ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<str_copy>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L2
  xor    %eax,%eax
  nopl   (%rax)
L4:
  mov    %dl,(%rsi,%rax,1)
  add    $0x1,%rax
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  jne    L4
L2:
  ret
```

assistant: ```c
void str_copy(unsigned char *from, unsigned char *to) {
  int i;
  for (i = 0; from[i] != '\0'; i++) {
    to[i] = from[i];
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<DoubleNAN>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D4(%rip),%rax
  movsd  (%rax),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double DoubleNAN(void) {
  typedef struct {
    double x;
    long l1;
    long l2;
  } L_nanDouble;
  static L_nanDouble l_nanDouble = {0.0, 0x7fffffff, 0x7fffffff};
  static double *d_nanDouble = (double *)&l_nanDouble.l1;
  return *d_nanDouble;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<DoubleNAN>:
  endbr64
  movsd  D2(%rip),%xmm0
  ret
```

assistant: ```c
double DoubleNAN(void) {
  typedef struct {
    double x;
    long l1;
    long l2;
  } L_nanDouble;
  static L_nanDouble l_nanDouble = {0.0, 0x7fffffff, 0x7fffffff};
  static double *d_nanDouble = (double *)&l_nanDouble.l1;
  return *d_nanDouble;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<DoubleNAN>:
  endbr64
  movsd  D8(%rip),%xmm0
  ret
```

assistant: ```c
double DoubleNAN(void) {
  typedef struct {
    double x;
    long l1;
    long l2;
  } L_nanDouble;
  static L_nanDouble l_nanDouble = {0.0, 0x7fffffff, 0x7fffffff};
  static double *d_nanDouble = (double *)&l_nanDouble.l1;
  return *d_nanDouble;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<DoubleNAN>:
  endbr64
  movsd  D7(%rip),%xmm0
  ret
```

assistant: ```c
double DoubleNAN(void) {
  typedef struct {
    double x;
    long l1;
    long l2;
  } L_nanDouble;
  static L_nanDouble l_nanDouble = {0.0, 0x7fffffff, 0x7fffffff};
  static double *d_nanDouble = (double *)&l_nanDouble.l1;
  return *d_nanDouble;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<alaw2linear>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x14(%rbp)
  xorb   $0x55,-0x14(%rbp)
  movzbl -0x14(%rbp),%eax
  shl    $0x4,%eax
  and    $0xff,%eax
  mov    %eax,-0x8(%rbp)
  movzbl -0x14(%rbp),%eax
  shr    $0x4,%eax
  and    $0x7,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L1
  cmpl   $0x1,-0x4(%rbp)
  je     L32
  jmp    L65
L1:
  addl   $0x8,-0x8(%rbp)
  jmp    L0
L32:
  addl   $0x108,-0x8(%rbp)
  jmp    L0
L65:
  addl   $0x108,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%ecx
  shll   %cl,-0x8(%rbp)
L0:
  movzbl -0x14(%rbp),%eax
  test   %al,%al
  js     L51
  mov    -0x8(%rbp),%eax
  neg    %eax
  jmp    L34
L51:
  mov    -0x8(%rbp),%eax
L34:
  pop    %rbp
  ret
```

assistant: ```c
int alaw2linear(unsigned char a_val) {
  int t;
  int seg;
  a_val ^= 0x55;
  t = (a_val & (0xf)) << 4;
  seg = ((unsigned)a_val & (0x70)) >> (4);
  switch (seg) {
  case 0:
    t += 8;
    break;
  case 1:
    t += 0x108;
    break;
  default:
    t += 0x108;
    t <<= seg - 1;
  }
  return ((a_val & (0x80)) ? t : -t);
}
```
user: Translate the below assembly code into C programming language:

```asm
<alaw2linear>:
  endbr64
  xor    $0x55,%edi
  movzbl %dil,%ecx
  mov    %ecx,%edx
  shl    $0x4,%edx
  movzbl %dl,%edx
  shr    $0x4,%ecx
  lea    0x8(%rdx),%eax
  and    $0x7,%ecx
  je     L2
  lea    0x108(%rdx),%eax
  cmp    $0x1,%ecx
  je     L2
  sub    $0x1,%ecx
  shl    %cl,%eax
L2:
  mov    %eax,%edx
  neg    %edx
  test   %dil,%dil
  cmovns %edx,%eax
  ret
```

assistant: ```c
int alaw2linear(unsigned char a_val) {
  int t;
  int seg;
  a_val ^= 0x55;
  t = (a_val & (0xf)) << 4;
  seg = ((unsigned)a_val & (0x70)) >> (4);
  switch (seg) {
  case 0:
    t += 8;
    break;
  case 1:
    t += 0x108;
    break;
  default:
    t += 0x108;
    t <<= seg - 1;
  }
  return ((a_val & (0x80)) ? t : -t);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<alaw2linear>:
  endbr64
  xor    $0x55,%edi
  movzbl %dil,%edx
  mov    %edx,%ecx
  shr    $0x4,%edx
  shl    $0x4,%ecx
  movzbl %cl,%ecx
  lea    0x8(%rcx),%eax
  and    $0x7,%edx
  je     L1
  lea    0x108(%rcx),%eax
  lea    -0x1(%rdx),%ecx
  mov    %eax,%esi
  shl    %cl,%eax
  cmp    $0x1,%edx
  cmove  %esi,%eax
L1:
  mov    %eax,%edx
  neg    %edx
  test   %dil,%dil
  cmovns %edx,%eax
  ret
```

assistant: ```c
int alaw2linear(unsigned char a_val) {
  int t;
  int seg;
  a_val ^= 0x55;
  t = (a_val & (0xf)) << 4;
  seg = ((unsigned)a_val & (0x70)) >> (4);
  switch (seg) {
  case 0:
    t += 8;
    break;
  case 1:
    t += 0x108;
    break;
  default:
    t += 0x108;
    t <<= seg - 1;
  }
  return ((a_val & (0x80)) ? t : -t);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<alaw2linear>:
  endbr64
  xor    $0x55,%edi
  movzbl %dil,%edx
  mov    %edx,%ecx
  shr    $0x4,%edx
  shl    $0x4,%ecx
  movzbl %cl,%ecx
  lea    0x8(%rcx),%eax
  and    $0x7,%edx
  je     L4
  lea    0x108(%rcx),%eax
  lea    -0x1(%rdx),%ecx
  mov    %eax,%esi
  shl    %cl,%eax
  cmp    $0x1,%edx
  cmove  %esi,%eax
L4:
  mov    %eax,%edx
  neg    %edx
  test   %dil,%dil
  cmovns %edx,%eax
  ret
```

assistant: ```c
int alaw2linear(unsigned char a_val) {
  int t;
  int seg;
  a_val ^= 0x55;
  t = (a_val & (0xf)) << 4;
  seg = ((unsigned)a_val & (0x70)) >> (4);
  switch (seg) {
  case 0:
    t += 8;
    break;
  case 1:
    t += 0x108;
    break;
  default:
    t += 0x108;
    t <<= seg - 1;
  }
  return ((a_val & (0x80)) ? t : -t);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<copy_file_to_mem>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %rdx,-0x58(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rdx
  mov    -0x48(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <stat@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <S_ISREG@plt>
  test   %rax,%rax
  je     L10
  mov    -0x20(%rbp),%rax
  test   %rax,%rax
  je     L10
  mov    D18(%rip),%rax
  mov    (%rax),%ecx
  mov    -0x48(%rbp),%rax
  mov    $0x0,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <open@plt>
  mov    %eax,-0x34(%rbp)
  cmpl   $0x0,-0x34(%rbp)
  jns    L26
  mov    $0xffffffff,%eax
  jmp    L12
L26:
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  mov    -0x58(%rbp),%rax
  mov    -0x30(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    -0x50(%rbp),%rax
  mov    %rdx,(%rax)
  mov    D10(%rip),%rax
  mov    (%rax),%ecx
  mov    D3(%rip),%rax
  mov    (%rax),%edx
  mov    -0x34(%rbp),%esi
  mov    -0x30(%rbp),%rax
  mov    $0x0,%r9d
  mov    %esi,%r8d
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <mmap@plt>
  mov    %rax,-0x28(%rbp)
  mov    -0x50(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <munmap@plt>
  mov    -0x34(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  mov    $0x0,%eax
  jmp    L12
L10:
  mov    $0xffffffff,%eax
L12:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L52
  call   <__stack_chk_fail@plt>
L52:
  leave
  ret
```

assistant: ```c
int copy_file_to_mem(const char *path, unsigned char **p_buf, size_t *p_sz) {
  struct stat fst;
  stat(path, &fst);
  if (S_ISREG(fst.st_mode) && fst.st_size > 0) {
    int fd = open(path, O_RDONLY, 0);
    if (fd < 0)
      return (-1);
    size_t fsz = fst.st_size;
    *p_sz = fsz;
    *p_buf = (unsigned char *)malloc(fsz);
    void *p = mmap(0, fsz, PROT_READ, MAP_PRIVATE, fd, 0);
    memcpy(*p_buf, p, fsz);
    munmap(p, fsz);
    close(fd);
    return (0);
  } else
    return (-1);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<copy_file_to_mem>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rdx,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <stat@plt>
  mov    0x8(%rsp),%edi
  call   <S_ISREG@plt>
  test   %rax,%rax
  je     L50
  cmpq   $0x0,(%rsp)
  je     L45
  mov    $0x0,%edx
  mov    D3(%rip),%rax
  mov    (%rax),%esi
  mov    %rbx,%rdi
  call   <open@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  js     L14
  mov    (%rsp),%r13
  mov    %r13,(%r12)
  mov    %r13,%rdi
  call   <malloc@plt>
  mov    %rax,0x0(%rbp)
  mov    $0x0,%r9d
  mov    %ebx,%r8d
  mov    D1(%rip),%rax
  mov    (%rax),%ecx
  mov    D17(%rip),%rax
  mov    (%rax),%edx
  mov    %r13,%rsi
  mov    $0x0,%edi
  call   <mmap@plt>
  mov    %rax,%r12
  mov    %r13,%rdx
  mov    %rax,%rsi
  mov    0x0(%rbp),%rdi
  call   <memcpy@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <munmap@plt>
  mov    %ebx,%edi
  call   <close@plt>
  mov    $0x0,%eax
L5:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L55
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L50:
  mov    $0xffffffff,%eax
  jmp    L5
L45:
  mov    $0xffffffff,%eax
  jmp    L5
L14:
  mov    $0xffffffff,%eax
  jmp    L5
L55:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int copy_file_to_mem(const char *path, unsigned char **p_buf, size_t *p_sz) {
  struct stat fst;
  stat(path, &fst);
  if (S_ISREG(fst.st_mode) && fst.st_size > 0) {
    int fd = open(path, O_RDONLY, 0);
    if (fd < 0)
      return (-1);
    size_t fsz = fst.st_size;
    *p_sz = fsz;
    *p_buf = (unsigned char *)malloc(fsz);
    void *p = mmap(0, fsz, PROT_READ, MAP_PRIVATE, fd, 0);
    memcpy(*p_buf, p, fsz);
    munmap(p, fsz);
    close(fd);
    return (0);
  } else
    return (-1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<copy_file_to_mem>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <stat@plt>
  mov    0x8(%rsp),%edi
  call   <S_ISREG@plt>
  test   %rax,%rax
  je     L17
  cmpq   $0x0,(%rsp)
  je     L17
  mov    D29(%rip),%rax
  mov    %rbp,%rdi
  xor    %edx,%edx
  mov    (%rax),%esi
  call   <open@plt>
  mov    %eax,%ebp
  test   %eax,%eax
  js     L17
  mov    (%rsp),%r12
  mov    %r12,0x0(%r13)
  mov    %r12,%rdi
  call   <malloc@plt>
  xor    %r9d,%r9d
  mov    %ebp,%r8d
  mov    %r12,%rsi
  mov    %rax,(%rbx)
  mov    D18(%rip),%rax
  xor    %edi,%edi
  mov    (%rax),%ecx
  mov    D7(%rip),%rax
  mov    (%rax),%edx
  call   <mmap@plt>
  mov    (%rbx),%rdi
  mov    %r12,%rdx
  mov    %rax,%r13
  mov    %rax,%rsi
  call   <memcpy@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <munmap@plt>
  mov    %ebp,%edi
  call   <close@plt>
  xor    %eax,%eax
L0:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L28
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  mov    $0xffffffff,%eax
  jmp    L0
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int copy_file_to_mem(const char *path, unsigned char **p_buf, size_t *p_sz) {
  struct stat fst;
  stat(path, &fst);
  if (S_ISREG(fst.st_mode) && fst.st_size > 0) {
    int fd = open(path, O_RDONLY, 0);
    if (fd < 0)
      return (-1);
    size_t fsz = fst.st_size;
    *p_sz = fsz;
    *p_buf = (unsigned char *)malloc(fsz);
    void *p = mmap(0, fsz, PROT_READ, MAP_PRIVATE, fd, 0);
    memcpy(*p_buf, p, fsz);
    munmap(p, fsz);
    close(fd);
    return (0);
  } else
    return (-1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<copy_file_to_mem>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <stat@plt>
  mov    0x8(%rsp),%edi
  call   <S_ISREG@plt>
  test   %rax,%rax
  je     L0
  cmpq   $0x0,(%rsp)
  je     L0
  mov    D6(%rip),%rax
  mov    %rbp,%rdi
  xor    %edx,%edx
  mov    (%rax),%esi
  call   <open@plt>
  mov    %eax,%ebp
  test   %eax,%eax
  js     L0
  mov    (%rsp),%r12
  mov    %r12,0x0(%r13)
  mov    %r12,%rdi
  call   <malloc@plt>
  xor    %r9d,%r9d
  mov    %ebp,%r8d
  mov    %r12,%rsi
  mov    %rax,(%rbx)
  mov    D26(%rip),%rax
  xor    %edi,%edi
  mov    (%rax),%ecx
  mov    D20(%rip),%rax
  mov    (%rax),%edx
  call   <mmap@plt>
  mov    (%rbx),%rdi
  mov    %r12,%rdx
  mov    %rax,%r13
  mov    %rax,%rsi
  call   <memcpy@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <munmap@plt>
  mov    %ebp,%edi
  call   <close@plt>
  xor    %eax,%eax
L21:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L48
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  mov    $0xffffffff,%eax
  jmp    L21
L48:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int copy_file_to_mem(const char *path, unsigned char **p_buf, size_t *p_sz) {
  struct stat fst;
  stat(path, &fst);
  if (S_ISREG(fst.st_mode) && fst.st_size > 0) {
    int fd = open(path, O_RDONLY, 0);
    if (fd < 0)
      return (-1);
    size_t fsz = fst.st_size;
    *p_sz = fsz;
    *p_buf = (unsigned char *)malloc(fsz);
    void *p = mmap(0, fsz, PROT_READ, MAP_PRIVATE, fd, 0);
    memcpy(*p_buf, p, fsz);
    munmap(p, fsz);
    close(fd);
    return (0);
  } else
    return (-1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<test_sudoku_solveSolved>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <sudoku_solve@plt>
  cmp    $0x1,%eax
  je     L6
  lea    D16(%rip),%rax
  jmp    L16
L6:
  mov    $0x0,%eax
L16:
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D16
datatype: string

tool: D16: e: test_sudoku_solveSolved()

assistant: ```c
char *test_sudoku_solveSolved() {
  static int grid[81] = {4, 9, 2, 3, 6, 8, 1, 7, 5, 8, 7, 6, 9, 5, 1, 2, 3, 4, 3, 1, 5, 2, 7, 4, 6, 8, 9, 9, 3, 4, 1, 2, 7, 8, 5, 6, 2, 5, 7, 8, 3, 6, 9, 4, 1, 1, 6, 8, 4, 9, 5, 3, 2, 7, 5, 2, 1, 6, 4, 3, 7, 9, 8, 7, 8, 3, 5, 1, 9, 4, 6, 2, 6, 4, 9, 7, 8, 2, 5, 1, 3};
  do {
    if (!(sudoku_solve(grid) == 1))
      return "e: test_sudoku_solveSolved()";
  } while (0);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<test_sudoku_solveSolved>:
  endbr64
  sub    $0x8,%rsp
  lea    D14(%rip),%rdi
  call   <sudoku_solve@plt>
  cmp    $0x1,%eax
  lea    D10(%rip),%rax
  mov    $0x0,%edx
  cmove  %rdx,%rax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D10
datatype: string

tool: D10: e: test_sudoku_solveSolved()

assistant: ```c
char *test_sudoku_solveSolved() {
  static int grid[81] = {4, 9, 2, 3, 6, 8, 1, 7, 5, 8, 7, 6, 9, 5, 1, 2, 3, 4, 3, 1, 5, 2, 7, 4, 6, 8, 9, 9, 3, 4, 1, 2, 7, 8, 5, 6, 2, 5, 7, 8, 3, 6, 9, 4, 1, 1, 6, 8, 4, 9, 5, 3, 2, 7, 5, 2, 1, 6, 4, 3, 7, 9, 8, 7, 8, 3, 5, 1, 9, 4, 6, 2, 6, 4, 9, 7, 8, 2, 5, 1, 3};
  do {
    if (!(sudoku_solve(grid) == 1))
      return "e: test_sudoku_solveSolved()";
  } while (0);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test_sudoku_solveSolved>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <sudoku_solve@plt>
  mov    $0x0,%edx
  cmp    $0x1,%eax
  lea    D16(%rip),%rax
  cmove  %rdx,%rax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D16
datatype: string

tool: D16: e: test_sudoku_solveSolved()

assistant: ```c
char *test_sudoku_solveSolved() {
  static int grid[81] = {4, 9, 2, 3, 6, 8, 1, 7, 5, 8, 7, 6, 9, 5, 1, 2, 3, 4, 3, 1, 5, 2, 7, 4, 6, 8, 9, 9, 3, 4, 1, 2, 7, 8, 5, 6, 2, 5, 7, 8, 3, 6, 9, 4, 1, 1, 6, 8, 4, 9, 5, 3, 2, 7, 5, 2, 1, 6, 4, 3, 7, 9, 8, 7, 8, 3, 5, 1, 9, 4, 6, 2, 6, 4, 9, 7, 8, 2, 5, 1, 3};
  do {
    if (!(sudoku_solve(grid) == 1))
      return "e: test_sudoku_solveSolved()";
  } while (0);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<test_sudoku_solveSolved>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <sudoku_solve@plt>
  mov    $0x0,%edx
  cmp    $0x1,%eax
  lea    D16(%rip),%rax
  cmove  %rdx,%rax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D16
datatype: string

tool: D16: e: test_sudoku_solveSolved()

assistant: ```c
char *test_sudoku_solveSolved() {
  static int grid[81] = {4, 9, 2, 3, 6, 8, 1, 7, 5, 8, 7, 6, 9, 5, 1, 2, 3, 4, 3, 1, 5, 2, 7, 4, 6, 8, 9, 9, 3, 4, 1, 2, 7, 8, 5, 6, 2, 5, 7, 8, 3, 6, 9, 4, 1, 1, 6, 8, 4, 9, 5, 3, 2, 7, 5, 2, 1, 6, 4, 3, 7, 9, 8, 7, 8, 3, 5, 1, 9, 4, 6, 2, 6, 4, 9, 7, 8, 2, 5, 1, 3};
  do {
    if (!(sudoku_solve(grid) == 1))
      return "e: test_sudoku_solveSolved()";
  } while (0);
  return 0;
}
```
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
  mov    $0x0,%eax
  call   <system_init@plt>
L8:
  mov    $0x0,%eax
  call   <ir_measure_cm@plt>
  movq   %xmm0,%rax
  lea    -0x20(%rbp),%rdx
  movq   %rax,%xmm0
  lea    D5(%rip),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <sprintf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <UART1_Write_Text@plt>
  mov    $0x1f4,%edi
  call   <Delay_ms@plt>
  jmp    L8
```

tool_calls: name: read_data
arguments: tag: D5
type: string

tool: D5: Distance is: %4.2f


assistant: ```c
void main() {
  char tmp_text[20];
  system_init();
  while (1) {
    sprintf(tmp_text, "Distance is: %4.2f\r\n", ir_measure_cm());
    UART1_Write_Text(tmp_text);
    Delay_ms(500);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <system_init@plt>
  lea    D9(%rip),%rbp
L3:
  mov    $0x0,%eax
  call   <ir_measure_cm@plt>
  mov    %rsp,%rbx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rdi
  call   <UART1_Write_Text@plt>
  mov    $0x1f4,%edi
  call   <Delay_ms@plt>
  jmp    L3
```

tool_calls: name: disassemble_data
arguments: data_label: D9
type_guessed: string

tool: D9: Distance is: %4.2f


assistant: ```c
void main() {
  char tmp_text[20];
  system_init();
  while (1) {
    sprintf(tmp_text, "Distance is: %4.2f\r\n", ir_measure_cm());
    UART1_Write_Text(tmp_text);
    Delay_ms(500);
  }
}
```
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
  lea    D7(%rip),%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <system_init@plt>
  nopl   0x0(%rax)
L5:
  xor    %eax,%eax
  call   <ir_measure_cm@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rdi
  call   <UART1_Write_Text@plt>
  mov    $0x1f4,%edi
  call   <Delay_ms@plt>
  jmp    L5
```

tool_calls: name: unpack_data
arguments: data_tag: D7
datatype: string

tool: D7: Distance is: %4.2f


assistant: ```c
void main() {
  char tmp_text[20];
  system_init();
  while (1) {
    sprintf(tmp_text, "Distance is: %4.2f\r\n", ir_measure_cm());
    UART1_Write_Text(tmp_text);
    Delay_ms(500);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <system_init@plt>
  nopl   0x0(%rax)
L4:
  xor    %eax,%eax
  call   <ir_measure_cm@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rdi
  call   <UART1_Write_Text@plt>
  mov    $0x1f4,%edi
  call   <Delay_ms@plt>
  jmp    L4
```

tool_calls: name: parse_data
arguments: label: D8
type_guessed: string

tool: D8: Distance is: %4.2f


assistant: ```c
void main() {
  char tmp_text[20];
  system_init();
  while (1) {
    sprintf(tmp_text, "Distance is: %4.2f\r\n", ir_measure_cm());
    UART1_Write_Text(tmp_text);
    Delay_ms(500);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%r12d
  mov    $0x0,%ebp
  mov    $0x1,%ebx
  jmp    L79
L31:
  movsbl %al,%eax
  lea    -0x30(%rax),%r13d
  jmp    L20
L14:
  movsbl %al,%eax
  sub    $0x30,%eax
  mov    %eax,%edx
  imul   %ebx,%edx
  add    %edx,%ebp
  sub    %eax,%edx
  add    %edx,%r12d
L20:
  add    $0x1,%ebx
L72:
  cmp    $0xb,%ebx
  jg     L5
L79:
  mov    $0x0,%eax
  call   <getchar@plt>
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  ja     L72
  cmp    $0xa,%ebx
  je     L31
  cmp    $0xb,%ebx
  jne    L14
  movsbl %al,%eax
  lea    -0x30(%rax),%r14d
  jmp    L20
L5:
  movslq %ebp,%rax
  imul   $0x2e8ba2e9,%rax,%rax
  sar    $0x21,%rax
  mov    %ebp,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  lea    (%rax,%rdx,2),%eax
  sub    %eax,%ebp
  movslq %ebp,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ebp,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  mov    %ebp,%eax
  sub    %edx,%eax
  lea    (%rax,%rax,8),%edx
  add    %r12d,%edx
  movslq %edx,%rcx
  imul   $0x2e8ba2e9,%rcx,%rcx
  sar    $0x21,%rcx
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%ecx
  lea    (%rcx,%rcx,4),%esi
  lea    (%rcx,%rsi,2),%ecx
  sub    %ecx,%edx
  movslq %edx,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%ecx
  lea    (%rcx,%rcx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  cmp    %r14d,%edx
  jne    L89
  cmp    %eax,%r13d
  jne    L89
  lea    D23(%rip),%rdi
  call   <printf@plt>
L66:
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L89:
  lea    D17(%rip),%rdi
  call   <printf@plt>
  jmp    L66
```

tool_calls: name: unpack_data
arguments: data_label: D19
type_infered: string
name: unpack_data
arguments: data_label: D23
type_infered: string
name: unpack_data
arguments: data_label: D17
type_infered: string

tool: D19: DIGITE O SEU CPF:

tool: D23: 
CPF VALIDO.

tool: D17: 
CPF INVALIDO.

assistant: ```c
int main(void) {
  int digitIndex = 1, term10sum = 0, term11sum = 0, digit10, digit11, term10, term11;
  char digit;
  printf("DIGITE O SEU CPF:");
  while (digitIndex <= 11) {
    digit = getchar();
    if (digit >= '0' && digit <= '9') {
      if (digitIndex == 10)
        digit10 = (int)(digit - '0');
      else if (digitIndex == 11)
        digit11 = (int)(digit - '0');
      else {
        term10sum += (int)(digit - '0') * (digitIndex);
        term11sum += (int)(digit - '0') * (digitIndex - 1);
      }
      digitIndex++;
    }
  }
  term10 = ((term10sum % 11) % 10);
  term11sum += term10 * 9;
  term11 = ((term11sum % 11) % 10);
  if ((term10 == digit10) && (term11 == digit11))
    printf("\nCPF VALIDO.");
  else
    printf("\nCPF INVALIDO.");
  return 0;
}
```