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

```asm
<strcmp>:
  endbr64
  movzbl (%rdi),%edx
  movzbl (%rsi),%ecx
  cmp    %cl,%dl
  jne    L17
  mov    $0x1,%eax
  jmp    L12
  nopl   (%rax)
L35:
  movzbl (%rdi,%rax,1),%edx
  add    $0x1,%rax
  movzbl -0x1(%rsi,%rax,1),%ecx
  cmp    %cl,%dl
  jne    L17
L12:
  test   %dl,%dl
  jne    L35
  xor    %eax,%eax
  ret
L17:
  xor    %eax,%eax
  cmp    %cl,%dl
  setg   %al
  lea    -0x1(%rax,%rax,1),%rax
  ret
```

assistant: ```c
char *strcmp(char *s1, char *s2) {
  int i, j;
  for (i = 0; s1[i] == s2[i]; i++)
    if (s1[i] == '\0' && s2[i] == '\0')
      return 0;
  if (s1[i] > s2[i])
    return 1;
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strcmp>:
  endbr64
  movzbl (%rdi),%edx
  movzbl (%rsi),%ecx
  cmp    %cl,%dl
  jne    L13
  mov    $0x1,%eax
  jmp    L17
  nopl   (%rax)
L21:
  movzbl (%rdi,%rax,1),%edx
  add    $0x1,%rax
  movzbl -0x1(%rsi,%rax,1),%ecx
  cmp    %cl,%dl
  jne    L13
L17:
  test   %dl,%dl
  jne    L21
  xor    %eax,%eax
  ret
L13:
  xor    %eax,%eax
  cmp    %cl,%dl
  setg   %al
  lea    -0x1(%rax,%rax,1),%rax
  ret
```

assistant: ```c
char *strcmp(char *s1, char *s2) {
  int i, j;
  for (i = 0; s1[i] == s2[i]; i++)
    if (s1[i] == '\0' && s2[i] == '\0')
      return 0;
  if (s1[i] > s2[i])
    return 1;
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<allocate_bitmap_with_margins>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  lea    0x2(%rax),%edx
  mov    -0x14(%rbp),%eax
  add    $0x2,%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <allocate_bitmap@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L4
  mov    $0x0,%eax
  jmp    L16
L4:
  movl   $0x0,-0xc(%rbp)
  jmp    L10
L22:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  add    $0x1,%rdx
  mov    %rdx,(%rax)
  addl   $0x1,-0xc(%rbp)
L10:
  mov    -0x18(%rbp),%eax
  add    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jle    L22
  addq   $0x8,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
L16:
  leave
  ret
```

assistant: ```c
unsigned char **allocate_bitmap_with_margins(int w, int h) {
  unsigned char **result = allocate_bitmap(w + 2, h + 2);
  int y;
  if (!result)
    return NULL;
  for (y = 0; y < h + 2; y++) {
    result[y]++;
  }
  result++;
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<allocate_bitmap_with_margins>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  lea    0x2(%rsi),%esi
  add    $0x2,%edi
  call   <allocate_bitmap@plt>
  test   %rax,%rax
  je     L11
  cmp    $0xffffffff,%ebx
  jl     L14
  mov    %rax,%rdx
  lea    0x1(%rbx),%ecx
  lea    0x8(%rax,%rcx,8),%rcx
L28:
  addq   $0x1,(%rdx)
  add    $0x8,%rdx
  cmp    %rcx,%rdx
  jne    L28
L14:
  add    $0x8,%rax
L11:
  pop    %rbx
  ret
```

assistant: ```c
unsigned char **allocate_bitmap_with_margins(int w, int h) {
  unsigned char **result = allocate_bitmap(w + 2, h + 2);
  int y;
  if (!result)
    return NULL;
  for (y = 0; y < h + 2; y++) {
    result[y]++;
  }
  result++;
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<allocate_bitmap_with_margins>:
  endbr64
  push   %rbx
  add    $0x2,%edi
  mov    %esi,%ebx
  add    $0x2,%esi
  call   <allocate_bitmap@plt>
  test   %rax,%rax
  je     L10
  cmp    $0xffffffff,%ebx
  jl     L22
  lea    0x1(%rbx),%ecx
  mov    %rax,%rdx
  lea    0x8(%rax,%rcx,8),%rcx
  nopw   0x0(%rax,%rax,1)
L20:
  addq   $0x1,(%rdx)
  add    $0x8,%rdx
  cmp    %rcx,%rdx
  jne    L20
L22:
  add    $0x8,%rax
L10:
  pop    %rbx
  ret
```

assistant: ```c
unsigned char **allocate_bitmap_with_margins(int w, int h) {
  unsigned char **result = allocate_bitmap(w + 2, h + 2);
  int y;
  if (!result)
    return NULL;
  for (y = 0; y < h + 2; y++) {
    result[y]++;
  }
  result++;
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<allocate_bitmap_with_margins>:
  endbr64
  push   %rbp
  lea    0x2(%rsi),%ebp
  add    $0x2,%edi
  push   %rbx
  mov    %esi,%ebx
  mov    %ebp,%esi
  sub    $0x8,%rsp
  call   <allocate_bitmap@plt>
  test   %rax,%rax
  je     L21
  cmp    $0xffffffff,%ebx
  jl     L3
  je     L44
  mov    %ebp,%ecx
  movdqa D7(%rip),%xmm1
  mov    %rax,%rdx
  shr    %ecx
  shl    $0x4,%rcx
  add    %rax,%rcx
  nopl   0x0(%rax,%rax,1)
L16:
  movdqu (%rdx),%xmm0
  add    $0x10,%rdx
  paddq  %xmm1,%xmm0
  movups %xmm0,-0x10(%rdx)
  cmp    %rcx,%rdx
  jne    L16
  mov    %ebp,%edx
  and    $0xfffffffe,%edx
  and    $0x1,%ebp
  je     L3
L5:
  movslq %edx,%rdx
  addq   $0x1,(%rax,%rdx,8)
L3:
  add    $0x8,%rax
L21:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L44:
  xor    %edx,%edx
  jmp    L5
```

tool_calls: name: parse_data
arguments: data_label: D7
type: byte[16]

tool: {"D7": [0x01,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
unsigned char **allocate_bitmap_with_margins(int w, int h) {
  unsigned char **result = allocate_bitmap(w + 2, h + 2);
  int y;
  if (!result)
    return NULL;
  for (y = 0; y < h + 2; y++) {
    result[y]++;
  }
  result++;
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x1,-0x4(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  1 % 1;
  1 / 1;
  int x = 1;
  1 / x;
  1 % x;
  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
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
  1 % 1;
  1 / 1;
  int x = 1;
  1 / x;
  1 % x;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  1 % 1;
  1 / 1;
  int x = 1;
  1 / x;
  1 % x;
  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
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  1 % 1;
  1 / 1;
  int x = 1;
  1 / x;
  1 % x;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<myAtoi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L79
  mov    $0x0,%eax
  jmp    L60
L79:
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L56
L106:
  addq   $0x1,-0x18(%rbp)
L56:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L106
  movl   $0x1,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  je     L128
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2b,%al
  jne    L23
L128:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  jne    L152
  mov    $0xffffffff,%eax
  jmp    L156
L152:
  mov    $0x1,%eax
L156:
  mov    %eax,-0xc(%rbp)
  addq   $0x1,-0x18(%rbp)
  jmp    L23
L59:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jle    L134
  mov    $0x7fffffff,%eax
  sub    -0x4(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  cmp    %eax,-0x10(%rbp)
  jle    L134
  mov    $0x7fffffff,%eax
  jmp    L60
L134:
  cmpl   $0x0,-0xc(%rbp)
  jns    L146
  mov    -0x10(%rbp),%eax
  neg    %eax
  movslq %eax,%rsi
  mov    -0x4(%rbp),%eax
  cltq
  lea    -0x80000000(%rax),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  cmp    %rax,%rsi
  jge    L146
  mov    $0x80000000,%eax
  jmp    L60
L146:
  mov    -0x10(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x10(%rbp)
  addq   $0x1,-0x18(%rbp)
L23:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L59
  mov    -0x10(%rbp),%eax
  imul   -0xc(%rbp),%eax
L60:
  leave
  ret
```

assistant: ```c
int myAtoi(char *s) {
  if (!s)
    return 0;
  int i = 0;
  int ret = 0;
  for (; isspace(*s); s++)
    ;
  int flag = 1;
  if (*s == '-' || *s == '+') {
    flag = (*s == '-') ? -1 : 1;
    s++;
  }
  for (; isdigit(*s); s++) {
    int digit = *s - '0';
    if (flag > 0 && ret > ((2147483647) - digit) / 10)
      return (2147483647);
    else if (flag < 0 && -ret < ((-2147483648) + digit) / 10)
      return (-2147483648);
    ret = ret * 10 + digit;
  }
  return ret * flag;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<myAtoi>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  test   %rdi,%rdi
  jne    L55
  mov    $0x0,%eax
  jmp    L61
L9:
  add    $0x1,%rbx
L55:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L9
  movzbl (%rbx),%eax
  lea    -0x2b(%rax),%edx
  mov    $0x1,%r12d
  test   $0xfd,%dl
  jne    L88
  cmp    $0x2d,%al
  setne  %r12b
  movzbl %r12b,%r12d
  lea    -0x1(%r12,%r12,1),%r12d
  add    $0x1,%rbx
L88:
  mov    $0x0,%ebp
  movabs $0x6666666666666667,%r13
  jmp    L106
L66:
  jns    L82
  mov    %ebp,%edi
  neg    %edi
  movslq %edi,%rdi
  movslq %ecx,%rsi
  add    $0xffffffff80000000,%rsi
  mov    %rsi,%rax
  imul   %r13
  sar    $0x2,%rdx
  sar    $0x3f,%rsi
  sub    %rsi,%rdx
  cmp    %rdx,%rdi
  jl     L28
L82:
  lea    0x0(%rbp,%rbp,4),%eax
  lea    (%rcx,%rax,2),%ebp
  add    $0x1,%rbx
L106:
  movsbl (%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L70
  movsbl (%rbx),%ecx
  sub    $0x30,%ecx
  test   %r12d,%r12d
  jle    L66
  mov    $0x7fffffff,%edx
  sub    %ecx,%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%edx
  sub    %edx,%eax
  cmp    %ebp,%eax
  jge    L82
  mov    $0x7fffffff,%eax
  jmp    L61
L70:
  mov    %ebp,%eax
  imul   %r12d,%eax
L61:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L28:
  mov    $0x80000000,%eax
  jmp    L61
```

assistant: ```c
int myAtoi(char *s) {
  if (!s)
    return 0;
  int i = 0;
  int ret = 0;
  for (; isspace(*s); s++)
    ;
  int flag = 1;
  if (*s == '-' || *s == '+') {
    flag = (*s == '-') ? -1 : 1;
    s++;
  }
  for (; isdigit(*s); s++) {
    int digit = *s - '0';
    if (flag > 0 && ret > ((2147483647) - digit) / 10)
      return (2147483647);
    else if (flag < 0 && -ret < ((-2147483648) + digit) / 10)
      return (-2147483648);
    ret = ret * 10 + digit;
  }
  return ret * flag;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<myAtoi>:
  endbr64
  test   %rdi,%rdi
  je     L31
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L76
L105:
  add    $0x1,%rbx
L76:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L105
  movsbl (%rbx),%edi
  mov    $0x1,%ebp
  lea    -0x2b(%rdi),%eax
  test   $0xfd,%al
  je     L95
L3:
  xor    %r15d,%r15d
  mov    $0x7fffffff,%r13d
  movabs $0x6666666666666667,%r14
  mov    $0xcccccccd,%r12d
  jmp    L29
  nop
L53:
  cmp    $0xffffffff,%ebp
  jne    L11
  movslq %ecx,%rsi
  mov    %r15d,%edi
  add    $0xffffffff80000000,%rsi
  neg    %edi
  mov    %rsi,%rax
  sar    $0x3f,%rsi
  movslq %edi,%rdi
  imul   %r14
  sar    $0x2,%rdx
  sub    %rsi,%rdx
  cmp    %rdx,%rdi
  jl     L17
L11:
  lea    (%r15,%r15,4),%eax
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  lea    (%rcx,%rax,2),%r15d
L29:
  call   <isdigit@plt>
  test   %rax,%rax
  je     L78
  movsbl (%rbx),%ecx
  sub    $0x30,%ecx
  cmp    $0x1,%ebp
  jne    L53
  mov    %r13d,%eax
  sub    %ecx,%eax
  imul   %r12,%rax
  shr    $0x23,%rax
  cmp    %r15d,%eax
  jge    L11
  add    $0x8,%rsp
  mov    $0x7fffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L95:
  xor    %ebp,%ebp
  cmp    $0x2d,%dil
  movsbl 0x1(%rbx),%edi
  setne  %bpl
  add    $0x1,%rbx
  lea    -0x1(%rbp,%rbp,1),%ebp
  jmp    L3
  nopl   0x0(%rax,%rax,1)
L78:
  add    $0x8,%rsp
  mov    %r15d,%eax
  pop    %rbx
  imul   %ebp,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L31:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L17:
  add    $0x8,%rsp
  mov    $0x80000000,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int myAtoi(char *s) {
  if (!s)
    return 0;
  int i = 0;
  int ret = 0;
  for (; isspace(*s); s++)
    ;
  int flag = 1;
  if (*s == '-' || *s == '+') {
    flag = (*s == '-') ? -1 : 1;
    s++;
  }
  for (; isdigit(*s); s++) {
    int digit = *s - '0';
    if (flag > 0 && ret > ((2147483647) - digit) / 10)
      return (2147483647);
    else if (flag < 0 && -ret < ((-2147483648) + digit) / 10)
      return (-2147483648);
    ret = ret * 10 + digit;
  }
  return ret * flag;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<myAtoi>:
  endbr64
  test   %rdi,%rdi
  je     L41
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  jmp    L6
  cs nopw 0x0(%rax,%rax,1)
L103:
  add    $0x1,%rbx
L6:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L103
  movzbl (%rbx),%eax
  xor    %ebp,%ebp
  lea    -0x2b(%rax),%edx
  and    $0xfd,%edx
  jne    L55
  movabs $0x6666666666666667,%r12
  add    $0x1,%rbx
  cmp    $0x2d,%al
  je     L38
L55:
  movsbl (%rbx),%edi
  mov    $0xcccccccd,%r12d
  call   <isdigit@plt>
  test   %rax,%rax
  je     L49
  nopw   0x0(%rax,%rax,1)
L88:
  movsbl (%rbx),%edx
  mov    $0x7fffffff,%eax
  sub    $0x30,%edx
  sub    %edx,%eax
  imul   %r12,%rax
  shr    $0x23,%rax
  cmp    %eax,%ebp
  jg     L98
  movsbl 0x1(%rbx),%edi
  lea    0x0(%rbp,%rbp,4),%eax
  add    $0x1,%rbx
  lea    (%rdx,%rax,2),%ebp
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L88
L49:
  mov    $0x1,%eax
  jmp    L5
  nopl   (%rax)
L39:
  movsbl (%rbx),%esi
  mov    %ebp,%edi
  neg    %edi
  sub    $0x30,%esi
  movslq %edi,%rdi
  movslq %esi,%rcx
  add    $0xffffffff80000000,%rcx
  mov    %rcx,%rax
  sar    $0x3f,%rcx
  imul   %r12
  sar    $0x2,%rdx
  sub    %rcx,%rdx
  cmp    %rdx,%rdi
  jl     L26
  lea    0x0(%rbp,%rbp,4),%eax
  add    $0x1,%rbx
  lea    (%rsi,%rax,2),%ebp
L38:
  movsbl (%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L39
  mov    $0xffffffff,%eax
L5:
  imul   %ebp,%eax
L37:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L41:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  pop    %rbx
  mov    $0x80000000,%eax
  pop    %rbp
  pop    %r12
  ret
L98:
  mov    $0x7fffffff,%eax
  jmp    L37
```

assistant: ```c
int myAtoi(char *s) {
  if (!s)
    return 0;
  int i = 0;
  int ret = 0;
  for (; isspace(*s); s++)
    ;
  int flag = 1;
  if (*s == '-' || *s == '+') {
    flag = (*s == '-') ? -1 : 1;
    s++;
  }
  for (; isdigit(*s); s++) {
    int digit = *s - '0';
    if (flag > 0 && ret > ((2147483647) - digit) / 10)
      return (2147483647);
    else if (flag < 0 && -ret < ((-2147483648) + digit) / 10)
      return (-2147483648);
    ret = ret * 10 + digit;
  }
  return ret * flag;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print_heap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L1
L0:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  test   %edx,%edx
  jne    L25
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L25:
  addl   $0x1,-0x4(%rbp)
L1:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jle    L0
  nop
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D7
type: string
name: disassemble_data
arguments: data_tag: D6
type: string

tool: D7: a[%d]=%d	

tool: D6: 


assistant: ```c
void print_heap(int *a) {
  int i;
  for (i = 1; i <= a[0]; i++) {
    printf("a[%d]=%d\t", i, a[i]);
    if (!(i % 10))
      puts("\n");
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<print_heap>:
  endbr64
  cmpl   $0x0,(%rdi)
  jle    L25
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    0x4(%rdi),%edx
  mov    $0x1,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x2,%ebx
  lea    D5(%rip),%r12
  lea    D22(%rip),%r13
  jmp    L36
L25:
  ret
L48:
  add    $0x1,%rbx
L36:
  mov    %ebx,%esi
  cmp    %ebx,0x0(%rbp)
  jl     L8
  mov    0x0(%rbp,%rbx,4),%edx
  mov    %r12,%rdi
  call   <printf@plt>
  movslq %ebx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  cmp    %ebx,%eax
  jne    L48
  mov    %r13,%rdi
  call   <puts@plt>
  jmp    L48
L8:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
data_type: string
name: extract_symbolic_data
arguments: data_label: D22
data_type: string

tool: D5: a[%d]=%d	

tool: D22: 


assistant: ```c
void print_heap(int *a) {
  int i;
  for (i = 1; i <= a[0]; i++) {
    printf("a[%d]=%d\t", i, a[i]);
    if (!(i % 10))
      puts("\n");
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<print_heap>:
  endbr64
  mov    (%rdi),%eax
  test   %eax,%eax
  jle    L9
  push   %r13
  mov    $0x1,%esi
  lea    D4(%rip),%r13
  push   %r12
  lea    D15(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    $0x2,%ebx
  sub    $0x8,%rsp
  mov    0x4(%rdi),%edx
  mov    %r12,%rdi
  call   <printf@plt>
  jmp    L29
  nopl   0x0(%rax)
L11:
  mov    0x0(%rbp,%rbx,4),%edx
  mov    %r12,%rdi
  call   <printf@plt>
  imul   $0xcccccccd,%ebx,%eax
  ror    %eax
  cmp    $0x19999999,%eax
  ja     L10
  mov    %r13,%rdi
  call   <puts@plt>
L10:
  add    $0x1,%rbx
L29:
  mov    %ebx,%esi
  cmp    %ebx,0x0(%rbp)
  jge    L11
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L9:
  ret
```

tool_calls: name: retrieve_data
arguments: label: D4
type_guessed: string
name: retrieve_data
arguments: label: D15
type_guessed: string

tool: D4: 


tool: D15: a[%d]=%d	

assistant: ```c
void print_heap(int *a) {
  int i;
  for (i = 1; i <= a[0]; i++) {
    printf("a[%d]=%d\t", i, a[i]);
    if (!(i % 10))
      puts("\n");
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<print_heap>:
  endbr64
  mov    (%rdi),%eax
  test   %eax,%eax
  jle    L44
  push   %r13
  mov    $0x1,%esi
  mov    %rdi,%r13
  push   %r12
  lea    D6(%rip),%r12
  push   %rbp
  lea    D3(%rip),%rbp
  push   %rbx
  mov    $0x2,%ebx
  sub    $0x8,%rsp
  mov    0x4(%rdi),%edx
  mov    %rbp,%rdi
  call   <printf@plt>
  cmpl   $0x1,0x0(%r13)
  jle    L33
  nopl   (%rax)
L42:
  mov    0x0(%r13,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  imul   $0xcccccccd,%ebx,%eax
  ror    %eax
  cmp    $0x19999999,%eax
  jbe    L19
  add    $0x1,%rbx
  cmp    %ebx,0x0(%r13)
  jge    L42
L33:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  mov    %r12,%rdi
  add    $0x1,%rbx
  call   <puts@plt>
  cmp    %ebx,0x0(%r13)
  jge    L42
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L44:
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
data_type: string
name: unpack_data
arguments: data_tag: D3
data_type: string

tool: D6: 


tool: D3: a[%d]=%d	

assistant: ```c
void print_heap(int *a) {
  int i;
  for (i = 1; i <= a[0]; i++) {
    printf("a[%d]=%d\t", i, a[i]);
    if (!(i % 10))
      puts("\n");
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<od_sub>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %rdi,-0x58(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %rdx,-0x68(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x50(%rbp),%rdx
  mov    -0x60(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <od_copy@plt>
  lea    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <od_minus@plt>
  mov    -0x68(%rbp),%rdx
  lea    -0x50(%rbp),%rcx
  mov    -0x58(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <od_add@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
void od_sub(const double *a, const double *b, double *c) {
  double minb[8];
  od_copy(b, minb);
  od_minus(minb);
  od_add(a, minb, c);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<od_sub>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x50,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  mov    %rdx,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  call   <od_copy@plt>
  mov    %rbp,%rdi
  call   <od_minus@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <od_add@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x50,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void od_sub(const double *a, const double *b, double *c) {
  double minb[8];
  od_copy(b, minb);
  od_minus(minb);
  od_add(a, minb, c);
}
```
user: Translate the below assembly code into C programming language:

```asm
<od_sub>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rsi
  call   <od_copy@plt>
  mov    %r12,%rdi
  call   <od_minus@plt>
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <od_add@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x50,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void od_sub(const double *a, const double *b, double *c) {
  double minb[8];
  od_copy(b, minb);
  od_minus(minb);
  od_add(a, minb, c);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<od_sub>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rsi
  call   <od_copy@plt>
  mov    %r12,%rdi
  call   <od_minus@plt>
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <od_add@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x50,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void od_sub(const double *a, const double *b, double *c) {
  double minb[8];
  od_copy(b, minb);
  od_minus(minb);
  od_add(a, minb, c);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<cmd_watch_rssi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1001,-0xc(%rbp)
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <watchoids@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
int cmd_watch_rssi(int argc, char **argv) {
  int oid = 0x1001;
  return watchoids(1, &oid);
}
```