ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r14
  pxor   %xmm0,%xmm0
  push   %r13
  lea    D13(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  movaps %xmm0,(%rsp)
  lea    0x28(%rsp),%r12
  movq   $0x0,0x20(%rsp)
  mov    %r14,%rbp
  mov    %r14,%rbx
  movaps %xmm0,0x10(%rsp)
  nopl   0x0(%rax,%rax,1)
L12:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L12
  mov    D27(%rip),%rax
  movdqa D27(%rip),%xmm0
  mov    %r14,%rbx
  mov    %rax,0x20(%rsp)
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  nopl   0x0(%rax,%rax,1)
L7:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L7
  mov    $0x28,%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  call   <memset@plt>
  nopw   0x0(%rax,%rax,1)
L10:
  mov    0x0(%rbp),%esi
  mov    %r13,%rdi
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %rbp,%r12
  jne    L10
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D13
type_guessed: string
name: unpack_data
arguments: data_tag: D27
type_guessed: byte[16]

tool: D13: %d

tool: {"D27": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int main(void) {
  int vetor[10] = {0};
  for (int i = 0; i < 10; i++)
    printf("%d", vetor[i]);
  for (int i = 0; i < 10; i++)
    vetor[i] = 1;
  for (int i = 0; i < 10; i++)
    printf("%d", vetor[i]);
  memset(vetor, 0, sizeof(vetor));
  for (int i = 0; i < 10; i++)
    printf("%d", vetor[i]);
}
```
user: Translate the below assembly code into C programming language:

```asm
<full_subtractor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %edx,-0x2c(%rbp)
  mov    $0x8,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <exor@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x2c(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <exor@plt>
  mov    %eax,-0x10(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  sete   %al
  movzbl %al,%eax
  and    -0x28(%rbp),%eax
  mov    %eax,%edx
  cmpl   $0x0,-0x14(%rbp)
  sete   %al
  movzbl %al,%eax
  and    -0x2c(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    -0x10(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *full_subtractor(int x, int y, int z) {
  int temp, diff, borrow;
  int *arr = malloc(sizeof(int) * 2);
  temp = exor(x, y);
  diff = exor(temp, z);
  borrow = (!x & y) | (z & !temp);
  *(arr + 0) = diff;
  *(arr + 1) = borrow;
  return arr;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<full_subtractor>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r14d
  mov    %esi,%r13d
  mov    %edx,%ebp
  mov    $0x8,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %r13d,%esi
  mov    %r14d,%edi
  call   <exor@plt>
  mov    %eax,%r12d
  mov    %ebp,%esi
  mov    %eax,%edi
  call   <exor@plt>
  mov    %eax,(%rbx)
  test   %r14d,%r14d
  sete   %al
  movzbl %al,%eax
  and    %r13d,%eax
  test   %r12d,%r12d
  sete   %dl
  movzbl %dl,%edx
  and    %ebp,%edx
  or     %edx,%eax
  mov    %eax,0x4(%rbx)
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int *full_subtractor(int x, int y, int z) {
  int temp, diff, borrow;
  int *arr = malloc(sizeof(int) * 2);
  temp = exor(x, y);
  diff = exor(temp, z);
  borrow = (!x & y) | (z & !temp);
  *(arr + 0) = diff;
  *(arr + 1) = borrow;
  return arr;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<full_subtractor>:
  endbr64
  push   %r14
  mov    %edi,%r14d
  mov    $0x8,%edi
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edx,%ebx
  call   <malloc@plt>
  mov    %r13d,%esi
  mov    %r14d,%edi
  mov    %rax,%r12
  call   <exor@plt>
  mov    %ebx,%esi
  mov    %eax,%edi
  mov    %eax,%ebp
  call   <exor@plt>
  mov    %eax,(%r12)
  xor    %eax,%eax
  test   %r14d,%r14d
  sete   %al
  xor    %edx,%edx
  and    %r13d,%eax
  test   %ebp,%ebp
  sete   %dl
  and    %ebx,%edx
  or     %edx,%eax
  mov    %eax,0x4(%r12)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int *full_subtractor(int x, int y, int z) {
  int temp, diff, borrow;
  int *arr = malloc(sizeof(int) * 2);
  temp = exor(x, y);
  diff = exor(temp, z);
  borrow = (!x & y) | (z & !temp);
  *(arr + 0) = diff;
  *(arr + 1) = borrow;
  return arr;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<full_subtractor>:
  endbr64
  push   %r14
  mov    %edi,%r14d
  mov    $0x8,%edi
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edx,%ebx
  call   <malloc@plt>
  mov    %r13d,%esi
  mov    %r14d,%edi
  mov    %rax,%r12
  call   <exor@plt>
  mov    %ebx,%esi
  mov    %eax,%edi
  mov    %eax,%ebp
  call   <exor@plt>
  xor    %edx,%edx
  test   %r14d,%r14d
  sete   %dl
  xor    %ecx,%ecx
  movd   %eax,%xmm0
  mov    %r12,%rax
  and    %r13d,%edx
  test   %ebp,%ebp
  sete   %cl
  and    %ebx,%ecx
  or     %ecx,%edx
  movd   %edx,%xmm1
  punpckldq %xmm1,%xmm0
  movq   %xmm0,(%r12)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int *full_subtractor(int x, int y, int z) {
  int temp, diff, borrow;
  int *arr = malloc(sizeof(int) * 2);
  temp = exor(x, y);
  diff = exor(temp, z);
  borrow = (!x & y) | (z & !temp);
  *(arr + 0) = diff;
  *(arr + 1) = borrow;
  return arr;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x6c,-0x8(%rbp)
  movl   $0x6f,-0x4(%rbp)
  mov    $0x48,%edi
  call   <putchar@plt>
  mov    $0x65,%edi
  call   <putchar@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    $0x57,%edi
  call   <putchar@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  mov    $0x72,%edi
  call   <putchar@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  mov    $0x64,%edi
  call   <putchar@plt>
  mov    $0x21,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int a = 108, b = 111;
  putchar(72);
  putchar(101);
  putchar(a);
  putchar(a);
  putchar(b);
  putchar(32);
  putchar(87);
  putchar(b);
  putchar(114);
  putchar(a);
  putchar(100);
  putchar(33);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x48,%edi
  call   <putchar@plt>
  mov    $0x65,%edi
  call   <putchar@plt>
  mov    $0x6c,%edi
  call   <putchar@plt>
  mov    $0x6c,%edi
  call   <putchar@plt>
  mov    $0x6f,%edi
  call   <putchar@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    $0x57,%edi
  call   <putchar@plt>
  mov    $0x6f,%edi
  call   <putchar@plt>
  mov    $0x72,%edi
  call   <putchar@plt>
  mov    $0x6c,%edi
  call   <putchar@plt>
  mov    $0x64,%edi
  call   <putchar@plt>
  mov    $0x21,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int a = 108, b = 111;
  putchar(72);
  putchar(101);
  putchar(a);
  putchar(a);
  putchar(b);
  putchar(32);
  putchar(87);
  putchar(b);
  putchar(114);
  putchar(a);
  putchar(100);
  putchar(33);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x48,%edi
  call   <putchar@plt>
  mov    $0x65,%edi
  call   <putchar@plt>
  mov    $0x6c,%edi
  call   <putchar@plt>
  mov    $0x6c,%edi
  call   <putchar@plt>
  mov    $0x6f,%edi
  call   <putchar@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    $0x57,%edi
  call   <putchar@plt>
  mov    $0x6f,%edi
  call   <putchar@plt>
  mov    $0x72,%edi
  call   <putchar@plt>
  mov    $0x6c,%edi
  call   <putchar@plt>
  mov    $0x64,%edi
  call   <putchar@plt>
  mov    $0x21,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int a = 108, b = 111;
  putchar(72);
  putchar(101);
  putchar(a);
  putchar(a);
  putchar(b);
  putchar(32);
  putchar(87);
  putchar(b);
  putchar(114);
  putchar(a);
  putchar(100);
  putchar(33);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x48,%edi
  call   <putchar@plt>
  mov    $0x65,%edi
  call   <putchar@plt>
  mov    $0x6c,%edi
  call   <putchar@plt>
  mov    $0x6c,%edi
  call   <putchar@plt>
  mov    $0x6f,%edi
  call   <putchar@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    $0x57,%edi
  call   <putchar@plt>
  mov    $0x6f,%edi
  call   <putchar@plt>
  mov    $0x72,%edi
  call   <putchar@plt>
  mov    $0x6c,%edi
  call   <putchar@plt>
  mov    $0x64,%edi
  call   <putchar@plt>
  mov    $0x21,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int a = 108, b = 111;
  putchar(72);
  putchar(101);
  putchar(a);
  putchar(a);
  putchar(b);
  putchar(32);
  putchar(87);
  putchar(b);
  putchar(114);
  putchar(a);
  putchar(100);
  putchar(33);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D6(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rsi
  lea    D24(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x18(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x1c(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x20(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x24(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  add    (%rsp),%eax
  add    0x8(%rsp),%eax
  add    0xc(%rsp),%eax
  add    0x10(%rsp),%eax
  add    0x14(%rsp),%eax
  add    0x18(%rsp),%eax
  add    0x1c(%rsp),%eax
  add    0x20(%rsp),%eax
  add    0x24(%rsp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  divsd  D7(%rip),%xmm0
  cvtsd2ss %xmm0,%xmm0
  cvtss2sd %xmm0,%xmm0
  lea    D39(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x24(%rsp),%eax
  push   %rax
  mov    0x28(%rsp),%eax
  push   %rax
  mov    0x2c(%rsp),%eax
  push   %rax
  mov    0x30(%rsp),%eax
  push   %rax
  mov    0x34(%rsp),%r9d
  mov    0x30(%rsp),%r8d
  mov    0x2c(%rsp),%ecx
  mov    0x28(%rsp),%edx
  mov    0x24(%rsp),%esi
  mov    0x20(%rsp),%edi
  call   <sort@plt>
  add    $0x20,%rsp
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D6
datatype: string
name: extract_symbolic_data
arguments: label: D24
datatype: string
name: extract_symbolic_data
arguments: label: D7
datatype: f64
name: extract_symbolic_data
arguments: label: D39
datatype: string

tool: D6: Please input 10 numbers:

tool: D24: %d

tool: D7: 10

tool: D39: Average: %f


assistant: ```c
int main() {
  int num1, num2, num3, num4, num5, num6;
  int num7, num8, num9, num10;
  float average;
  printf("Please input 10 numbers:\n");
  scanf("%d", &num1);
  scanf("%d", &num2);
  scanf("%d", &num3);
  scanf("%d", &num4);
  scanf("%d", &num5);
  scanf("%d", &num6);
  scanf("%d", &num7);
  scanf("%d", &num8);
  scanf("%d", &num9);
  scanf("%d", &num10);
  average = (num1 + num2 + num3 + num4 + num5 + num6 + num7 + num8 + num9 + num10) / 10.0;
  printf("Average: %f\n", average);
  sort(num1, num2, num3, num4, num5, num6, num7, num8, num9, num10);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D22(%rip),%rdi
  lea    D31(%rip),%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x18(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x1c(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x20(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x24(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  pxor   %xmm0,%xmm0
  add    (%rsp),%eax
  add    0x8(%rsp),%eax
  add    0xc(%rsp),%eax
  lea    D18(%rip),%rdi
  add    0x10(%rsp),%eax
  add    0x14(%rsp),%eax
  add    0x18(%rsp),%eax
  add    0x1c(%rsp),%eax
  add    0x20(%rsp),%eax
  add    0x24(%rsp),%eax
  cvtsi2sd %eax,%xmm0
  mov    $0x1,%eax
  divsd  D21(%rip),%xmm0
  cvtsd2ss %xmm0,%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x24(%rsp),%eax
  push   %rax
  mov    0x28(%rsp),%eax
  push   %rax
  mov    0x2c(%rsp),%eax
  push   %rax
  mov    0x30(%rsp),%eax
  push   %rax
  mov    0x34(%rsp),%r9d
  mov    0x30(%rsp),%r8d
  mov    0x2c(%rsp),%ecx
  mov    0x28(%rsp),%edx
  mov    0x24(%rsp),%esi
  mov    0x20(%rsp),%edi
  call   <sort@plt>
  add    $0x20,%rsp
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D22
data_type: string
name: disassemble_data
arguments: label: D31
data_type: string
name: disassemble_data
arguments: label: D18
data_type: string
name: disassemble_data
arguments: label: D21
data_type: f64

tool: D22: Please input 10 numbers:

tool: D31: %d

tool: D18: Average: %f


tool: D21: 10

assistant: ```c
int main() {
  int num1, num2, num3, num4, num5, num6;
  int num7, num8, num9, num10;
  float average;
  printf("Please input 10 numbers:\n");
  scanf("%d", &num1);
  scanf("%d", &num2);
  scanf("%d", &num3);
  scanf("%d", &num4);
  scanf("%d", &num5);
  scanf("%d", &num6);
  scanf("%d", &num7);
  scanf("%d", &num8);
  scanf("%d", &num9);
  scanf("%d", &num10);
  average = (num1 + num2 + num3 + num4 + num5 + num6 + num7 + num8 + num9 + num10) / 10.0;
  printf("Average: %f\n", average);
  sort(num1, num2, num3, num4, num5, num6, num7, num8, num9, num10);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D34(%rip),%rdi
  lea    D14(%rip),%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x18(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x1c(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x20(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x24(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  pxor   %xmm0,%xmm0
  add    (%rsp),%eax
  add    0x8(%rsp),%eax
  add    0xc(%rsp),%eax
  lea    D32(%rip),%rdi
  add    0x10(%rsp),%eax
  add    0x14(%rsp),%eax
  add    0x18(%rsp),%eax
  add    0x1c(%rsp),%eax
  add    0x20(%rsp),%eax
  add    0x24(%rsp),%eax
  cvtsi2sd %eax,%xmm0
  mov    $0x1,%eax
  divsd  D20(%rip),%xmm0
  cvtsd2ss %xmm0,%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x24(%rsp),%eax
  push   %rax
  mov    0x28(%rsp),%eax
  push   %rax
  mov    0x2c(%rsp),%eax
  push   %rax
  mov    0x30(%rsp),%eax
  push   %rax
  mov    0x34(%rsp),%r9d
  mov    0x30(%rsp),%r8d
  mov    0x2c(%rsp),%ecx
  mov    0x28(%rsp),%edx
  mov    0x24(%rsp),%esi
  mov    0x20(%rsp),%edi
  call   <sort@plt>
  add    $0x20,%rsp
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D34
data_type: string
name: read_data
arguments: data_label: D14
data_type: string
name: read_data
arguments: data_label: D32
data_type: string
name: read_data
arguments: data_label: D20
data_type: f64

tool: D34: Please input 10 numbers:

tool: D14: %d

tool: D32: Average: %f


tool: D20: 10

assistant: ```c
int main() {
  int num1, num2, num3, num4, num5, num6;
  int num7, num8, num9, num10;
  float average;
  printf("Please input 10 numbers:\n");
  scanf("%d", &num1);
  scanf("%d", &num2);
  scanf("%d", &num3);
  scanf("%d", &num4);
  scanf("%d", &num5);
  scanf("%d", &num6);
  scanf("%d", &num7);
  scanf("%d", &num8);
  scanf("%d", &num9);
  scanf("%d", &num10);
  average = (num1 + num2 + num3 + num4 + num5 + num6 + num7 + num8 + num9 + num10) / 10.0;
  printf("Average: %f\n", average);
  sort(num1, num2, num3, num4, num5, num6, num7, num8, num9, num10);
}
```
user: Translate the below assembly code into C programming language:

```asm
<print_matrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L6
L31:
  movl   $0x0,-0x8(%rbp)
  jmp    L17
L25:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x8(%rbp)
L17:
  mov    -0x8(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L25
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x4(%rbp)
L6:
  mov    -0x4(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L31
  nop
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D19
type_guessed: string
name: disassemble_data
arguments: label: D10
type_guessed: string

tool: D19: %s:


tool: D10: %d 

assistant: ```c
void print_matrix(char *description, int **M, int nx, int ny) {
  int i, j;
  printf("%s:\n", description);
  for (j = 0; j < ny; j++) {
    for (i = 0; i < nx; i++) {
      printf("%d ", M[i][j]);
    }
    printf("\n");
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<print_matrix>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,%r14
  mov    %edx,%r15d
  mov    %ecx,%ebx
  mov    %rdi,%rsi
  lea    D10(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %ebx,%ebx
  jle    L8
  mov    %ebx,%ebx
  lea    0x0(,%rbx,4),%rax
  mov    %rax,0x8(%rsp)
  lea    -0x1(%r15),%eax
  lea    0x8(%r14,%rax,8),%r13
  mov    $0x0,%ebp
  lea    D0(%rip),%r12
  jmp    L31
L40:
  mov    (%rbx),%rax
  mov    (%rax,%rbp,1),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x8,%rbx
  cmp    %r13,%rbx
  jne    L40
L16:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x4,%rbp
  cmp    0x8(%rsp),%rbp
  je     L8
L31:
  mov    %r14,%rbx
  test   %r15d,%r15d
  jg     L40
  jmp    L16
L8:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: tag: D10
type: string
name: read_data
arguments: tag: D0
type: string

tool: D10: %s:


tool: D0: %d 

assistant: ```c
void print_matrix(char *description, int **M, int nx, int ny) {
  int i, j;
  printf("%s:\n", description);
  for (j = 0; j < ny; j++) {
    for (i = 0; i < nx; i++) {
      printf("%d ", M[i][j]);
    }
    printf("\n");
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print_matrix>:
  endbr64
  push   %r15
  xor    %eax,%eax
  mov    %edx,%r15d
  push   %r14
  mov    %rsi,%r14
  mov    %rdi,%rsi
  lea    D2(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  movslq %ecx,%rbx
  sub    $0x18,%rsp
  call   <printf@plt>
  test   %ebx,%ebx
  jle    L31
  lea    0x0(,%rbx,4),%rax
  lea    D18(%rip),%rbp
  xor    %ebx,%ebx
  mov    %rax,0x8(%rsp)
  lea    -0x1(%r15),%eax
  lea    0x8(%r14,%rax,8),%r12
  nop
L6:
  mov    %r14,%r13
  test   %r15d,%r15d
  jle    L2
  nopl   0x0(%rax,%rax,1)
L25:
  mov    0x0(%r13),%rax
  mov    %rbp,%rdi
  add    $0x8,%r13
  mov    (%rax,%rbx,1),%esi
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %r12,%r13
  jne    L25
L2:
  mov    $0xa,%edi
  add    $0x4,%rbx
  call   <putchar@plt>
  cmp    0x8(%rsp),%rbx
  jne    L6
L31:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: data_label: D2
type_infered: string
name: parse_data
arguments: data_label: D18
type_infered: string

tool: D2: %s:


tool: D18: %d 

assistant: ```c
void print_matrix(char *description, int **M, int nx, int ny) {
  int i, j;
  printf("%s:\n", description);
  for (j = 0; j < ny; j++) {
    for (i = 0; i < nx; i++) {
      printf("%d ", M[i][j]);
    }
    printf("\n");
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<print_matrix>:
  endbr64
  push   %r15
  xor    %eax,%eax
  mov    %edx,%r15d
  push   %r14
  mov    %rsi,%r14
  mov    %rdi,%rsi
  lea    D2(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  movslq %ecx,%rbx
  sub    $0x18,%rsp
  call   <printf@plt>
  test   %ebx,%ebx
  jle    L22
  lea    0x0(,%rbx,4),%rax
  lea    D9(%rip),%rbp
  xor    %ebx,%ebx
  mov    %rax,0x8(%rsp)
  lea    -0x1(%r15),%eax
  lea    0x8(%r14,%rax,8),%r12
  nop
L16:
  mov    %r14,%r13
  test   %r15d,%r15d
  jle    L18
  nopl   0x0(%rax,%rax,1)
L1:
  mov    0x0(%r13),%rax
  mov    %rbp,%rdi
  add    $0x8,%r13
  mov    (%rax,%rbx,1),%esi
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %r13,%r12
  jne    L1
L18:
  mov    $0xa,%edi
  add    $0x4,%rbx
  call   <putchar@plt>
  cmp    0x8(%rsp),%rbx
  jne    L16
L22:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D2
type_infered: string
name: disassemble_data
arguments: tag: D9
type_infered: string

tool: D2: %s:


tool: D9: %d 

assistant: ```c
void print_matrix(char *description, int **M, int nx, int ny) {
  int i, j;
  printf("%s:\n", description);
  for (j = 0; j < ny; j++) {
    for (i = 0; i < nx; i++) {
      printf("%d ", M[i][j]);
    }
    printf("\n");
  }
}
```