ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<page32>:
  endbr64
  sub    $0x8,%rsp
  lea    D47(%rip),%rdi
  call   <printf@plt>
  lea    D12(%rip),%rdi
  call   <printf@plt>
  lea    D32(%rip),%rdi
  call   <printf@plt>
  lea    D39(%rip),%rdi
  call   <printf@plt>
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    $0x16,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D47
type_guessed: string
name: disassemble_data
arguments: label: D12
type_guessed: string
name: disassemble_data
arguments: label: D32
type_guessed: string
name: disassemble_data
arguments: label: D39
type_guessed: string
name: disassemble_data
arguments: label: D17
type_guessed: string

tool: D47: Finally it's your big chance to prove that you're as good a troubleshooter


tool: D12: as your previous clone.  You walk briskly to mission briefing and pick up your


tool: D32: previous clone's personal effects and notepad.  After reviewing the notes you


tool: D39: know what has to be done.  You catch the purple line to Goods Distribution Hall


tool: D17: 7-beta and begin to search for the blast door.


assistant: ```c
page32() {
  printf("Finally it\'s your big chance to prove that you\'re as good a troubleshooter\n");
  printf("as your previous clone.  You walk briskly to mission briefing and pick up your\n");
  printf("previous clone\'s personal effects and notepad.  After reviewing the notes you\n");
  printf("know what has to be done.  You catch the purple line to Goods Distribution Hall\n");
  printf("7-beta and begin to search for the blast door.\n");
  return 22;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<page32>:
  endbr64
  sub    $0x8,%rsp
  lea    D25(%rip),%rdi
  call   <printf@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  lea    D40(%rip),%rdi
  call   <printf@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    $0x16,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D25
datatype: string
name: read_data
arguments: data_label: D1
datatype: string
name: read_data
arguments: data_label: D40
datatype: string
name: read_data
arguments: data_label: D0
datatype: string
name: read_data
arguments: data_label: D14
datatype: string

tool: D25: Finally it's your big chance to prove that you're as good a troubleshooter


tool: D1: as your previous clone.  You walk briskly to mission briefing and pick up your


tool: D40: previous clone's personal effects and notepad.  After reviewing the notes you


tool: D0: know what has to be done.  You catch the purple line to Goods Distribution Hall


tool: D14: 7-beta and begin to search for the blast door.


assistant: ```c
page32() {
  printf("Finally it\'s your big chance to prove that you\'re as good a troubleshooter\n");
  printf("as your previous clone.  You walk briskly to mission briefing and pick up your\n");
  printf("previous clone\'s personal effects and notepad.  After reviewing the notes you\n");
  printf("know what has to be done.  You catch the purple line to Goods Distribution Hall\n");
  printf("7-beta and begin to search for the blast door.\n");
  return 22;
}
```
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
  push   %rbp
  push   %rbx
  sub    $0xb8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xa8(%rsp)
  xor    %eax,%eax
  movl   $0x676e696c,(%rsp)
  movb   $0x0,0x4(%rsp)
  movl   $0x6979,0x5(%rsp)
  movb   $0x0,0x9(%rsp)
  movl   $0x7265,0xa(%rsp)
  movb   $0x0,0xe(%rsp)
  movl   $0x6e6173,0xf(%rsp)
  movb   $0x0,0x13(%rsp)
  movl   $0x6973,0x14(%rsp)
  movb   $0x0,0x18(%rsp)
  movl   $0x7577,0x19(%rsp)
  movb   $0x0,0x1d(%rsp)
  movl   $0x75696c,0x1e(%rsp)
  movb   $0x0,0x22(%rsp)
  movl   $0x6971,0x23(%rsp)
  movb   $0x0,0x27(%rsp)
  movl   $0x6162,0x28(%rsp)
  movb   $0x0,0x2c(%rsp)
  movl   $0x75696a,0x2d(%rsp)
  movb   $0x0,0x31(%rsp)
  lea    0x40(%rsp),%rsi
  mov    $0xc,%ecx
  mov    %rsi,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x0,(%rdi)
  movl   $0x0,0x3b(%rsp)
  movb   $0x0,0x3f(%rsp)
  lea    D12(%rip),%rdi
  call   <scanf@plt>
  movzbl 0x40(%rsp),%edx
  test   %dl,%dl
  je     L72
  lea    0x41(%rsp),%rcx
  mov    $0x0,%eax
L89:
  movsbl %dl,%edx
  lea    -0x30(%rax,%rdx,1),%eax
  add    $0x1,%rcx
  movzbl -0x1(%rcx),%edx
  test   %dl,%dl
  jne    L89
  test   %eax,%eax
  je     L72
  lea    0x3b(%rsp),%rsi
L1:
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  add    $0x30,%eax
  mov    %al,(%rsi)
  mov    %edx,%eax
  add    $0x1,%rsi
  test   %edx,%edx
  jne    L1
L72:
  mov    $0x4,%ebx
  lea    D12(%rip),%rbp
  jmp    L41
L15:
  mov    $0x20,%edi
  call   <putchar@plt>
L69:
  sub    $0x1,%rbx
  cmp    $0xffffffffffffffff,%rbx
  je     L8
L41:
  movzbl 0x3b(%rsp,%rbx,1),%eax
  test   %al,%al
  je     L69
  movsbl %al,%eax
  sub    $0x30,%eax
  cltq
  lea    (%rax,%rax,4),%rax
  lea    (%rsp,%rax,1),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %ebx,%ebx
  jne    L15
L8:
  mov    0xa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  mov    $0x0,%eax
  add    $0xb8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D12
data_type: string

tool: D12: %s

assistant: ```c
int main(void) {
  char num[10][5] = {"ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu"};
  char input[100];
  char output[5];
  int sum = 0;
  int i = 0;
  int j = 0;
  memset(input, 0, sizeof(input) / sizeof(char));
  memset(output, 0, sizeof(output) / sizeof(char));
  scanf("%s", input);
  while (input[i] != '\0') {
    sum += input[i] - '0';
    i++;
  }
  while (sum != 0) {
    output[j] = sum % 10 + '0';
    sum /= 10;
    j++;
  }
  for (int j = 4; j >= 0; j--) {
    if (output[j] != '\0') {
      printf("%s", num[output[j] - '0']);
      if (j != 0) {
        printf(" ");
      }
    }
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_sys_uptime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  lea    -0x10(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <clock_gettime@plt>
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
const long get_sys_uptime() {
  struct timespec sp;
  (void)clock_gettime(CLOCK_MONOTONIC, &sp);
  return (sp.tv_sec);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_sys_uptime>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    D6(%rip),%rax
  mov    (%rax),%edi
  call   <clock_gettime@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
const long get_sys_uptime() {
  struct timespec sp;
  (void)clock_gettime(CLOCK_MONOTONIC, &sp);
  return (sp.tv_sec);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<get_sys_uptime>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D9(%rip),%rax
  mov    %rsp,%rsi
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
const long get_sys_uptime() {
  struct timespec sp;
  (void)clock_gettime(CLOCK_MONOTONIC, &sp);
  return (sp.tv_sec);
}
```
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_sys_uptime>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D7(%rip),%rax
  mov    %rsp,%rsi
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
const long get_sys_uptime() {
  struct timespec sp;
  (void)clock_gettime(CLOCK_MONOTONIC, &sp);
  return (sp.tv_sec);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<invert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L4
L6:
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  add    %edx,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
L4:
  cmpl   $0x0,-0x14(%rbp)
  jne    L6
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int invert(int a) {
  int r = 0;
  while (a != 0) {
    r = r * 10;
    r = r + a % 10;
    a = a / 10;
  }
  return r;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<invert>:
  endbr64
  test   %edi,%edi
  je     L11
  mov    $0x0,%edx
L2:
  lea    (%rdx,%rdx,4),%ecx
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  lea    (%rdi,%rcx,2),%edx
  mov    %eax,%edi
  test   %eax,%eax
  jne    L2
L17:
  mov    %edx,%eax
  ret
L11:
  mov    %edi,%edx
  jmp    L17
```

assistant: ```c
int invert(int a) {
  int r = 0;
  while (a != 0) {
    r = r * 10;
    r = r + a % 10;
    a = a / 10;
  }
  return r;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<invert>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L11
  nopl   0x0(%rax,%rax,1)
L1:
  movslq %edi,%rax
  mov    %edi,%edx
  lea    (%r8,%r8,4),%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  lea    (%rdi,%rcx,2),%r8d
  mov    %eax,%edi
  test   %eax,%eax
  jne    L1
L11:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int invert(int a) {
  int r = 0;
  while (a != 0) {
    r = r * 10;
    r = r + a % 10;
    a = a / 10;
  }
  return r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<invert>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L19
  nopl   0x0(%rax,%rax,1)
L15:
  movslq %edi,%rax
  mov    %edi,%edx
  lea    (%r8,%r8,4),%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  lea    (%rdi,%rcx,2),%r8d
  mov    %eax,%edi
  test   %eax,%eax
  jne    L15
L19:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int invert(int a) {
  int r = 0;
  while (a != 0) {
    r = r * 10;
    r = r + a % 10;
    a = a / 10;
  }
  return r;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<chkword>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <word@plt>
  test   %rax,%rax
  je     L2
  mov    $0x0,%eax
  call   <synbad@plt>
L2:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void chkword(void) {
  if (word())
    synbad();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<chkword>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <word@plt>
  test   %rax,%rax
  jne    L17
L4:
  add    $0x8,%rsp
  ret
L17:
  mov    $0x0,%eax
  call   <synbad@plt>
  jmp    L4
```

assistant: ```c
void chkword(void) {
  if (word())
    synbad();
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<chkword>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <word@plt>
  test   %rax,%rax
  jne    L8
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax)
L8:
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <synbad@plt>
```

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

```asm
<chkword>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <word@plt>
  test   %rax,%rax
  jne    L3
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax)
L3:
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <synbad@plt>
```

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

```asm
<Hfun>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x10(%rbp)
  jmp    L15
L4:
  shll   $0x4,-0x14(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0xf0000000,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  je     L11
  mov    -0xc(%rbp),%eax
  shr    $0x18,%eax
  xor    -0xc(%rbp),%eax
  xor    %eax,-0x14(%rbp)
L11:
  addq   $0x1,-0x8(%rbp)
  subl   $0x1,-0x10(%rbp)
L15:
  cmpl   $0x0,-0x10(%rbp)
  jne    L4
  mov    -0x2c(%rbp),%ecx
  mov    -0x14(%rbp),%eax
  mov    $0x0,%edx
  div    %ecx
  mov    %edx,%eax
  leave
  ret
```

assistant: ```c
int Hfun(char *s, int tablen) {
  char *p;
  unsigned int h = 0, g;
  int i;
  for (p = s, i = strlen(s); i; p++, i--) {
    h <<= 4;
    h += *p;
    if ((g = (h & 0xF0000000L)))
      h ^= g ^ (g >> 24);
  }
  return h % tablen;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Hfun>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  call   <strlen@plt>
  test   %eax,%eax
  je     L36
  mov    %eax,%eax
  lea    (%rbx,%rax,1),%rcx
  mov    $0x0,%eax
  jmp    L10
L13:
  add    $0x1,%rbx
  cmp    %rcx,%rbx
  je     L23
L10:
  shl    $0x4,%eax
  movsbl (%rbx),%edx
  add    %edx,%eax
  mov    %eax,%edx
  and    $0xf0000000,%edx
  je     L13
  and    $0xfffffff,%eax
  shr    $0x18,%edx
  xor    %edx,%eax
  jmp    L13
L36:
  mov    $0x0,%eax
L23:
  mov    $0x0,%edx
  div    %ebp
  mov    %edx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int Hfun(char *s, int tablen) {
  char *p;
  unsigned int h = 0, g;
  int i;
  for (p = s, i = strlen(s); i; p++, i--) {
    h <<= 4;
    h += *p;
    if ((g = (h & 0xF0000000L)))
      h ^= g ^ (g >> 24);
  }
  return h % tablen;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Hfun>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%r8d
  test   %eax,%eax
  je     L28
  mov    %eax,%ecx
  xor    %eax,%eax
  add    %rbx,%rcx
  nopw   0x0(%rax,%rax,1)
L21:
  movsbl (%rbx),%edx
  shl    $0x4,%eax
  add    %edx,%eax
  mov    %eax,%edx
  and    $0xf0000000,%edx
  je     L0
  and    $0xfffffff,%eax
  shr    $0x18,%edx
  xor    %edx,%eax
L0:
  add    $0x1,%rbx
  cmp    %rcx,%rbx
  jne    L21
  xor    %edx,%edx
  div    %ebp
  mov    %edx,%r8d
L28:
  add    $0x8,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int Hfun(char *s, int tablen) {
  char *p;
  unsigned int h = 0, g;
  int i;
  for (p = s, i = strlen(s); i; p++, i--) {
    h <<= 4;
    h += *p;
    if ((g = (h & 0xF0000000L)))
      h ^= g ^ (g >> 24);
  }
  return h % tablen;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Hfun>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%r8d
  test   %eax,%eax
  je     L4
  mov    %eax,%ecx
  xor    %eax,%eax
  add    %rbx,%rcx
  nopw   0x0(%rax,%rax,1)
L19:
  movsbl (%rbx),%edx
  shl    $0x4,%eax
  add    %edx,%eax
  mov    %eax,%edx
  and    $0xf0000000,%edx
  je     L26
  and    $0xfffffff,%eax
  shr    $0x18,%edx
  xor    %edx,%eax
L26:
  add    $0x1,%rbx
  cmp    %rcx,%rbx
  jne    L19
  xor    %edx,%edx
  div    %ebp
  mov    %edx,%r8d
L4:
  add    $0x8,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int Hfun(char *s, int tablen) {
  char *p;
  unsigned int h = 0, g;
  int i;
  for (p = s, i = strlen(s); i; p++, i--) {
    h <<= 4;
    h += *p;
    if ((g = (h & 0xF0000000L)))
      h ^= g ^ (g >> 24);
  }
  return h % tablen;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<writeblock_fd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %rdx,-0x10(%rbp)
  mov    -0x8(%rbp),%eax
  lea    (%rax,%rax,1),%ecx
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <fdc_write_sector@plt>
  mov    -0x10(%rbp),%rax
  lea    0x200(%rax),%rdx
  mov    -0x8(%rbp),%eax
  add    %eax,%eax
  lea    0x1(%rax),%ecx
  mov    -0x4(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <fdc_write_sector@plt>
  nop
  leave
  ret
```

assistant: ```c
void writeblock_fd(int device, int blockno, char *buffer) {
  fdc_write_sector(device, blockno * 2, buffer);
  fdc_write_sector(device, blockno * 2 + 1, buffer + 512);
}
```
user: Translate the below assembly code into C programming language:

```asm
<writeblock_fd>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %rdx,%rbp
  lea    (%rsi,%rsi,1),%ebx
  mov    %ebx,%esi
  call   <fdc_write_sector@plt>
  lea    0x200(%rbp),%rdx
  lea    0x1(%rbx),%esi
  mov    %r12d,%edi
  call   <fdc_write_sector@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void writeblock_fd(int device, int blockno, char *buffer) {
  fdc_write_sector(device, blockno * 2, buffer);
  fdc_write_sector(device, blockno * 2 + 1, buffer + 512);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<writeblock_fd>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  lea    (%rsi,%rsi,1),%ebx
  mov    %ebx,%esi
  call   <fdc_write_sector@plt>
  lea    0x200(%rbp),%rdx
  lea    0x1(%rbx),%esi
  mov    %r12d,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1050 <fdc_write_sector@plt>
```

assistant: ```c
void writeblock_fd(int device, int blockno, char *buffer) {
  fdc_write_sector(device, blockno * 2, buffer);
  fdc_write_sector(device, blockno * 2 + 1, buffer + 512);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<writeblock_fd>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  lea    (%rsi,%rsi,1),%ebx
  mov    %ebx,%esi
  call   <fdc_write_sector@plt>
  lea    0x200(%rbp),%rdx
  lea    0x1(%rbx),%esi
  mov    %r12d,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1050 <fdc_write_sector@plt>
```

assistant: ```c
void writeblock_fd(int device, int blockno, char *buffer) {
  fdc_write_sector(device, blockno * 2, buffer);
  fdc_write_sector(device, blockno * 2 + 1, buffer + 512);
}
```
user: What is the c source code of the assembly code below?

```asm
<mpz_EGlpNumReadStr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movb   $0x1,-0x7(%rbp)
  movb   $0x0,-0x6(%rbp)
  movb   $0x0,-0x5(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x5(%rbp)
  mov    -0x14(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <mpz_set_ui@plt>
  jmp    L34
L124:
  movsbl -0x5(%rbp),%eax
  cmp    $0x39,%eax
  jg     L7
  cmp    $0x30,%eax
  jge    L67
  cmp    $0x2b,%eax
  je     L36
  cmp    $0x2d,%eax
  je     L85
  jmp    L7
L67:
  mov    -0x14(%rbp),%ecx
  mov    -0x14(%rbp),%eax
  mov    $0xa,%edx
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <mpz_mul_ui@plt>
  movsbl -0x5(%rbp),%eax
  sub    $0x30,%eax
  movslq %eax,%rdx
  mov    -0x14(%rbp),%ecx
  mov    -0x14(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <mpz_add_ui@plt>
  movb   $0x0,-0x7(%rbp)
  jmp    L7
L85:
  movb   $0x1,-0x6(%rbp)
L36:
  movb   $0x0,-0x7(%rbp)
  nop
L7:
  addl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x5(%rbp)
L34:
  cmpb   $0x2f,-0x5(%rbp)
  jle    L44
  cmpb   $0x39,-0x5(%rbp)
  jle    L124
L44:
  cmpb   $0x0,-0x7(%rbp)
  je     L71
  cmpb   $0x2b,-0x5(%rbp)
  je     L124
  cmpb   $0x2d,-0x5(%rbp)
  je     L124
L71:
  cmpb   $0x0,-0x6(%rbp)
  je     L78
  mov    -0x14(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <mpz_neg@plt>
L78:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int mpz_EGlpNumReadStr(mpz_t var, char const *str) {
  char unsigned a_sgn = 1;
  char unsigned sgn = 0;
  char c = 0;
  int n_char = 0;
  c = str[n_char];
  mpz_set_ui(var, (unsigned long int)0);
  while ((('0' <= c) && (c <= '9')) || (a_sgn && (c == '+' || c == '-'))) {
    switch (c) {
    case '0':
    case '1':
    case '2':
    case '3':
    case '4':
    case '5':
    case '6':
    case '7':
    case '8':
    case '9':
      mpz_mul_ui(var, var, (unsigned long int)10);
      mpz_add_ui(var, var, (unsigned long int)(c - '0'));
      a_sgn = 0;
      break;
    case '-':
      sgn = 1;
    case '+':
      a_sgn = 0;
      break;
    }
    c = str[++n_char];
  }
  if (sgn)
    mpz_neg(var, var);
  return n_char;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<mpz_EGlpNumReadStr>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r12d
  mov    %rsi,%r13
  movzbl (%rsi),%ebx
  mov    $0x0,%esi
  call   <mpz_set_ui@plt>
  mov    $0x0,%ebp
  mov    $0x0,%r15d
  mov    $0x1,%edx
  jmp    L79
L41:
  cmp    $0x2d,%bl
  je     L93
  jg     L13
  cmp    $0x2b,%bl
  mov    $0x0,%eax
  cmove  %eax,%edx
L17:
  movzbl 0x1(%r13,%rbp,1),%ebx
  add    $0x1,%rbp
L79:
  mov    %ebp,%r14d
  mov    %ebx,%ecx
  lea    -0x30(%rbx),%eax
  cmp    $0x9,%al
  jbe    L41
  test   %dl,%dl
  je     L103
  sub    $0x2b,%ecx
  test   $0xfd,%cl
  je     L41
L103:
  test   %r15b,%r15b
  jne    L95
L16:
  mov    %r14d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L13:
  cmp    $0x9,%al
  ja     L17
  mov    $0xa,%edx
  mov    %r12d,%esi
  mov    %r12d,%edi
  call   <mpz_mul_ui@plt>
  movsbl %bl,%edx
  sub    $0x30,%edx
  movslq %edx,%rdx
  mov    %r12d,%esi
  mov    %r12d,%edi
  call   <mpz_add_ui@plt>
  mov    $0x0,%edx
  jmp    L17
L93:
  mov    $0x1,%r15d
  mov    $0x0,%edx
  jmp    L17
L95:
  mov    %r12d,%esi
  mov    %r12d,%edi
  call   <mpz_neg@plt>
  jmp    L16
```

assistant: ```c
int mpz_EGlpNumReadStr(mpz_t var, char const *str) {
  char unsigned a_sgn = 1;
  char unsigned sgn = 0;
  char c = 0;
  int n_char = 0;
  c = str[n_char];
  mpz_set_ui(var, (unsigned long int)0);
  while ((('0' <= c) && (c <= '9')) || (a_sgn && (c == '+' || c == '-'))) {
    switch (c) {
    case '0':
    case '1':
    case '2':
    case '3':
    case '4':
    case '5':
    case '6':
    case '7':
    case '8':
    case '9':
      mpz_mul_ui(var, var, (unsigned long int)10);
      mpz_add_ui(var, var, (unsigned long int)(c - '0'));
      a_sgn = 0;
      break;
    case '-':
      sgn = 1;
    case '+':
      a_sgn = 0;
      break;
    }
    c = str[++n_char];
  }
  if (sgn)
    mpz_neg(var, var);
  return n_char;
}
```