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

```asm
<main>:
  endbr64
  push   %r14
  lea    D18(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D50(%rip),%rdi
  call   <scanf_s@plt>
  mov    0x4(%rsp),%ebx
  test   %ebx,%ebx
  jle    L51
  xor    %ebp,%ebp
  lea    D43(%rip),%r13
  lea    D23(%rip),%r14
  lea    D36(%rip),%r12
  nopl   0x0(%rax)
L1:
  sub    %ebp,%ebx
  sub    $0x1,%ebx
  test   %ebx,%ebx
  jle    L52
  nopl   0x0(%rax)
L3:
  mov    %r12,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L3
L52:
  xor    %ebx,%ebx
  nop
L0:
  mov    %r13,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebp,%ebx
  jle    L0
  mov    %r14,%rdi
  add    $0x1,%ebp
  call   <printf@plt>
  mov    0x4(%rsp),%ebx
  cmp    %ebp,%ebx
  jg     L1
L51:
  lea    D26(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L59
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L59:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D18
datatype: string
name: read_data
arguments: label: D50
datatype: string
name: read_data
arguments: label: D43
datatype: string
name: read_data
arguments: label: D23
datatype: string
name: read_data
arguments: label: D36
datatype: string
name: read_data
arguments: label: D26
datatype: string

tool: D18: Enter X: 

tool: D50: %d

tool: D43: *

tool: D23: 


tool: D36:  

tool: D26: pause

assistant: ```c
void main() {
  int i, j, k, x;
  printf("Enter X: ");
  scanf_s("%d", &x);
  for (i = 0; i < x; i++) {
    for (k = x - i - 1; k > 0; k--) {
      printf(" ");
    }
    for (j = 0; j < i + 1; j++) {
      printf("*");
    }
    printf("\n");
  }
  system("pause");
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r14
  lea    D18(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D49(%rip),%rdi
  call   <scanf_s@plt>
  mov    0x4(%rsp),%ebx
  test   %ebx,%ebx
  jle    L55
  xor    %ebp,%ebp
  lea    D21(%rip),%r13
  lea    D11(%rip),%r14
  lea    D6(%rip),%r12
  nopl   0x0(%rax)
L18:
  sub    %ebp,%ebx
  sub    $0x1,%ebx
  test   %ebx,%ebx
  jle    L37
  nopl   0x0(%rax)
L46:
  mov    %r12,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L46
L37:
  xor    %ebx,%ebx
  nop
L9:
  mov    %r13,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebp,%ebx
  jle    L9
  mov    %r14,%rdi
  add    $0x1,%ebp
  call   <printf@plt>
  mov    0x4(%rsp),%ebx
  cmp    %ebp,%ebx
  jg     L18
L55:
  lea    D38(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D18
type_guessed: string
name: parse_data
arguments: label: D49
type_guessed: string
name: parse_data
arguments: label: D21
type_guessed: string
name: parse_data
arguments: label: D11
type_guessed: string
name: parse_data
arguments: label: D6
type_guessed: string
name: parse_data
arguments: label: D38
type_guessed: string

tool: D18: Enter X: 

tool: D49: %d

tool: D21: *

tool: D11: 


tool: D6:  

tool: D38: pause

assistant: ```c
void main() {
  int i, j, k, x;
  printf("Enter X: ");
  scanf_s("%d", &x);
  for (i = 0; i < x; i++) {
    for (k = x - i - 1; k > 0; k--) {
      printf(" ");
    }
    for (j = 0; j < i + 1; j++) {
      printf("*");
    }
    printf("\n");
  }
  system("pause");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<Print>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <getch@plt>
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <system@plt>
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <getch@plt>
  cmp    $0x1b,%eax
  jne    L7
  mov    $0x0,%edi
  call   <exit@plt>
L7:
  nop
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D25
type_infered: string
name: retrieve_data
arguments: data_label: D1
type_infered: string
name: retrieve_data
arguments: data_label: D12
type_infered: string
name: retrieve_data
arguments: data_label: D33
type_infered: string
name: retrieve_data
arguments: data_label: D38
type_infered: string

tool: D25: 	Welcome to the mini Snake game.(press any key to continue)


tool: D1: cls

tool: D12: 	Game instructions:


tool: D33: 
-> Use arrow keys to move the snake.

-> You will be provided foods at the several coordinates of the screen which you have to eat. Everytime you eat a food the length of the snake will be increased by 1 element and thus the score.

-> Here you are provided with three lives. Your life will decrease as you hit the wall or snake's body.

-> YOu can pause the game in its middle by pressing any key. To continue the paused game press any other key once again

-> If you want to exit press esc. 


tool: D38: 

Press any key to play game...

assistant: ```c
void Print() {
  printf("\tWelcome to the mini Snake game.(press any key to continue)\n");
  getch();
  system("cls");
  printf("\tGame instructions:\n");
  printf("\n-> Use arrow keys to move the snake.\n\n-> You will be provided foods at the several coordinates of the screen which you have to eat. Everytime you eat a food the length of the snake will be increased by 1 element and thus the score.\n\n-> Here you are provided with three lives. Your life will decrease as you hit the wall or snake's body.\n\n-> YOu can pause the game in its middle by pressing any key. To continue the paused game press any other key once again\n\n-> If you want to exit press esc. \n");
  printf("\n\nPress any key to play game...");
  if (getch() == 27)
    exit(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
<Print>:
  endbr64
  sub    $0x8,%rsp
  lea    D48(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <getch@plt>
  lea    D17(%rip),%rdi
  call   <system@plt>
  lea    D21(%rip),%rdi
  call   <printf@plt>
  lea    D42(%rip),%rdi
  call   <printf@plt>
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <getch@plt>
  cmp    $0x1b,%eax
  je     L14
L9:
  add    $0x8,%rsp
  ret
L14:
  mov    $0x0,%edi
  call   <exit@plt>
  jmp    L9
```

tool_calls: name: read_data
arguments: label: D48
datatype: string
name: read_data
arguments: label: D17
datatype: string
name: read_data
arguments: label: D21
datatype: string
name: read_data
arguments: label: D42
datatype: string
name: read_data
arguments: label: D14
datatype: string

tool: D48: 	Welcome to the mini Snake game.(press any key to continue)


tool: D17: cls

tool: D21: 	Game instructions:


tool: D42: 
-> Use arrow keys to move the snake.

-> You will be provided foods at the several coordinates of the screen which you have to eat. Everytime you eat a food the length of the snake will be increased by 1 element and thus the score.

-> Here you are provided with three lives. Your life will decrease as you hit the wall or snake's body.

-> YOu can pause the game in its middle by pressing any key. To continue the paused game press any other key once again

-> If you want to exit press esc. 


tool: D14: 

Press any key to play game...

assistant: ```c
void Print() {
  printf("\tWelcome to the mini Snake game.(press any key to continue)\n");
  getch();
  system("cls");
  printf("\tGame instructions:\n");
  printf("\n-> Use arrow keys to move the snake.\n\n-> You will be provided foods at the several coordinates of the screen which you have to eat. Everytime you eat a food the length of the snake will be increased by 1 element and thus the score.\n\n-> Here you are provided with three lives. Your life will decrease as you hit the wall or snake's body.\n\n-> YOu can pause the game in its middle by pressing any key. To continue the paused game press any other key once again\n\n-> If you want to exit press esc. \n");
  printf("\n\nPress any key to play game...");
  if (getch() == 27)
    exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<Print>:
  endbr64
  sub    $0x8,%rsp
  lea    D20(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <getch@plt>
  lea    D23(%rip),%rdi
  call   <system@plt>
  lea    D48(%rip),%rdi
  call   <printf@plt>
  lea    D31(%rip),%rdi
  call   <printf@plt>
  lea    D24(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <getch@plt>
  cmp    $0x1b,%eax
  je     L8
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax)
L8:
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    10b0 <exit@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D20
data_type: string
name: parse_data
arguments: data_tag: D23
data_type: string
name: parse_data
arguments: data_tag: D48
data_type: string
name: parse_data
arguments: data_tag: D31
data_type: string
name: parse_data
arguments: data_tag: D24
data_type: string

tool: D20: 	Welcome to the mini Snake game.(press any key to continue)


tool: D23: cls

tool: D48: 	Game instructions:


tool: D31: 
-> Use arrow keys to move the snake.

-> You will be provided foods at the several coordinates of the screen which you have to eat. Everytime you eat a food the length of the snake will be increased by 1 element and thus the score.

-> Here you are provided with three lives. Your life will decrease as you hit the wall or snake's body.

-> YOu can pause the game in its middle by pressing any key. To continue the paused game press any other key once again

-> If you want to exit press esc. 


tool: D24: 

Press any key to play game...

assistant: ```c
void Print() {
  printf("\tWelcome to the mini Snake game.(press any key to continue)\n");
  getch();
  system("cls");
  printf("\tGame instructions:\n");
  printf("\n-> Use arrow keys to move the snake.\n\n-> You will be provided foods at the several coordinates of the screen which you have to eat. Everytime you eat a food the length of the snake will be increased by 1 element and thus the score.\n\n-> Here you are provided with three lives. Your life will decrease as you hit the wall or snake's body.\n\n-> YOu can pause the game in its middle by pressing any key. To continue the paused game press any other key once again\n\n-> If you want to exit press esc. \n");
  printf("\n\nPress any key to play game...");
  if (getch() == 27)
    exit(0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<Print>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <getch@plt>
  lea    D39(%rip),%rdi
  call   <system@plt>
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    D37(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <getch@plt>
  cmp    $0x1b,%eax
  je     L7
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax)
L7:
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    10b0 <exit@plt>
```

tool_calls: name: read_data
arguments: tag: D3
type_infered: string
name: read_data
arguments: tag: D39
type_infered: string
name: read_data
arguments: tag: D4
type_infered: string
name: read_data
arguments: tag: D24
type_infered: string
name: read_data
arguments: tag: D37
type_infered: string

tool: D3: 	Welcome to the mini Snake game.(press any key to continue)


tool: D39: cls

tool: D4: 	Game instructions:


tool: D24: 
-> Use arrow keys to move the snake.

-> You will be provided foods at the several coordinates of the screen which you have to eat. Everytime you eat a food the length of the snake will be increased by 1 element and thus the score.

-> Here you are provided with three lives. Your life will decrease as you hit the wall or snake's body.

-> YOu can pause the game in its middle by pressing any key. To continue the paused game press any other key once again

-> If you want to exit press esc. 


tool: D37: 

Press any key to play game...

assistant: ```c
void Print() {
  printf("\tWelcome to the mini Snake game.(press any key to continue)\n");
  getch();
  system("cls");
  printf("\tGame instructions:\n");
  printf("\n-> Use arrow keys to move the snake.\n\n-> You will be provided foods at the several coordinates of the screen which you have to eat. Everytime you eat a food the length of the snake will be increased by 1 element and thus the score.\n\n-> Here you are provided with three lives. Your life will decrease as you hit the wall or snake's body.\n\n-> YOu can pause the game in its middle by pressing any key. To continue the paused game press any other key once again\n\n-> If you want to exit press esc. \n");
  printf("\n\nPress any key to play game...");
  if (getch() == 27)
    exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<vec_abs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movss  (%rax),%xmm0
  movss  %xmm0,-0x4(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L9
L24:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm1
  pxor   %xmm0,%xmm0
  comiss %xmm1,%xmm0
  jbe    L12
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  movss  D1(%rip),%xmm1
  xorps  %xmm1,%xmm0
  jmp    L1
L12:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
L1:
  movss  -0x4(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L9:
  cmpl   $0x4,-0x8(%rbp)
  jle    L24
  movss  -0x4(%rbp),%xmm0
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D1
type_guessed: byte[4]

tool: {"D1": [0x00,0x00,0x00,0x80]}

assistant: ```c
float vec_abs(float v[]) {
  int i;
  float r = v[0];
  for (i = 1; i < 5; i++)
    r += ((v[i]) < 0.0 ? (-v[i]) : (v[i]));
  return r;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<vec_abs>:
  endbr64
  movss  (%rdi),%xmm1
  lea    0x4(%rdi),%rax
  add    $0x14,%rdi
  pxor   %xmm2,%xmm2
  jmp    L30
L9:
  addss  %xmm0,%xmm1
  add    $0x4,%rax
  cmp    %rdi,%rax
  je     L22
L30:
  movss  (%rax),%xmm0
  comiss %xmm0,%xmm2
  jbe    L9
  xorps  D7(%rip),%xmm0
  jmp    L9
L22:
  movaps %xmm1,%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
type_infered: byte[16]

tool: {"D7": [0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
float vec_abs(float v[]) {
  int i;
  float r = v[0];
  for (i = 1; i < 5; i++)
    r += ((v[i]) < 0.0 ? (-v[i]) : (v[i]));
  return r;
}
```
user: Translate the below assembly code into C programming language:

```asm
<vec_abs>:
  endbr64
  movss  (%rdi),%xmm1
  lea    0x4(%rdi),%rax
  pxor   %xmm2,%xmm2
  add    $0x14,%rdi
  movss  D8(%rip),%xmm3
L6:
  movss  (%rax),%xmm0
  comiss %xmm0,%xmm2
  jbe    L13
  xorps  %xmm3,%xmm0
L13:
  add    $0x4,%rax
  addss  %xmm0,%xmm1
  cmp    %rdi,%rax
  jne    L6
  movaps %xmm1,%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
type_infered: byte[4]

tool: {"D8": [0x00,0x00,0x00,0x80]}

assistant: ```c
float vec_abs(float v[]) {
  int i;
  float r = v[0];
  for (i = 1; i < 5; i++)
    r += ((v[i]) < 0.0 ? (-v[i]) : (v[i]));
  return r;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<vec_abs>:
  endbr64
  movups 0x4(%rdi),%xmm2
  pxor   %xmm3,%xmm3
  movaps D3(%rip),%xmm0
  movaps %xmm2,%xmm1
  xorps  %xmm2,%xmm0
  cmpltps %xmm3,%xmm1
  andps  %xmm1,%xmm0
  andnps %xmm2,%xmm1
  orps   %xmm0,%xmm1
  movss  (%rdi),%xmm0
  movaps %xmm1,%xmm2
  addss  %xmm1,%xmm0
  shufps $0x55,%xmm1,%xmm2
  addss  %xmm2,%xmm0
  movaps %xmm1,%xmm2
  unpckhps %xmm1,%xmm2
  shufps $0xff,%xmm1,%xmm1
  addss  %xmm2,%xmm0
  addss  %xmm1,%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D3
type: byte[16]

tool: {"D3": [0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x80]}

assistant: ```c
float vec_abs(float v[]) {
  int i;
  float r = v[0];
  for (i = 1; i < 5; i++)
    r += ((v[i]) < 0.0 ? (-v[i]) : (v[i]));
  return r;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<find_maximum_column>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x3c(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L21:
  cmp    %rdx,%rsp
  je     L15
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L21
L15:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L29
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L29:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x20(%rbp)
  movl   $0x0,-0x2c(%rbp)
  jmp    L38
L4:
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x40(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%ecx
  mov    -0x20(%rbp),%rax
  mov    -0x2c(%rbp),%edx
  movslq %edx,%rdx
  mov    %ecx,(%rax,%rdx,4)
  addl   $0x1,-0x2c(%rbp)
L38:
  mov    -0x2c(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jl     L4
  mov    -0x3c(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <find_max_subsequence@plt>
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L51
  call   <__stack_chk_fail@plt>
L51:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int find_maximum_column(int **matrix, int size, int column) {
  int values[size];
  for (int i = 0; i < size; i++)
    values[i] = matrix[i][column];
  return find_max_subsequence(values, size);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<find_maximum_column>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %esi,%r9d
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movslq %esi,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rcx
  sub    %rax,%rcx
L21:
  cmp    %rcx,%rsp
  je     L25
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L21
L25:
  mov    %rsi,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L4
  orq    $0x0,-0x8(%rsp,%rax,1)
L4:
  mov    %rsp,%r8
  test   %r9d,%r9d
  jle    L29
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    %r9d,%esi
  mov    $0x0,%eax
L44:
  mov    (%rdi,%rax,8),%rdx
  mov    (%rdx,%rcx,1),%edx
  mov    %edx,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L44
L29:
  mov    %r9d,%esi
  mov    %r8,%rdi
  call   <find_max_subsequence@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L38
  leave
  ret
L38:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int find_maximum_column(int **matrix, int size, int column) {
  int values[size];
  for (int i = 0; i < size; i++)
    values[i] = matrix[i][column];
  return find_max_subsequence(values, size);
}
```
user: What is the c source code of the assembly code below:

```asm
<find_maximum_column>:
  endbr64
  push   %rbp
  movslq %esi,%rsi
  mov    %rsi,%r9
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    0xf(,%rsi,4),%rax
  mov    %rsp,%r8
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%r8
  and    $0xfffffffffffffff0,%rcx
  cmp    %r8,%rsp
  je     L39
L15:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %r8,%rsp
  jne    L15
L39:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L16
L37:
  mov    %rsp,%r8
  test   %r9d,%r9d
  jle    L25
  movslq %edx,%rdx
  xor    %eax,%eax
  lea    0x0(,%rdx,4),%rcx
  nopl   0x0(%rax)
L18:
  mov    (%rdi,%rax,8),%rdx
  mov    (%rdx,%rcx,1),%edx
  mov    %edx,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L18
L25:
  mov    %r9d,%esi
  mov    %r8,%rdi
  call   <find_max_subsequence@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L45
  leave
  ret
L16:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L37
L45:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int find_maximum_column(int **matrix, int size, int column) {
  int values[size];
  for (int i = 0; i < size; i++)
    values[i] = matrix[i][column];
  return find_max_subsequence(values, size);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<find_maximum_column>:
  endbr64
  push   %rbp
  movslq %esi,%rsi
  mov    %rsi,%r9
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    0xf(,%rsi,4),%rax
  mov    %rsp,%r8
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%r8
  and    $0xfffffffffffffff0,%rcx
  cmp    %r8,%rsp
  je     L35
L48:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %r8,%rsp
  jne    L48
L35:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L38
L8:
  mov    %rsp,%r8
  test   %r9d,%r9d
  jle    L46
  movslq %edx,%rdx
  xor    %eax,%eax
  lea    0x0(,%rdx,4),%rcx
  nopl   0x0(%rax)
L12:
  mov    (%rdi,%rax,8),%rdx
  mov    (%rdx,%rcx,1),%edx
  mov    %edx,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L12
L46:
  mov    %r9d,%esi
  mov    %r8,%rdi
  call   <find_max_subsequence@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L41
  leave
  ret
L38:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L8
L41:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int find_maximum_column(int **matrix, int size, int column) {
  int values[size];
  for (int i = 0; i < size; i++)
    values[i] = matrix[i][column];
  return find_max_subsequence(values, size);
}
```