ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x14(%rbp),%rcx
  lea    -0x18(%rbp),%rdx
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movss  -0x14(%rbp),%xmm1
  movss  -0x1c(%rbp),%xmm0
  mulss  %xmm0,%xmm1
  movss  -0x18(%rbp),%xmm0
  mulss  %xmm1,%xmm0
  movss  %xmm0,-0x10(%rbp)
  pxor   %xmm2,%xmm2
  cvtss2sd -0x10(%rbp),%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movss  -0x18(%rbp),%xmm1
  movss  -0x10(%rbp),%xmm0
  divss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  pxor   %xmm3,%xmm3
  cvtss2sd -0xc(%rbp),%xmm3
  movq   %xmm3,%rax
  movq   %rax,%xmm0
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D3
type: string
name: unpack_data
arguments: tag: D24
type: string
name: unpack_data
arguments: tag: D30
type: string
name: unpack_data
arguments: tag: D38
type: string

tool: D3: Enter velocity,time and force

tool: D24: %f%f%f

tool: D30: Work=
%f

tool: D38: Power=
%f

assistant: ```c
int main() {
  float v, t, f, w, p;
  printf("Enter velocity,time and force\n");
  scanf("%f%f%f", &v, &t, &f);
  w = f * v * t;
  printf("Work=\n%f", w);
  p = w / t;
  printf("Power=\n%f", p);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D31(%rip),%rdi
  call   <puts@plt>
  lea    0x24(%rsp),%rcx
  lea    0x20(%rsp),%rdx
  lea    0x1c(%rsp),%rsi
  lea    D22(%rip),%rdi
  call   <scanf@plt>
  movss  0x24(%rsp),%xmm0
  mulss  0x1c(%rsp),%xmm0
  mulss  0x20(%rsp),%xmm0
  movss  %xmm0,0xc(%rsp)
  cvtss2sd %xmm0,%xmm0
  lea    D8(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movss  0xc(%rsp),%xmm0
  divss  0x20(%rsp),%xmm0
  cvtss2sd %xmm0,%xmm0
  lea    D35(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x38,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D31
data_type: string
name: unpack_data
arguments: data_label: D22
data_type: string
name: unpack_data
arguments: data_label: D8
data_type: string
name: unpack_data
arguments: data_label: D35
data_type: string

tool: D31: Enter velocity,time and force

tool: D22: %f%f%f

tool: D8: Work=
%f

tool: D35: Power=
%f

assistant: ```c
int main() {
  float v, t, f, w, p;
  printf("Enter velocity,time and force\n");
  scanf("%f%f%f", &v, &t, &f);
  w = f * v * t;
  printf("Work=\n%f", w);
  p = w / t;
  printf("Power=\n%f", p);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  lea    D36(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x24(%rsp),%rcx
  lea    0x20(%rsp),%rdx
  lea    0x1c(%rsp),%rsi
  lea    D12(%rip),%rdi
  call   <scanf@plt>
  movss  0x24(%rsp),%xmm1
  mulss  0x1c(%rsp),%xmm1
  lea    D17(%rip),%rdi
  mulss  0x20(%rsp),%xmm1
  pxor   %xmm0,%xmm0
  mov    $0x1,%eax
  cvtss2sd %xmm1,%xmm0
  movss  %xmm1,0xc(%rsp)
  call   <printf@plt>
  pxor   %xmm0,%xmm0
  mov    $0x1,%eax
  lea    D39(%rip),%rdi
  movss  0xc(%rsp),%xmm1
  divss  0x20(%rsp),%xmm1
  cvtss2sd %xmm1,%xmm0
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D36
type_guessed: string
name: unpack_data
arguments: data_tag: D12
type_guessed: string
name: unpack_data
arguments: data_tag: D17
type_guessed: string
name: unpack_data
arguments: data_tag: D39
type_guessed: string

tool: D36: Enter velocity,time and force

tool: D12: %f%f%f

tool: D17: Work=
%f

tool: D39: Power=
%f

assistant: ```c
int main() {
  float v, t, f, w, p;
  printf("Enter velocity,time and force\n");
  scanf("%f%f%f", &v, &t, &f);
  w = f * v * t;
  printf("Work=\n%f", w);
  p = w / t;
  printf("Power=\n%f", p);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  lea    D30(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x24(%rsp),%rcx
  lea    0x20(%rsp),%rdx
  lea    0x1c(%rsp),%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  movss  0x24(%rsp),%xmm1
  mulss  0x1c(%rsp),%xmm1
  lea    D26(%rip),%rdi
  mulss  0x20(%rsp),%xmm1
  pxor   %xmm0,%xmm0
  mov    $0x1,%eax
  cvtss2sd %xmm1,%xmm0
  movss  %xmm1,0xc(%rsp)
  call   <printf@plt>
  pxor   %xmm0,%xmm0
  mov    $0x1,%eax
  lea    D2(%rip),%rdi
  movss  0xc(%rsp),%xmm1
  divss  0x20(%rsp),%xmm1
  cvtss2sd %xmm1,%xmm0
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D30
type_infered: string
name: read_data
arguments: data_tag: D14
type_infered: string
name: read_data
arguments: data_tag: D26
type_infered: string
name: read_data
arguments: data_tag: D2
type_infered: string

tool: D30: Enter velocity,time and force

tool: D14: %f%f%f

tool: D26: Work=
%f

tool: D2: Power=
%f

assistant: ```c
int main() {
  float v, t, f, w, p;
  printf("Enter velocity,time and force\n");
  scanf("%f%f%f", &v, &t, &f);
  w = f * v * t;
  printf("Work=\n%f", w);
  p = w / t;
  printf("Power=\n%f", p);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
int main(int argc, char **argv) {
  h();
  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    $0x0,%eax
  call   <h@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  h();
  return 0;
}
```
user: What is the c source code of the assembly code below?

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

assistant: ```c
int main(int argc, char **argv) {
  h();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

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

assistant: ```c
int main(int argc, char **argv) {
  h();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<yas_bma150_power_down>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0xa,%edi
  call   <yas_bma150_read_reg_byte@plt>
  or     $0x1,%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <yas_bma150_write_reg_byte@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int yas_bma150_power_down(void) {
  yas_bma150_write_reg_byte(0x0a, ((yas_bma150_read_reg_byte(0x0a) & ~0x01) | ((1) << 0)));
  return (0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<yas_bma150_power_down>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%edi
  call   <yas_bma150_read_reg_byte@plt>
  or     $0x1,%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <yas_bma150_write_reg_byte@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int yas_bma150_power_down(void) {
  yas_bma150_write_reg_byte(0x0a, ((yas_bma150_read_reg_byte(0x0a) & ~0x01) | ((1) << 0)));
  return (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<yas_bma150_power_down>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%edi
  call   <yas_bma150_read_reg_byte@plt>
  mov    $0xa,%edi
  or     $0x1,%eax
  mov    %eax,%esi
  call   <yas_bma150_write_reg_byte@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int yas_bma150_power_down(void) {
  yas_bma150_write_reg_byte(0x0a, ((yas_bma150_read_reg_byte(0x0a) & ~0x01) | ((1) << 0)));
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<yas_bma150_power_down>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%edi
  call   <yas_bma150_read_reg_byte@plt>
  mov    $0xa,%edi
  or     $0x1,%eax
  mov    %eax,%esi
  call   <yas_bma150_write_reg_byte@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int yas_bma150_power_down(void) {
  yas_bma150_write_reg_byte(0x0a, ((yas_bma150_read_reg_byte(0x0a) & ~0x01) | ((1) << 0)));
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<getlcookie>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movq   $0x0,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  lea    D0(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strstr@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L29
  mov    -0x10(%rbp),%rax
  add    $0x2,%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    $0x26,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L12
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
L12:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L29
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  movq   $0x0,-0x8(%rbp)
L29:
  mov    -0x18(%rbp),%rax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D0
type_infered: string

tool: D0: n=

assistant: ```c
char *getlcookie(char *cookie) {
  char *tmp;
  char *tmpend;
  char *login_cookie = NULL;
  tmpend = strstr(cookie, "n=");
  if (tmpend) {
    tmp = strdup(tmpend + 2);
    tmpend = strchr(tmp, '&');
    if (tmpend)
      *tmpend = '\0';
    login_cookie = strdup(tmp);
    if (tmp) {
      free(tmp);
      tmp = NULL;
    };
  }
  return login_cookie;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getlcookie>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    D1(%rip),%rsi
  call   <strstr@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L1
  lea    0x2(%rax),%rdi
  call   <strdup@plt>
  mov    %rax,%rbp
  mov    $0x26,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L14
  movb   $0x0,(%rax)
L14:
  mov    %rbp,%rdi
  call   <strdup@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdi
  call   <free@plt>
L1:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D1
datatype: string

tool: D1: n=

assistant: ```c
char *getlcookie(char *cookie) {
  char *tmp;
  char *tmpend;
  char *login_cookie = NULL;
  tmpend = strstr(cookie, "n=");
  if (tmpend) {
    tmp = strdup(tmpend + 2);
    tmpend = strchr(tmp, '&');
    if (tmpend)
      *tmpend = '\0';
    login_cookie = strdup(tmp);
    if (tmp) {
      free(tmp);
      tmp = NULL;
    };
  }
  return login_cookie;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getlcookie>:
  endbr64
  push   %r12
  lea    D9(%rip),%rsi
  push   %rbp
  sub    $0x8,%rsp
  call   <strstr@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L10
  lea    0x2(%rax),%rdi
  call   <strdup@plt>
  mov    $0x26,%esi
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <strchr@plt>
  test   %rax,%rax
  je     L16
  movb   $0x0,(%rax)
L16:
  mov    %rbp,%rdi
  call   <strdup@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <free@plt>
L10:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D9
datatype: string

tool: D9: n=

assistant: ```c
char *getlcookie(char *cookie) {
  char *tmp;
  char *tmpend;
  char *login_cookie = NULL;
  tmpend = strstr(cookie, "n=");
  if (tmpend) {
    tmp = strdup(tmpend + 2);
    tmpend = strchr(tmp, '&');
    if (tmpend)
      *tmpend = '\0';
    login_cookie = strdup(tmp);
    if (tmp) {
      free(tmp);
      tmp = NULL;
    };
  }
  return login_cookie;
}
```
user: What is the c source code of the assembly code below?

```asm
<getlcookie>:
  endbr64
  push   %r12
  lea    D6(%rip),%rsi
  push   %rbp
  sub    $0x8,%rsp
  call   <strstr@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L18
  lea    0x2(%rax),%rdi
  call   <strdup@plt>
  mov    $0x26,%esi
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <strchr@plt>
  test   %rax,%rax
  je     L4
  movb   $0x0,(%rax)
L4:
  mov    %rbp,%rdi
  call   <strdup@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <free@plt>
L18:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: label: D6
type_infered: string

tool: D6: n=

assistant: ```c
char *getlcookie(char *cookie) {
  char *tmp;
  char *tmpend;
  char *login_cookie = NULL;
  tmpend = strstr(cookie, "n=");
  if (tmpend) {
    tmp = strdup(tmpend + 2);
    tmpend = strchr(tmp, '&');
    if (tmpend)
      *tmpend = '\0';
    login_cookie = strdup(tmp);
    if (tmp) {
      free(tmp);
      tmp = NULL;
    };
  }
  return login_cookie;
}
```
user: What is the c source code of the assembly code below:

```asm
<xfree>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L0
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L0:
  nop
  leave
  ret
```

assistant: ```c
void xfree(void *ptr) {
  if (ptr) {
    free(ptr);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<xfree>:
  endbr64
  test   %rdi,%rdi
  je     L9
  sub    $0x8,%rsp
  call   <free@plt>
  add    $0x8,%rsp
  ret
L9:
  ret
```

assistant: ```c
void xfree(void *ptr) {
  if (ptr) {
    free(ptr);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<xfree>:
  endbr64
  test   %rdi,%rdi
  je     L4
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L4:
  ret
```

assistant: ```c
void xfree(void *ptr) {
  if (ptr) {
    free(ptr);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<xfree>:
  endbr64
  test   %rdi,%rdi
  je     L6
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L6:
  ret
```

assistant: ```c
void xfree(void *ptr) {
  if (ptr) {
    free(ptr);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L4:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L13:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L4
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int sum(int data[], int length) {
  int result = data[0];
  for (int i = 0; i < length; i++)
    result += data[i];
  return result;
}
```
user: What is the c source code of the assembly code below:

```asm
<sum>:
  endbr64
  mov    (%rdi),%edx
  test   %esi,%esi
  jle    L4
  mov    %rdi,%rax
  lea    -0x1(%rsi),%ecx
  lea    0x4(%rdi,%rcx,4),%rcx
L17:
  add    (%rax),%edx
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L17
L4:
  mov    %edx,%eax
  ret
```

assistant: ```c
int sum(int data[], int length) {
  int result = data[0];
  for (int i = 0; i < length; i++)
    result += data[i];
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sum>:
  endbr64
  mov    (%rdi),%eax
  test   %esi,%esi
  jle    L18
  lea    -0x1(%rsi),%edx
  add    $0x4,%rdi
  lea    (%rdi,%rdx,4),%rcx
  mov    %eax,%edx
  jmp    L20
  nopl   0x0(%rax)
L11:
  mov    (%rdi),%edx
  add    $0x4,%rdi
L20:
  add    %edx,%eax
  cmp    %rcx,%rdi
  jne    L11
L18:
  ret
```

assistant: ```c
int sum(int data[], int length) {
  int result = data[0];
  for (int i = 0; i < length; i++)
    result += data[i];
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<sum>:
  endbr64
  mov    (%rdi),%r8d
  mov    %rdi,%rdx
  test   %esi,%esi
  jle    L11
  lea    -0x1(%rsi),%eax
  cmp    $0x2,%eax
  jbe    L22
  mov    %esi,%ecx
  mov    %rdi,%rax
  pxor   %xmm0,%xmm0
  shr    $0x2,%ecx
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rdi,%rcx,1),%rcx
  cs nopw 0x0(%rax,%rax,1)
L27:
  movdqu (%rax),%xmm2
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  cmp    %rcx,%rax
  jne    L27
  movdqa %xmm0,%xmm1
  psrldq $0x8,%xmm1
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  add    %eax,%r8d
  mov    %esi,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%sil
  je     L11
L66:
  movslq %eax,%rcx
  lea    0x0(,%rcx,4),%rdi
  add    (%rdx,%rcx,4),%r8d
  lea    0x1(%rax),%ecx
  cmp    %ecx,%esi
  jle    L11
  add    $0x2,%eax
  add    0x4(%rdx,%rdi,1),%r8d
  cmp    %eax,%esi
  jle    L11
  add    0x8(%rdx,%rdi,1),%r8d
L11:
  mov    %r8d,%eax
  ret
L22:
  xor    %eax,%eax
  jmp    L66
```

assistant: ```c
int sum(int data[], int length) {
  int result = data[0];
  for (int i = 0; i < length; i++)
    result += data[i];
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<muldivu64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  cltd
  idivl  -0x1c(%rbp)
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  cltd
  idivl  -0x1c(%rbp)
  mov    %edx,-0x4(%rbp)
  jmp    L17
L5:
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L37
  mov    -0x8(%rbp),%eax
  add    %eax,-0x10(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L37
  addl   $0x1,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    %eax,-0xc(%rbp)
L37:
  sarl   -0x14(%rbp)
  shll   -0x8(%rbp)
  shll   -0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L17
  addl   $0x1,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    %eax,-0x4(%rbp)
L17:
  cmpl   $0x0,-0x14(%rbp)
  jne    L5
  mov    -0x10(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
uint64_t muldivu64(uint64_t a, uint64_t b, uint64_t c) {
  uint64_t quotient = 0;
  uint64_t remainder = 0;
  uint64_t cur_quotient = b / c;
  uint64_t cur_remainder = b % c;
  while (a) {
    if (a & 1) {
      quotient += cur_quotient;
      remainder += cur_remainder;
      if (remainder >= c) {
        quotient++;
        remainder -= c;
      }
    }
    a >>= 1;
    cur_quotient <<= 1;
    cur_remainder <<= 1;
    if (cur_remainder >= c) {
      cur_quotient++;
      cur_remainder -= c;
    }
  }
  return quotient;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<muldivu64>:
  endbr64
  mov    %esi,%eax
  mov    %edx,%ecx
  cltd
  idiv   %ecx
  test   %edi,%edi
  je     L28
  mov    $0x0,%esi
  mov    $0x0,%r8d
  jmp    L59
L24:
  sar    %edi
  add    %eax,%eax
  add    %edx,%edx
  cmp    %edx,%ecx
  jg     L18
  add    $0x1,%eax
  sub    %ecx,%edx
L18:
  test   %edi,%edi
  je     L33
L59:
  test   $0x1,%dil
  je     L24
  add    %eax,%r8d
  add    %edx,%esi
  cmp    %esi,%ecx
  jg     L24
  add    $0x1,%r8d
  sub    %ecx,%esi
  jmp    L24
L28:
  mov    %edi,%r8d
L33:
  mov    %r8d,%eax
  ret
```

assistant: ```c
uint64_t muldivu64(uint64_t a, uint64_t b, uint64_t c) {
  uint64_t quotient = 0;
  uint64_t remainder = 0;
  uint64_t cur_quotient = b / c;
  uint64_t cur_remainder = b % c;
  while (a) {
    if (a & 1) {
      quotient += cur_quotient;
      remainder += cur_remainder;
      if (remainder >= c) {
        quotient++;
        remainder -= c;
      }
    }
    a >>= 1;
    cur_quotient <<= 1;
    cur_remainder <<= 1;
    if (cur_remainder >= c) {
      cur_quotient++;
      cur_remainder -= c;
    }
  }
  return quotient;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<muldivu64>:
  endbr64
  mov    %esi,%eax
  mov    %edx,%ecx
  cltd
  idiv   %ecx
  test   %edi,%edi
  je     L3
  xor    %esi,%esi
  xor    %r8d,%r8d
  nopl   0x0(%rax)
L2:
  test   $0x1,%dil
  je     L20
  add    %edx,%esi
  add    %eax,%r8d
  cmp    %esi,%ecx
  jg     L20
  add    $0x1,%r8d
  sub    %ecx,%esi
L20:
  add    %edx,%edx
  sar    %edi
  add    %eax,%eax
  cmp    %edx,%ecx
  jg     L40
  add    $0x1,%eax
  sub    %ecx,%edx
L40:
  test   %edi,%edi
  jne    L2
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L3:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
uint64_t muldivu64(uint64_t a, uint64_t b, uint64_t c) {
  uint64_t quotient = 0;
  uint64_t remainder = 0;
  uint64_t cur_quotient = b / c;
  uint64_t cur_remainder = b % c;
  while (a) {
    if (a & 1) {
      quotient += cur_quotient;
      remainder += cur_remainder;
      if (remainder >= c) {
        quotient++;
        remainder -= c;
      }
    }
    a >>= 1;
    cur_quotient <<= 1;
    cur_remainder <<= 1;
    if (cur_remainder >= c) {
      cur_quotient++;
      cur_remainder -= c;
    }
  }
  return quotient;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<muldivu64>:
  endbr64
  mov    %esi,%eax
  mov    %edx,%ecx
  cltd
  idiv   %ecx
  test   %edi,%edi
  je     L42
  xor    %esi,%esi
  xor    %r8d,%r8d
  nopl   0x0(%rax)
L47:
  test   $0x1,%dil
  je     L23
  add    %edx,%esi
  add    %eax,%r8d
  cmp    %esi,%ecx
  jg     L23
  add    $0x1,%r8d
  sub    %ecx,%esi
L23:
  add    %edx,%edx
  sar    %edi
  add    %eax,%eax
  cmp    %edx,%ecx
  jg     L48
  add    $0x1,%eax
  sub    %ecx,%edx
L48:
  test   %edi,%edi
  jne    L47
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L42:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
uint64_t muldivu64(uint64_t a, uint64_t b, uint64_t c) {
  uint64_t quotient = 0;
  uint64_t remainder = 0;
  uint64_t cur_quotient = b / c;
  uint64_t cur_remainder = b % c;
  while (a) {
    if (a & 1) {
      quotient += cur_quotient;
      remainder += cur_remainder;
      if (remainder >= c) {
        quotient++;
        remainder -= c;
      }
    }
    a >>= 1;
    cur_quotient <<= 1;
    cur_remainder <<= 1;
    if (cur_remainder >= c) {
      cur_quotient++;
      cur_remainder -= c;
    }
  }
  return quotient;
}
```