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

```asm
<__vsqrt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %r8,-0x48(%rbp)
  movq   $0x0,-0x20(%rbp)
  jmp    L17
L0:
  mov    -0x28(%rbp),%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x30(%rbp),%rax
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x40(%rbp),%rax
  shl    $0x3,%rax
  add    %rax,-0x28(%rbp)
  mov    -0x48(%rbp),%rax
  shl    $0x3,%rax
  add    %rax,-0x30(%rbp)
  addq   $0x1,-0x20(%rbp)
L17:
  mov    -0x20(%rbp),%rax
  cmp    -0x38(%rbp),%rax
  jl     L0
  nop
  nop
  leave
  ret
```

assistant: ```c
void __vsqrt(double *x, double *y, long count, long stridex, long stridey) {
  long i;
  double u;
  double z;
  double result;
  for (i = 0; i < count; i++) {
    u = z = *x;
    result = sqrt(u);
    *y = result;
    x += stridex;
    y += stridey;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<__vsqrt>:
  endbr64
  test   %rdx,%rdx
  jle    L2
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rdx,%r15
  lea    0x0(,%rcx,8),%r14
  lea    0x0(,%r8,8),%r13
  mov    $0x0,%r12d
L34:
  movsd  (%rbx),%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L6
  sqrtsd %xmm0,%xmm0
L32:
  movsd  %xmm0,0x0(%rbp)
  add    %r14,%rbx
  add    %r13,%rbp
  add    $0x1,%r12
  cmp    %r12,%r15
  jne    L34
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L6:
  call   <sqrt@plt>
  jmp    L32
L2:
  ret
```

assistant: ```c
void __vsqrt(double *x, double *y, long count, long stridex, long stridey) {
  long i;
  double u;
  double z;
  double result;
  for (i = 0; i < count; i++) {
    u = z = *x;
    result = sqrt(u);
    *y = result;
    x += stridex;
    y += stridey;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<__vsqrt>:
  endbr64
  test   %rdx,%rdx
  jle    L25
  push   %r12
  shl    $0x3,%rcx
  shl    $0x3,%r8
  xor    %r12d,%r12d
  pxor   %xmm1,%xmm1
  sub    $0x30,%rsp
  xchg   %ax,%ax
L17:
  movsd  (%rdi),%xmm0
  ucomisd %xmm0,%xmm1
  ja     L33
  sqrtsd %xmm0,%xmm0
L16:
  add    $0x1,%r12
  movsd  %xmm0,(%rsi)
  add    %rcx,%rdi
  add    %r8,%rsi
  cmp    %r12,%rdx
  jne    L17
  add    $0x30,%rsp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L25:
  ret
L33:
  mov    %rdx,0x28(%rsp)
  mov    %rsi,0x20(%rsp)
  mov    %rdi,0x18(%rsp)
  mov    %r8,0x10(%rsp)
  mov    %rcx,0x8(%rsp)
  call   <sqrt@plt>
  mov    0x28(%rsp),%rdx
  mov    0x20(%rsp),%rsi
  pxor   %xmm1,%xmm1
  mov    0x18(%rsp),%rdi
  mov    0x10(%rsp),%r8
  mov    0x8(%rsp),%rcx
  jmp    L16
```

assistant: ```c
void __vsqrt(double *x, double *y, long count, long stridex, long stridey) {
  long i;
  double u;
  double z;
  double result;
  for (i = 0; i < count; i++) {
    u = z = *x;
    result = sqrt(u);
    *y = result;
    x += stridex;
    y += stridey;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<__vsqrt>:
  endbr64
  test   %rdx,%rdx
  jle    L15
  push   %r12
  shl    $0x3,%rcx
  shl    $0x3,%r8
  xor    %r12d,%r12d
  pxor   %xmm1,%xmm1
  sub    $0x30,%rsp
  xchg   %ax,%ax
L33:
  movsd  (%rdi),%xmm0
  ucomisd %xmm0,%xmm1
  ja     L28
  sqrtsd %xmm0,%xmm0
  add    $0x1,%r12
  add    %rcx,%rdi
  movsd  %xmm0,(%rsi)
  add    %r8,%rsi
  cmp    %r12,%rdx
  jne    L33
L35:
  add    $0x30,%rsp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  ret
L28:
  mov    %rdx,0x28(%rsp)
  add    $0x1,%r12
  mov    %rsi,0x20(%rsp)
  mov    %rdi,0x18(%rsp)
  mov    %r8,0x10(%rsp)
  mov    %rcx,0x8(%rsp)
  call   <sqrt@plt>
  mov    0x20(%rsp),%rsi
  mov    0x8(%rsp),%rcx
  pxor   %xmm1,%xmm1
  mov    0x18(%rsp),%rdi
  mov    0x10(%rsp),%r8
  mov    0x28(%rsp),%rdx
  movsd  %xmm0,(%rsi)
  add    %rcx,%rdi
  add    %r8,%rsi
  cmp    %r12,%rdx
  jne    L33
  jmp    L35
```

assistant: ```c
void __vsqrt(double *x, double *y, long count, long stridex, long stridey) {
  long i;
  double u;
  double z;
  double result;
  for (i = 0; i < count; i++) {
    u = z = *x;
    result = sqrt(u);
    *y = result;
    x += stridex;
    y += stridey;
  }
}
```
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
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D74(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  cmp    %eax,%edx
  jne    L66
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L49
L66:
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  cmp    %eax,%edx
  jge    L16
  jmp    L57
L27:
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  mov    %eax,%esi
  lea    D69(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L57:
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  cmp    %eax,%edx
  jle    L27
  jmp    L49
L24:
  mov    -0x10(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  mov    %eax,%esi
  lea    D69(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L16:
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  cmp    %eax,%edx
  jge    L24
L49:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
type: string
name: extract_symbolic_data
arguments: data_tag: D2
type: string
name: extract_symbolic_data
arguments: data_tag: D3
type: string
name: extract_symbolic_data
arguments: data_tag: D74
type: string
name: extract_symbolic_data
arguments: data_tag: D40
type: string
name: extract_symbolic_data
arguments: data_tag: D69
type: string

tool: D0: Para saber os inteiros entre dois numeros siga os passos abaixo: 

tool: D2: Digite o primeiro numero: 

tool: D3: %d

tool: D74: Digite o segundo numero: 

tool: D40: NUMEROS IGUAIS!

tool: D69: %d


assistant: ```c
int main() {
  int n1, n2;
  printf("Para saber os inteiros entre dois numeros siga os passos abaixo: \n");
  printf("Digite o primeiro numero: ");
  scanf("%d", &n1);
  printf("Digite o segundo numero: ");
  scanf("%d", &n2);
  if (n1 == n2)
    printf("NUMEROS IGUAIS!\n");
  else if (n1 < n2)
    while (n1 <= n2)
      printf("%d\n", n1++);
  else
    while (n1 >= n2)
      printf("%d\n", n1--);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D41(%rip),%rdi
  call   <puts@plt>
  lea    D58(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D19(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D66(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    (%rsp),%esi
  mov    0x4(%rsp),%eax
  cmp    %eax,%esi
  je     L44
  lea    D48(%rip),%rbx
  jge    L0
  lea    D48(%rip),%rbx
L7:
  lea    0x1(%rsi),%eax
  mov    %eax,(%rsp)
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    (%rsp),%esi
  cmp    0x4(%rsp),%esi
  jle    L7
L19:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L49
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L44:
  lea    D45(%rip),%rdi
  call   <puts@plt>
  jmp    L19
L0:
  lea    -0x1(%rsi),%eax
  mov    %eax,(%rsp)
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    (%rsp),%esi
  cmp    0x4(%rsp),%esi
  jge    L0
  jmp    L19
L49:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D41
data_type: string
name: disassemble_data
arguments: data_tag: D58
data_type: string
name: disassemble_data
arguments: data_tag: D19
data_type: string
name: disassemble_data
arguments: data_tag: D66
data_type: string
name: disassemble_data
arguments: data_tag: D48
data_type: string
name: disassemble_data
arguments: data_tag: D45
data_type: string

tool: D41: Para saber os inteiros entre dois numeros siga os passos abaixo: 

tool: D58: Digite o primeiro numero: 

tool: D19: %d

tool: D66: Digite o segundo numero: 

tool: D48: %d


tool: D45: NUMEROS IGUAIS!

assistant: ```c
int main() {
  int n1, n2;
  printf("Para saber os inteiros entre dois numeros siga os passos abaixo: \n");
  printf("Digite o primeiro numero: ");
  scanf("%d", &n1);
  printf("Digite o segundo numero: ");
  scanf("%d", &n2);
  if (n1 == n2)
    printf("NUMEROS IGUAIS!\n");
  else if (n1 < n2)
    while (n1 <= n2)
      printf("%d\n", n1++);
  else
    while (n1 >= n2)
      printf("%d\n", n1--);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D47(%rip),%rdi
  lea    D42(%rip),%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D30(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D40(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    (%rsp),%esi
  cmp    0x4(%rsp),%esi
  je     L41
  lea    D45(%rip),%rbx
  jge    L30
  nopl   0x0(%rax,%rax,1)
L7:
  lea    0x1(%rsi),%eax
  mov    %rbx,%rdi
  mov    %eax,(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rsp),%esi
  cmp    0x4(%rsp),%esi
  jle    L7
L40:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L30:
  lea    -0x1(%rsi),%eax
  mov    %rbx,%rdi
  mov    %eax,(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rsp),%esi
  cmp    0x4(%rsp),%esi
  jge    L30
  jmp    L40
L41:
  lea    D57(%rip),%rdi
  call   <puts@plt>
  jmp    L40
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D47
type_guessed: string
name: retrieve_data
arguments: data_tag: D42
type_guessed: string
name: retrieve_data
arguments: data_tag: D30
type_guessed: string
name: retrieve_data
arguments: data_tag: D40
type_guessed: string
name: retrieve_data
arguments: data_tag: D45
type_guessed: string
name: retrieve_data
arguments: data_tag: D57
type_guessed: string

tool: D47: Para saber os inteiros entre dois numeros siga os passos abaixo: 

tool: D42: %d

tool: D30: Digite o primeiro numero: 

tool: D40: Digite o segundo numero: 

tool: D45: %d


tool: D57: NUMEROS IGUAIS!

assistant: ```c
int main() {
  int n1, n2;
  printf("Para saber os inteiros entre dois numeros siga os passos abaixo: \n");
  printf("Digite o primeiro numero: ");
  scanf("%d", &n1);
  printf("Digite o segundo numero: ");
  scanf("%d", &n2);
  if (n1 == n2)
    printf("NUMEROS IGUAIS!\n");
  else if (n1 < n2)
    while (n1 <= n2)
      printf("%d\n", n1++);
  else
    while (n1 >= n2)
      printf("%d\n", n1--);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D38(%rip),%rdi
  lea    D40(%rip),%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D22(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D43(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    (%rsp),%esi
  cmp    0x4(%rsp),%esi
  je     L19
  lea    D23(%rip),%rbx
  jge    L17
  nopl   0x0(%rax,%rax,1)
L30:
  lea    0x1(%rsi),%eax
  mov    %rbx,%rdi
  mov    %eax,(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rsp),%esi
  cmp    0x4(%rsp),%esi
  jle    L30
L31:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  lea    -0x1(%rsi),%eax
  mov    %rbx,%rdi
  mov    %eax,(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rsp),%esi
  cmp    0x4(%rsp),%esi
  jge    L17
  jmp    L31
L19:
  lea    D29(%rip),%rdi
  call   <puts@plt>
  jmp    L31
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D38
data_type: string
name: read_data
arguments: data_label: D40
data_type: string
name: read_data
arguments: data_label: D22
data_type: string
name: read_data
arguments: data_label: D43
data_type: string
name: read_data
arguments: data_label: D23
data_type: string
name: read_data
arguments: data_label: D29
data_type: string

tool: D38: Para saber os inteiros entre dois numeros siga os passos abaixo: 

tool: D40: %d

tool: D22: Digite o primeiro numero: 

tool: D43: Digite o segundo numero: 

tool: D23: %d


tool: D29: NUMEROS IGUAIS!

assistant: ```c
int main() {
  int n1, n2;
  printf("Para saber os inteiros entre dois numeros siga os passos abaixo: \n");
  printf("Digite o primeiro numero: ");
  scanf("%d", &n1);
  printf("Digite o segundo numero: ");
  scanf("%d", &n2);
  if (n1 == n2)
    printf("NUMEROS IGUAIS!\n");
  else if (n1 < n2)
    while (n1 <= n2)
      printf("%d\n", n1++);
  else
    while (n1 >= n2)
      printf("%d\n", n1--);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<tmpl_insert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movsd  %xmm0,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x1c(%rbp),%edx
  mov    -0x28(%rbp),%rsi
  mov    -0x18(%rbp),%rcx
  lea    -0x10(%rbp),%rax
  movq   %rsi,%xmm0
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <asprintf@plt>
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
char *tmpl_insert(char *tmpl, int n, double val) {
  char *s;
  asprintf(&s, tmpl, val, n);
  return s;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<tmpl_insert>:
  endbr64
  sub    $0x18,%rsp
  mov    %esi,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rdi,%rsi
  mov    %rax,%rdi
  call   <asprintf@plt>
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *tmpl_insert(char *tmpl, int n, double val) {
  char *s;
  asprintf(&s, tmpl, val, n);
  return s;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<tmpl_insert>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,%r8
  mov    %esi,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    %r8,%rsi
  call   <asprintf@plt>
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *tmpl_insert(char *tmpl, int n, double val) {
  char *s;
  asprintf(&s, tmpl, val, n);
  return s;
}
```
user: Translate the below assembly code into C programming language:

```asm
<tmpl_insert>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,%r8
  mov    %esi,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    %r8,%rsi
  call   <asprintf@plt>
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *tmpl_insert(char *tmpl, int n, double val) {
  char *s;
  asprintf(&s, tmpl, val, n);
  return s;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<particion>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x2c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x30(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L9
L23:
  addl   $0x1,-0x14(%rbp)
L14:
  mov    -0x14(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jg     L12
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jge    L23
  jmp    L12
L20:
  subl   $0x1,-0x10(%rbp)
L12:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jg     L37
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jge    L20
L37:
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jge    L60
  movl   $0x1,-0xc(%rbp)
  jmp    L9
L60:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x14(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
L9:
  cmpl   $0x0,-0xc(%rbp)
  je     L14
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x2c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int particion(int lista[], int f, int l) {
  int pivote = lista[f];
  int izq = f + 1;
  int der = l;
  int temp;
  int done = 0;
  while (!done) {
    while (izq <= der && lista[izq] <= pivote) {
      izq = izq + 1;
    }
    while (lista[der] >= pivote && der >= izq) {
      der = der - 1;
    }
    if (der < izq) {
      done = 1;
    } else {
      temp = lista[izq];
      lista[izq] = lista[der];
      lista[der] = temp;
    }
  }
  temp = lista[f];
  lista[f] = lista[der];
  lista[der] = temp;
  return der;
}
```
user: What is the c source code of the assembly code below?

```asm
<particion>:
  endbr64
  mov    %rdi,%r9
  movslq %esi,%rax
  lea    (%rdi,%rax,4),%r11
  mov    (%r11),%edi
  lea    0x1(%rsi),%eax
  jmp    L95
L39:
  movslq %edx,%rcx
  lea    (%r9,%rcx,4),%r8
  mov    (%r8),%esi
  cmp    %edx,%eax
  jg     L32
  cmp    %esi,%edi
  jg     L32
  lea    -0x4(%r9,%rcx,4),%rcx
L84:
  sub    $0x1,%edx
  mov    %rcx,%r8
  mov    (%rcx),%esi
  sub    $0x4,%rcx
  cmp    %edi,%esi
  jl     L32
  cmp    %eax,%edx
  jge    L84
L32:
  cmp    %eax,%edx
  jl     L42
  movslq %eax,%rcx
  lea    (%r9,%rcx,4),%rcx
  mov    (%rcx),%r10d
  mov    %esi,(%rcx)
  mov    %r10d,(%r8)
L95:
  cmp    %edx,%eax
  jg     L42
  movslq %eax,%rcx
  lea    (%r9,%rcx,4),%rcx
L23:
  cmp    %edi,(%rcx)
  jg     L39
  add    $0x1,%eax
  add    $0x4,%rcx
  cmp    %edx,%eax
  jle    L23
  movslq %edx,%rcx
  lea    (%r9,%rcx,4),%r8
  mov    (%r8),%esi
  jmp    L32
L42:
  mov    (%r11),%ecx
  movslq %edx,%rax
  lea    (%r9,%rax,4),%rax
  mov    (%rax),%esi
  mov    %esi,(%r11)
  mov    %ecx,(%rax)
  mov    %edx,%eax
  ret
```

assistant: ```c
int particion(int lista[], int f, int l) {
  int pivote = lista[f];
  int izq = f + 1;
  int der = l;
  int temp;
  int done = 0;
  while (!done) {
    while (izq <= der && lista[izq] <= pivote) {
      izq = izq + 1;
    }
    while (lista[der] >= pivote && der >= izq) {
      der = der - 1;
    }
    if (der < izq) {
      done = 1;
    } else {
      temp = lista[izq];
      lista[izq] = lista[der];
      lista[der] = temp;
    }
  }
  temp = lista[f];
  lista[f] = lista[der];
  lista[der] = temp;
  return der;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<particion>:
  endbr64
  movslq %esi,%rax
  mov    %rdi,%r9
  lea    0x1(%rsi),%ecx
  lea    (%rdi,%rax,4),%r11
  mov    %edx,%eax
  movslq %eax,%r10
  mov    (%r11),%edi
  lea    (%r9,%r10,4),%r8
  mov    (%r8),%esi
  cmp    %eax,%ecx
  jg     L75
  nopl   0x0(%rax)
L33:
  movslq %ecx,%rdx
  lea    (%r9,%rdx,4),%rdx
  jmp    L32
  nopl   0x0(%rax)
L45:
  add    $0x1,%ecx
  add    $0x4,%rdx
  cmp    %eax,%ecx
  jg     L43
L32:
  cmp    %edi,(%rdx)
  jle    L45
  cmp    %esi,%edi
  jg     L97
  cmp    %eax,%ecx
  jg     L75
  lea    -0x4(%r9,%r10,4),%rdx
  jmp    L109
  cs nopw 0x0(%rax,%rax,1)
L112:
  sub    $0x4,%rdx
  cmp    %ecx,%eax
  jl     L75
L109:
  mov    (%rdx),%esi
  sub    $0x1,%eax
  mov    %rdx,%r8
  cmp    %edi,%esi
  jge    L112
L97:
  cmp    %eax,%ecx
  jg     L76
  movslq %ecx,%rdx
  lea    (%r9,%rdx,4),%rdx
  mov    (%rdx),%r10d
  mov    %esi,(%rdx)
  mov    %r10d,(%r8)
  movslq %eax,%r10
  lea    (%r9,%r10,4),%r8
  mov    (%r8),%esi
  cmp    %eax,%ecx
  jle    L33
L75:
  mov    (%r11),%edx
  mov    %esi,(%r11)
  mov    %edx,(%r8)
  ret
  nop
L43:
  cmp    %esi,%edi
  jle    L75
L76:
  mov    (%r8),%esi
  jmp    L75
```

assistant: ```c
int particion(int lista[], int f, int l) {
  int pivote = lista[f];
  int izq = f + 1;
  int der = l;
  int temp;
  int done = 0;
  while (!done) {
    while (izq <= der && lista[izq] <= pivote) {
      izq = izq + 1;
    }
    while (lista[der] >= pivote && der >= izq) {
      der = der - 1;
    }
    if (der < izq) {
      done = 1;
    } else {
      temp = lista[izq];
      lista[izq] = lista[der];
      lista[der] = temp;
    }
  }
  temp = lista[f];
  lista[f] = lista[der];
  lista[der] = temp;
  return der;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<particion>:
  endbr64
  movslq %esi,%rax
  mov    %rdi,%r9
  lea    0x1(%rsi),%ecx
  lea    (%rdi,%rax,4),%r11
  mov    %edx,%eax
  movslq %eax,%r10
  mov    (%r11),%edi
  lea    (%r9,%r10,4),%r8
  mov    (%r8),%esi
  cmp    %eax,%ecx
  jg     L103
  nopl   0x0(%rax)
L79:
  movslq %ecx,%rdx
  lea    (%r9,%rdx,4),%rdx
  jmp    L107
  nopl   0x0(%rax)
L90:
  add    $0x1,%ecx
  add    $0x4,%rdx
  cmp    %eax,%ecx
  jg     L103
L107:
  cmp    %edi,(%rdx)
  jle    L90
  cmp    %esi,%edi
  jg     L86
  cmp    %eax,%ecx
  jg     L103
  lea    -0x4(%r9,%r10,4),%rdx
  jmp    L35
  cs nopw 0x0(%rax,%rax,1)
L21:
  sub    $0x4,%rdx
  cmp    %ecx,%eax
  jl     L103
L35:
  mov    (%rdx),%esi
  sub    $0x1,%eax
  mov    %rdx,%r8
  cmp    %edi,%esi
  jge    L21
L86:
  cmp    %eax,%ecx
  jg     L103
  movslq %ecx,%rdx
  lea    (%r9,%rdx,4),%rdx
  mov    (%rdx),%r10d
  mov    %esi,(%rdx)
  mov    %r10d,(%r8)
  movslq %eax,%r10
  lea    (%r9,%r10,4),%r8
  mov    (%r8),%esi
  cmp    %eax,%ecx
  jle    L79
L103:
  mov    (%r11),%edx
  mov    %esi,(%r11)
  mov    %edx,(%r8)
  ret
```

assistant: ```c
int particion(int lista[], int f, int l) {
  int pivote = lista[f];
  int izq = f + 1;
  int der = l;
  int temp;
  int done = 0;
  while (!done) {
    while (izq <= der && lista[izq] <= pivote) {
      izq = izq + 1;
    }
    while (lista[der] >= pivote && der >= izq) {
      der = der - 1;
    }
    if (der < izq) {
      done = 1;
    } else {
      temp = lista[izq];
      lista[izq] = lista[der];
      lista[der] = temp;
    }
  }
  temp = lista[f];
  lista[f] = lista[der];
  lista[der] = temp;
  return der;
}
```