ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<reverse_endianness>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  mov    %rsi,%rdi
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  call   <safe_malloc@plt>
  mov    %r13,%rcx
  mov    %rax,%rbp
  lea    -0x1(%rax,%r12,1),%rax
  test   %r12,%r12
  je     L10
  nopw   0x0(%rax,%rax,1)
L3:
  movzbl (%rcx),%r8d
  add    $0x1,%rcx
  mov    %r8b,(%rax)
  mov    %rax,%r8
  sub    $0x1,%rax
  cmp    %rbp,%r8
  jne    L3
L10:
  mov    %r13,%rdi
  mov    %r12,%rdx
  mov    %rbp,%rsi
  call   <memcpy@plt>
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <free@plt>
```

assistant: ```c
void reverse_endianness(void *ptr, size_t buffsize) {
  char *buffer;
  char *temp;
  int i;
  buffer = (char *)ptr;
  temp = safe_malloc(buffsize);
  for (i = 0; i < buffsize; i++) {
    temp[(buffsize - 1) - i] = buffer[i];
  }
  memcpy(buffer, temp, buffsize);
  free(temp);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<reverse_endianness>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  mov    %rsi,%rdi
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  call   <safe_malloc@plt>
  mov    %r13,%rcx
  mov    %rax,%rbp
  lea    -0x1(%rax,%r12,1),%rax
  test   %r12,%r12
  je     L10
  nopw   0x0(%rax,%rax,1)
L15:
  movzbl (%rcx),%r8d
  add    $0x1,%rcx
  mov    %r8b,(%rax)
  mov    %rax,%r8
  sub    $0x1,%rax
  cmp    %rbp,%r8
  jne    L15
L10:
  mov    %r13,%rdi
  mov    %r12,%rdx
  mov    %rbp,%rsi
  call   <memcpy@plt>
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <free@plt>
```

assistant: ```c
void reverse_endianness(void *ptr, size_t buffsize) {
  char *buffer;
  char *temp;
  int i;
  buffer = (char *)ptr;
  temp = safe_malloc(buffsize);
  for (i = 0; i < buffsize; i++) {
    temp[(buffsize - 1) - i] = buffer[i];
  }
  memcpy(buffer, temp, buffsize);
  free(temp);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xff7a7a7a,-0x10(%rbp)
  movl   $0xff7a7a7a,-0xc(%rbp)
  movl   $0xff7b7b7b,-0x18(%rbp)
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <Predictor@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jne    L5
  mov    $0x0,%eax
  jmp    L18
L5:
  mov    $0x1,%eax
L18:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

assistant: ```c
int main(void) {
  const unsigned top[2] = {0xff7a7a7a, 0xff7a7a7a};
  const unsigned left = 0xff7b7b7b;
  const unsigned pred = Predictor(left, top);
  if (pred == left)
    return 0;
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0xff7a7a7a,(%rsp)
  movl   $0xff7a7a7a,0x4(%rsp)
  mov    %rsp,%rsi
  mov    $0xff7b7b7b,%edi
  call   <Predictor@plt>
  cmp    $0xff7b7b7b,%eax
  setne  %al
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  movzbl %al,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  const unsigned top[2] = {0xff7a7a7a, 0xff7a7a7a};
  const unsigned left = 0xff7b7b7b;
  const unsigned pred = Predictor(left, top);
  if (pred == left)
    return 0;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    $0xff7b7b7b,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  movabs $0xff7a7a7aff7a7a7a,%rax
  mov    %rsp,%rsi
  mov    %rax,(%rsp)
  call   <Predictor@plt>
  cmp    $0xff7b7b7b,%eax
  setne  %al
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  movzbl %al,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  const unsigned top[2] = {0xff7a7a7a, 0xff7a7a7a};
  const unsigned left = 0xff7b7b7b;
  const unsigned pred = Predictor(left, top);
  if (pred == left)
    return 0;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    $0xff7b7b7b,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D9(%rip),%rax
  mov    %rsp,%rsi
  mov    %rax,(%rsp)
  call   <Predictor@plt>
  cmp    $0xff7b7b7b,%eax
  setne  %al
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  movzbl %al,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D9
type_guessed: byte[8]

tool: {"D9": [0x7a,0x7a,0x7a,0xff,0x7a,0x7a,0x7a,0xff]}

assistant: ```c
int main(void) {
  const unsigned top[2] = {0xff7a7a7a, 0xff7a7a7a};
  const unsigned left = 0xff7b7b7b;
  const unsigned pred = Predictor(left, top);
  if (pred == left)
    return 0;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```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   <rand@plt>
  movslq %eax,%rdx
  imul   $0xffffffff88888889,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x3,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,-0xc(%rbp)
  mov    -0xc(%rbp),%ecx
  mov    %ecx,%edx
  shl    $0x4,%edx
  sub    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0xa,-0x14(%rbp)
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L18:
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jne    L58
  movl   $0x1,-0x10(%rbp)
L58:
  subl   $0x1,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jle    L33
  cmpl   $0x0,-0x10(%rbp)
  je     L18
L33:
  cmpl   $0x1,-0x10(%rbp)
  jne    L46
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L36
L46:
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L36:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L45
  call   <__stack_chk_fail@plt>
L45:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D24
type: string
name: retrieve_data
arguments: data_label: D21
type: string
name: retrieve_data
arguments: data_label: D17
type: string
name: retrieve_data
arguments: data_label: D41
type: string
name: retrieve_data
arguments: data_label: D27
type: string

tool: D24: il pc ha pensato un numero da 1 a 14

tool: D21: 
Indovinalo con 10 tentativi:

tool: D17: %d

tool: D41: Hai vinto

tool: D27: Hai perso

assistant: ```c
int main() {
  int n_u;
  int t;
  int n_c;
  int ind;
  n_c = rand() % 15;
  ind = 0;
  t = 10;
  printf("il pc ha pensato un numero da 1 a 14");
  do {
    printf("\nIndovinalo con 10 tentativi:");
    scanf("%d", &n_u);
    if (n_u == n_c) {
      ind = 1;
    }
    t--;
  } while (t > 0 && ind == 0);
  if (ind == 1) {
    printf("Hai vinto");
  } else {
    printf("Hai perso");
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rbp
  imul   $0xffffffff88888889,%rbp,%rbp
  shr    $0x20,%rbp
  add    %eax,%ebp
  sar    $0x3,%ebp
  cltd
  sub    %edx,%ebp
  mov    %ebp,%edx
  shl    $0x4,%edx
  sub    %ebp,%edx
  sub    %edx,%eax
  mov    %eax,%ebp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0xa,%ebx
  lea    D49(%rip),%r14
  lea    0x4(%rsp),%r13
  lea    D22(%rip),%r12
L2:
  mov    %r14,%rdi
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  sub    $0x1,%ebx
  test   %ebx,%ebx
  jle    L32
  cmp    %ebp,%eax
  jne    L2
L32:
  cmp    %ebp,%eax
  je     L27
  lea    D6(%rip),%rdi
  call   <printf@plt>
L16:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L27:
  lea    D34(%rip),%rdi
  call   <printf@plt>
  jmp    L16
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D0
data_type: string
name: read_data
arguments: data_label: D49
data_type: string
name: read_data
arguments: data_label: D22
data_type: string
name: read_data
arguments: data_label: D6
data_type: string
name: read_data
arguments: data_label: D34
data_type: string

tool: D0: il pc ha pensato un numero da 1 a 14

tool: D49: 
Indovinalo con 10 tentativi:

tool: D22: %d

tool: D6: Hai perso

tool: D34: Hai vinto

assistant: ```c
int main() {
  int n_u;
  int t;
  int n_c;
  int ind;
  n_c = rand() % 15;
  ind = 0;
  t = 10;
  printf("il pc ha pensato un numero da 1 a 14");
  do {
    printf("\nIndovinalo con 10 tentativi:");
    scanf("%d", &n_u);
    if (n_u == n_c) {
      ind = 1;
    }
    t--;
  } while (t > 0 && ind == 0);
  if (ind == 1) {
    printf("Hai vinto");
  } else {
    printf("Hai perso");
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  lea    D25(%rip),%r14
  push   %r13
  push   %r12
  lea    D3(%rip),%r12
  push   %rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
  call   <rand@plt>
  lea    D10(%rip),%rdi
  movslq %eax,%rbp
  cltd
  imul   $0xffffffff88888889,%rbp,%rbp
  shr    $0x20,%rbp
  add    %eax,%ebp
  sar    $0x3,%ebp
  sub    %edx,%ebp
  mov    %ebp,%edx
  shl    $0x4,%edx
  sub    %ebp,%edx
  sub    %edx,%eax
  mov    %eax,%ebp
  call   <printf@plt>
  jmp    L9
  nopl   0x0(%rax)
L44:
  sub    $0x1,%ebx
  je     L36
L9:
  mov    %r14,%rdi
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmp    %ebp,0x4(%rsp)
  jne    L44
  lea    D7(%rip),%rdi
  call   <printf@plt>
  jmp    L14
L36:
  lea    D6(%rip),%rdi
  call   <printf@plt>
L14:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D25
datatype: string
name: parse_data
arguments: label: D3
datatype: string
name: parse_data
arguments: label: D10
datatype: string
name: parse_data
arguments: label: D7
datatype: string
name: parse_data
arguments: label: D6
datatype: string

tool: D25: 
Indovinalo con 10 tentativi:

tool: D3: %d

tool: D10: il pc ha pensato un numero da 1 a 14

tool: D7: Hai vinto

tool: D6: Hai perso

assistant: ```c
int main() {
  int n_u;
  int t;
  int n_c;
  int ind;
  n_c = rand() % 15;
  ind = 0;
  t = 10;
  printf("il pc ha pensato un numero da 1 a 14");
  do {
    printf("\nIndovinalo con 10 tentativi:");
    scanf("%d", &n_u);
    if (n_u == n_c) {
      ind = 1;
    }
    t--;
  } while (t > 0 && ind == 0);
  if (ind == 1) {
    printf("Hai vinto");
  } else {
    printf("Hai perso");
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r14
  lea    D21(%rip),%r14
  push   %r13
  push   %r12
  lea    D38(%rip),%r12
  push   %rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
  call   <rand@plt>
  lea    D15(%rip),%rdi
  movslq %eax,%rbp
  cltd
  imul   $0xffffffff88888889,%rbp,%rbp
  shr    $0x20,%rbp
  add    %eax,%ebp
  sar    $0x3,%ebp
  sub    %edx,%ebp
  mov    %ebp,%edx
  shl    $0x4,%edx
  sub    %ebp,%edx
  sub    %edx,%eax
  mov    %eax,%ebp
  call   <printf@plt>
  jmp    L9
  nopl   0x0(%rax)
L37:
  sub    $0x1,%ebx
  je     L27
L9:
  mov    %r14,%rdi
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmp    %ebp,0x4(%rsp)
  jne    L37
  lea    D34(%rip),%rdi
  call   <printf@plt>
  jmp    L4
L27:
  lea    D19(%rip),%rdi
  call   <printf@plt>
L4:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L36:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D21
type_guessed: string
name: unpack_data
arguments: label: D38
type_guessed: string
name: unpack_data
arguments: label: D15
type_guessed: string
name: unpack_data
arguments: label: D34
type_guessed: string
name: unpack_data
arguments: label: D19
type_guessed: string

tool: D21: 
Indovinalo con 10 tentativi:

tool: D38: %d

tool: D15: il pc ha pensato un numero da 1 a 14

tool: D34: Hai vinto

tool: D19: Hai perso

assistant: ```c
int main() {
  int n_u;
  int t;
  int n_c;
  int ind;
  n_c = rand() % 15;
  ind = 0;
  t = 10;
  printf("il pc ha pensato un numero da 1 a 14");
  do {
    printf("\nIndovinalo con 10 tentativi:");
    scanf("%d", &n_u);
    if (n_u == n_c) {
      ind = 1;
    }
    t--;
  } while (t > 0 && ind == 0);
  if (ind == 1) {
    printf("Hai vinto");
  } else {
    printf("Hai perso");
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<endian2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%ecx
  mov    %esi,%eax
  mov    %edx,-0xc(%rbp)
  mov    %ecx,%edx
  mov    %dl,-0x4(%rbp)
  mov    %al,-0x8(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  je     L18
  movzbl -0x4(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x8(%rbp),%eax
  or     %edx,%eax
  jmp    L0
L18:
  movzbl -0x8(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x4(%rbp),%eax
  or     %edx,%eax
L0:
  pop    %rbp
  ret
```

assistant: ```c
ulong endian2(unsigned char u1, unsigned char u2, int endian) {
  if (endian)
    return ((u1 << 8) & 0xFF00) | ((u2)&0xFF);
  else
    return ((u2 << 8) & 0xFF00) | ((u1)&0xFF);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<endian2>:
  endbr64
  test   %edx,%edx
  mov    %edi,%eax
  cmovne %esi,%eax
  ret
```

assistant: ```c
ulong endian2(unsigned char u1, unsigned char u2, int endian) {
  if (endian)
    return ((u1 << 8) & 0xFF00) | ((u2)&0xFF);
  else
    return ((u2 << 8) & 0xFF00) | ((u1)&0xFF);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<endian2>:
  endbr64
  test   %edx,%edx
  mov    %edi,%eax
  cmovne %esi,%eax
  ret
```

assistant: ```c
ulong endian2(unsigned char u1, unsigned char u2, int endian) {
  if (endian)
    return ((u1 << 8) & 0xFF00) | ((u2)&0xFF);
  else
    return ((u2 << 8) & 0xFF00) | ((u1)&0xFF);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<endian2>:
  endbr64
  test   %edx,%edx
  mov    %edi,%eax
  cmovne %esi,%eax
  ret
```

assistant: ```c
ulong endian2(unsigned char u1, unsigned char u2, int endian) {
  if (endian)
    return ((u1 << 8) & 0xFF00) | ((u2)&0xFF);
  else
    return ((u2 << 8) & 0xFF00) | ((u1)&0xFF);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ignore_render>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void ignore_render(const char *to_render, void *data) {
  (void)data;
  (void)to_render;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ignore_render>:
  endbr64
  ret
```

assistant: ```c
void ignore_render(const char *to_render, void *data) {
  (void)data;
  (void)to_render;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ignore_render>:
  endbr64
  ret
```

assistant: ```c
void ignore_render(const char *to_render, void *data) {
  (void)data;
  (void)to_render;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ignore_render>:
  endbr64
  ret
```

assistant: ```c
void ignore_render(const char *to_render, void *data) {
  (void)data;
  (void)to_render;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<max_diff>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,-0x14(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movl   $0x1,-0x20(%rbp)
  jmp    L27
L41:
  movl   $0x1,-0x1c(%rbp)
  jmp    L20
L37:
  movl   $0x1,-0x18(%rbp)
  jmp    L25
L49:
  mov    -0x20(%rbp),%eax
  imul   -0x34(%rbp),%eax
  imul   -0x34(%rbp),%eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  imul   -0x34(%rbp),%eax
  add    %eax,%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x20(%rbp),%eax
  imul   -0x34(%rbp),%eax
  imul   -0x34(%rbp),%eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  imul   -0x34(%rbp),%eax
  add    %eax,%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  subsd  %xmm1,%xmm0
  movq   D7(%rip),%xmm1
  andpd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  comisd -0x10(%rbp),%xmm0
  jbe    L68
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
L68:
  addl   $0x1,-0x18(%rbp)
L25:
  mov    -0x14(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jge    L49
  addl   $0x1,-0x1c(%rbp)
L20:
  mov    -0x14(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L37
  addl   $0x1,-0x20(%rbp)
L27:
  mov    -0x14(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L41
  movsd  -0x10(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D7
datatype: qword

tool: D7: 9223372036854775807

assistant: ```c
double max_diff(double *array1, double *array2, int sz) {
  double tmpdiff, diff;
  int i, j, k;
  int n = sz - 2;
  diff = 0.0;
  for (i = 1; i < n + 1; i++) {
    for (j = 1; j < n + 1; j++) {
      for (k = 1; k < n + 1; k++) {
        tmpdiff = fabs(array1[i * sz * sz + j * sz + k] - array2[i * sz * sz + j * sz + k]);
        if (tmpdiff > diff)
          diff = tmpdiff;
      }
    }
  }
  return diff;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<max_diff>:
  endbr64
  cmp    $0x2,%edx
  jle    L28
  push   %rbp
  push   %rbx
  mov    %rdi,%rcx
  mov    %edx,%r8d
  imul   %edx,%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rbx
  movslq %r8d,%r10
  lea    -0x3(%r8),%eax
  lea    0x2(%rdx,%r10,1),%r11
  add    %rax,%r11
  shl    $0x3,%r11
  sub    $0x1,%r8d
  shl    $0x3,%r10
  mov    $0x1,%ebp
  pxor   %xmm1,%xmm1
  not    %rax
  lea    0x0(,%rax,8),%r9
  movq   D3(%rip),%xmm2
L6:
  mov    %r11,%rdx
  mov    $0x1,%edi
L39:
  lea    (%r9,%rdx,1),%rax
L34:
  movsd  (%rcx,%rax,1),%xmm0
  subsd  (%rsi,%rax,1),%xmm0
  andpd  %xmm2,%xmm0
  maxsd  %xmm1,%xmm0
  movapd %xmm0,%xmm1
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L34
  add    $0x1,%edi
  add    %r10,%rdx
  cmp    %r8d,%edi
  jne    L39
  add    $0x1,%ebp
  add    %rbx,%r11
  cmp    %r8d,%ebp
  jne    L6
  pop    %rbx
  pop    %rbp
  ret
L28:
  pxor   %xmm1,%xmm1
  movapd %xmm1,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D3
datatype: qword

tool: D3: 9223372036854775807

assistant: ```c
double max_diff(double *array1, double *array2, int sz) {
  double tmpdiff, diff;
  int i, j, k;
  int n = sz - 2;
  diff = 0.0;
  for (i = 1; i < n + 1; i++) {
    for (j = 1; j < n + 1; j++) {
      for (k = 1; k < n + 1; k++) {
        tmpdiff = fabs(array1[i * sz * sz + j * sz + k] - array2[i * sz * sz + j * sz + k]);
        if (tmpdiff > diff)
          diff = tmpdiff;
      }
    }
  }
  return diff;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<max_diff>:
  endbr64
  cmp    $0x2,%edx
  jle    L14
  mov    %edx,%r8d
  imul   %edx,%edx
  movq   D1(%rip),%xmm2
  push   %rbp
  movslq %r8d,%r10
  lea    -0x3(%r8),%eax
  push   %rbx
  mov    %rdi,%rcx
  sub    $0x1,%r8d
  mov    $0x1,%ebp
  pxor   %xmm1,%xmm1
  movslq %edx,%rdx
  lea    0x2(%r10,%rdx,1),%r11
  lea    0x0(,%rdx,8),%rbx
  shl    $0x3,%r10
  add    %rax,%r11
  not    %rax
  shl    $0x3,%r11
  lea    0x0(,%rax,8),%r9
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %r11,%rdx
  mov    $0x1,%edi
  nopl   0x0(%rax,%rax,1)
L38:
  lea    (%r9,%rdx,1),%rax
  nopl   0x0(%rax)
L9:
  movsd  (%rcx,%rax,1),%xmm0
  subsd  (%rsi,%rax,1),%xmm0
  add    $0x8,%rax
  andpd  %xmm2,%xmm0
  maxsd  %xmm1,%xmm0
  movapd %xmm0,%xmm1
  cmp    %rax,%rdx
  jne    L9
  add    $0x1,%edi
  add    %r10,%rdx
  cmp    %r8d,%edi
  jne    L38
  add    $0x1,%ebp
  add    %rbx,%r11
  cmp    %r8d,%ebp
  jne    L1
  pop    %rbx
  pop    %rbp
  ret
L14:
  pxor   %xmm1,%xmm1
  movapd %xmm1,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D1
datatype: qword

tool: D1: 9223372036854775807

assistant: ```c
double max_diff(double *array1, double *array2, int sz) {
  double tmpdiff, diff;
  int i, j, k;
  int n = sz - 2;
  diff = 0.0;
  for (i = 1; i < n + 1; i++) {
    for (j = 1; j < n + 1; j++) {
      for (k = 1; k < n + 1; k++) {
        tmpdiff = fabs(array1[i * sz * sz + j * sz + k] - array2[i * sz * sz + j * sz + k]);
        if (tmpdiff > diff)
          diff = tmpdiff;
      }
    }
  }
  return diff;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<max_diff>:
  endbr64
  cmp    $0x2,%edx
  jle    L33
  mov    %edx,%r8d
  imul   %edx,%edx
  movq   D4(%rip),%xmm2
  push   %rbp
  movslq %r8d,%r10
  lea    -0x3(%r8),%eax
  push   %rbx
  mov    %rdi,%rcx
  sub    $0x1,%r8d
  mov    $0x1,%ebp
  pxor   %xmm1,%xmm1
  movslq %edx,%rdx
  lea    0x2(%r10,%rdx,1),%r11
  lea    0x0(,%rdx,8),%rbx
  shl    $0x3,%r10
  add    %rax,%r11
  not    %rax
  shl    $0x3,%r11
  lea    0x0(,%rax,8),%r9
  nopw   0x0(%rax,%rax,1)
L19:
  mov    %r11,%rdx
  mov    $0x1,%edi
  nopl   0x0(%rax,%rax,1)
L35:
  lea    (%r9,%rdx,1),%rax
  nopl   0x0(%rax)
L14:
  movsd  (%rcx,%rax,1),%xmm0
  subsd  (%rsi,%rax,1),%xmm0
  add    $0x8,%rax
  andpd  %xmm2,%xmm0
  maxsd  %xmm1,%xmm0
  movapd %xmm0,%xmm1
  cmp    %rdx,%rax
  jne    L14
  add    $0x1,%edi
  lea    (%rax,%r10,1),%rdx
  cmp    %r8d,%edi
  jne    L35
  add    $0x1,%ebp
  add    %rbx,%r11
  cmp    %r8d,%ebp
  jne    L19
  pop    %rbx
  pop    %rbp
  ret
L33:
  pxor   %xmm1,%xmm1
  movapd %xmm1,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: tag: D4
type: qword

tool: D4: 9223372036854775807

assistant: ```c
double max_diff(double *array1, double *array2, int sz) {
  double tmpdiff, diff;
  int i, j, k;
  int n = sz - 2;
  diff = 0.0;
  for (i = 1; i < n + 1; i++) {
    for (j = 1; j < n + 1; j++) {
      for (k = 1; k < n + 1; k++) {
        tmpdiff = fabs(array1[i * sz * sz + j * sz + k] - array2[i * sz * sz + j * sz + k]);
        if (tmpdiff > diff)
          diff = tmpdiff;
      }
    }
  }
  return diff;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<write_rmb>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x4,%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,-0x4(%rbp)
  mov    -0x4(%rbp),%ecx
  mov    %ecx,%edx
  shl    $0x2,%edx
  add    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%esi
  mov    $0x1,%edi
  call   <write_mail_box@plt>
  nop
  leave
  ret
```

assistant: ```c
void write_rmb(int index) {
  int right_index = (index + (5 - 1)) % 5;
  write_mail_box(true, right_index, true);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<write_rmb>:
  endbr64
  sub    $0x8,%rsp
  lea    0x4(%rdi),%esi
  movslq %esi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x21,%rax
  mov    %esi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  sub    %eax,%esi
  mov    $0x1,%edx
  mov    $0x1,%edi
  call   <write_mail_box@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void write_rmb(int index) {
  int right_index = (index + (5 - 1)) % 5;
  write_mail_box(true, right_index, true);
}
```