ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<cam16i_r>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    %r8,-0x28(%rbp)
  mov    %r9d,-0x2c(%rbp)
  movw   $0x0,-0x2(%rbp)
  jmp    L8
L15:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  lea    0x2(%rax),%rcx
  mov    -0x28(%rbp),%rdx
  mov    %rcx,(%rdx)
  mov    %eax,%edi
  mov    -0x20(%rbp),%ecx
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%esi
  mov    -0x14(%rbp),%eax
  mov    %edi,%r8d
  mov    %eax,%edi
  call   <cam16i@plt>
  movzwl -0x2(%rbp),%eax
  add    $0x1,%eax
  mov    %ax,-0x2(%rbp)
L8:
  movzwl -0x2(%rbp),%eax
  cmp    %eax,-0x2c(%rbp)
  jg     L15
  nop
  nop
  leave
  ret
```

assistant: ```c
void cam16i_r(const int c, const int n, const int a, const int f, unsigned short **d, const int r) {
  unsigned short i;
  for (i = 0; i < r; i++)
    cam16i(c, n, a, f, (*d)++);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<cam16i_r>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  test   %r9d,%r9d
  jle    L0
  mov    %esi,%r13d
  mov    %edx,%r14d
  mov    %ecx,%r15d
  mov    %r8,%rbp
  mov    %r9d,%r12d
  mov    $0x0,%ebx
L17:
  mov    0x0(%rbp),%r8
  lea    0x2(%r8),%rax
  mov    %rax,0x0(%rbp)
  mov    %r15d,%ecx
  mov    %r14d,%edx
  mov    %r13d,%esi
  mov    0xc(%rsp),%edi
  call   <cam16i@plt>
  add    $0x1,%ebx
  movzwl %bx,%eax
  cmp    %r12d,%eax
  jl     L17
L0:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void cam16i_r(const int c, const int n, const int a, const int f, unsigned short **d, const int r) {
  unsigned short i;
  for (i = 0; i < r; i++)
    cam16i(c, n, a, f, (*d)++);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cam16i_r>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %r9d,0xc(%rsp)
  test   %r9d,%r9d
  jle    L17
  mov    %edi,%r12d
  mov    %esi,%r13d
  mov    %edx,%r14d
  mov    %ecx,%r15d
  mov    %r8,%rbp
  xor    %ebx,%ebx
  nopl   (%rax)
L18:
  mov    0x0(%rbp),%r8
  mov    %r15d,%ecx
  mov    %r14d,%edx
  mov    %r13d,%esi
  mov    %r12d,%edi
  add    $0x1,%ebx
  lea    0x2(%r8),%rax
  mov    %rax,0x0(%rbp)
  call   <cam16i@plt>
  movzwl %bx,%eax
  cmp    0xc(%rsp),%eax
  jl     L18
L17:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void cam16i_r(const int c, const int n, const int a, const int f, unsigned short **d, const int r) {
  unsigned short i;
  for (i = 0; i < r; i++)
    cam16i(c, n, a, f, (*d)++);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<cam16i_r>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %r9d,0xc(%rsp)
  test   %r9d,%r9d
  jle    L19
  mov    %edi,%r12d
  mov    %esi,%r13d
  mov    %edx,%r14d
  mov    %ecx,%r15d
  mov    %r8,%rbp
  xor    %ebx,%ebx
  nopl   (%rax)
L15:
  mov    0x0(%rbp),%r8
  mov    %r15d,%ecx
  mov    %r14d,%edx
  mov    %r13d,%esi
  mov    %r12d,%edi
  add    $0x1,%ebx
  lea    0x2(%r8),%rax
  mov    %rax,0x0(%rbp)
  call   <cam16i@plt>
  movzwl %bx,%eax
  cmp    0xc(%rsp),%eax
  jl     L15
L19:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void cam16i_r(const int c, const int n, const int a, const int f, unsigned short **d, const int r) {
  unsigned short i;
  for (i = 0; i < r; i++)
    cam16i(c, n, a, f, (*d)++);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<autohelperowl_attackpat159>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  mov    %edx,%ecx
  mov    %eax,%edx
  mov    $0xfffffffe,%esi
  mov    $0xffffffff,%edi
  call   <offset@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <countlib@plt>
  cmp    $0x2,%eax
  sete   %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
int autohelperowl_attackpat159(struct pattern *patt, int transformation, int move, int color, int action) {
  int A;
  (void)patt;
  (void)color;
  (void)action;
  A = offset(-1, -2, move, transformation);
  return countlib(A) == 2;
}
```
user: What is the c source code of the assembly code below?

```asm
<autohelperowl_attackpat159>:
  endbr64
  sub    $0x8,%rsp
  mov    %esi,%ecx
  mov    $0xfffffffe,%esi
  mov    $0xffffffff,%edi
  call   <offset@plt>
  mov    %eax,%edi
  call   <countlib@plt>
  cmp    $0x2,%eax
  sete   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int autohelperowl_attackpat159(struct pattern *patt, int transformation, int move, int color, int action) {
  int A;
  (void)patt;
  (void)color;
  (void)action;
  A = offset(-1, -2, move, transformation);
  return countlib(A) == 2;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<autohelperowl_attackpat159>:
  endbr64
  sub    $0x8,%rsp
  mov    %esi,%ecx
  mov    $0xffffffff,%edi
  mov    $0xfffffffe,%esi
  call   <offset@plt>
  mov    %eax,%edi
  call   <countlib@plt>
  cmp    $0x2,%eax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int autohelperowl_attackpat159(struct pattern *patt, int transformation, int move, int color, int action) {
  int A;
  (void)patt;
  (void)color;
  (void)action;
  A = offset(-1, -2, move, transformation);
  return countlib(A) == 2;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<autohelperowl_attackpat159>:
  endbr64
  sub    $0x8,%rsp
  mov    %esi,%ecx
  mov    $0xffffffff,%edi
  mov    $0xfffffffe,%esi
  call   <offset@plt>
  mov    %eax,%edi
  call   <countlib@plt>
  cmp    $0x2,%eax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int autohelperowl_attackpat159(struct pattern *patt, int transformation, int move, int color, int action) {
  int A;
  (void)patt;
  (void)color;
  (void)action;
  A = offset(-1, -2, move, transformation);
  return countlib(A) == 2;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    D49(%rip),%rdi
  call   <printf@plt>
  lea    -0x3c(%rbp),%rsi
  lea    D45(%rip),%rdi
  call   <scanf@plt>
  mov    -0x3c(%rbp),%edx
  movslq %edx,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rcx
  sub    %rax,%rcx
L4:
  cmp    %rcx,%rsp
  je     L59
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L4
L59:
  mov    %rsi,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L50
  orq    $0x0,-0x8(%rsp,%rax,1)
L50:
  mov    %rsp,%r13
  test   %edx,%edx
  jle    L30
  mov    %r13,%r12
  mov    $0x0,%ebx
  lea    D38(%rip),%r15
  lea    D45(%rip),%r14
L24:
  add    $0x1,%ebx
  mov    %ebx,%esi
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  add    $0x4,%r12
  cmp    -0x3c(%rbp),%ebx
  jl     L24
L30:
  lea    D33(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x3c(%rbp),%esi
  test   %esi,%esi
  jle    L55
  mov    $0x0,%ebx
  lea    D61(%rip),%r12
L53:
  mov    0x0(%r13,%rbx,4),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x3c(%rbp),%esi
  add    $0x1,%rbx
  cmp    %ebx,%esi
  jg     L53
L55:
  mov    %r13,%rdi
  call   <sort@plt>
  lea    D44(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0x0,-0x3c(%rbp)
  jle    L13
  mov    $0x0,%ebx
  lea    D61(%rip),%r12
L71:
  mov    0x0(%r13,%rbx,4),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %ebx,-0x3c(%rbp)
  jg     L71
L13:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L89
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L89:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D49
type_guessed: string
name: read_data
arguments: label: D45
type_guessed: string
name: read_data
arguments: label: D38
type_guessed: string
name: read_data
arguments: label: D33
type_guessed: string
name: read_data
arguments: label: D61
type_guessed: string
name: read_data
arguments: label: D44
type_guessed: string

tool: D49: Enter the number of elements: 

tool: D45: %d

tool: D38: Enter element(%d): 

tool: D33: 

The array is: 

tool: D61: %d 

tool: D44: 

The sorted array is: 

assistant: ```c
void main() {
  int element_index, number_of_elements, i = 0;
  printf("Enter the number of elements: ");
  scanf("%d", &number_of_elements);
  int elements[number_of_elements];
  while (i < number_of_elements) {
    printf("Enter element(%d): ", i + 1);
    scanf("%d", &elements[i]);
    i++;
  }
  printf("\n\nThe array is: ");
  for (i = 0; i < number_of_elements; i++) {
    printf("%d ", elements[i]);
  }
  sort(elements, number_of_elements);
  printf("\n\nThe sorted array is: ");
  for (i = 0; i < number_of_elements; i++) {
    printf("%d ", elements[i]);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x38(%rbp)
  lea    -0x3c(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x3c(%rbp),%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x64,-0x30(%rbp)
  movl   $0x32,-0x2c(%rbp)
  movl   $0x14,-0x28(%rbp)
  movl   $0xa,-0x24(%rbp)
  movl   $0x5,-0x20(%rbp)
  movl   $0x2,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L6
L16:
  mov    -0x3c(%rbp),%eax
  mov    -0x38(%rbp),%edx
  movslq %edx,%rdx
  mov    -0x30(%rbp,%rdx,4),%ecx
  cltd
  idiv   %ecx
  mov    %eax,-0x34(%rbp)
  mov    -0x3c(%rbp),%eax
  mov    -0x38(%rbp),%edx
  movslq %edx,%rdx
  mov    -0x30(%rbp,%rdx,4),%ecx
  cltd
  idiv   %ecx
  mov    %edx,%eax
  mov    %eax,-0x3c(%rbp)
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%edx
  mov    -0x34(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x38(%rbp)
L6:
  cmpl   $0x6,-0x38(%rbp)
  jle    L16
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D26
data_type: string
name: read_data
arguments: data_tag: D23
data_type: string
name: read_data
arguments: data_tag: D2
data_type: string

tool: D26: %d

tool: D23: %d


tool: D2: %d nota(s) de R$ %d,00


assistant: ```c
int main() {
  int n, temp, i = 0;
  scanf("%d", &n);
  printf("%d\n", n);
  int a[7] = {100, 50, 20, 10, 5, 2, 1};
  while (i < 7) {
    temp = n / a[i];
    n = n % a[i];
    printf("%d nota(s) de R$ %d,00\n", temp, a[i]);
    i++;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  lea    D26(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x64,0x10(%rsp)
  movl   $0x32,0x14(%rsp)
  movl   $0x14,0x18(%rsp)
  movl   $0xa,0x1c(%rsp)
  movl   $0x5,0x20(%rsp)
  movl   $0x2,0x24(%rsp)
  movl   $0x1,0x28(%rsp)
  lea    0x10(%rsp),%rbx
  lea    0x2c(%rsp),%r12
  lea    D2(%rip),%rbp
L16:
  mov    0xc(%rsp),%eax
  mov    (%rbx),%ecx
  cltd
  idiv   %ecx
  mov    %edx,0xc(%rsp)
  mov    %ecx,%edx
  mov    %eax,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L16
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D16
type: string
name: retrieve_data
arguments: data_label: D26
type: string
name: retrieve_data
arguments: data_label: D2
type: string

tool: D16: %d

tool: D26: %d


tool: D2: %d nota(s) de R$ %d,00


assistant: ```c
int main() {
  int n, temp, i = 0;
  scanf("%d", &n);
  printf("%d\n", n);
  int a[7] = {100, 50, 20, 10, 5, 2, 1};
  while (i < 7) {
    temp = n / a[i];
    n = n % a[i];
    printf("%d nota(s) de R$ %d,00\n", temp, a[i]);
    i++;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D15(%rip),%rdi
  push   %rbp
  lea    D22(%rip),%rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    0x14(%rsp),%rbx
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  xor    %eax,%eax
  lea    0x2c(%rsp),%r12
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x64,%r8d
  movabs $0x1400000032,%rax
  mov    %rax,0x14(%rsp)
  movabs $0x50000000a,%rax
  mov    %rax,0x1c(%rsp)
  movabs $0x100000002,%rax
  mov    %rax,0x24(%rsp)
  jmp    L4
  nopl   (%rax)
L17:
  mov    (%rbx),%r8d
  add    $0x4,%rbx
L4:
  mov    0xc(%rsp),%eax
  mov    %rbp,%rdi
  cltd
  idiv   %r8d
  mov    %edx,0xc(%rsp)
  mov    %eax,%esi
  mov    %r8d,%edx
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L17
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D15
type_infered: string
name: retrieve_data
arguments: tag: D22
type_infered: string
name: retrieve_data
arguments: tag: D4
type_infered: string

tool: D15: %d

tool: D22: %d nota(s) de R$ %d,00


tool: D4: %d


assistant: ```c
int main() {
  int n, temp, i = 0;
  scanf("%d", &n);
  printf("%d\n", n);
  int a[7] = {100, 50, 20, 10, 5, 2, 1};
  while (i < 7) {
    temp = n / a[i];
    n = n % a[i];
    printf("%d nota(s) de R$ %d,00\n", temp, a[i]);
    i++;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  lea    D5(%rip),%rdi
  push   %rbp
  lea    D14(%rip),%rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    0x10(%rsp),%rbx
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  xor    %eax,%eax
  lea    0x2c(%rsp),%r12
  lea    D42(%rip),%rdi
  call   <printf@plt>
  mov    D35(%rip),%rax
  movdqa D29(%rip),%xmm0
  movl   $0x1,0x28(%rsp)
  mov    %rax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  nopl   0x0(%rax)
L8:
  mov    0xc(%rsp),%eax
  mov    (%rbx),%r8d
  mov    %rbp,%rdi
  add    $0x4,%rbx
  cltd
  idiv   %r8d
  mov    %edx,0xc(%rsp)
  mov    %eax,%esi
  mov    %r8d,%edx
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L8
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D5
type_guessed: string
name: extract_symbolic_data
arguments: label: D14
type_guessed: string
name: extract_symbolic_data
arguments: label: D42
type_guessed: string
name: extract_symbolic_data
arguments: label: D35
type_guessed: byte[8]
name: extract_symbolic_data
arguments: label: D29
type_guessed: byte[16]

tool: D5: %d

tool: D14: %d nota(s) de R$ %d,00


tool: D42: %d


tool: {"D35": [0x05,0x00,0x00,0x00,0x02,0x00,0x00,0x00]}

tool: {"D29": [0x64,0x00,0x00,0x00,0x32,0x00,0x00,0x00,0x14,0x00,0x00,0x00,0x0a,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int n, temp, i = 0;
  scanf("%d", &n);
  printf("%d\n", n);
  int a[7] = {100, 50, 20, 10, 5, 2, 1};
  while (i < 7) {
    temp = n / a[i];
    n = n % a[i];
    printf("%d nota(s) de R$ %d,00\n", temp, a[i]);
    i++;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <daemonize@plt>
L6:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L6
```

assistant: ```c
main() {
  daemonize();
  while (1)
    sleep(1);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <daemonize@plt>
L2:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L2
```

assistant: ```c
main() {
  daemonize();
  while (1)
    sleep(1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <daemonize@plt>
  nop
L6:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L6
```

assistant: ```c
main() {
  daemonize();
  while (1)
    sleep(1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <daemonize@plt>
  nop
L9:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L9
```

assistant: ```c
main() {
  daemonize();
  while (1)
    sleep(1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<malloc_double_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %edx,-0x2c(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L3
  mov    $0x0,%eax
  call   <error_and_exit@plt>
L3:
  movl   $0x0,-0x20(%rbp)
  jmp    L61
L57:
  mov    -0x28(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x18(%rbp),%rax
  lea    (%rcx,%rax,1),%rbx
  mov    %edx,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L25
  mov    $0x0,%eax
  call   <error_and_exit@plt>
L25:
  movl   $0x0,-0x1c(%rbp)
  jmp    L49
L18:
  mov    -0x2c(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rcx
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    (%rcx,%rax,1),%rbx
  mov    %edx,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L53
  mov    $0x0,%eax
  call   <error_and_exit@plt>
L53:
  addl   $0x1,-0x1c(%rbp)
L49:
  mov    -0x1c(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L18
  addl   $0x1,-0x20(%rbp)
L61:
  mov    -0x20(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L57
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
double ***malloc_double_array(int dim1, int dim2, int dim3) {
  int i, j;
  double ***array;
  array = malloc(sizeof(double **) * dim1);
  if (array == NULL)
    error_and_exit();
  for (i = 0; i < dim1; i++) {
    array[i] = malloc(sizeof(double *) * dim2);
    if (array[i] == NULL)
      error_and_exit();
    for (j = 0; j < dim2; j++) {
      array[i][j] = malloc(sizeof(double) * dim3);
      if (array[i][j] == NULL)
        error_and_exit();
    }
  }
  return array;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<malloc_double_array>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,%ebx
  mov    %esi,0x8(%rsp)
  mov    %edx,%r13d
  lea    0x0(,%rdi,8),%edi
  call   <malloc@plt>
  mov    %rax,0x18(%rsp)
  test   %rax,%rax
  je     L105
L42:
  test   %ebx,%ebx
  jle    L97
  mov    0x8(%rsp),%eax
  lea    0x0(,%rax,8),%edx
  mov    %edx,0xc(%rsp)
  shl    $0x3,%r13d
  mov    0x18(%rsp),%rcx
  mov    %rcx,%r15
  mov    %ebx,%ebx
  lea    (%rcx,%rbx,8),%rsi
  mov    %rsi,0x10(%rsp)
  lea    -0x1(%rax),%eax
  lea    0x8(,%rax,8),%r14
  jmp    L41
L105:
  call   <error_and_exit@plt>
  jmp    L42
L21:
  call   <error_and_exit@plt>
  jmp    L116
L14:
  add    $0x8,%rbx
  cmp    %r14,%rbx
  je     L4
L71:
  mov    %rbx,%r12
  add    0x0(%rbp),%r12
  mov    %r13d,%edi
  call   <malloc@plt>
  mov    %rax,(%r12)
  mov    0x0(%rbp),%rax
  cmpq   $0x0,(%rax,%rbx,1)
  jne    L14
  mov    $0x0,%eax
  call   <error_and_exit@plt>
  jmp    L14
L4:
  add    $0x8,%r15
  cmp    0x10(%rsp),%r15
  je     L97
L41:
  mov    0xc(%rsp),%edi
  call   <malloc@plt>
  mov    %r15,%rbp
  mov    %rax,(%r15)
  test   %rax,%rax
  je     L21
L116:
  cmpl   $0x0,0x8(%rsp)
  jle    L4
  mov    $0x0,%ebx
  jmp    L71
L97:
  mov    0x18(%rsp),%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
double ***malloc_double_array(int dim1, int dim2, int dim3) {
  int i, j;
  double ***array;
  array = malloc(sizeof(double **) * dim1);
  if (array == NULL)
    error_and_exit();
  for (i = 0; i < dim1; i++) {
    array[i] = malloc(sizeof(double *) * dim2);
    if (array[i] == NULL)
      error_and_exit();
    for (j = 0; j < dim2; j++) {
      array[i][j] = malloc(sizeof(double) * dim3);
      if (array[i][j] == NULL)
        error_and_exit();
    }
  }
  return array;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<malloc_double_array>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  lea    0x0(,%rbp,8),%edi
  mov    %edx,%ebx
  sub    $0x28,%rsp
  mov    %esi,0xc(%rsp)
  call   <malloc@plt>
  mov    %rax,0x18(%rsp)
  test   %rax,%rax
  je     L78
L37:
  test   %ebp,%ebp
  jle    L49
  mov    0x18(%rsp),%rcx
  mov    0xc(%rsp),%eax
  shl    $0x3,%ebx
  lea    (%rcx,%rbp,8),%rdx
  lea    0x0(,%rax,8),%r14d
  sub    $0x1,%eax
  mov    %rcx,%r15
  mov    %rdx,0x10(%rsp)
  lea    0x8(,%rax,8),%rbp
  nopw   0x0(%rax,%rax,1)
L84:
  mov    %r14d,%edi
  call   <malloc@plt>
  mov    %rax,(%r15)
  test   %rax,%rax
  je     L0
L5:
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L59
  xor    %r13d,%r13d
  jmp    L24
  nopl   (%rax)
L48:
  add    $0x8,%r13
  cmp    %r13,%rbp
  je     L59
L24:
  mov    (%r15),%r12
  mov    %ebx,%edi
  call   <malloc@plt>
  add    %r13,%r12
  mov    %rax,(%r12)
  mov    (%r15),%rax
  cmpq   $0x0,(%rax,%r13,1)
  jne    L48
  xor    %eax,%eax
  add    $0x8,%r13
  call   <error_and_exit@plt>
  cmp    %r13,%rbp
  jne    L24
  nopl   0x0(%rax)
L59:
  add    $0x8,%r15
  cmp    0x10(%rsp),%r15
  jne    L84
L49:
  mov    0x18(%rsp),%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  call   <error_and_exit@plt>
  jmp    L5
L78:
  call   <error_and_exit@plt>
  jmp    L37
```

assistant: ```c
double ***malloc_double_array(int dim1, int dim2, int dim3) {
  int i, j;
  double ***array;
  array = malloc(sizeof(double **) * dim1);
  if (array == NULL)
    error_and_exit();
  for (i = 0; i < dim1; i++) {
    array[i] = malloc(sizeof(double *) * dim2);
    if (array[i] == NULL)
      error_and_exit();
    for (j = 0; j < dim2; j++) {
      array[i][j] = malloc(sizeof(double) * dim3);
      if (array[i][j] == NULL)
        error_and_exit();
    }
  }
  return array;
}
```