tool: {"D45": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

tool: D38: Roots are %.3lf + %.3lfi 

tool: D36: and %.3lf - %.3lfi

tool: D13: 
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-


assistant: ```c
void imagine_root(int CoeffA, int CoeffB, double discrim) {
  double root;
  double root_b;
  root = (pow(abs(discrim), 0.5)) / (2.0 * abs(CoeffA));
  root_b = (-1.0 * CoeffB) / (2.0 * CoeffA);
  printf("Roots are %.3lf + %.3lfi ", root_b, root);
  printf("and %.3lf - %.3lfi", root_b, root);
  printf("\n-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-\n\n");
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<imagine_root>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %esi,%r12d
  cvttsd2si %xmm0,%edi
  call   <abs@plt>
  movsd  D37(%rip),%xmm1
  call   <pow@plt>
  movq   %xmm0,%rbx
  mov    %ebp,%edi
  call   <abs@plt>
  addsd  %xmm0,%xmm0
  movq   %rbx,%xmm3
  divsd  %xmm0,%xmm3
  movq   %xmm3,%rbx
  pxor   %xmm2,%xmm2
  cvtsi2sd %r12d,%xmm2
  xorpd  D6(%rip),%xmm2
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebp,%xmm0
  addsd  %xmm0,%xmm0
  divsd  %xmm0,%xmm2
  movq   %xmm2,%rbp
  movapd %xmm3,%xmm1
  movapd %xmm2,%xmm0
  lea    D45(%rip),%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  movq   %rbx,%xmm1
  movq   %rbp,%xmm0
  lea    D24(%rip),%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  lea    D38(%rip),%rdi
  call   <puts@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: label: D37
type: f64
name: retrieve_data
arguments: label: D6
type: byte[16]
name: retrieve_data
arguments: label: D45
type: string
name: retrieve_data
arguments: label: D24
type: string
name: retrieve_data
arguments: label: D38
type: string

tool: D37: 0.5

tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D45: Roots are %.3lf + %.3lfi 

tool: D24: and %.3lf - %.3lfi

tool: D38: 
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-


assistant: ```c
void imagine_root(int CoeffA, int CoeffB, double discrim) {
  double root;
  double root_b;
  root = (pow(abs(discrim), 0.5)) / (2.0 * abs(CoeffA));
  root_b = (-1.0 * CoeffB) / (2.0 * CoeffA);
  printf("Roots are %.3lf + %.3lfi ", root_b, root);
  printf("and %.3lf - %.3lfi", root_b, root);
  printf("\n-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-\n\n");
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<imagine_root>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  cvttsd2si %xmm0,%edi
  sub    $0x18,%rsp
  call   <abs@plt>
  movsd  D40(%rip),%xmm1
  call   <pow@plt>
  mov    %ebx,%edi
  movsd  %xmm0,(%rsp)
  call   <abs@plt>
  pxor   %xmm2,%xmm2
  movsd  (%rsp),%xmm1
  lea    D37(%rip),%rdi
  cvtsi2sd %ebx,%xmm2
  addsd  %xmm0,%xmm0
  mov    $0x2,%eax
  divsd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebp,%xmm0
  addsd  %xmm2,%xmm2
  xorpd  D20(%rip),%xmm0
  divsd  %xmm2,%xmm0
  movsd  %xmm1,0x8(%rsp)
  movsd  %xmm0,(%rsp)
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm1
  movsd  (%rsp),%xmm0
  lea    D1(%rip),%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  add    $0x18,%rsp
  lea    D17(%rip),%rdi
  pop    %rbx
  pop    %rbp
  jmp    1080 <puts@plt>
```

tool_calls: name: parse_data
arguments: tag: D40
type_infered: f64
name: parse_data
arguments: tag: D37
type_infered: string
name: parse_data
arguments: tag: D20
type_infered: byte[16]
name: parse_data
arguments: tag: D1
type_infered: string
name: parse_data
arguments: tag: D17
type_infered: string

tool: D40: 0.5

tool: D37: Roots are %.3lf + %.3lfi 

tool: {"D20": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D1: and %.3lf - %.3lfi

tool: D17: 
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-


assistant: ```c
void imagine_root(int CoeffA, int CoeffB, double discrim) {
  double root;
  double root_b;
  root = (pow(abs(discrim), 0.5)) / (2.0 * abs(CoeffA));
  root_b = (-1.0 * CoeffB) / (2.0 * CoeffA);
  printf("Roots are %.3lf + %.3lfi ", root_b, root);
  printf("and %.3lf - %.3lfi", root_b, root);
  printf("\n-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-\n\n");
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<imagine_root>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  cvttsd2si %xmm0,%edi
  sub    $0x18,%rsp
  call   <abs@plt>
  movsd  D6(%rip),%xmm1
  call   <pow@plt>
  mov    %ebx,%edi
  movsd  %xmm0,(%rsp)
  call   <abs@plt>
  pxor   %xmm2,%xmm2
  movsd  (%rsp),%xmm1
  lea    D36(%rip),%rdi
  cvtsi2sd %ebx,%xmm2
  addsd  %xmm0,%xmm0
  mov    $0x2,%eax
  divsd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebp,%xmm0
  addsd  %xmm2,%xmm2
  xorpd  D45(%rip),%xmm0
  divsd  %xmm2,%xmm0
  movsd  %xmm1,0x8(%rsp)
  movsd  %xmm0,(%rsp)
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm1
  movsd  (%rsp),%xmm0
  lea    D10(%rip),%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  add    $0x18,%rsp
  lea    D32(%rip),%rdi
  pop    %rbx
  pop    %rbp
  jmp    1080 <puts@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D6
type_guessed: f64
name: retrieve_data
arguments: tag: D36
type_guessed: string
name: retrieve_data
arguments: tag: D45
type_guessed: byte[16]
name: retrieve_data
arguments: tag: D10
type_guessed: string
name: retrieve_data
arguments: tag: D32
type_guessed: string

tool: D6: 0.5

tool: D36: Roots are %.3lf + %.3lfi 

tool: {"D45": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D10: and %.3lf - %.3lfi

tool: D32: 
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-


assistant: ```c
void imagine_root(int CoeffA, int CoeffB, double discrim) {
  double root;
  double root_b;
  root = (pow(abs(discrim), 0.5)) / (2.0 * abs(CoeffA));
  root_b = (-1.0 * CoeffB) / (2.0 * CoeffA);
  printf("Roots are %.3lf + %.3lfi ", root_b, root);
  printf("and %.3lf - %.3lfi", root_b, root);
  printf("\n-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-\n\n");
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<I2CMasterErr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  and    $0x1,%eax
  test   %rax,%rax
  je     L15
  mov    $0x0,%eax
  jmp    L3
L15:
  mov    -0x8(%rbp),%rax
  and    $0x2,%eax
  test   %rax,%rax
  je     L6
  mov    -0x8(%rbp),%rax
  and    $0x1c,%eax
  jmp    L3
L6:
  mov    $0x0,%eax
L3:
  pop    %rbp
  ret
```

assistant: ```c
unsigned long I2CMasterErr(unsigned long ulBase) {
  unsigned long ulErr;
  ;
  ulErr = (*((volatile unsigned long *)(ulBase + 0x00000004)));
  if (ulErr & 0x00000001) {
    return (0);
  }
  if (ulErr & 0x00000002) {
    return (ulErr & (0x00000010 | 0x00000008 | 0x00000004));
  } else {
    return (0);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<I2CMasterErr>:
  endbr64
  mov    0x4(%rdi),%rdx
  mov    $0x0,%eax
  test   $0x1,%dl
  jne    L3
  mov    %rdx,%rcx
  and    $0x1c,%ecx
  mov    %rdx,%rax
  and    $0x2,%eax
  cmovne %rcx,%rax
L3:
  ret
```

assistant: ```c
unsigned long I2CMasterErr(unsigned long ulBase) {
  unsigned long ulErr;
  ;
  ulErr = (*((volatile unsigned long *)(ulBase + 0x00000004)));
  if (ulErr & 0x00000001) {
    return (0);
  }
  if (ulErr & 0x00000002) {
    return (ulErr & (0x00000010 | 0x00000008 | 0x00000004));
  } else {
    return (0);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<I2CMasterErr>:
  endbr64
  mov    0x4(%rdi),%rdx
  xor    %eax,%eax
  test   $0x1,%dl
  jne    L6
  mov    %rdx,%rcx
  mov    %rdx,%rax
  and    $0x1c,%ecx
  and    $0x2,%eax
  cmovne %rcx,%rax
L6:
  ret
```

assistant: ```c
unsigned long I2CMasterErr(unsigned long ulBase) {
  unsigned long ulErr;
  ;
  ulErr = (*((volatile unsigned long *)(ulBase + 0x00000004)));
  if (ulErr & 0x00000001) {
    return (0);
  }
  if (ulErr & 0x00000002) {
    return (ulErr & (0x00000010 | 0x00000008 | 0x00000004));
  } else {
    return (0);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<I2CMasterErr>:
  endbr64
  mov    0x4(%rdi),%rdx
  xor    %eax,%eax
  test   $0x1,%dl
  jne    L3
  mov    %rdx,%rcx
  mov    %rdx,%rax
  and    $0x1c,%ecx
  and    $0x2,%eax
  cmovne %rcx,%rax
L3:
  ret
```

assistant: ```c
unsigned long I2CMasterErr(unsigned long ulBase) {
  unsigned long ulErr;
  ;
  ulErr = (*((volatile unsigned long *)(ulBase + 0x00000004)));
  if (ulErr & 0x00000001) {
    return (0);
  }
  if (ulErr & 0x00000002) {
    return (ulErr & (0x00000010 | 0x00000008 | 0x00000004));
  } else {
    return (0);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<write_le32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <write_le16@plt>
  mov    -0xc(%rbp),%eax
  sar    $0x10,%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%rax
  mov    $0x4,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <write_at_le16@plt>
  nop
  leave
  ret
```

assistant: ```c
void write_le32(void *dest, uint32_t val) {
  write_le16(dest, val >> 0);
  write_at_le16(dest, val >> 16, sizeof(uint16_t));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<write_le32>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  call   <write_le16@plt>
  sar    $0x10,%ebx
  mov    %ebx,%esi
  mov    $0x4,%edx
  mov    %rbp,%rdi
  call   <write_at_le16@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void write_le32(void *dest, uint32_t val) {
  write_le16(dest, val >> 0);
  write_at_le16(dest, val >> 16, sizeof(uint16_t));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<write_le32>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  sar    $0x10,%ebx
  sub    $0x8,%rsp
  call   <write_le16@plt>
  add    $0x8,%rsp
  mov    %ebx,%esi
  mov    %rbp,%rdi
  pop    %rbx
  mov    $0x4,%edx
  pop    %rbp
  jmp    1070 <write_at_le16@plt>
```

assistant: ```c
void write_le32(void *dest, uint32_t val) {
  write_le16(dest, val >> 0);
  write_at_le16(dest, val >> 16, sizeof(uint16_t));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<write_le32>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  sar    $0x10,%ebx
  sub    $0x8,%rsp
  call   <write_le16@plt>
  add    $0x8,%rsp
  mov    %ebx,%esi
  mov    %rbp,%rdi
  pop    %rbx
  mov    $0x4,%edx
  pop    %rbp
  jmp    1070 <write_at_le16@plt>
```

assistant: ```c
void write_le32(void *dest, uint32_t val) {
  write_le16(dest, val >> 0);
  write_at_le16(dest, val >> 16, sizeof(uint16_t));
}
```
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
  movsd  D13(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  movsd  -0x10(%rbp),%xmm0
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D6(%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: retrieve_data
arguments: label: D13
type_guessed: f64
name: retrieve_data
arguments: label: D6
type_guessed: string

tool: D13: 3.1415926

tool: D6: x = %f, y = %f


assistant: ```c
int main(void) {
  double x = 3.1415926, y;
  int *p;
  p = (int *)&x;
  y = *p;
  printf("x = %f, y = %f\n", x, y);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  movsd  D16(%rip),%xmm1
  movsd  D4(%rip),%xmm0
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D16
type: byte[8]
name: retrieve_data
arguments: data_label: D4
type: f64
name: retrieve_data
arguments: data_label: D12
type: string

tool: {"D16": [0x00,0x00,0x80,0x12,0xb6,0x44,0xd3,0x41]}

tool: D4: 3.1415926

tool: D12: x = %f, y = %f


assistant: ```c
int main(void) {
  double x = 3.1415926, y;
  int *p;
  p = (int *)&x;
  y = *p;
  printf("x = %f, y = %f\n", x, y);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  movsd  D19(%rip),%xmm1
  movsd  D3(%rip),%xmm0
  lea    D11(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D19
data_type: byte[8]
name: extract_symbolic_data
arguments: label: D3
data_type: f64
name: extract_symbolic_data
arguments: label: D11
data_type: string

tool: {"D19": [0x00,0x00,0x80,0x12,0xb6,0x44,0xd3,0x41]}

tool: D3: 3.1415926

tool: D11: x = %f, y = %f


assistant: ```c
int main(void) {
  double x = 3.1415926, y;
  int *p;
  p = (int *)&x;
  y = *p;
  printf("x = %f, y = %f\n", x, y);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  movsd  D29(%rip),%xmm1
  movsd  D21(%rip),%xmm0
  lea    D22(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D29
datatype: byte[8]
name: parse_data
arguments: label: D21
datatype: f64
name: parse_data
arguments: label: D22
datatype: string

tool: {"D29": [0x00,0x00,0x80,0x12,0xb6,0x44,0xd3,0x41]}

tool: D21: 3.1415926

tool: D22: x = %f, y = %f


assistant: ```c
int main(void) {
  double x = 3.1415926, y;
  int *p;
  p = (int *)&x;
  y = *p;
  printf("x = %f, y = %f\n", x, y);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ipfw_del_chain>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  lea    -0xc(%rbp),%rax
  mov    $0x0,%edx
  mov    $0x200,%esi
  mov    %rax,%rdi
  call   <add_command@plt>
  mov    -0xc(%rbp),%edx
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  mov    $0x0,%r8d
  mov    %eax,%edi
  call   <ipfw_init@plt>
  mov    D0(%rip),%rax
  movq   $0x0,(%rax)
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
int ipfw_del_chain(int args, char **arglist, const char *chain) {
  unsigned int command = 0;
  add_command(&command, 0x0200U, 0x0000U);
  ipfw_init(args, arglist, command, chain, 0);
  optind = 0;
  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
<ipfw_del_chain>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  mov    %rdx,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    0x4(%rsp),%rdi
  mov    $0x0,%edx
  mov    $0x200,%esi
  call   <add_command@plt>
  mov    $0x0,%r8d
  mov    %r12,%rcx
  mov    0x4(%rsp),%edx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <ipfw_init@plt>
  mov    D1(%rip),%rax
  movq   $0x0,(%rax)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ipfw_del_chain(int args, char **arglist, const char *chain) {
  unsigned int command = 0;
  add_command(&command, 0x0200U, 0x0000U);
  ipfw_init(args, arglist, command, chain, 0);
  optind = 0;
  return (0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ipfw_del_chain>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  xor    %edx,%edx
  push   %r12
  mov    %rsi,%r12
  mov    $0x200,%esi
  push   %rbp
  mov    %edi,%ebp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  movl   $0x0,0x4(%rsp)
  call   <add_command@plt>
  mov    0x4(%rsp),%edx
  xor    %r8d,%r8d
  mov    %r13,%rcx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <ipfw_init@plt>
  mov    D1(%rip),%rax
  movq   $0x0,(%rax)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ipfw_del_chain(int args, char **arglist, const char *chain) {
  unsigned int command = 0;
  add_command(&command, 0x0200U, 0x0000U);
  ipfw_init(args, arglist, command, chain, 0);
  optind = 0;
  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
<ipfw_del_chain>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  xor    %edx,%edx
  push   %r12
  mov    %rsi,%r12
  mov    $0x200,%esi
  push   %rbp
  mov    %edi,%ebp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  movl   $0x0,0x4(%rsp)
  call   <add_command@plt>
  mov    0x4(%rsp),%edx
  xor    %r8d,%r8d
  mov    %r13,%rcx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <ipfw_init@plt>
  mov    D1(%rip),%rax
  movq   $0x0,(%rax)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ipfw_del_chain(int args, char **arglist, const char *chain) {
  unsigned int command = 0;
  add_command(&command, 0x0200U, 0x0000U);
  ipfw_init(args, arglist, command, chain, 0);
  optind = 0;
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<firstBadVersion>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x1,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jne    L44
  mov    $0x1,%eax
  jmp    L4
L28:
  mov    -0x8(%rbp),%eax
  sub    -0xc(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <isBadVersion@plt>
  test   %rax,%rax
  je     L21
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L44
L21:
  mov    -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L44:
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  cmp    %eax,-0x8(%rbp)
  jg     L28
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <isBadVersion@plt>
  test   %rax,%rax
  je     L0
  mov    -0xc(%rbp),%eax
  jmp    L4
L0:
  mov    -0x8(%rbp),%eax
L4:
  leave
  ret
```

assistant: ```c
int firstBadVersion(int n) {
  int left = 1, right = n, mid;
  if (n == 1)
    return 1;
  while (left + 1 < right) {
    mid = left + (right - left) / 2;
    if (isBadVersion(mid))
      right = mid;
    else
      left = mid;
  }
  return isBadVersion(left) ? left : right;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<firstBadVersion>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  cmp    $0x1,%edi
  je     L30
  mov    %edi,%r12d
  cmp    $0x2,%edi
  jle    L29
  mov    $0x1,%ebp
L39:
  mov    %r12d,%eax
  sub    %ebp,%eax
  mov    %eax,%ebx
  shr    $0x1f,%ebx
  add    %eax,%ebx
  sar    %ebx
  add    %ebp,%ebx
  mov    %ebx,%edi
  call   <isBadVersion@plt>
  test   %rax,%rax
  cmove  %ebx,%ebp
  cmovne %ebx,%r12d
  lea    0x1(%rbp),%eax
  cmp    %r12d,%eax
  jl     L39
L12:
  mov    %ebp,%edi
  call   <isBadVersion@plt>
  test   %rax,%rax
  cmove  %r12d,%ebp
L30:
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L29:
  mov    $0x1,%ebp
  jmp    L12
```

assistant: ```c
int firstBadVersion(int n) {
  int left = 1, right = n, mid;
  if (n == 1)
    return 1;
  while (left + 1 < right) {
    mid = left + (right - left) / 2;
    if (isBadVersion(mid))
      right = mid;
    else
      left = mid;
  }
  return isBadVersion(left) ? left : right;
}
```