ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<threebyte_memmem>:
  endbr64
  mov    %rdi,%r8
  movzbl 0x1(%rdx),%eax
  movzbl (%rdx),%edi
  shl    $0x10,%eax
  shl    $0x18,%edi
  or     %eax,%edi
  movzbl 0x2(%rdx),%eax
  movzbl (%r8),%edx
  shl    $0x8,%eax
  shl    $0x18,%edx
  or     %eax,%edi
  movzbl 0x1(%r8),%eax
  shl    $0x10,%eax
  or     %eax,%edx
  movzbl 0x2(%r8),%eax
  shl    $0x8,%eax
  or     %eax,%edx
  lea    0x3(%r8),%rax
  cmp    $0x3,%rsi
  je     L19
  cmp    %edx,%edi
  je     L29
  lea    0x4(%r8),%rcx
  add    %rsi,%r8
  jmp    L30
  nopl   0x0(%rax)
L36:
  add    $0x1,%rcx
  cmp    %edx,%edi
  je     L49
L30:
  movzbl -0x1(%rcx),%esi
  mov    %rcx,%rax
  or     %esi,%edx
  shl    $0x8,%edx
  cmp    %r8,%rcx
  jne    L36
L19:
  sub    $0x3,%rax
  cmp    %edx,%edi
  mov    $0x0,%edx
  cmovne %rdx,%rax
  ret
  nopl   0x0(%rax)
L49:
  sub    $0x3,%rax
  ret
L29:
  mov    %r8,%rax
  ret
```

assistant: ```c
char *threebyte_memmem(const unsigned char *h, size_t k, const unsigned char *n) {
  uint32_t nw = (uint32_t)n[0] << 24 | n[1] << 16 | n[2] << 8;
  uint32_t hw = (uint32_t)h[0] << 24 | h[1] << 16 | h[2] << 8;
  for (h += 3, k -= 3; k; k--, hw = (hw | *h++) << 8)
    if (hw == nw)
      return (char *)h - 3;
  return hw == nw ? (char *)h - 3 : 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<threebyte_memmem>:
  endbr64
  mov    %rdi,%r8
  movzbl 0x1(%rdx),%eax
  movzbl (%rdx),%edi
  shl    $0x10,%eax
  shl    $0x18,%edi
  or     %eax,%edi
  movzbl 0x2(%rdx),%eax
  movzbl (%r8),%edx
  shl    $0x8,%eax
  shl    $0x18,%edx
  or     %eax,%edi
  movzbl 0x1(%r8),%eax
  shl    $0x10,%eax
  or     %eax,%edx
  movzbl 0x2(%r8),%eax
  shl    $0x8,%eax
  or     %eax,%edx
  lea    0x3(%r8),%rax
  cmp    $0x3,%rsi
  je     L7
  cmp    %edx,%edi
  je     L33
  lea    0x4(%r8),%rcx
  add    %rsi,%r8
  jmp    L22
  nopl   0x0(%rax)
L6:
  add    $0x1,%rcx
  cmp    %edx,%edi
  je     L15
L22:
  movzbl -0x1(%rcx),%esi
  mov    %rcx,%rax
  or     %esi,%edx
  shl    $0x8,%edx
  cmp    %rcx,%r8
  jne    L6
L7:
  sub    $0x3,%rax
  cmp    %edx,%edi
  mov    $0x0,%edx
  cmovne %rdx,%rax
  ret
  nopl   0x0(%rax)
L15:
  sub    $0x3,%rax
  ret
L33:
  mov    %r8,%rax
  ret
```

assistant: ```c
char *threebyte_memmem(const unsigned char *h, size_t k, const unsigned char *n) {
  uint32_t nw = (uint32_t)n[0] << 24 | n[1] << 16 | n[2] << 8;
  uint32_t hw = (uint32_t)h[0] << 24 | h[1] << 16 | h[2] << 8;
  for (h += 3, k -= 3; k; k--, hw = (hw | *h++) << 8)
    if (hw == nw)
      return (char *)h - 3;
  return hw == nw ? (char *)h - 3 : 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<_cycalloc_atomic>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <GC_malloc_atomic@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L9
  mov    $0x0,%eax
  call   <_throw_badalloc@plt>
L9:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *_cycalloc_atomic(int n) {
  void *ans = (void *)GC_malloc_atomic(n);
  if (!ans)
    _throw_badalloc();
  return ans;
}
```
user: What is the c source code of the assembly code below:

```asm
<_cycalloc_atomic>:
  endbr64
  push   %rbx
  call   <GC_malloc_atomic@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L7
L12:
  mov    %rbx,%rax
  pop    %rbx
  ret
L7:
  mov    $0x0,%eax
  call   <_throw_badalloc@plt>
  jmp    L12
```

assistant: ```c
void *_cycalloc_atomic(int n) {
  void *ans = (void *)GC_malloc_atomic(n);
  if (!ans)
    _throw_badalloc();
  return ans;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<_cycalloc_atomic>:
  endbr64
  push   %r12
  call   <GC_malloc_atomic@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L2
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L2:
  xor    %eax,%eax
  call   <_throw_badalloc@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
void *_cycalloc_atomic(int n) {
  void *ans = (void *)GC_malloc_atomic(n);
  if (!ans)
    _throw_badalloc();
  return ans;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<_cycalloc_atomic>:
  endbr64
  push   %r12
  call   <GC_malloc_atomic@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L5
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L5:
  xor    %eax,%eax
  call   <_throw_badalloc@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
void *_cycalloc_atomic(int n) {
  void *ans = (void *)GC_malloc_atomic(n);
  if (!ans)
    _throw_badalloc();
  return ans;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<find_count_of_char>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
L16:
  movsbl -0x1c(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L17
  addl   $0x1,-0xc(%rbp)
  addq   $0x1,-0x8(%rbp)
L17:
  cmpq   $0x0,-0x8(%rbp)
  jne    L16
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
int find_count_of_char(const char *str, char ch) {
  const char *find_ptr = str;
  int times = 0;
  do {
    find_ptr = strchr(find_ptr, ch);
    if (find_ptr) {
      times++;
      find_ptr++;
    }
  } while (find_ptr != NULL);
  return times;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<find_count_of_char>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%ebx
  movsbl %sil,%ebp
L16:
  mov    %ebp,%esi
  call   <strchr@plt>
  test   %rax,%rax
  je     L0
  add    $0x1,%ebx
  lea    0x1(%rax),%rdi
  jmp    L16
L0:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int find_count_of_char(const char *str, char ch) {
  const char *find_ptr = str;
  int times = 0;
  do {
    find_ptr = strchr(find_ptr, ch);
    if (find_ptr) {
      times++;
      find_ptr++;
    }
  } while (find_ptr != NULL);
  return times;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<find_count_of_char>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbx
  movsbl %sil,%ebx
  mov    %ebx,%esi
  sub    $0x8,%rsp
  call   <strchr@plt>
  test   %rax,%rax
  je     L12
  xchg   %ax,%ax
L8:
  lea    0x1(%rax),%rdi
  mov    %ebx,%esi
  add    $0x1,%r12d
  call   <strchr@plt>
  test   %rax,%rax
  jne    L8
L12:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
int find_count_of_char(const char *str, char ch) {
  const char *find_ptr = str;
  int times = 0;
  do {
    find_ptr = strchr(find_ptr, ch);
    if (find_ptr) {
      times++;
      find_ptr++;
    }
  } while (find_ptr != NULL);
  return times;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<find_count_of_char>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbx
  movsbl %sil,%ebx
  mov    %ebx,%esi
  sub    $0x8,%rsp
  call   <strchr@plt>
  test   %rax,%rax
  je     L7
  xchg   %ax,%ax
L9:
  lea    0x1(%rax),%rdi
  mov    %ebx,%esi
  add    $0x1,%r12d
  call   <strchr@plt>
  test   %rax,%rax
  jne    L9
L7:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
int find_count_of_char(const char *str, char ch) {
  const char *find_ptr = str;
  int times = 0;
  do {
    find_ptr = strchr(find_ptr, ch);
    if (find_ptr) {
      times++;
      find_ptr++;
    }
  } while (find_ptr != NULL);
  return times;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<step>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void step(unsigned int *x, unsigned int *y, unsigned int *mv) {
  y[*mv] = *y;
  x[*mv] = *x;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<step>:
  endbr64
  mov    (%rsi),%ecx
  mov    (%rdx),%eax
  mov    %ecx,(%rsi,%rax,4)
  mov    (%rdi),%ecx
  mov    (%rdx),%eax
  mov    %ecx,(%rdi,%rax,4)
  ret
```

assistant: ```c
void step(unsigned int *x, unsigned int *y, unsigned int *mv) {
  y[*mv] = *y;
  x[*mv] = *x;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<step>:
  endbr64
  mov    (%rsi),%ecx
  mov    (%rdx),%eax
  mov    %ecx,(%rsi,%rax,4)
  mov    (%rdi),%ecx
  mov    (%rdx),%eax
  mov    %ecx,(%rdi,%rax,4)
  ret
```

assistant: ```c
void step(unsigned int *x, unsigned int *y, unsigned int *mv) {
  y[*mv] = *y;
  x[*mv] = *x;
}
```
user: Translate the below assembly code into C programming language:

```asm
<step>:
  endbr64
  mov    (%rsi),%ecx
  mov    (%rdx),%eax
  mov    %ecx,(%rsi,%rax,4)
  mov    (%rdi),%ecx
  mov    (%rdx),%eax
  mov    %ecx,(%rdi,%rax,4)
  ret
```

assistant: ```c
void step(unsigned int *x, unsigned int *y, unsigned int *mv) {
  y[*mv] = *y;
  x[*mv] = *x;
}
```
user: Translate the below assembly code into C programming language:

```asm
<dcd11>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x30(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x2a,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0xf,-0x14(%rbp)
  jbe    L17
  mov    $0x0,%eax
  jmp    L22
L17:
  mov    -0x14(%rbp),%eax
  shl    $0x1e,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  shl    $0x1c,%eax
  and    $0xc0000000,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L32
L59:
  movl   $0x0,-0x18(%rbp)
  jmp    L55
L11:
  mov    -0x1c(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %eax,%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x3b,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x3f,-0x14(%rbp)
  jbe    L73
  mov    $0x0,%eax
  jmp    L22
L73:
  mov    -0x1c(%rbp),%eax
  cltq
  mov    -0x10(%rbp,%rax,4),%esi
  mov    -0x18(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%ecx
  shl    %cl,%eax
  or     %eax,%esi
  mov    %esi,%edx
  mov    -0x1c(%rbp),%eax
  cltq
  mov    %edx,-0x10(%rbp,%rax,4)
  addl   $0x1,-0x18(%rbp)
L55:
  cmpl   $0x4,-0x18(%rbp)
  jle    L11
  addl   $0x1,-0x1c(%rbp)
L32:
  cmpl   $0x1,-0x1c(%rbp)
  jle    L59
  lea    -0x10(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    $0x8,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    $0x1,%eax
L22:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L86
  call   <__stack_chk_fail@plt>
L86:
  leave
  ret
```

assistant: ```c
int dcd11(double *to, const char *s) {
  unsigned int qw[2], k = s[0] - 42u;
  if (k > 15u)
    return 0;
  qw[0] = (k & 3u) << 30;
  qw[1] = (k & 12u) << 28;
  int i, j;
  for (i = 0; i < 2; i++) {
    for (j = 0; j < 5; j++) {
      if ((k = s[5 * i + j + 1] - 59) > 63u)
        return 0;
      else
        qw[i] |= k << (6 * j);
    }
  }
  return memcpy(to, qw, 8), 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<dcd11>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movsbl (%rsi),%eax
  sub    $0x2a,%eax
  mov    $0x0,%edx
  cmp    $0xf,%eax
  ja     L23
  mov    %eax,%r8d
  shl    $0x1e,%r8d
  shl    $0x1c,%eax
  and    $0xc0000000,%eax
  mov    %eax,0x4(%rsp)
  lea    0x1(%rsi),%rdx
  mov    $0x0,%ecx
L26:
  movsbl (%rdx),%eax
  sub    $0x3b,%eax
  cmp    $0x3f,%eax
  ja     L19
  shl    %cl,%eax
  or     %eax,%r8d
  add    $0x1,%rdx
  add    $0x6,%ecx
  cmp    $0x1e,%ecx
  jne    L26
  mov    %r8d,(%rsp)
  add    $0x6,%rsi
  mov    $0x0,%ecx
L31:
  movsbl (%rsi),%eax
  sub    $0x3b,%eax
  cmp    $0x3f,%eax
  ja     L19
  shl    %cl,%eax
  or     %eax,0x4(%rsp)
  add    $0x1,%rsi
  add    $0x6,%ecx
  cmp    $0x1e,%ecx
  jne    L31
  mov    %rsp,%rsi
  mov    $0x8,%edx
  call   <memcpy@plt>
  mov    $0x1,%edx
L23:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  mov    %edx,%eax
  add    $0x18,%rsp
  ret
L19:
  mov    $0x0,%edx
  jmp    L23
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dcd11(double *to, const char *s) {
  unsigned int qw[2], k = s[0] - 42u;
  if (k > 15u)
    return 0;
  qw[0] = (k & 3u) << 30;
  qw[1] = (k & 12u) << 28;
  int i, j;
  for (i = 0; i < 2; i++) {
    for (j = 0; j < 5; j++) {
      if ((k = s[5 * i + j + 1] - 59) > 63u)
        return 0;
      else
        qw[i] |= k << (6 * j);
    }
  }
  return memcpy(to, qw, 8), 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dcd11>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movsbl (%rsi),%eax
  sub    $0x2a,%eax
  cmp    $0xf,%eax
  ja     L55
  mov    %eax,%r8d
  shl    $0x1c,%eax
  lea    0x1(%rsi),%rdx
  xor    %ecx,%ecx
  and    $0xc0000000,%eax
  shl    $0x1e,%r8d
  mov    %eax,0x4(%rsp)
L2:
  movsbl (%rdx),%eax
  sub    $0x3b,%eax
  cmp    $0x3f,%eax
  ja     L55
  shl    %cl,%eax
  add    $0x6,%ecx
  add    $0x1,%rdx
  or     %eax,%r8d
  cmp    $0x1e,%ecx
  jne    L2
  mov    %r8d,(%rsp)
  add    $0x6,%rsi
  xor    %ecx,%ecx
L50:
  movsbl (%rsi),%eax
  sub    $0x3b,%eax
  cmp    $0x3f,%eax
  ja     L55
  shl    %cl,%eax
  add    $0x6,%ecx
  or     %eax,0x4(%rsp)
  add    $0x1,%rsi
  cmp    $0x1e,%ecx
  jne    L50
  mov    %rsp,%rsi
  mov    $0x8,%edx
  call   <memcpy@plt>
  mov    $0x1,%eax
  jmp    L31
  nopl   0x0(%rax,%rax,1)
L55:
  xor    %eax,%eax
L31:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L51
  add    $0x18,%rsp
  ret
L51:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dcd11(double *to, const char *s) {
  unsigned int qw[2], k = s[0] - 42u;
  if (k > 15u)
    return 0;
  qw[0] = (k & 3u) << 30;
  qw[1] = (k & 12u) << 28;
  int i, j;
  for (i = 0; i < 2; i++) {
    for (j = 0; j < 5; j++) {
      if ((k = s[5 * i + j + 1] - 59) > 63u)
        return 0;
      else
        qw[i] |= k << (6 * j);
    }
  }
  return memcpy(to, qw, 8), 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<dcd11>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rdx
  mov    %rdx,0x8(%rsp)
  xor    %edx,%edx
  movsbl (%rsi),%edx
  sub    $0x2a,%edx
  cmp    $0xf,%edx
  ja     L117
  mov    %rsi,%rax
  mov    %edx,%esi
  shl    $0x1c,%edx
  movsbl 0x1(%rax),%ecx
  and    $0xc0000000,%edx
  shl    $0x1e,%esi
  mov    %edx,0x4(%rsp)
  sub    $0x3b,%ecx
  cmp    $0x3f,%ecx
  ja     L117
  or     %ecx,%esi
  movsbl 0x2(%rax),%ecx
  sub    $0x3b,%ecx
  cmp    $0x3f,%ecx
  ja     L117
  shl    $0x6,%ecx
  or     %esi,%ecx
  movsbl 0x3(%rax),%esi
  sub    $0x3b,%esi
  cmp    $0x3f,%esi
  ja     L117
  shl    $0xc,%esi
  or     %ecx,%esi
  movsbl 0x4(%rax),%ecx
  sub    $0x3b,%ecx
  cmp    $0x3f,%ecx
  ja     L117
  shl    $0x12,%ecx
  or     %esi,%ecx
  movsbl 0x5(%rax),%esi
  sub    $0x3b,%esi
  cmp    $0x3f,%esi
  ja     L117
  shl    $0x18,%esi
  or     %ecx,%esi
  mov    %esi,(%rsp)
  movsbl 0x6(%rax),%esi
  sub    $0x3b,%esi
  cmp    $0x3f,%esi
  ja     L117
  movsbl 0x7(%rax),%ecx
  sub    $0x3b,%ecx
  cmp    $0x3f,%ecx
  ja     L117
  shl    $0x6,%ecx
  or     %edx,%ecx
  movsbl 0x8(%rax),%edx
  or     %esi,%ecx
  sub    $0x3b,%edx
  cmp    $0x3f,%edx
  ja     L117
  shl    $0xc,%edx
  or     %ecx,%edx
  movsbl 0x9(%rax),%ecx
  sub    $0x3b,%ecx
  cmp    $0x3f,%ecx
  ja     L117
  movsbl 0xa(%rax),%eax
  shl    $0x12,%ecx
  or     %edx,%ecx
  sub    $0x3b,%eax
  cmp    $0x3f,%eax
  ja     L117
  shl    $0x18,%eax
  mov    %rsp,%rsi
  mov    $0x8,%edx
  or     %ecx,%eax
  mov    %eax,0x4(%rsp)
  call   <memcpy@plt>
  mov    $0x1,%eax
  jmp    L127
  nopl   0x0(%rax,%rax,1)
L117:
  xor    %eax,%eax
L127:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L77
  add    $0x18,%rsp
  ret
L77:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dcd11(double *to, const char *s) {
  unsigned int qw[2], k = s[0] - 42u;
  if (k > 15u)
    return 0;
  qw[0] = (k & 3u) << 30;
  qw[1] = (k & 12u) << 28;
  int i, j;
  for (i = 0; i < 2; i++) {
    for (j = 0; j < 5; j++) {
      if ((k = s[5 * i + j + 1] - 59) > 63u)
        return 0;
      else
        qw[i] |= k << (6 * j);
    }
  }
  return memcpy(to, qw, 8), 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<Se>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <Ue@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jle    L8
  mov    -0x4(%rbp),%eax
  and    $0x1,%eax
  add    %eax,-0x8(%rbp)
L8:
  mov    -0x4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L5
  negl   -0x8(%rbp)
L5:
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int Se(unsigned char *pBuff, unsigned int nLen, unsigned int *nStartBit) {
  int UeVal = Ue(pBuff, nLen, nStartBit);
  int nValue = UeVal / 2;
  if (nValue > 0)
    nValue += UeVal & 0x01;
  if (UeVal % 2 == 0)
    nValue = -nValue;
  return nValue;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<Se>:
  endbr64
  sub    $0x8,%rsp
  call   <Ue@plt>
  mov    %eax,%ecx
  shr    $0x1f,%eax
  add    %ecx,%eax
  sar    %eax
  mov    %ecx,%edx
  and    $0x1,%edx
  add    %eax,%edx
  cmp    $0x1,%ecx
  cmovg  %edx,%eax
  mov    %eax,%edx
  neg    %edx
  test   $0x1,%cl
  cmove  %edx,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int Se(unsigned char *pBuff, unsigned int nLen, unsigned int *nStartBit) {
  int UeVal = Ue(pBuff, nLen, nStartBit);
  int nValue = UeVal / 2;
  if (nValue > 0)
    nValue += UeVal & 0x01;
  if (UeVal % 2 == 0)
    nValue = -nValue;
  return nValue;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Se>:
  endbr64
  sub    $0x8,%rsp
  call   <Ue@plt>
  mov    %eax,%ecx
  mov    %eax,%esi
  shr    $0x1f,%ecx
  and    $0x1,%esi
  add    %eax,%ecx
  sar    %ecx
  cmp    $0x1,%eax
  lea    (%rcx,%rsi,1),%edx
  mov    %ecx,%eax
  cmovg  %edx,%eax
  mov    %eax,%edx
  neg    %edx
  test   %esi,%esi
  cmove  %edx,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int Se(unsigned char *pBuff, unsigned int nLen, unsigned int *nStartBit) {
  int UeVal = Ue(pBuff, nLen, nStartBit);
  int nValue = UeVal / 2;
  if (nValue > 0)
    nValue += UeVal & 0x01;
  if (UeVal % 2 == 0)
    nValue = -nValue;
  return nValue;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<Se>:
  endbr64
  sub    $0x8,%rsp
  call   <Ue@plt>
  mov    %eax,%ecx
  mov    %eax,%esi
  shr    $0x1f,%ecx
  and    $0x1,%esi
  add    %eax,%ecx
  sar    %ecx
  cmp    $0x1,%eax
  lea    (%rcx,%rsi,1),%edx
  mov    %ecx,%eax
  cmovg  %edx,%eax
  mov    %eax,%edx
  neg    %edx
  test   %esi,%esi
  cmove  %edx,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int Se(unsigned char *pBuff, unsigned int nLen, unsigned int *nStartBit) {
  int UeVal = Ue(pBuff, nLen, nStartBit);
  int nValue = UeVal / 2;
  if (nValue > 0)
    nValue += UeVal & 0x01;
  if (UeVal % 2 == 0)
    nValue = -nValue;
  return nValue;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<test_x>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x5,%edi
  call   <x@plt>
  mov    $0x5,%edi
  call   <x2@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void test_x() {
  x(5);
  x2(5);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<test_x>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%edi
  call   <x@plt>
  mov    $0x5,%edi
  call   <x2@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void test_x() {
  x(5);
  x2(5);
}
```
user: What is the c source code of the assembly code below?

```asm
<test_x>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%edi
  call   <x@plt>
  mov    $0x5,%edi
  add    $0x8,%rsp
  jmp    1070 <x2@plt>
```

assistant: ```c
void test_x() {
  x(5);
  x2(5);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<test_x>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%edi
  call   <x@plt>
  mov    $0x5,%edi
  add    $0x8,%rsp
  jmp    1070 <x2@plt>
```

assistant: ```c
void test_x() {
  x(5);
  x2(5);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x1300,-0x18(%rbp)
  movq   $0xd82,-0x10(%rbp)
  lea    -0x30(%rbp),%rdi
  lea    -0x20(%rbp),%rcx
  lea    -0x28(%rbp),%rdx
  mov    -0x10(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <extended_euclid@plt>
  mov    -0x30(%rbp),%rcx
  mov    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D6
type: string

tool: D6: x = %ld y = %ld d = %ld


assistant: ```c
int main(void) {
  long a = 4864, b = 3458, d, x, y;
  extended_euclid(a, b, &x, &y, &d);
  printf("x = %ld y = %ld d = %ld\n", x, y, d);
  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
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  mov    %rsp,%r8
  mov    $0xd82,%esi
  mov    $0x1300,%edi
  call   <extended_euclid@plt>
  mov    (%rsp),%rcx
  mov    0x10(%rsp),%rdx
  mov    0x8(%rsp),%rsi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D6
type_infered: string

tool: D6: x = %ld y = %ld d = %ld


assistant: ```c
int main(void) {
  long a = 4864, b = 3458, d, x, y;
  extended_euclid(a, b, &x, &y, &d);
  printf("x = %ld y = %ld d = %ld\n", x, y, d);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    $0xd82,%esi
  mov    $0x1300,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  mov    %rsp,%r8
  call   <extended_euclid@plt>
  mov    (%rsp),%rcx
  mov    0x10(%rsp),%rdx
  lea    D1(%rip),%rdi
  mov    0x8(%rsp),%rsi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D1
data_type: string

tool: D1: x = %ld y = %ld d = %ld


assistant: ```c
int main(void) {
  long a = 4864, b = 3458, d, x, y;
  extended_euclid(a, b, &x, &y, &d);
  printf("x = %ld y = %ld d = %ld\n", x, y, d);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    $0xd82,%esi
  mov    $0x1300,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  mov    %rsp,%r8
  call   <extended_euclid@plt>
  mov    (%rsp),%rcx
  mov    0x10(%rsp),%rdx
  lea    D8(%rip),%rdi
  mov    0x8(%rsp),%rsi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D8
type_infered: string

tool: D8: x = %ld y = %ld d = %ld


assistant: ```c
int main(void) {
  long a = 4864, b = 3458, d, x, y;
  extended_euclid(a, b, &x, &y, &d);
  printf("x = %ld y = %ld d = %ld\n", x, y, d);
  return 0;
}
```