ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<strncpy_no_overflow2>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x4,%esi
  jg     L27
L15:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L27:
  mov    %rdi,%rbp
  mov    %esi,%ebx
  call   <strlen@plt>
  cmp    $0x3,%eax
  jne    L15
  lea    0x4(%rsp),%rdi
  mov    %ebx,%edx
  mov    %rbp,%rsi
  call   <strncpy@plt>
  jmp    L15
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void strncpy_no_overflow2(char *y, int n) {
  if (n <= 4)
    return;
  char x[4];
  if (strlen(y) == 3)
    strncpy(x, y, n);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strncpy_no_overflow2>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x4,%esi
  jg     L14
L21:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L14:
  mov    %rdi,%rbp
  mov    %esi,%r12d
  call   <strlen@plt>
  cmp    $0x3,%eax
  jne    L21
  lea    0x4(%rsp),%rdi
  mov    %r12d,%edx
  mov    %rbp,%rsi
  call   <strncpy@plt>
  jmp    L21
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void strncpy_no_overflow2(char *y, int n) {
  if (n <= 4)
    return;
  char x[4];
  if (strlen(y) == 3)
    strncpy(x, y, n);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strncpy_no_overflow2>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x4,%esi
  jg     L34
L7:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L34:
  mov    %rdi,%rbp
  mov    %esi,%r12d
  call   <strlen@plt>
  cmp    $0x3,%eax
  jne    L7
  lea    0x4(%rsp),%rdi
  mov    %r12d,%edx
  mov    %rbp,%rsi
  call   <strncpy@plt>
  jmp    L7
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void strncpy_no_overflow2(char *y, int n) {
  if (n <= 4)
    return;
  char x[4];
  if (strlen(y) == 3)
    strncpy(x, y, n);
}
```
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
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  movl   $0xa,-0x2c(%rbp)
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  movslq %eax,%rdx
  mov    %rdx,%rsi
  mov    $0x0,%edi
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%ecx
  mov    $0x0,%edx
  div    %rcx
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L22:
  cmp    %rdx,%rsp
  je     L53
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L22
L53:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L57
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L57:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x20(%rbp)
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x38(%rbp)
  jmp    L67
L12:
  mov    -0x38(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x38(%rbp)
L67:
  mov    -0x38(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L12
  movl   $0x0,-0x34(%rbp)
  jmp    L74
L21:
  mov    -0x20(%rbp),%rdx
  mov    -0x34(%rbp),%eax
  cltq
  movzbl (%rdx,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x34(%rbp)
L74:
  mov    -0x34(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L21
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x2c(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <maopao@plt>
  movl   $0x0,-0x30(%rbp)
  jmp    L88
L84:
  mov    -0x20(%rbp),%rdx
  mov    -0x30(%rbp),%eax
  cltq
  movzbl (%rdx,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x30(%rbp)
L88:
  mov    -0x30(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L84
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rbx,%rsp
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D36
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D28
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D9
type_guessed: string

tool: D36: array elements:

tool: D28: %c

tool: D9: %c	

assistant: ```c
int main() {
  int num = 10;
  char array[num];
  printf("array elements:\n");
  for (int i = 0; i < num; i++)
    scanf("%c", &array[i]);
  for (int i = 0; i < num; i++)
    printf("%c\t", array[i]);
  printf("\n");
  maopao(array, num);
  for (int i = 0; i < num; i++)
    printf("%c\t", array[i]);
  printf("\n");
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D14(%rip),%rdi
  call   <puts@plt>
  lea    0xe(%rsp),%rbx
  lea    0x18(%rsp),%r12
  mov    %rbx,%rbp
  lea    D8(%rip),%r13
L22:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x1,%rbp
  cmp    %r12,%rbp
  jne    L22
  mov    %rbx,%rbp
  lea    D30(%rip),%r13
L24:
  movsbl 0x0(%rbp),%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbp
  cmp    %r12,%rbp
  jne    L24
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    0xe(%rsp),%rdi
  mov    $0xa,%esi
  call   <maopao@plt>
  lea    D30(%rip),%rbp
L28:
  movsbl (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L28
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D14
type_guessed: string
name: retrieve_data
arguments: data_label: D8
type_guessed: string
name: retrieve_data
arguments: data_label: D30
type_guessed: string

tool: D14: array elements:

tool: D8: %c

tool: D30: %c	

assistant: ```c
int main() {
  int num = 10;
  char array[num];
  printf("array elements:\n");
  for (int i = 0; i < num; i++)
    scanf("%c", &array[i]);
  for (int i = 0; i < num; i++)
    printf("%c\t", array[i]);
  printf("\n");
  maopao(array, num);
  for (int i = 0; i < num; i++)
    printf("%c\t", array[i]);
  printf("\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r14
  lea    D6(%rip),%rdi
  push   %r13
  lea    D28(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%r14
  lea    0x18(%rsp),%r12
  call   <puts@plt>
  mov    %r14,%rbx
  mov    %r14,%rbp
  nopl   0x0(%rax,%rax,1)
L23:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x1,%rbp
  call   <scanf@plt>
  cmp    %r12,%rbp
  jne    L23
  mov    %r14,%rbp
  lea    D10(%rip),%r13
  cs nopw 0x0(%rax,%rax,1)
L32:
  movsbl 0x0(%rbp),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <printf@plt>
  cmp    %r12,%rbp
  jne    L32
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0xa,%esi
  mov    %r14,%rdi
  call   <maopao@plt>
  xchg   %ax,%ax
L2:
  movsbl (%rbx),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L2
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D6
datatype: string
name: retrieve_data
arguments: label: D28
datatype: string
name: retrieve_data
arguments: label: D10
datatype: string

tool: D6: array elements:

tool: D28: %c

tool: D10: %c	

assistant: ```c
int main() {
  int num = 10;
  char array[num];
  printf("array elements:\n");
  for (int i = 0; i < num; i++)
    scanf("%c", &array[i]);
  for (int i = 0; i < num; i++)
    printf("%c\t", array[i]);
  printf("\n");
  maopao(array, num);
  for (int i = 0; i < num; i++)
    printf("%c\t", array[i]);
  printf("\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  lea    D8(%rip),%rdi
  push   %r13
  lea    D13(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%r14
  lea    0x18(%rsp),%r12
  call   <puts@plt>
  mov    %r14,%rbx
  mov    %r14,%rbp
  nopl   0x0(%rax,%rax,1)
L4:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x1,%rbp
  call   <scanf@plt>
  cmp    %r12,%rbp
  jne    L4
  mov    %r14,%rbp
  lea    D17(%rip),%r13
  cs nopw 0x0(%rax,%rax,1)
L35:
  movsbl 0x0(%rbp),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <printf@plt>
  cmp    %r12,%rbp
  jne    L35
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0xa,%esi
  mov    %r14,%rdi
  call   <maopao@plt>
  xchg   %ax,%ax
L25:
  movsbl (%rbx),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L25
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D8
datatype: string
name: read_data
arguments: tag: D13
datatype: string
name: read_data
arguments: tag: D17
datatype: string

tool: D8: array elements:

tool: D13: %c

tool: D17: %c	

assistant: ```c
int main() {
  int num = 10;
  char array[num];
  printf("array elements:\n");
  for (int i = 0; i < num; i++)
    scanf("%c", &array[i]);
  for (int i = 0; i < num; i++)
    printf("%c\t", array[i]);
  printf("\n");
  maopao(array, num);
  for (int i = 0; i < num; i++)
    printf("%c\t", array[i]);
  printf("\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<a52_resample_MONO_to_5_C>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L14
L12:
  mov    -0x1c(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  cltq
  add    $0x3,%rax
  lea    (%rax,%rax,1),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  movw   $0x0,(%rdx)
  mov    -0x1c(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  cltq
  add    $0x2,%rax
  lea    (%rax,%rax,1),%rcx
  mov    -0x30(%rbp),%rax
  add    %rax,%rcx
  movzwl (%rdx),%eax
  mov    %ax,(%rcx)
  mov    -0x1c(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  cltq
  add    $0x1,%rax
  lea    (%rax,%rax,1),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  movzwl (%rcx),%eax
  mov    %ax,(%rdx)
  mov    -0x1c(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  cltq
  lea    (%rax,%rax,1),%rcx
  mov    -0x30(%rbp),%rax
  add    %rax,%rcx
  movzwl (%rdx),%eax
  mov    %ax,(%rcx)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x1c(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  cltq
  add    $0x4,%rax
  lea    (%rax,%rax,1),%rcx
  mov    -0x30(%rbp),%rax
  lea    (%rcx,%rax,1),%rbx
  mov    %edx,%edi
  call   <convert@plt>
  mov    %ax,(%rbx)
  addl   $0x1,-0x1c(%rbp)
L14:
  cmpl   $0xff,-0x1c(%rbp)
  jle    L12
  mov    $0x500,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int a52_resample_MONO_to_5_C(float *_f, signed short *s16) {
  int i;
  signed int *f = (signed int *)_f;
  for (i = 0; i < 256; i++) {
    s16[5 * i] = s16[5 * i + 1] = s16[5 * i + 2] = s16[5 * i + 3] = 0;
    s16[5 * i + 4] = convert(f[i]);
  }
  return 5 * 256;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<a52_resample_MONO_to_5_C>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    %rdi,%rbp
  lea    0xa00(%rsi),%r12
L5:
  movw   $0x0,0x6(%rbx)
  movw   $0x0,0x4(%rbx)
  movw   $0x0,0x2(%rbx)
  movw   $0x0,(%rbx)
  mov    0x0(%rbp),%edi
  call   <convert@plt>
  mov    %ax,0x8(%rbx)
  add    $0xa,%rbx
  add    $0x4,%rbp
  cmp    %r12,%rbx
  jne    L5
  mov    $0x500,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int a52_resample_MONO_to_5_C(float *_f, signed short *s16) {
  int i;
  signed int *f = (signed int *)_f;
  for (i = 0; i < 256; i++) {
    s16[5 * i] = s16[5 * i + 1] = s16[5 * i + 2] = s16[5 * i + 3] = 0;
    s16[5 * i + 4] = convert(f[i]);
  }
  return 5 * 256;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<a52_resample_MONO_to_5_C>:
  endbr64
  push   %r12
  lea    0xa00(%rsi),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  nopl   (%rax)
L0:
  xor    %eax,%eax
  xor    %edx,%edx
  xor    %ecx,%ecx
  xor    %esi,%esi
  mov    %ax,0x6(%rbx)
  mov    0x0(%rbp),%edi
  add    $0xa,%rbx
  add    $0x4,%rbp
  mov    %dx,-0x6(%rbx)
  mov    %cx,-0x8(%rbx)
  mov    %si,-0xa(%rbx)
  call   <convert@plt>
  mov    %ax,-0x2(%rbx)
  cmp    %r12,%rbx
  jne    L0
  pop    %rbx
  mov    $0x500,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int a52_resample_MONO_to_5_C(float *_f, signed short *s16) {
  int i;
  signed int *f = (signed int *)_f;
  for (i = 0; i < 256; i++) {
    s16[5 * i] = s16[5 * i + 1] = s16[5 * i + 2] = s16[5 * i + 3] = 0;
    s16[5 * i + 4] = convert(f[i]);
  }
  return 5 * 256;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<a52_resample_MONO_to_5_C>:
  endbr64
  push   %r12
  lea    0xa00(%rsi),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  nopl   (%rax)
L6:
  movq   $0x0,(%rbx)
  mov    0x0(%rbp),%edi
  add    $0xa,%rbx
  add    $0x4,%rbp
  call   <convert@plt>
  mov    %ax,-0x2(%rbx)
  cmp    %rbx,%r12
  jne    L6
  pop    %rbx
  mov    $0x500,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int a52_resample_MONO_to_5_C(float *_f, signed short *s16) {
  int i;
  signed int *f = (signed int *)_f;
  for (i = 0; i < 256; i++) {
    s16[5 * i] = s16[5 * i + 1] = s16[5 * i + 2] = s16[5 * i + 3] = 0;
    s16[5 * i + 4] = convert(f[i]);
  }
  return 5 * 256;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <initArray@plt>
  mov    $0x0,%eax
  call   <printArray@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char const *argv[]) {
  initArray();
  printArray();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <initArray@plt>
  mov    $0x0,%eax
  call   <printArray@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char const *argv[]) {
  initArray();
  printArray();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <initArray@plt>
  xor    %eax,%eax
  call   <printArray@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char const *argv[]) {
  initArray();
  printArray();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <initArray@plt>
  xor    %eax,%eax
  call   <printArray@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char const *argv[]) {
  initArray();
  printArray();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<casoCuatro>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x38(%rbp)
  movsd  %xmm1,-0x40(%rbp)
  movsd  %xmm2,-0x48(%rbp)
  movsd  -0x38(%rbp),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x40(%rbp),%xmm1
  movsd  D2(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  mulsd  -0x48(%rbp),%xmm0
  movsd  -0x38(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x28(%rbp)
  movsd  -0x40(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x20(%rbp),%xmm0
  subsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  addsd  -0x40(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x20(%rbp),%xmm0
  comisd -0x28(%rbp),%xmm0
  jbe    L5
  movsd  -0x20(%rbp),%xmm0
  addsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  jmp    L8
L5:
  movsd  -0x20(%rbp),%xmm1
  movsd  D45(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x28(%rbp),%xmm1
  movsd  D29(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x28(%rbp)
  movsd  -0x40(%rbp),%xmm1
  movsd  D2(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  -0x8(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jbe    L7
  movsd  -0x20(%rbp),%xmm0
  addsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  jmp    L8
L7:
  movsd  -0x20(%rbp),%xmm1
  movsd  D45(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x28(%rbp),%xmm0
  movsd  D29(%rip),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x28(%rbp)
L8:
  movsd  -0x28(%rbp),%xmm0
  mulsd  -0x10(%rbp),%xmm0
  addsd  -0x20(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D2
type_guessed: f64
name: read_data
arguments: label: D45
type_guessed: f64
name: read_data
arguments: label: D29
type_guessed: byte[8]

tool: D2: 5

tool: D45: 2.75

tool: {"D29": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double casoCuatro(double a, double b, double c) {
  double f;
  double d = (a * b) + 5;
  f = a + (d * c);
  double r = b * b;
  double k = r - f;
  double s = k + b;
  if (r > f) {
    r = r + f;
  } else {
    r = r * 2.75;
    f++;
    if (s > 5 * b) {
      r = r + f;
    } else {
      r = r * 2.75;
      f--;
    }
  }
  return f * k + r;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<casoCuatro>:
  endbr64
  movapd %xmm0,%xmm3
  mulsd  %xmm1,%xmm0
  addsd  D31(%rip),%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm3,%xmm0
  movapd %xmm1,%xmm2
  mulsd  %xmm1,%xmm2
  movapd %xmm2,%xmm3
  subsd  %xmm0,%xmm3
  comisd %xmm0,%xmm2
  jbe    L32
  addsd  %xmm0,%xmm2
L16:
  mulsd  %xmm3,%xmm0
  addsd  %xmm2,%xmm0
  ret
L32:
  mulsd  D25(%rip),%xmm2
  addsd  D0(%rip),%xmm0
  movapd %xmm1,%xmm4
  mulsd  D31(%rip),%xmm4
  addsd  %xmm3,%xmm1
  comisd %xmm4,%xmm1
  jbe    L21
  addsd  %xmm0,%xmm2
  jmp    L16
L21:
  mulsd  D25(%rip),%xmm2
  subsd  D0(%rip),%xmm0
  jmp    L16
```

tool_calls: name: unpack_data
arguments: data_label: D31
type_guessed: f64
name: unpack_data
arguments: data_label: D25
type_guessed: f64
name: unpack_data
arguments: data_label: D0
type_guessed: byte[8]

tool: D31: 5

tool: D25: 2.75

tool: {"D0": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double casoCuatro(double a, double b, double c) {
  double f;
  double d = (a * b) + 5;
  f = a + (d * c);
  double r = b * b;
  double k = r - f;
  double s = k + b;
  if (r > f) {
    r = r + f;
  } else {
    r = r * 2.75;
    f++;
    if (s > 5 * b) {
      r = r + f;
    } else {
      r = r * 2.75;
      f--;
    }
  }
  return f * k + r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<casoCuatro>:
  endbr64
  movapd %xmm0,%xmm3
  movsd  D27(%rip),%xmm4
  mulsd  %xmm1,%xmm0
  addsd  %xmm4,%xmm0
  mulsd  %xmm2,%xmm0
  movapd %xmm1,%xmm2
  mulsd  %xmm1,%xmm2
  addsd  %xmm3,%xmm0
  movapd %xmm2,%xmm3
  comisd %xmm0,%xmm2
  subsd  %xmm0,%xmm3
  ja     L17
  mulsd  %xmm1,%xmm4
  addsd  %xmm3,%xmm1
  movsd  D20(%rip),%xmm6
  movsd  D6(%rip),%xmm5
  mulsd  %xmm6,%xmm2
  addsd  %xmm5,%xmm0
  comisd %xmm4,%xmm1
  jbe    L19
L17:
  addsd  %xmm0,%xmm2
  mulsd  %xmm3,%xmm0
  addsd  %xmm2,%xmm0
  ret
  nopl   0x0(%rax)
L19:
  subsd  %xmm5,%xmm0
  mulsd  %xmm6,%xmm2
  mulsd  %xmm3,%xmm0
  addsd  %xmm2,%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D27
type_guessed: f64
name: extract_symbolic_data
arguments: label: D20
type_guessed: f64
name: extract_symbolic_data
arguments: label: D6
type_guessed: byte[8]

tool: D27: 5

tool: D20: 2.75

tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double casoCuatro(double a, double b, double c) {
  double f;
  double d = (a * b) + 5;
  f = a + (d * c);
  double r = b * b;
  double k = r - f;
  double s = k + b;
  if (r > f) {
    r = r + f;
  } else {
    r = r * 2.75;
    f++;
    if (s > 5 * b) {
      r = r + f;
    } else {
      r = r * 2.75;
      f--;
    }
  }
  return f * k + r;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<casoCuatro>:
  endbr64
  movapd %xmm0,%xmm3
  movsd  D19(%rip),%xmm4
  mulsd  %xmm1,%xmm0
  addsd  %xmm4,%xmm0
  mulsd  %xmm2,%xmm0
  movapd %xmm1,%xmm2
  mulsd  %xmm1,%xmm2
  addsd  %xmm3,%xmm0
  movapd %xmm2,%xmm3
  comisd %xmm0,%xmm2
  subsd  %xmm0,%xmm3
  ja     L9
  mulsd  %xmm1,%xmm4
  addsd  %xmm3,%xmm1
  movsd  D11(%rip),%xmm6
  movsd  D7(%rip),%xmm5
  mulsd  %xmm6,%xmm2
  addsd  %xmm5,%xmm0
  comisd %xmm4,%xmm1
  jbe    L19
L9:
  addsd  %xmm0,%xmm2
  mulsd  %xmm3,%xmm0
  addsd  %xmm2,%xmm0
  ret
  nopl   0x0(%rax)
L19:
  subsd  %xmm5,%xmm0
  mulsd  %xmm6,%xmm2
  mulsd  %xmm3,%xmm0
  addsd  %xmm2,%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D19
type_guessed: f64
name: extract_symbolic_data
arguments: tag: D11
type_guessed: f64
name: extract_symbolic_data
arguments: tag: D7
type_guessed: byte[8]

tool: D19: 5

tool: D11: 2.75

tool: {"D7": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double casoCuatro(double a, double b, double c) {
  double f;
  double d = (a * b) + 5;
  f = a + (d * c);
  double r = b * b;
  double k = r - f;
  double s = k + b;
  if (r > f) {
    r = r + f;
  } else {
    r = r * 2.75;
    f++;
    if (s > 5 * b) {
      r = r + f;
    } else {
      r = r * 2.75;
      f--;
    }
  }
  return f * k + r;
}
```
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
  mov    $0x0,%eax
  call   <simple@plt>
  mov    $0x0,%eax
  call   <loop@plt>
  mov    $0x0,%eax
  call   <undef@plt>
  mov    $0x0,%eax
  call   <cond_incl@plt>
  mov    $0x0,%eax
  call   <const_expr@plt>
  mov    $0x0,%eax
  call   <defined@plt>
  mov    $0x0,%eax
  call   <ifdef@plt>
  mov    $0x0,%eax
  call   <funclike@plt>
  mov    $0x0,%eax
  call   <empty@plt>
  mov    $0x0,%eax
  call   <noarg@plt>
  mov    $0x0,%eax
  call   <null@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  simple();
  loop();
  undef();
  cond_incl();
  const_expr();
  defined();
  ifdef();
  funclike();
  empty();
  noarg();
  null();
  return 0;
}
```