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

```asm
<strerrno>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L17
  lea    D6(%rip),%rax
  jmp    L9
L17:
  lea    D1(%rip),%rax
L9:
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D6
type_guessed: string
name: read_data
arguments: label: D1
type_guessed: string

tool: D6: OK

tool: D1: E??

assistant: ```c
const char *strerrno(int err) {
  switch (err) {
  case 0:
    return "OK";
  default:
    return "E??";
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strerrno>:
  endbr64
  test   %edi,%edi
  lea    D10(%rip),%rax
  lea    D6(%rip),%rdx
  cmovne %rdx,%rax
  ret
```

tool_calls: name: parse_data
arguments: label: D10
datatype: string
name: parse_data
arguments: label: D6
datatype: string

tool: D10: OK

tool: D6: E??

assistant: ```c
const char *strerrno(int err) {
  switch (err) {
  case 0:
    return "OK";
  default:
    return "E??";
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<strerrno>:
  endbr64
  test   %edi,%edi
  lea    D13(%rip),%rdx
  lea    D1(%rip),%rax
  cmovne %rdx,%rax
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D13
data_type: string
name: unpack_data
arguments: data_tag: D1
data_type: string

tool: D13: E??

tool: D1: OK

assistant: ```c
const char *strerrno(int err) {
  switch (err) {
  case 0:
    return "OK";
  default:
    return "E??";
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<strerrno>:
  endbr64
  test   %edi,%edi
  lea    D8(%rip),%rdx
  lea    D0(%rip),%rax
  cmovne %rdx,%rax
  ret
```

tool_calls: name: parse_data
arguments: tag: D8
type_guessed: string
name: parse_data
arguments: tag: D0
type_guessed: string

tool: D8: E??

tool: D0: OK

assistant: ```c
const char *strerrno(int err) {
  switch (err) {
  case 0:
    return "OK";
  default:
    return "E??";
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<opMinLongGASPI>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x3c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L1
L7:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  cmovg  %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  addq   $0x8,-0x10(%rbp)
  addq   $0x8,-0x8(%rbp)
  addq   $0x8,-0x18(%rbp)
  addl   $0x1,-0x1c(%rbp)
L1:
  mov    -0x1c(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jl     L7
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void opMinLongGASPI(void *res, void *localVal, void *dstVal, const int cnt) {
  int i;
  long *rv = (long *)res;
  long *lv = (long *)localVal;
  long *dv = (long *)dstVal;
  for (i = 0; i < cnt; i++) {
    *rv = (((*lv) > (*dv)) ? (*dv) : (*lv));
    lv++;
    dv++;
    rv++;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<opMinLongGASPI>:
  endbr64
  test   %ecx,%ecx
  jle    L14
  mov    %ecx,%ecx
  mov    $0x0,%eax
L15:
  mov    (%rsi,%rax,8),%r8
  mov    (%rdx,%rax,8),%r9
  cmp    %r9,%r8
  cmovg  %r9,%r8
  mov    %r8,(%rdi,%rax,8)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L15
L14:
  ret
```

assistant: ```c
void opMinLongGASPI(void *res, void *localVal, void *dstVal, const int cnt) {
  int i;
  long *rv = (long *)res;
  long *lv = (long *)localVal;
  long *dv = (long *)dstVal;
  for (i = 0; i < cnt; i++) {
    *rv = (((*lv) > (*dv)) ? (*dv) : (*lv));
    lv++;
    dv++;
    rv++;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<opMinLongGASPI>:
  endbr64
  test   %ecx,%ecx
  jle    L14
  movslq %ecx,%rcx
  xor    %eax,%eax
  nopl   (%rax)
L4:
  mov    (%rsi,%rax,8),%r8
  mov    (%rdx,%rax,8),%r9
  cmp    %r9,%r8
  cmovg  %r9,%r8
  mov    %r8,(%rdi,%rax,8)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L4
L14:
  ret
```

assistant: ```c
void opMinLongGASPI(void *res, void *localVal, void *dstVal, const int cnt) {
  int i;
  long *rv = (long *)res;
  long *lv = (long *)localVal;
  long *dv = (long *)dstVal;
  for (i = 0; i < cnt; i++) {
    *rv = (((*lv) > (*dv)) ? (*dv) : (*lv));
    lv++;
    dv++;
    rv++;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<opMinLongGASPI>:
  endbr64
  test   %ecx,%ecx
  jle    L0
  movslq %ecx,%rcx
  xor    %eax,%eax
  nopl   (%rax)
L16:
  mov    (%rsi,%rax,8),%r8
  mov    (%rdx,%rax,8),%r9
  cmp    %r9,%r8
  cmovg  %r9,%r8
  mov    %r8,(%rdi,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L16
L0:
  ret
```

assistant: ```c
void opMinLongGASPI(void *res, void *localVal, void *dstVal, const int cnt) {
  int i;
  long *rv = (long *)res;
  long *lv = (long *)localVal;
  long *dv = (long *)dstVal;
  for (i = 0; i < cnt; i++) {
    *rv = (((*lv) > (*dv)) ? (*dv) : (*lv));
    lv++;
    dv++;
    rv++;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<u8_unescape>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x20(%rbp)
  jmp    L13
L76:
  mov    -0x48(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5c,%al
  jne    L47
  addq   $0x1,-0x48(%rbp)
  lea    -0x28(%rbp),%rdx
  mov    -0x48(%rbp),%rax
  mov    $0x3e8,%esi
  mov    %rax,%rdi
  call   <u8_read_escape_sequence@plt>
  mov    %rax,-0x18(%rbp)
  jmp    L77
L47:
  mov    -0x48(%rbp),%rax
  movzbl (%rax),%eax
  movsbq %al,%rax
  mov    %rax,-0x28(%rbp)
  movq   $0x1,-0x18(%rbp)
L77:
  mov    -0x18(%rbp),%rax
  add    %rax,-0x48(%rbp)
  mov    -0x28(%rbp),%rdx
  lea    -0xc(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <u8_wc_toutf8@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x40(%rbp),%rax
  sub    -0x20(%rbp),%rax
  cmp    %rax,-0x18(%rbp)
  ja     L79
  mov    -0x38(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x18(%rbp),%rdx
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    -0x18(%rbp),%rax
  add    %rax,-0x20(%rbp)
L13:
  mov    -0x48(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L33
  mov    -0x20(%rbp),%rax
  cmp    -0x40(%rbp),%rax
  jb     L76
  jmp    L33
L79:
  nop
L33:
  mov    -0x20(%rbp),%rax
  cmp    -0x40(%rbp),%rax
  jae    L22
  mov    -0x38(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L22:
  mov    -0x20(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L72
  call   <__stack_chk_fail@plt>
L72:
  leave
  ret
```

assistant: ```c
size_t u8_unescape(char *buf, size_t sz, const char *src) {
  size_t c = 0, amt;
  u_int32_t ch;
  char temp[4];
  while (*src && c < sz) {
    if (*src == '\\') {
      src++;
      amt = u8_read_escape_sequence(src, 1000, &ch);
    } else {
      ch = (u_int32_t)*src;
      amt = 1;
    }
    src += amt;
    amt = u8_wc_toutf8(temp, ch);
    if (amt > sz - c)
      break;
    memcpy(&buf[c], temp, amt);
    c += amt;
  }
  if (c < sz)
    buf[c] = '\0';
  return c;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<u8_unescape>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%r14
  mov    %rsi,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movzbl (%rdx),%eax
  test   %rsi,%rsi
  je     L88
  mov    %rdx,%r12
  test   %al,%al
  je     L88
  mov    $0x0,%ebp
  mov    $0x1,%r15d
  jmp    L99
L6:
  add    $0x1,%r12
  lea    0x8(%rsp),%rdx
  mov    $0x3e8,%esi
  mov    %r12,%rdi
  call   <u8_read_escape_sequence@plt>
L25:
  add    %rax,%r12
  lea    0x14(%rsp),%rdi
  mov    0x8(%rsp),%rsi
  call   <u8_wc_toutf8@plt>
  mov    %rax,%rbx
  mov    %r13,%rax
  sub    %rbp,%rax
  cmp    %rbx,%rax
  jb     L91
  lea    0x14(%rsp),%rsi
  lea    (%r14,%rbp,1),%rdi
  mov    %rbx,%rdx
  call   <memcpy@plt>
  add    %rbx,%rbp
  movzbl (%r12),%eax
  cmp    %rbp,%r13
  jbe    L91
  test   %al,%al
  je     L91
L99:
  cmp    $0x5c,%al
  je     L6
  movsbq %al,%rax
  mov    %rax,0x8(%rsp)
  mov    %r15,%rax
  jmp    L25
L88:
  mov    $0x0,%ebp
L91:
  cmp    %rbp,%r13
  jbe    L15
  movb   $0x0,(%r14,%rbp,1)
L15:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L51
  mov    %rbp,%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L51:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t u8_unescape(char *buf, size_t sz, const char *src) {
  size_t c = 0, amt;
  u_int32_t ch;
  char temp[4];
  while (*src && c < sz) {
    if (*src == '\\') {
      src++;
      amt = u8_read_escape_sequence(src, 1000, &ch);
    } else {
      ch = (u_int32_t)*src;
      amt = 1;
    }
    src += amt;
    amt = u8_wc_toutf8(temp, ch);
    if (amt > sz - c)
      break;
    memcpy(&buf[c], temp, amt);
    c += amt;
  }
  if (c < sz)
    buf[c] = '\0';
  return c;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<u8_unescape>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x28,%rsp
  movsbq (%rdx),%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %sil,%sil
  je     L71
  test   %rbp,%rbp
  je     L66
  mov    %rdx,%rbx
  xor    %r15d,%r15d
  lea    0x14(%rsp),%r12
  jmp    L36
  nopl   0x0(%rax,%rax,1)
L107:
  mov    %rsi,0x8(%rsp)
  mov    $0x1,%eax
L6:
  mov    %r12,%rdi
  add    %rax,%rbx
  call   <u8_wc_toutf8@plt>
  lea    0x0(%r13,%r15,1),%rdi
  mov    %rax,%r14
  mov    %rbp,%rax
  sub    %r15,%rax
  cmp    %r14,%rax
  jb     L8
  mov    %r12,%rsi
  mov    %r14,%rdx
  add    %r14,%r15
  call   <memcpy@plt>
  movsbq (%rbx),%rsi
  test   %sil,%sil
  je     L78
  cmp    %r15,%rbp
  jbe    L42
L36:
  cmp    $0x5c,%sil
  jne    L107
  add    $0x1,%rbx
  mov    $0x3e8,%esi
  lea    0x8(%rsp),%rdx
  mov    %rbx,%rdi
  call   <u8_read_escape_sequence@plt>
  mov    0x8(%rsp),%rsi
  jmp    L6
L71:
  xor    %r15d,%r15d
  nopw   0x0(%rax,%rax,1)
L78:
  cmp    %rbp,%r15
  jae    L42
L8:
  movb   $0x0,0x0(%r13,%r15,1)
L42:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x28,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L66:
  xor    %r15d,%r15d
  jmp    L42
L21:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t u8_unescape(char *buf, size_t sz, const char *src) {
  size_t c = 0, amt;
  u_int32_t ch;
  char temp[4];
  while (*src && c < sz) {
    if (*src == '\\') {
      src++;
      amt = u8_read_escape_sequence(src, 1000, &ch);
    } else {
      ch = (u_int32_t)*src;
      amt = 1;
    }
    src += amt;
    amt = u8_wc_toutf8(temp, ch);
    if (amt > sz - c)
      break;
    memcpy(&buf[c], temp, amt);
    c += amt;
  }
  if (c < sz)
    buf[c] = '\0';
  return c;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<u8_unescape>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x28,%rsp
  movsbq (%rdx),%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %sil,%sil
  je     L73
  test   %rbp,%rbp
  je     L14
  mov    %rdx,%rbx
  xor    %r15d,%r15d
  lea    0x14(%rsp),%r12
  jmp    L50
  nopl   0x0(%rax,%rax,1)
L105:
  mov    %rsi,0x8(%rsp)
  mov    $0x1,%eax
L101:
  mov    %r12,%rdi
  add    %rax,%rbx
  call   <u8_wc_toutf8@plt>
  lea    0x0(%r13,%r15,1),%rdi
  mov    %rax,%r14
  mov    %rbp,%rax
  sub    %r15,%rax
  cmp    %r14,%rax
  jb     L41
  mov    %r12,%rsi
  mov    %r14,%rdx
  add    %r14,%r15
  call   <memcpy@plt>
  movsbq (%rbx),%rsi
  test   %sil,%sil
  je     L88
  cmp    %r15,%rbp
  jbe    L96
L50:
  cmp    $0x5c,%sil
  jne    L105
  add    $0x1,%rbx
  mov    $0x3e8,%esi
  lea    0x8(%rsp),%rdx
  mov    %rbx,%rdi
  call   <u8_read_escape_sequence@plt>
  mov    0x8(%rsp),%rsi
  jmp    L101
L73:
  xor    %r15d,%r15d
  nopw   0x0(%rax,%rax,1)
L88:
  cmp    %rbp,%r15
  jae    L96
L41:
  movb   $0x0,0x0(%r13,%r15,1)
L96:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L109
  add    $0x28,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L14:
  xor    %r15d,%r15d
  jmp    L96
L109:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t u8_unescape(char *buf, size_t sz, const char *src) {
  size_t c = 0, amt;
  u_int32_t ch;
  char temp[4];
  while (*src && c < sz) {
    if (*src == '\\') {
      src++;
      amt = u8_read_escape_sequence(src, 1000, &ch);
    } else {
      ch = (u_int32_t)*src;
      amt = 1;
    }
    src += amt;
    amt = u8_wc_toutf8(temp, ch);
    if (amt > sz - c)
      break;
    memcpy(&buf[c], temp, amt);
    c += amt;
  }
  if (c < sz)
    buf[c] = '\0';
  return c;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<lseek>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    D13(%rip),%rax
  mov    (%rax),%edx
  mov    D14(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
off_t lseek(int fd, off_t off, int whence) {
  (void)fd;
  (void)off;
  (void)whence;
  errno = ENODEV;
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<lseek>:
  endbr64
  mov    D17(%rip),%rax
  mov    (%rax),%edx
  mov    D2(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
off_t lseek(int fd, off_t off, int whence) {
  (void)fd;
  (void)off;
  (void)whence;
  errno = ENODEV;
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<lseek>:
  endbr64
  mov    D17(%rip),%rax
  mov    (%rax),%edx
  mov    D10(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
off_t lseek(int fd, off_t off, int whence) {
  (void)fd;
  (void)off;
  (void)whence;
  errno = ENODEV;
  return -1;
}
```
user: What is the c source code of the assembly code below?

```asm
<lseek>:
  endbr64
  mov    D17(%rip),%rax
  mov    (%rax),%edx
  mov    D10(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
off_t lseek(int fd, off_t off, int whence) {
  (void)fd;
  (void)off;
  (void)whence;
  errno = ENODEV;
  return -1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<rop3_229_0_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  movzbl -0x1c(%rbp),%eax
  and    -0x20(%rbp),%al
  mov    %al,-0x3(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  or     -0x3(%rbp),%al
  mov    %al,-0x2(%rbp)
  movzbl -0x20(%rbp),%eax
  xor    -0x2(%rbp),%al
  mov    %al,-0x1(%rbp)
  notb   -0x1(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl -0x1(%rbp),%edx
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rop3_229_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = S & T;
  stk2 = *D | stk3;
  stk1 = T ^ stk2;
  stk1 = ~stk1;
  *D = stk1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rop3_229_0_0>:
  endbr64
  and    %edx,%esi
  or     (%rdi),%sil
  xor    %edx,%esi
  not    %esi
  mov    %sil,(%rdi)
  ret
```

assistant: ```c
void rop3_229_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = S & T;
  stk2 = *D | stk3;
  stk1 = T ^ stk2;
  stk1 = ~stk1;
  *D = stk1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<rop3_229_0_0>:
  endbr64
  and    %edx,%esi
  or     (%rdi),%sil
  xor    %edx,%esi
  not    %esi
  mov    %sil,(%rdi)
  ret
```

assistant: ```c
void rop3_229_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = S & T;
  stk2 = *D | stk3;
  stk1 = T ^ stk2;
  stk1 = ~stk1;
  *D = stk1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<rop3_229_0_0>:
  endbr64
  and    %edx,%esi
  or     (%rdi),%sil
  xor    %edx,%esi
  not    %esi
  mov    %sil,(%rdi)
  ret
```

assistant: ```c
void rop3_229_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = S & T;
  stk2 = *D | stk3;
  stk1 = T ^ stk2;
  stk1 = ~stk1;
  *D = stk1;
}
```
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
  sub    $0x50,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x2,-0x44(%rbp)
  je     L8
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
L8:
  mov    -0x50(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x2c(%rbp)
  lea    -0x28(%rbp),%rdx
  lea    -0x3c(%rbp),%rcx
  mov    -0x2c(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <solveNQueens@plt>
  mov    %rax,-0x20(%rbp)
  movl   $0x0,-0x38(%rbp)
  jmp    L15
L58:
  mov    -0x38(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x34(%rbp)
  jmp    L49
L45:
  mov    -0x34(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x30(%rbp)
  jmp    L11
L73:
  mov    -0x30(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0x30(%rbp)
L11:
  mov    -0x30(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L73
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x34(%rbp)
L49:
  mov    -0x34(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L45
  mov    -0x38(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x38(%rbp)
L15:
  mov    -0x3c(%rbp),%eax
  cmp    %eax,-0x38(%rbp)
  jl     L58
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D5
type_guessed: string
name: read_data
arguments: data_tag: D18
type_guessed: string

tool: D5: Usage: ./queen 8

tool: D18: The %dth solution.


assistant: ```c
int main(int argc, char **argv) {
  int i, n, row, col, num_of_solution;
  if (argc != 2) {
    printf("Usage: ./queen 8\n");
    exit(-1);
  }
  n = atoi(argv[1]);
  int *col_sizes;
  char ***solutions = solveNQueens(n, &num_of_solution, &col_sizes);
  for (i = 0; i < num_of_solution; i++) {
    char **solution = solutions[i];
    for (row = 0; row < n; row++) {
      char *line = solution[row];
      for (col = 0; col < n; col++) {
        putchar(line[col]);
      }
      putchar('\n');
    }
    printf("The %dth solution.\n", i + 1);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  cmp    $0x2,%edi
  jne    L11
L9:
  mov    0x8(%rbx),%rdi
  call   <atoi@plt>
  mov    %eax,%ebx
  mov    %eax,0x14(%rsp)
  lea    0x30(%rsp),%rdx
  lea    0x2c(%rsp),%rsi
  mov    %eax,%edi
  call   <solveNQueens@plt>
  cmpl   $0x0,0x2c(%rsp)
  jle    L60
  mov    %rax,0x8(%rsp)
  lea    -0x1(%rbx),%r14d
  mov    $0x0,%r15d
  add    $0x1,%r14
  lea    0x0(,%r14,8),%rax
  mov    %rax,0x18(%rsp)
L52:
  mov    0x8(%rsp),%rax
  mov    (%rax),%r13
  cmpl   $0x0,0x14(%rsp)
  jle    L65
  mov    %r13,%r12
  add    0x18(%rsp),%r13
L29:
  mov    (%r12),%rbx
  lea    (%r14,%rbx,1),%rbp
L37:
  movsbl (%rbx),%edi
  call   <putchar@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L37
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%r12
  cmp    %r13,%r12
  jne    L29
L65:
  add    $0x1,%r15d
  mov    %r15d,%esi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addq   $0x8,0x8(%rsp)
  cmp    0x2c(%rsp),%r15d
  jl     L52
L60:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L46
  mov    $0x0,%eax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L11:
  lea    D17(%rip),%rdi
  call   <puts@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  jmp    L9
L46:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
datatype: string
name: extract_symbolic_data
arguments: data_label: D17
datatype: string

tool: D9: The %dth solution.


tool: D17: Usage: ./queen 8

assistant: ```c
int main(int argc, char **argv) {
  int i, n, row, col, num_of_solution;
  if (argc != 2) {
    printf("Usage: ./queen 8\n");
    exit(-1);
  }
  n = atoi(argv[1]);
  int *col_sizes;
  char ***solutions = solveNQueens(n, &num_of_solution, &col_sizes);
  for (i = 0; i < num_of_solution; i++) {
    char **solution = solutions[i];
    for (row = 0; row < n; row++) {
      char *line = solution[row];
      for (col = 0; col < n; col++) {
        putchar(line[col]);
      }
      putchar('\n');
    }
    printf("The %dth solution.\n", i + 1);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  cmp    $0x2,%edi
  je     L49
  lea    D10(%rip),%rdi
  call   <puts@plt>
  or     $0xffffffff,%edi
  call   <exit@plt>
L49:
  mov    0x8(%rbx),%rdi
  call   <atoi@plt>
  lea    0x20(%rsp),%rdx
  lea    0x1c(%rsp),%rsi
  mov    %eax,%edi
  mov    %eax,0xc(%rsp)
  mov    %eax,%ebx
  call   <solveNQueens@plt>
  mov    0x1c(%rsp),%edx
  test   %edx,%edx
  jle    L20
  mov    %rax,(%rsp)
  lea    -0x1(%rbx),%r12d
  xor    %r14d,%r14d
  add    $0x1,%r12
  nopl   0x0(%rax)
L54:
  mov    (%rsp),%rax
  mov    (%rax),%rbp
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L33
  lea    0x0(%rbp,%r12,8),%r13
  nopl   0x0(%rax)
L7:
  mov    0x0(%rbp),%r15
  lea    (%r12,%r15,1),%rbx
L5:
  movsbl (%r15),%edi
  add    $0x1,%r15
  call   <putchar@plt>
  cmp    %r15,%rbx
  jne    L5
  mov    $0xa,%edi
  add    $0x8,%rbp
  call   <putchar@plt>
  cmp    %r13,%rbp
  jne    L7
L33:
  add    $0x1,%r14d
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  mov    %r14d,%esi
  call   <printf@plt>
  addq   $0x8,(%rsp)
  cmp    0x1c(%rsp),%r14d
  jl     L54
L20:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D10
type_guessed: string
name: retrieve_data
arguments: data_label: D9
type_guessed: string

tool: D10: Usage: ./queen 8

tool: D9: The %dth solution.


assistant: ```c
int main(int argc, char **argv) {
  int i, n, row, col, num_of_solution;
  if (argc != 2) {
    printf("Usage: ./queen 8\n");
    exit(-1);
  }
  n = atoi(argv[1]);
  int *col_sizes;
  char ***solutions = solveNQueens(n, &num_of_solution, &col_sizes);
  for (i = 0; i < num_of_solution; i++) {
    char **solution = solutions[i];
    for (row = 0; row < n; row++) {
      char *line = solution[row];
      for (col = 0; col < n; col++) {
        putchar(line[col]);
      }
      putchar('\n');
    }
    printf("The %dth solution.\n", i + 1);
  }
  return 0;
}
```