ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<appendext>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x3c(%rbp)
  mov    -0x38(%rbp),%rbx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <pathlast@plt>
  cltq
  mov    %rax,%r13
  mov    $0x0,%r12d
L44:
  cmp    %r13,-0x28(%rbp)
  jb     L25
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  jne    L25
  mov    %rbx,%r12
L25:
  mov    -0x28(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x28(%rbp)
  mov    %rbx,%rax
  lea    0x1(%rax),%rbx
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L44
  sub    $0x1,%rbx
  test   %r12,%r12
  je     L38
  cmpl   $0x0,-0x3c(%rbp)
  je     L62
  mov    %r12,%rbx
  jmp    L38
L62:
  mov    $0x0,%eax
  jmp    L55
L38:
  mov    -0x30(%rbp),%rax
  mov    %rax,%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <mystrcpy@plt>
  cltq
  mov    %rax,%rbx
  mov    %rbx,%rax
  sub    -0x38(%rbp),%rax
L55:
  add    $0x28,%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
```

assistant: ```c
int appendext(path, ext, result, force)
char *path, *ext, *result;
int force;
{
  register char *p, *q, *r;
  p = result;
  q = pathlast(path);
  r = (char *)0;
  do {
    if (path >= q && *path == '.')
      r = p;
  } while ((*p++ = *path++) != 0);
  p--;
  if (r != (char *)0) {
    if (force)
      p = r;
    else
      return 0;
  }
  p = mystrcpy(p, ext);
  return p - result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<appendext>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r13
  mov    %rdx,%rbp
  mov    %ecx,%r12d
  mov    $0x0,%eax
  call   <pathlast@plt>
  cltq
  lea    0x1(%rbx),%rdx
  lea    0x1(%rbp),%rcx
  mov    $0x0,%r8d
  jmp    L38
L43:
  movzbl -0x1(%rdx),%esi
  mov    %sil,-0x1(%rcx)
  test   %sil,%sil
  je     L39
L8:
  add    $0x1,%rdx
  add    $0x1,%rcx
L38:
  lea    -0x1(%rcx),%rdi
  lea    -0x1(%rdx),%rsi
  cmp    %rsi,%rax
  ja     L43
  cmpb   $0x2e,-0x1(%rdx)
  jne    L43
  movb   $0x2e,-0x1(%rcx)
  mov    %rdi,%r8
  jmp    L8
L39:
  test   %r8,%r8
  je     L47
  mov    %r12d,%eax
  test   %r12d,%r12d
  je     L24
L2:
  mov    %r13,%rsi
  mov    %r8,%rdi
  mov    $0x0,%eax
  call   <mystrcpy@plt>
  sub    %ebp,%eax
L24:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L47:
  mov    %rdi,%r8
  jmp    L2
```

assistant: ```c
int appendext(path, ext, result, force)
char *path, *ext, *result;
int force;
{
  register char *p, *q, *r;
  p = result;
  q = pathlast(path);
  r = (char *)0;
  do {
    if (path >= q && *path == '.')
      r = p;
  } while ((*p++ = *path++) != 0);
  p--;
  if (r != (char *)0) {
    if (force)
      p = r;
    else
      return 0;
  }
  p = mystrcpy(p, ext);
  return p - result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<appendext>:
  endbr64
  push   %r13
  xor    %eax,%eax
  mov    %rsi,%r13
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <pathlast@plt>
  mov    %rbx,%rdi
  mov    %rbp,%rsi
  xor    %r8d,%r8d
  cltq
  jmp    L79
  xchg   %ax,%ax
L47:
  mov    %dl,-0x1(%rcx)
  test   %dl,%dl
  je     L15
L73:
  add    $0x1,%rdi
  mov    %rcx,%rsi
L79:
  movzbl (%rdi),%edx
  lea    0x1(%rsi),%rcx
  cmp    %rdi,%rax
  ja     L47
  cmp    $0x2e,%dl
  jne    L47
  movb   $0x2e,-0x1(%rcx)
  mov    %rsi,%r8
  jmp    L73
  nopl   0x0(%rax,%rax,1)
L15:
  test   %r8,%r8
  je     L27
  xor    %eax,%eax
  test   %r12d,%r12d
  je     L46
L22:
  mov    %r13,%rsi
  mov    %r8,%rdi
  xor    %eax,%eax
  call   <mystrcpy@plt>
  sub    %ebp,%eax
L46:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L27:
  mov    %rsi,%r8
  jmp    L22
```

assistant: ```c
int appendext(path, ext, result, force)
char *path, *ext, *result;
int force;
{
  register char *p, *q, *r;
  p = result;
  q = pathlast(path);
  r = (char *)0;
  do {
    if (path >= q && *path == '.')
      r = p;
  } while ((*p++ = *path++) != 0);
  p--;
  if (r != (char *)0) {
    if (force)
      p = r;
    else
      return 0;
  }
  p = mystrcpy(p, ext);
  return p - result;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<appendext>:
  endbr64
  push   %r13
  xor    %eax,%eax
  mov    %rsi,%r13
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <pathlast@plt>
  mov    %rbx,%rdi
  mov    %rbp,%rsi
  xor    %r8d,%r8d
  cltq
  jmp    L29
  xchg   %ax,%ax
L43:
  mov    %dl,-0x1(%rcx)
  test   %dl,%dl
  je     L9
L10:
  add    $0x1,%rdi
  mov    %rcx,%rsi
L29:
  movzbl (%rdi),%edx
  lea    0x1(%rsi),%rcx
  cmp    %rdi,%rax
  ja     L43
  cmp    $0x2e,%dl
  jne    L43
  movb   $0x2e,-0x1(%rcx)
  mov    %rsi,%r8
  jmp    L10
  nopl   0x0(%rax,%rax,1)
L9:
  test   %r8,%r8
  je     L5
  xor    %eax,%eax
  test   %r12d,%r12d
  je     L69
L11:
  mov    %r13,%rsi
  mov    %r8,%rdi
  xor    %eax,%eax
  call   <mystrcpy@plt>
  sub    %ebp,%eax
L69:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L5:
  mov    %rsi,%r8
  jmp    L11
```

assistant: ```c
int appendext(path, ext, result, force)
char *path, *ext, *result;
int force;
{
  register char *p, *q, *r;
  p = result;
  q = pathlast(path);
  r = (char *)0;
  do {
    if (path >= q && *path == '.')
      r = p;
  } while ((*p++ = *path++) != 0);
  p--;
  if (r != (char *)0) {
    if (force)
      p = r;
    else
      return 0;
  }
  p = mystrcpy(p, ext);
  return p - result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<dir>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xa0,%rsp
  mov    %rdi,-0x98(%rbp)
  mov    %esi,-0x9c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <tmpnam@plt>
  cmpl   $0x0,-0x9c(%rbp)
  je     L6
  mov    -0x98(%rbp),%rdx
  lea    -0x90(%rbp),%rax
  lea    D59(%rip),%rcx
  lea    D39(%rip),%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  jmp    L23
L6:
  mov    -0x98(%rbp),%rdx
  lea    -0x90(%rbp),%rax
  lea    D59(%rip),%rcx
  lea    D16(%rip),%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
L23:
  lea    -0x90(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <system@plt>
  lea    D59(%rip),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D39
datatype: string
name: read_data
arguments: label: D16
datatype: string

tool: D39: ls -l %s > %s

tool: D16: ls %s > %s

assistant: ```c
char *dir(path, full)
char *path;
int full;
{
  char cmd[128];
  static char name[32];
  tmpnam(name);
  if (full)
    sprintf(cmd, "ls -l %s > %s", path, name);
  else
    sprintf(cmd, "ls %s > %s", path, name);
  system(cmd);
  return (name);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<dir>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  lea    D0(%rip),%rdi
  call   <tmpnam@plt>
  test   %ebp,%ebp
  je     L21
  mov    %rsp,%rdi
  lea    D0(%rip),%rcx
  mov    %rbx,%rdx
  lea    D26(%rip),%rsi
  mov    $0x0,%eax
  call   <sprintf@plt>
L4:
  mov    %rsp,%rdi
  mov    $0x0,%eax
  call   <system@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  lea    D0(%rip),%rax
  add    $0x98,%rsp
  pop    %rbx
  pop    %rbp
  ret
L21:
  mov    %rsp,%rdi
  lea    D0(%rip),%rcx
  mov    %rbx,%rdx
  lea    D31(%rip),%rsi
  mov    $0x0,%eax
  call   <sprintf@plt>
  jmp    L4
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D26
type: string
name: read_data
arguments: tag: D31
type: string

tool: D26: ls -l %s > %s

tool: D31: ls %s > %s

assistant: ```c
char *dir(path, full)
char *path;
int full;
{
  char cmd[128];
  static char name[32];
  tmpnam(name);
  if (full)
    sprintf(cmd, "ls -l %s > %s", path, name);
  else
    sprintf(cmd, "ls %s > %s", path, name);
  system(cmd);
  return (name);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<dir>:
  endbr64
  push   %r13
  lea    D27(%rip),%r13
  push   %r12
  mov    %rdi,%r12
  mov    %r13,%rdi
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <tmpnam@plt>
  mov    %r13,%rcx
  mov    %r12,%rdx
  test   %ebx,%ebx
  je     L28
  lea    D3(%rip),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <sprintf@plt>
L7:
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <system@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x98,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L28:
  lea    D21(%rip),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <sprintf@plt>
  jmp    L7
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D3
type_infered: string
name: parse_data
arguments: data_label: D21
type_infered: string

tool: D3: ls -l %s > %s

tool: D21: ls %s > %s

assistant: ```c
char *dir(path, full)
char *path;
int full;
{
  char cmd[128];
  static char name[32];
  tmpnam(name);
  if (full)
    sprintf(cmd, "ls -l %s > %s", path, name);
  else
    sprintf(cmd, "ls %s > %s", path, name);
  system(cmd);
  return (name);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<dir>:
  endbr64
  push   %r13
  lea    D7(%rip),%r13
  push   %r12
  mov    %rdi,%r12
  mov    %r13,%rdi
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <tmpnam@plt>
  mov    %r13,%rcx
  mov    %r12,%rdx
  test   %ebx,%ebx
  je     L11
  lea    D14(%rip),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <sprintf@plt>
L12:
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <system@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x98,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L11:
  lea    D18(%rip),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <sprintf@plt>
  jmp    L12
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D14
datatype: string
name: unpack_data
arguments: label: D18
datatype: string

tool: D14: ls -l %s > %s

tool: D18: ls %s > %s

assistant: ```c
char *dir(path, full)
char *path;
int full;
{
  char cmd[128];
  static char name[32];
  tmpnam(name);
  if (full)
    sprintf(cmd, "ls -l %s > %s", path, name);
  else
    sprintf(cmd, "ls %s > %s", path, name);
  system(cmd);
  return (name);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<integer_overflow_by_addition_l2_Bad>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L1
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L1:
  movl   $0x77359400,-0x4(%rbp)
L17:
  shll   -0x4(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void integer_overflow_by_addition_l2_Bad(int x) {
  int32_t y;
  if (x) {
    y = 0;
  } else {
    y = 2000000000;
  }
  y = y + y;
}
```
user: What is the c source code of the assembly code below:

```asm
<integer_overflow_by_addition_l2_Bad>:
  endbr64
  ret
```

assistant: ```c
void integer_overflow_by_addition_l2_Bad(int x) {
  int32_t y;
  if (x) {
    y = 0;
  } else {
    y = 2000000000;
  }
  y = y + y;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<integer_overflow_by_addition_l2_Bad>:
  endbr64
  ret
```

assistant: ```c
void integer_overflow_by_addition_l2_Bad(int x) {
  int32_t y;
  if (x) {
    y = 0;
  } else {
    y = 2000000000;
  }
  y = y + y;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<isPossible>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  movl   $0x3,-0x14(%rbp)
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x38e38e39,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  mov    %eax,%esi
  sar    $0x1f,%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %ecx,%edx
  shl    $0x3,%edx
  add    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x38e38e39,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  cltd
  idivl  -0x14(%rbp)
  mov    -0x10(%rbp),%eax
  sub    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  cltd
  idivl  -0x14(%rbp)
  mov    -0xc(%rbp),%eax
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L8
L29:
  mov    -0xc(%rbp),%edx
  mov    -0x18(%rbp),%ecx
  mov    -0x28(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <getAt@plt>
  cmp    %eax,-0x30(%rbp)
  je     L35
  mov    -0x18(%rbp),%edx
  mov    -0x10(%rbp),%ecx
  mov    -0x28(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <getAt@plt>
  cmp    %eax,-0x30(%rbp)
  je     L35
  mov    -0x18(%rbp),%eax
  cltd
  idivl  -0x14(%rbp)
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  lea    (%rdx,%rax,1),%esi
  mov    -0x18(%rbp),%eax
  cltd
  idivl  -0x14(%rbp)
  mov    -0x8(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x28(%rbp),%rax
  mov    %esi,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <getAt@plt>
  cmp    %eax,-0x30(%rbp)
  jne    L19
L35:
  mov    $0x0,%eax
  jmp    L46
L19:
  addl   $0x1,-0x18(%rbp)
L8:
  cmpl   $0x8,-0x18(%rbp)
  jle    L29
  mov    $0x1,%eax
L46:
  leave
  ret
```

assistant: ```c
int isPossible(int *grid, int index, int value) {
  int boxN = 9 / 3;
  int x = index % 9;
  int y = index / 9;
  int boxX = x - (x % boxN);
  int boxY = y - (y % boxN);
  int i;
  for (i = 0; i < 9; i++) {
    if (getAt(grid, i, y) == value || getAt(grid, x, i) == value || getAt(grid, boxX + (int)(i % boxN), boxY + (int)(i / boxN)) == value) {
      return 0;
    }
  }
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<isPossible>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r12
  mov    %edx,%ebp
  movslq %esi,%r13
  imul   $0x38e38e39,%r13,%r13
  sar    $0x21,%r13
  mov    %esi,%eax
  sar    $0x1f,%eax
  sub    %eax,%r13d
  mov    %r13d,%r14d
  lea    0x0(%r13,%r13,8),%eax
  sub    %eax,%esi
  mov    %esi,%r13d
  movslq %esi,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %esi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,2),%eax
  mov    %eax,0xc(%rsp)
  movslq %r14d,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %r14d,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,2),%r15d
  mov    $0x0,%ebx
L42:
  mov    %r14d,%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <getAt@plt>
  cmp    %ebp,%eax
  je     L7
  mov    %ebx,%edx
  mov    %r13d,%esi
  mov    %r12,%rdi
  call   <getAt@plt>
  cmp    %ebp,%eax
  je     L17
  movslq %ebx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%r15,1),%edx
  lea    (%rax,%rax,2),%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  add    0xc(%rsp),%esi
  mov    %r12,%rdi
  call   <getAt@plt>
  cmp    %ebp,%eax
  je     L24
  add    $0x1,%ebx
  cmp    $0x9,%ebx
  jne    L42
  mov    $0x1,%eax
L2:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L7:
  mov    $0x0,%eax
  jmp    L2
L17:
  mov    $0x0,%eax
  jmp    L2
L24:
  mov    $0x0,%eax
  jmp    L2
```

assistant: ```c
int isPossible(int *grid, int index, int value) {
  int boxN = 9 / 3;
  int x = index % 9;
  int y = index / 9;
  int boxX = x - (x % boxN);
  int boxY = y - (y % boxN);
  int i;
  for (i = 0; i < 9; i++) {
    if (getAt(grid, i, y) == value || getAt(grid, x, i) == value || getAt(grid, boxX + (int)(i % boxN), boxY + (int)(i / boxN)) == value) {
      return 0;
    }
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<isPossible>:
  endbr64
  push   %r15
  mov    %esi,%eax
  push   %r14
  sar    $0x1f,%eax
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  movslq %esi,%r12
  imul   $0x38e38e39,%r12,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %edx,%ebx
  sar    $0x21,%r12
  sub    %eax,%r12d
  sub    $0x18,%rsp
  lea    (%r12,%r12,8),%eax
  mov    %r12d,%r13d
  sub    %eax,%esi
  movslq %esi,%rax
  mov    %rax,%r12
  imul   $0x55555556,%rax,%rax
  mov    %r12d,%edx
  sar    $0x1f,%edx
  shr    $0x20,%rax
  sub    %edx,%eax
  mov    %r13d,%edx
  lea    (%rax,%rax,2),%r15d
  movslq %r13d,%rax
  sar    $0x1f,%edx
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,2),%eax
  mov    %eax,0xc(%rsp)
  nopl   0x0(%rax,%rax,1)
L32:
  mov    %r13d,%edx
  mov    %r14d,%esi
  mov    %rbp,%rdi
  call   <getAt@plt>
  cmp    %ebx,%eax
  je     L15
  mov    %r14d,%edx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <getAt@plt>
  cmp    %ebx,%eax
  je     L15
  mov    $0xaaaaaaab,%ecx
  mov    %r14d,%eax
  mov    %r14d,%esi
  mov    %rbp,%rdi
  imul   %rcx,%rax
  mov    0xc(%rsp),%ecx
  shr    $0x21,%rax
  lea    (%rax,%rcx,1),%edx
  lea    (%rax,%rax,2),%eax
  sub    %eax,%esi
  add    %r15d,%esi
  call   <getAt@plt>
  cmp    %ebx,%eax
  je     L15
  add    $0x1,%r14d
  cmp    $0x9,%r14d
  jne    L32
  add    $0x18,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int isPossible(int *grid, int index, int value) {
  int boxN = 9 / 3;
  int x = index % 9;
  int y = index / 9;
  int boxX = x - (x % boxN);
  int boxY = y - (y % boxN);
  int i;
  for (i = 0; i < 9; i++) {
    if (getAt(grid, i, y) == value || getAt(grid, x, i) == value || getAt(grid, boxX + (int)(i % boxN), boxY + (int)(i / boxN)) == value) {
      return 0;
    }
  }
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<isPossible>:
  endbr64
  push   %r15
  mov    %esi,%eax
  push   %r14
  sar    $0x1f,%eax
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  movslq %esi,%r12
  imul   $0x38e38e39,%r12,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %edx,%ebx
  sar    $0x21,%r12
  sub    %eax,%r12d
  sub    $0x18,%rsp
  lea    (%r12,%r12,8),%eax
  mov    %r12d,%r13d
  sub    %eax,%esi
  movslq %esi,%rax
  mov    %rax,%r12
  imul   $0x55555556,%rax,%rax
  mov    %r12d,%edx
  sar    $0x1f,%edx
  shr    $0x20,%rax
  sub    %edx,%eax
  mov    %r13d,%edx
  lea    (%rax,%rax,2),%r15d
  movslq %r13d,%rax
  sar    $0x1f,%edx
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,2),%eax
  mov    %eax,0xc(%rsp)
  nopl   0x0(%rax,%rax,1)
L11:
  mov    %r13d,%edx
  mov    %r14d,%esi
  mov    %rbp,%rdi
  call   <getAt@plt>
  cmp    %ebx,%eax
  je     L38
  mov    %r14d,%edx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <getAt@plt>
  cmp    %ebx,%eax
  je     L38
  mov    $0xaaaaaaab,%ecx
  mov    %r14d,%eax
  mov    %r14d,%esi
  mov    %rbp,%rdi
  imul   %rcx,%rax
  mov    0xc(%rsp),%ecx
  shr    $0x21,%rax
  lea    (%rax,%rcx,1),%edx
  lea    (%rax,%rax,2),%eax
  sub    %eax,%esi
  add    %r15d,%esi
  call   <getAt@plt>
  cmp    %ebx,%eax
  je     L38
  add    $0x1,%r14d
  cmp    $0x9,%r14d
  jne    L11
  add    $0x18,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L38:
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int isPossible(int *grid, int index, int value) {
  int boxN = 9 / 3;
  int x = index % 9;
  int y = index / 9;
  int boxX = x - (x % boxN);
  int boxY = y - (y % boxN);
  int i;
  for (i = 0; i < 9; i++) {
    if (getAt(grid, i, y) == value || getAt(grid, x, i) == value || getAt(grid, boxX + (int)(i % boxN), boxY + (int)(i / boxN)) == value) {
      return 0;
    }
  }
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_lexer_encode_unicode_char>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %al,D14(%rip)
  movb   $0x0,D5(%rip)
  lea    D14(%rip),%rax
  pop    %rbp
  ret
```

assistant: ```c
const char *_lexer_encode_unicode_char(long inCodePoint) {
  static char short_string[2];
  short_string[0] = inCodePoint;
  short_string[1] = 0;
  return short_string;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_lexer_encode_unicode_char>:
  endbr64
  mov    %dil,D25(%rip)
  movb   $0x0,D21(%rip)
  lea    D25(%rip),%rax
  ret
```

assistant: ```c
const char *_lexer_encode_unicode_char(long inCodePoint) {
  static char short_string[2];
  short_string[0] = inCodePoint;
  short_string[1] = 0;
  return short_string;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<_lexer_encode_unicode_char>:
  endbr64
  mov    %dil,D27(%rip)
  lea    D27(%rip),%rax
  movb   $0x0,D19(%rip)
  ret
```

assistant: ```c
const char *_lexer_encode_unicode_char(long inCodePoint) {
  static char short_string[2];
  short_string[0] = inCodePoint;
  short_string[1] = 0;
  return short_string;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_lexer_encode_unicode_char>:
  endbr64
  mov    %dil,D7(%rip)
  lea    D7(%rip),%rax
  movb   $0x0,D29(%rip)
  ret
```

assistant: ```c
const char *_lexer_encode_unicode_char(long inCodePoint) {
  static char short_string[2];
  short_string[0] = inCodePoint;
  short_string[1] = 0;
  return short_string;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D14
type: string
name: unpack_data
arguments: tag: D41
type: string
name: unpack_data
arguments: tag: D53
type: string
name: unpack_data
arguments: tag: D35
type: string
name: unpack_data
arguments: tag: D49
type: string
name: unpack_data
arguments: tag: D0
type: string

tool: D14: %s [-hf] [-s string] [-i interger] [-d float]


tool: D41:    -h            print help/usage 

tool: D53:    -f            set the flag

tool: D35:    -s string     pass string to the s-flag

tool: D49:    -i int        passing the interger to the i-flag

tool: D0:    -d float      passing floating point number to the d-flag (d for double)

assistant: ```c
void usage(char *p) {
  printf("%s [-hf] [-s string] [-i interger] [-d float]\n", p);
  printf("   -h            print help/usage \n");
  printf("   -f            set the flag\n");
  printf("   -s string     pass string to the s-flag\n");
  printf("   -i int        passing the interger to the i-flag\n");
  printf("   -d float      passing floating point number to the d-flag (d for double)\n");
}
```