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

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D18(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D46(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D47(%rip),%rdi
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%esi
  mov    0x14(%rsp),%edx
  cmp    %edx,%esi
  je     L26
  mov    0xc(%rsp),%edi
  call   <hanoi@plt>
  mov    $0x0,%eax
L25:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L19
  add    $0x20,%rsp
  pop    %rbx
  ret
L26:
  lea    D20(%rip),%rdi
  call   <printf@plt>
  jmp    L25
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D18
type_guessed: string
name: extract_symbolic_data
arguments: tag: D46
type_guessed: string
name: extract_symbolic_data
arguments: tag: D47
type_guessed: string
name: extract_symbolic_data
arguments: tag: D0
type_guessed: string
name: extract_symbolic_data
arguments: tag: D20
type_guessed: string

tool: D18: How Many Disk? : 

tool: D46: %d

tool: D47: From Where? [1-3] : 

tool: D0: To where? [1-3] : 

tool: D20: Have the same start and finish


assistant: ```c
int main() {
  int n, from, to;
  printf("How Many Disk? : ");
  scanf("%d", &n);
  printf("From Where? [1-3] : ");
  scanf("%d", &from);
  printf("To where? [1-3] : ");
  scanf("%d", &to);
  if (from == to)
    return printf("Have the same start and finish\n");
  else
    hanoi(n, from, to);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D15(%rip),%rdi
  lea    D29(%rip),%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D12(%rip),%rdi
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D13(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%esi
  mov    0x14(%rsp),%edx
  cmp    %edx,%esi
  je     L14
  mov    0xc(%rsp),%edi
  call   <hanoi@plt>
  xor    %eax,%eax
L21:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L10
  add    $0x20,%rsp
  pop    %rbp
  ret
L14:
  lea    D22(%rip),%rdi
  call   <printf@plt>
  jmp    L21
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D15
type_guessed: string
name: extract_symbolic_data
arguments: label: D29
type_guessed: string
name: extract_symbolic_data
arguments: label: D12
type_guessed: string
name: extract_symbolic_data
arguments: label: D13
type_guessed: string
name: extract_symbolic_data
arguments: label: D22
type_guessed: string

tool: D15: How Many Disk? : 

tool: D29: %d

tool: D12: From Where? [1-3] : 

tool: D13: To where? [1-3] : 

tool: D22: Have the same start and finish


assistant: ```c
int main() {
  int n, from, to;
  printf("How Many Disk? : ");
  scanf("%d", &n);
  printf("From Where? [1-3] : ");
  scanf("%d", &from);
  printf("To where? [1-3] : ");
  scanf("%d", &to);
  if (from == to)
    return printf("Have the same start and finish\n");
  else
    hanoi(n, from, to);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D31(%rip),%rdi
  lea    D12(%rip),%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D33(%rip),%rdi
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D23(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%esi
  mov    0x14(%rsp),%edx
  cmp    %edx,%esi
  je     L27
  mov    0xc(%rsp),%edi
  call   <hanoi@plt>
  xor    %eax,%eax
L15:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L25
  add    $0x20,%rsp
  pop    %rbp
  ret
L27:
  lea    D17(%rip),%rdi
  call   <printf@plt>
  jmp    L15
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D31
type: string
name: read_data
arguments: data_label: D12
type: string
name: read_data
arguments: data_label: D33
type: string
name: read_data
arguments: data_label: D23
type: string
name: read_data
arguments: data_label: D17
type: string

tool: D31: How Many Disk? : 

tool: D12: %d

tool: D33: From Where? [1-3] : 

tool: D23: To where? [1-3] : 

tool: D17: Have the same start and finish


assistant: ```c
int main() {
  int n, from, to;
  printf("How Many Disk? : ");
  scanf("%d", &n);
  printf("From Where? [1-3] : ");
  scanf("%d", &from);
  printf("To where? [1-3] : ");
  scanf("%d", &to);
  if (from == to)
    return printf("Have the same start and finish\n");
  else
    hanoi(n, from, to);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <InitApp@plt>
  mov    $0x0,%eax
  call   <Flash_LED@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void main(void) {
  InitApp();
  Flash_LED();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

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

assistant: ```c
void main(void) {
  InitApp();
  Flash_LED();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <InitApp@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <Flash_LED@plt>
```

assistant: ```c
void main(void) {
  InitApp();
  Flash_LED();
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <InitApp@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <Flash_LED@plt>
```

assistant: ```c
void main(void) {
  InitApp();
  Flash_LED();
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<divvv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  movss  (%rax),%xmm0
  mov    -0x10(%rbp),%rax
  movss  (%rax),%xmm1
  divss  %xmm1,%xmm0
  mov    -0x8(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm1
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  divss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm0
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm1
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  divss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void divvv(float *result, const float *s1) {
  result[0] /= s1[0];
  result[1] /= s1[1];
  result[2] /= s1[2];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<divvv>:
  endbr64
  movss  (%rdi),%xmm0
  divss  (%rsi),%xmm0
  movss  %xmm0,(%rdi)
  movss  0x4(%rdi),%xmm0
  divss  0x4(%rsi),%xmm0
  movss  %xmm0,0x4(%rdi)
  movss  0x8(%rdi),%xmm0
  divss  0x8(%rsi),%xmm0
  movss  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void divvv(float *result, const float *s1) {
  result[0] /= s1[0];
  result[1] /= s1[1];
  result[2] /= s1[2];
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<divvv>:
  endbr64
  movss  (%rdi),%xmm0
  divss  (%rsi),%xmm0
  movss  %xmm0,(%rdi)
  movss  0x4(%rdi),%xmm0
  divss  0x4(%rsi),%xmm0
  movss  %xmm0,0x4(%rdi)
  movss  0x8(%rdi),%xmm0
  divss  0x8(%rsi),%xmm0
  movss  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void divvv(float *result, const float *s1) {
  result[0] /= s1[0];
  result[1] /= s1[1];
  result[2] /= s1[2];
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<divvv>:
  endbr64
  movss  (%rdi),%xmm0
  divss  (%rsi),%xmm0
  movss  %xmm0,(%rdi)
  movss  0x4(%rdi),%xmm0
  divss  0x4(%rsi),%xmm0
  movss  %xmm0,0x4(%rdi)
  movss  0x8(%rdi),%xmm0
  divss  0x8(%rsi),%xmm0
  movss  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void divvv(float *result, const float *s1) {
  result[0] /= s1[0];
  result[1] /= s1[1];
  result[2] /= s1[2];
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<get_pid_time>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    D3(%rip),%eax
  test   %eax,%eax
  jns    L11
  mov    -0x4(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <custom_pid_time@plt>
  mov    %eax,D29(%rip)
L11:
  cmpl   $0x1,-0x8(%rbp)
  jne    L27
  mov    D29(%rip),%eax
  jmp    L23
L27:
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <custom_pid_time@plt>
L23:
  leave
  ret
```

assistant: ```c
int get_pid_time(int procfd, pid_t pid) {
  static int init_pid_time = -1;
  if (init_pid_time < 0) {
    init_pid_time = custom_pid_time(procfd, 1);
  }
  if (pid == 1)
    return init_pid_time;
  return custom_pid_time(procfd, pid);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<get_pid_time>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
  cmpl   $0x0,D8(%rip)
  js     L24
L15:
  mov    D8(%rip),%eax
  cmp    $0x1,%ebx
  je     L23
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <custom_pid_time@plt>
L23:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L24:
  mov    $0x1,%esi
  call   <custom_pid_time@plt>
  mov    %eax,D5(%rip)
  jmp    L15
```

assistant: ```c
int get_pid_time(int procfd, pid_t pid) {
  static int init_pid_time = -1;
  if (init_pid_time < 0) {
    init_pid_time = custom_pid_time(procfd, 1);
  }
  if (pid == 1)
    return init_pid_time;
  return custom_pid_time(procfd, pid);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<get_pid_time>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  sub    $0x10,%rsp
  mov    D0(%rip),%eax
  test   %eax,%eax
  js     L24
L10:
  cmp    $0x1,%ebp
  je     L12
  add    $0x10,%rsp
  mov    %ebp,%esi
  pop    %rbp
  jmp    1050 <custom_pid_time@plt>
  cs nopw 0x0(%rax,%rax,1)
L12:
  mov    D0(%rip),%eax
  add    $0x10,%rsp
  pop    %rbp
  ret
  nopl   0x0(%rax)
L24:
  mov    $0x1,%esi
  mov    %edi,0xc(%rsp)
  call   <custom_pid_time@plt>
  mov    0xc(%rsp),%edi
  mov    %eax,D24(%rip)
  jmp    L10
```

assistant: ```c
int get_pid_time(int procfd, pid_t pid) {
  static int init_pid_time = -1;
  if (init_pid_time < 0) {
    init_pid_time = custom_pid_time(procfd, 1);
  }
  if (pid == 1)
    return init_pid_time;
  return custom_pid_time(procfd, pid);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_pid_time>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  sub    $0x10,%rsp
  mov    D6(%rip),%eax
  test   %eax,%eax
  js     L15
L10:
  cmp    $0x1,%ebp
  je     L12
  add    $0x10,%rsp
  mov    %ebp,%esi
  pop    %rbp
  jmp    1050 <custom_pid_time@plt>
  cs nopw 0x0(%rax,%rax,1)
L12:
  add    $0x10,%rsp
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L15:
  mov    $0x1,%esi
  mov    %edi,0xc(%rsp)
  call   <custom_pid_time@plt>
  mov    0xc(%rsp),%edi
  mov    %eax,D10(%rip)
  jmp    L10
```

assistant: ```c
int get_pid_time(int procfd, pid_t pid) {
  static int init_pid_time = -1;
  if (init_pid_time < 0) {
    init_pid_time = custom_pid_time(procfd, 1);
  }
  if (pid == 1)
    return init_pid_time;
  return custom_pid_time(procfd, pid);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<store>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x11(%rax),%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  lea    0x15(%rax),%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  lea    0x19(%rax),%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  lea    0x1d(%rax),%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void store(char *p, int k) {
  *(int *)(p + 17) = k;
  *(int *)(p + 21) = k;
  *(int *)(p + 25) = k;
  *(int *)(p + 29) = k;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<store>:
  endbr64
  mov    %esi,0x11(%rdi)
  mov    %esi,0x15(%rdi)
  mov    %esi,0x19(%rdi)
  mov    %esi,0x1d(%rdi)
  ret
```

assistant: ```c
void store(char *p, int k) {
  *(int *)(p + 17) = k;
  *(int *)(p + 21) = k;
  *(int *)(p + 25) = k;
  *(int *)(p + 29) = k;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<store>:
  endbr64
  mov    %esi,0x11(%rdi)
  mov    %esi,0x15(%rdi)
  mov    %esi,0x19(%rdi)
  mov    %esi,0x1d(%rdi)
  ret
```

assistant: ```c
void store(char *p, int k) {
  *(int *)(p + 17) = k;
  *(int *)(p + 21) = k;
  *(int *)(p + 25) = k;
  *(int *)(p + 29) = k;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<store>:
  endbr64
  movd   %esi,%xmm1
  pshufd $0x0,%xmm1,%xmm0
  movups %xmm0,0x11(%rdi)
  ret
```

assistant: ```c
void store(char *p, int k) {
  *(int *)(p + 17) = k;
  *(int *)(p + 21) = k;
  *(int *)(p + 25) = k;
  *(int *)(p + 29) = k;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<read_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L37
L13:
  cmpb   $0xa,-0x5(%rbp)
  je     L26
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x5(%rbp),%eax
  mov    %al,(%rdx)
L37:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x5(%rbp)
  mov    D3(%rip),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x5(%rbp)
  je     L1
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jl     L13
  jmp    L1
L26:
  nop
L1:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int read_str(char str[], int *size) {
  char c;
  int i = 0;
  while (((c = getchar()) != EOF) && (i < *size)) {
    if (c == '\n') {
      break;
    }
    str[i++] = c;
  }
  return i;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<read_str>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    $0x0,%ebx
L11:
  mov    %ebx,%r13d
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D1(%rip),%rdx
  cmp    %al,(%rdx)
  je     L34
  cmp    %ebx,0x0(%rbp)
  jle    L34
  cmp    $0xa,%al
  je     L34
  mov    %al,(%r12,%rbx,1)
  add    $0x1,%rbx
  jmp    L11
L34:
  mov    %r13d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int read_str(char str[], int *size) {
  char c;
  int i = 0;
  while (((c = getchar()) != EOF) && (i < *size)) {
    if (c == '\n') {
      break;
    }
    str[i++] = c;
  }
  return i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<read_str>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    D9(%rip),%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L38
  xchg   %ax,%ax
L28:
  cmp    %r14d,(%rbx)
  jle    L29
  cmp    $0xa,%al
  je     L29
  mov    %al,0x0(%rbp,%r14,1)
  add    $0x1,%r14
L38:
  xor    %eax,%eax
  mov    %r14d,%r12d
  call   <getchar@plt>
  cmp    %al,0x0(%r13)
  jne    L28
L29:
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int read_str(char str[], int *size) {
  char c;
  int i = 0;
  while (((c = getchar()) != EOF) && (i < *size)) {
    if (c == '\n') {
      break;
    }
    str[i++] = c;
  }
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<read_str>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    D9(%rip),%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L32
  xchg   %ax,%ax
L2:
  cmp    %r14d,(%rbx)
  jle    L15
  cmp    $0xa,%al
  je     L15
  mov    %al,0x0(%rbp,%r14,1)
  add    $0x1,%r14
L32:
  xor    %eax,%eax
  mov    %r14d,%r12d
  call   <getchar@plt>
  cmp    %al,0x0(%r13)
  jne    L2
L15:
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int read_str(char str[], int *size) {
  char c;
  int i = 0;
  while (((c = getchar()) != EOF) && (i < *size)) {
    if (c == '\n') {
      break;
    }
    str[i++] = c;
  }
  return i;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<sqliterkParseNumber>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpq   $0x0,-0x28(%rbp)
  je     L47
  cmpq   $0x0,-0x38(%rbp)
  jne    L54
L47:
  mov    $0x2,%eax
  jmp    L19
L54:
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L42
L2:
  mov    $0x7,%eax
  sub    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x1c(%rbp),%eax
  cltq
  mov    %dl,-0x10(%rbp,%rax,1)
  addl   $0x1,-0x1c(%rbp)
L42:
  cmpl   $0x7,-0x1c(%rbp)
  jle    L2
  lea    -0x10(%rbp),%rcx
  mov    -0x38(%rbp),%rax
  mov    $0x8,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    $0x0,%eax
L19:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L52
  call   <__stack_chk_fail@plt>
L52:
  leave
  ret
```

assistant: ```c
int sqliterkParseNumber(const unsigned char *data, int offset, double *value) {
  if (!data || !value) {
    return 2;
  }
  unsigned char out[8];
  const unsigned char *begin = data + offset;
  int i;
  for (i = 0; i < 8; i++) {
    out[i] = begin[8 - 1 - i];
  }
  memcpy(value, out, 8);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sqliterkParseNumber>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,%rax
  mov    %rdx,%rdi
  mov    %fs:0x28,%rdx
  mov    %rdx,0x8(%rsp)
  xor    %edx,%edx
  test   %rax,%rax
  je     L27
  test   %rdi,%rdi
  je     L27
  movslq %esi,%rsi
  lea    0x7(%rax,%rsi,1),%rcx
  mov    %rsp,%rax
  lea    0x8(%rsp),%r8
L21:
  movzbl (%rcx),%esi
  mov    %sil,(%rax)
  sub    $0x1,%rcx
  add    $0x1,%rax
  cmp    %r8,%rax
  jne    L21
  mov    %rsp,%rsi
  mov    $0x8,%edx
  call   <memcpy@plt>
  mov    $0x0,%eax
L30:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L49
  add    $0x18,%rsp
  ret
L27:
  mov    $0x2,%eax
  jmp    L30
L49:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int sqliterkParseNumber(const unsigned char *data, int offset, double *value) {
  if (!data || !value) {
    return 2;
  }
  unsigned char out[8];
  const unsigned char *begin = data + offset;
  int i;
  for (i = 0; i < 8; i++) {
    out[i] = begin[8 - 1 - i];
  }
  memcpy(value, out, 8);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<sqliterkParseNumber>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,%rax
  mov    %rdx,%rdi
  mov    %fs:0x28,%rdx
  mov    %rdx,0x8(%rsp)
  xor    %edx,%edx
  test   %rax,%rax
  je     L17
  test   %rdi,%rdi
  je     L17
  movslq %esi,%rsi
  lea    0x8(%rsp),%r8
  lea    0x7(%rax,%rsi,1),%rcx
  mov    %rsp,%rsi
  mov    %rsi,%rax
  nopl   0x0(%rax,%rax,1)
L0:
  movzbl (%rcx),%edx
  add    $0x1,%rax
  sub    $0x1,%rcx
  mov    %dl,-0x1(%rax)
  cmp    %r8,%rax
  jne    L0
  mov    $0x8,%edx
  call   <memcpy@plt>
  xor    %eax,%eax
L48:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L39
  add    $0x18,%rsp
  ret
  nopl   0x0(%rax)
L17:
  mov    $0x2,%eax
  jmp    L48
L39:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int sqliterkParseNumber(const unsigned char *data, int offset, double *value) {
  if (!data || !value) {
    return 2;
  }
  unsigned char out[8];
  const unsigned char *begin = data + offset;
  int i;
  for (i = 0; i < 8; i++) {
    out[i] = begin[8 - 1 - i];
  }
  memcpy(value, out, 8);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<sqliterkParseNumber>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L19
  mov    %rdx,%rdi
  test   %rdx,%rdx
  je     L19
  movslq %esi,%rdx
  movzbl (%rcx,%rdx,1),%eax
  movzbl 0x1(%rcx,%rdx,1),%esi
  shl    $0x8,%rax
  or     %rsi,%rax
  movzbl 0x2(%rcx,%rdx,1),%esi
  shl    $0x8,%rax
  or     %rsi,%rax
  movzbl 0x3(%rcx,%rdx,1),%esi
  shl    $0x8,%rax
  or     %rsi,%rax
  movzbl 0x4(%rcx,%rdx,1),%esi
  shl    $0x8,%rax
  or     %rsi,%rax
  movzbl 0x5(%rcx,%rdx,1),%esi
  shl    $0x8,%rax
  or     %rsi,%rax
  movzbl 0x6(%rcx,%rdx,1),%esi
  movzbl 0x7(%rcx,%rdx,1),%edx
  shl    $0x8,%rax
  or     %rsi,%rax
  mov    %rsp,%rsi
  shl    $0x8,%rax
  or     %rdx,%rax
  mov    $0x8,%edx
  mov    %rax,(%rsp)
  call   <memcpy@plt>
  xor    %eax,%eax
L23:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  add    $0x18,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  mov    $0x2,%eax
  jmp    L23
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int sqliterkParseNumber(const unsigned char *data, int offset, double *value) {
  if (!data || !value) {
    return 2;
  }
  unsigned char out[8];
  const unsigned char *begin = data + offset;
  int i;
  for (i = 0; i < 8; i++) {
    out[i] = begin[8 - 1 - i];
  }
  memcpy(value, out, 8);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rotate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    %r8,-0x28(%rbp)
  mov    %r9,-0x30(%rbp)
  movsd  %xmm0,-0x38(%rbp)
  movsd  %xmm1,-0x40(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x30(%rbp),%rax
  movsd  %xmm0,(%rax)
  mov    -0x20(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x28(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    0x10(%rbp),%rax
  movsd  %xmm0,(%rax)
  mov    -0x30(%rbp),%rax
  movsd  (%rax),%xmm0
  mov    0x10(%rbp),%rax
  movsd  (%rax),%xmm2
  mov    -0x30(%rbp),%rax
  movsd  (%rax),%xmm1
  mulsd  -0x40(%rbp),%xmm1
  addsd  %xmm2,%xmm1
  mulsd  -0x38(%rbp),%xmm1
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    0x10(%rbp),%rax
  movsd  (%rax),%xmm2
  mov    -0x30(%rbp),%rax
  movsd  (%rax),%xmm0
  mov    0x10(%rbp),%rax
  movsd  (%rax),%xmm1
  mulsd  -0x40(%rbp),%xmm1
  subsd  %xmm1,%xmm0
  mulsd  -0x38(%rbp),%xmm0
  mov    -0x20(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x28(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  addsd  %xmm2,%xmm0
  movsd  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rotate(double **a, long i, long j, long k, long l, double *g, double *h, double s, double tau) {
  *g = a[i][j];
  *h = a[k][l];
  a[i][j] = *g - s * (*h + *g * tau);
  a[k][l] = *h + s * (*g - *h * tau);
}
```