ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <adainit@plt>
  xor    %eax,%eax
  call   <greetings@plt>
  lea    D28(%rip),%rdi
  mov    %eax,%r12d
  call   <printf@plt>
  lea    D47(%rip),%rdi
  call   <printf@plt>
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    D21(%rip),%rdi
  call   <printf@plt>
  lea    D50(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D55(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  je     L52
  cmp    $0x1,%eax
  je     L67
  cmp    $0x2,%eax
  je     L24
L37:
  xor    %eax,%eax
  call   <adafinal@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
L52:
  call   <standard_scaler@plt>
  mov    %eax,%r12d
  jmp    L37
L24:
  xor    %eax,%eax
  call   <quaddobl_scaler@plt>
  mov    %eax,%r12d
  jmp    L37
L67:
  xor    %eax,%eax
  call   <dobldobl_scaler@plt>
  mov    %eax,%r12d
  jmp    L37
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D28
type: string
name: read_data
arguments: label: D47
type: string
name: read_data
arguments: label: D4
type: string
name: read_data
arguments: label: D21
type: string
name: read_data
arguments: label: D50
type: string
name: read_data
arguments: label: D55
type: string

tool: D28: 
MENU for the precision of the scalers :


tool: D47:   0. use standard double precision arithmetic; or


tool: D4:   1. use double double precision arithmetic; or


tool: D21:   2. use quad double precision arithmetic.


tool: D50: Type 1, 2, or 3 to make your choice : 

tool: D55: %d

assistant: ```c
int main(int argc, char *argv[]) {
  int fail, choice;
  adainit();
  fail = greetings();
  printf("\nMENU for the precision of the scalers :\n");
  printf("  0. use standard double precision arithmetic; or\n");
  printf("  1. use double double precision arithmetic; or\n");
  printf("  2. use quad double precision arithmetic.\n");
  printf("Type 1, 2, or 3 to make your choice : ");
  scanf("%d", &choice);
  if (choice == 0)
    fail = standard_scaler();
  else if (choice == 1)
    fail = dobldobl_scaler();
  else if (choice == 2)
    fail = quaddobl_scaler();
  adafinal();
  return fail;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <adainit@plt>
  xor    %eax,%eax
  call   <greetings@plt>
  lea    D25(%rip),%rdi
  mov    %eax,%r12d
  call   <printf@plt>
  lea    D13(%rip),%rdi
  call   <printf@plt>
  lea    D27(%rip),%rdi
  call   <printf@plt>
  lea    D36(%rip),%rdi
  call   <printf@plt>
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D43(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  je     L29
  cmp    $0x1,%eax
  je     L1
  cmp    $0x2,%eax
  je     L4
L19:
  xor    %eax,%eax
  call   <adafinal@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
L29:
  call   <standard_scaler@plt>
  mov    %eax,%r12d
  jmp    L19
L4:
  xor    %eax,%eax
  call   <quaddobl_scaler@plt>
  mov    %eax,%r12d
  jmp    L19
L1:
  xor    %eax,%eax
  call   <dobldobl_scaler@plt>
  mov    %eax,%r12d
  jmp    L19
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D25
type_infered: string
name: retrieve_data
arguments: data_tag: D13
type_infered: string
name: retrieve_data
arguments: data_tag: D27
type_infered: string
name: retrieve_data
arguments: data_tag: D36
type_infered: string
name: retrieve_data
arguments: data_tag: D24
type_infered: string
name: retrieve_data
arguments: data_tag: D43
type_infered: string

tool: D25: 
MENU for the precision of the scalers :


tool: D13:   0. use standard double precision arithmetic; or


tool: D27:   1. use double double precision arithmetic; or


tool: D36:   2. use quad double precision arithmetic.


tool: D24: Type 1, 2, or 3 to make your choice : 

tool: D43: %d

assistant: ```c
int main(int argc, char *argv[]) {
  int fail, choice;
  adainit();
  fail = greetings();
  printf("\nMENU for the precision of the scalers :\n");
  printf("  0. use standard double precision arithmetic; or\n");
  printf("  1. use double double precision arithmetic; or\n");
  printf("  2. use quad double precision arithmetic.\n");
  printf("Type 1, 2, or 3 to make your choice : ");
  scanf("%d", &choice);
  if (choice == 0)
    fail = standard_scaler();
  else if (choice == 1)
    fail = dobldobl_scaler();
  else if (choice == 2)
    fail = quaddobl_scaler();
  adafinal();
  return fail;
}
```
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
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x78(%rbp)
  jmp    L28
L12:
  movl   $0x0,-0x74(%rbp)
  jmp    L47
L9:
  mov    -0x78(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %eax,%edx
  mov    -0x74(%rbp),%eax
  add    %edx,%eax
  lea    0x1(%rax),%ecx
  mov    -0x74(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x78(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rsi,%rax
  mov    %ecx,-0x70(%rbp,%rax,4)
  addl   $0x1,-0x74(%rbp)
L47:
  cmpl   $0x4,-0x74(%rbp)
  jle    L9
  addl   $0x1,-0x78(%rbp)
L28:
  cmpl   $0x4,-0x78(%rbp)
  jle    L12
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <fun@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <pr@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L35
  call   <__stack_chk_fail@plt>
L35:
  leave
  ret
```

assistant: ```c
int main() {
  int a[5][5], i, j;
  for (i = 0; i < 5; i++)
    for (j = 0; j < 5; j++)
      a[i][j] = i * 5 + j + 1;
  fun(a);
  pr(a);
  getchar();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    $0x6,%ecx
L11:
  lea    -0x5(%rcx),%eax
  mov    %rsi,%rdx
L19:
  mov    %eax,(%rdx)
  add    $0x1,%eax
  add    $0x4,%rdx
  cmp    %ecx,%eax
  jne    L19
  add    $0x14,%rsi
  add    $0x5,%ecx
  cmp    $0x1f,%ecx
  jne    L11
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <fun@plt>
  mov    %rbx,%rdi
  call   <pr@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  mov    $0x0,%eax
  add    $0x70,%rsp
  pop    %rbx
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int a[5][5], i, j;
  for (i = 0; i < 5; i++)
    for (j = 0; j < 5; j++)
      a[i][j] = i * 5 + j + 1;
  fun(a);
  pr(a);
  getchar();
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x6,%ecx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
L19:
  lea    -0x5(%rcx),%eax
  mov    %rsi,%rdx
L0:
  mov    %eax,(%rdx)
  add    $0x1,%eax
  add    $0x4,%rdx
  cmp    %ecx,%eax
  jne    L0
  add    $0x14,%rsi
  lea    0x5(%rax),%ecx
  cmp    $0x1a,%eax
  jne    L19
  mov    %rbp,%rdi
  call   <fun@plt>
  mov    %rbp,%rdi
  call   <pr@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int a[5][5], i, j;
  for (i = 0; i < 5; i++)
    for (j = 0; j < 5; j++)
      a[i][j] = i * 5 + j + 1;
  fun(a);
  pr(a);
  getchar();
  return 0;
}
```
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
  sub    $0x70,%rsp
  movdqa D40(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movl   $0x19,0x60(%rsp)
  movaps %xmm0,(%rsp)
  movdqa D51(%rip),%xmm0
  mov    %rbp,%rdi
  movaps %xmm0,0x10(%rsp)
  movdqa D0(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  movdqa D20(%rip),%xmm0
  movaps %xmm0,0x30(%rsp)
  movdqa D2(%rip),%xmm0
  movaps %xmm0,0x40(%rsp)
  movdqa D57(%rip),%xmm0
  movaps %xmm0,0x50(%rsp)
  call   <fun@plt>
  mov    %rbp,%rdi
  call   <pr@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D40
type_guessed: byte[16]
name: retrieve_data
arguments: tag: D51
type_guessed: byte[16]
name: retrieve_data
arguments: tag: D0
type_guessed: byte[16]
name: retrieve_data
arguments: tag: D20
type_guessed: byte[16]
name: retrieve_data
arguments: tag: D2
type_guessed: byte[16]
name: retrieve_data
arguments: tag: D57
type_guessed: byte[16]

tool: {"D40": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D51": [0x05,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x08,0x00,0x00,0x00]}

tool: {"D0": [0x09,0x00,0x00,0x00,0x0a,0x00,0x00,0x00,0x0b,0x00,0x00,0x00,0x0c,0x00,0x00,0x00]}

tool: {"D20": [0x0d,0x00,0x00,0x00,0x0e,0x00,0x00,0x00,0x0f,0x00,0x00,0x00,0x10,0x00,0x00,0x00]}

tool: {"D2": [0x11,0x00,0x00,0x00,0x12,0x00,0x00,0x00,0x13,0x00,0x00,0x00,0x14,0x00,0x00,0x00]}

tool: {"D57": [0x15,0x00,0x00,0x00,0x16,0x00,0x00,0x00,0x17,0x00,0x00,0x00,0x18,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int a[5][5], i, j;
  for (i = 0; i < 5; i++)
    for (j = 0; j < 5; j++)
      a[i][j] = i * 5 + j + 1;
  fun(a);
  pr(a);
  getchar();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<bug>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x4000(%rsp),%r11
L5:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L5
  sub    $0x20,%rsp
  mov    %rdi,-0x4018(%rbp)
  mov    %esi,-0x401c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x4010(%rbp),%rax
  movabs $0x5542205d2a2a2a5b,%rcx
  mov    %rcx,(%rax)
  movl   $0x203a47,0x8(%rax)
  lea    -0x4010(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    -0x4010(%rbp),%rdx
  lea    (%rdx,%rax,1),%rcx
  mov    -0x401c(%rbp),%edx
  mov    -0x4018(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  lea    -0x4010(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <log_string@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

assistant: ```c
void bug(const char *str, int param) {
  char buf[16384];
  strcpy(buf, "[***] BUG: ");
  sprintf(buf + strlen(buf), str, param);
  log_string(buf);
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<bug>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x4000(%rsp),%r11
L13:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L13
  sub    $0x10,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x4008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  movabs $0x5542205d2a2a2a5b,%rax
  mov    %rax,(%rsp)
  movl   $0x203a47,0x8(%rsp)
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    (%rbx,%rax,1),%rdi
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    $0x0,%eax
  call   <sprintf@plt>
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <log_string@plt>
  mov    0x4008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x4010,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void bug(const char *str, int param) {
  char buf[16384];
  strcpy(buf, "[***] BUG: ");
  sprintf(buf + strlen(buf), str, param);
  log_string(buf);
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<bug>:
  endbr64
  push   %rbp
  lea    -0x4000(%rsp),%r11
L4:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L4
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x4008(%rsp)
  xor    %eax,%eax
  mov    %rdi,%r8
  mov    %esi,%edx
  lea    0xb(%rsp),%rdi
  movabs $0x5542205d2a2a2a5b,%rax
  mov    %rsp,%rbp
  mov    %r8,%rsi
  movl   $0x203a47,0x8(%rsp)
  mov    %rax,(%rsp)
  xor    %eax,%eax
  call   <sprintf@plt>
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <log_string@plt>
  mov    0x4008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x4010,%rsp
  pop    %rbp
  ret
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void bug(const char *str, int param) {
  char buf[16384];
  strcpy(buf, "[***] BUG: ");
  sprintf(buf + strlen(buf), str, param);
  log_string(buf);
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<bug>:
  endbr64
  push   %rbp
  lea    -0x4000(%rsp),%r11
L8:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L8
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x4008(%rsp)
  xor    %eax,%eax
  mov    %rdi,%r8
  mov    %esi,%edx
  lea    0xb(%rsp),%rdi
  movabs $0x5542205d2a2a2a5b,%rax
  mov    %rsp,%rbp
  mov    %r8,%rsi
  movl   $0x203a47,0x8(%rsp)
  mov    %rax,(%rsp)
  xor    %eax,%eax
  call   <sprintf@plt>
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <log_string@plt>
  mov    0x4008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x4010,%rsp
  pop    %rbp
  ret
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void bug(const char *str, int param) {
  char buf[16384];
  strcpy(buf, "[***] BUG: ");
  sprintf(buf + strlen(buf), str, param);
  log_string(buf);
  return;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<schrl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
L126:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L91
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x9(%rbp)
  je     L10
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L62
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x9(%rbp)
  je     L84
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L146
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x9(%rbp)
  je     L69
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L100
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x9(%rbp)
  je     L79
  addq   $0x1,-0x8(%rbp)
  jmp    L126
L91:
  nop
  jmp    L63
L10:
  nop
  jmp    L63
L62:
  nop
  jmp    L63
L84:
  nop
  jmp    L63
L146:
  nop
  jmp    L63
L69:
  nop
  jmp    L63
L100:
  nop
  jmp    L63
L79:
  nop
L63:
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
int schrl(const char *s, int c) {
  char ch;
  const char *t;
  ch = c;
  t = s;
  for (;;) {
    if (!*t)
      break;
    if (*t == ch)
      break;
    ++t;
    if (!*t)
      break;
    if (*t == ch)
      break;
    ++t;
    if (!*t)
      break;
    if (*t == ch)
      break;
    ++t;
    if (!*t)
      break;
    if (*t == ch)
      break;
    ++t;
  }
  return t - s;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<schrl>:
  endbr64
  mov    %esi,%edx
  movzbl (%rdi),%eax
  cmp    %sil,%al
  je     L72
  test   %al,%al
  je     L72
  mov    %rdi,%rax
L2:
  movzbl 0x1(%rax),%ecx
  cmp    %dl,%cl
  je     L41
  test   %cl,%cl
  je     L41
  movzbl 0x2(%rax),%ecx
  cmp    %dl,%cl
  je     L6
  test   %cl,%cl
  je     L6
  movzbl 0x3(%rax),%ecx
  cmp    %dl,%cl
  je     L19
  test   %cl,%cl
  je     L19
  add    $0x4,%rax
  movzbl (%rax),%ecx
  cmp    %dl,%cl
  je     L22
  test   %cl,%cl
  jne    L2
  jmp    L22
L41:
  add    $0x1,%rax
L22:
  sub    %edi,%eax
  ret
L6:
  add    $0x2,%rax
  jmp    L22
L19:
  add    $0x3,%rax
  jmp    L22
L72:
  mov    %rdi,%rax
  jmp    L22
```

assistant: ```c
int schrl(const char *s, int c) {
  char ch;
  const char *t;
  ch = c;
  t = s;
  for (;;) {
    if (!*t)
      break;
    if (*t == ch)
      break;
    ++t;
    if (!*t)
      break;
    if (*t == ch)
      break;
    ++t;
    if (!*t)
      break;
    if (*t == ch)
      break;
    ++t;
    if (!*t)
      break;
    if (*t == ch)
      break;
    ++t;
  }
  return t - s;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<schrl>:
  endbr64
  movzbl (%rdi),%eax
  mov    %esi,%edx
  cmp    %sil,%al
  je     L101
  test   %al,%al
  je     L101
  mov    %rdi,%rax
  jmp    L58
  nopw   0x0(%rax,%rax,1)
L14:
  cmp    %dl,%cl
  je     L38
  movzbl 0x2(%rax),%ecx
  cmp    %dl,%cl
  je     L3
  test   %cl,%cl
  je     L3
  movzbl 0x3(%rax),%ecx
  cmp    %dl,%cl
  je     L47
  test   %cl,%cl
  je     L47
  movzbl 0x4(%rax),%ecx
  add    $0x4,%rax
  cmp    %dl,%cl
  je     L63
  test   %cl,%cl
  je     L63
L58:
  movzbl 0x1(%rax),%ecx
  test   %cl,%cl
  jne    L14
L38:
  add    $0x1,%rax
  sub    %edi,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  add    $0x2,%rax
  sub    %edi,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L47:
  add    $0x3,%rax
  sub    %edi,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L63:
  sub    %edi,%eax
  ret
L101:
  xor    %eax,%eax
  ret
```

assistant: ```c
int schrl(const char *s, int c) {
  char ch;
  const char *t;
  ch = c;
  t = s;
  for (;;) {
    if (!*t)
      break;
    if (*t == ch)
      break;
    ++t;
    if (!*t)
      break;
    if (*t == ch)
      break;
    ++t;
    if (!*t)
      break;
    if (*t == ch)
      break;
    ++t;
    if (!*t)
      break;
    if (*t == ch)
      break;
    ++t;
  }
  return t - s;
}
```
user: What is the c source code of the assembly code below?

```asm
<schrl>:
  endbr64
  movzbl (%rdi),%eax
  mov    %esi,%edx
  cmp    %sil,%al
  je     L51
  test   %al,%al
  je     L51
  mov    %rdi,%rax
  jmp    L88
  nopw   0x0(%rax,%rax,1)
L12:
  cmp    %dl,%cl
  je     L68
  movzbl 0x2(%rax),%ecx
  cmp    %dl,%cl
  je     L43
  test   %cl,%cl
  je     L43
  movzbl 0x3(%rax),%ecx
  cmp    %dl,%cl
  je     L46
  test   %cl,%cl
  je     L46
  movzbl 0x4(%rax),%ecx
  add    $0x4,%rax
  cmp    %dl,%cl
  je     L55
  test   %cl,%cl
  je     L55
L88:
  movzbl 0x1(%rax),%ecx
  test   %cl,%cl
  jne    L12
L68:
  add    $0x1,%rax
  sub    %edi,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L43:
  add    $0x2,%rax
  sub    %edi,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L46:
  add    $0x3,%rax
  sub    %edi,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L55:
  sub    %edi,%eax
  ret
L51:
  xor    %eax,%eax
  ret
```

assistant: ```c
int schrl(const char *s, int c) {
  char ch;
  const char *t;
  ch = c;
  t = s;
  for (;;) {
    if (!*t)
      break;
    if (*t == ch)
      break;
    ++t;
    if (!*t)
      break;
    if (*t == ch)
      break;
    ++t;
    if (!*t)
      break;
    if (*t == ch)
      break;
    ++t;
    if (!*t)
      break;
    if (*t == ch)
      break;
    ++t;
  }
  return t - s;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<test_and_clear_bit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  cltd
  shr    $0x1d,%edx
  add    %edx,%eax
  and    $0x7,%eax
  sub    %edx,%eax
  mov    %eax,-0x8(%rbp)
  cmpl   $0x7,-0x14(%rbp)
  jle    L6
  mov    -0x14(%rbp),%eax
  lea    0x7(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x3,%eax
  cltq
  shl    $0x2,%rax
  add    %rax,-0x20(%rbp)
L6:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  and    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  not    %eax
  and    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int test_and_clear_bit(int nr, uint8_t *addr) {
  uint8_t bit_place = nr % 8;
  if (nr > 7) {
    addr += (nr / 8);
  }
  int oldbit = *addr & (0x1 << bit_place);
  *addr &= ~(0x1 << bit_place);
  return oldbit;
}
```
user: Translate the below assembly code into C programming language:

```asm
<test_and_clear_bit>:
  endbr64
  mov    %edi,%ecx
  cmp    $0x7,%edi
  jle    L8
  lea    0x7(%rdi),%eax
  test   %edi,%edi
  cmovns %edi,%eax
  sar    $0x3,%eax
  cltq
  lea    (%rsi,%rax,4),%rsi
L8:
  mov    (%rsi),%eax
  and    $0x7,%ecx
  mov    $0x1,%edx
  shl    %cl,%edx
  mov    %edx,%ecx
  not    %ecx
  and    %eax,%ecx
  mov    %ecx,(%rsi)
  and    %edx,%eax
  ret
```

assistant: ```c
int test_and_clear_bit(int nr, uint8_t *addr) {
  uint8_t bit_place = nr % 8;
  if (nr > 7) {
    addr += (nr / 8);
  }
  int oldbit = *addr & (0x1 << bit_place);
  *addr &= ~(0x1 << bit_place);
  return oldbit;
}
```
user: What is the c source code of the assembly code below:

```asm
<test_and_clear_bit>:
  endbr64
  mov    %edi,%ecx
  cmp    $0x7,%edi
  jle    L5
  mov    %edi,%eax
  sar    $0x3,%eax
  cltq
  lea    (%rsi,%rax,4),%rsi
L5:
  and    $0x7,%ecx
  mov    $0x1,%edx
  mov    (%rsi),%eax
  shl    %cl,%edx
  mov    %edx,%ecx
  not    %ecx
  and    %eax,%ecx
  and    %edx,%eax
  mov    %ecx,(%rsi)
  ret
```

assistant: ```c
int test_and_clear_bit(int nr, uint8_t *addr) {
  uint8_t bit_place = nr % 8;
  if (nr > 7) {
    addr += (nr / 8);
  }
  int oldbit = *addr & (0x1 << bit_place);
  *addr &= ~(0x1 << bit_place);
  return oldbit;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test_and_clear_bit>:
  endbr64
  mov    %edi,%ecx
  cmp    $0x7,%edi
  jle    L3
  mov    %edi,%eax
  sar    $0x3,%eax
  cltq
  lea    (%rsi,%rax,4),%rsi
L3:
  and    $0x7,%ecx
  mov    $0x1,%edx
  mov    (%rsi),%eax
  shl    %cl,%edx
  mov    %edx,%ecx
  not    %ecx
  and    %eax,%ecx
  and    %edx,%eax
  mov    %ecx,(%rsi)
  ret
```

assistant: ```c
int test_and_clear_bit(int nr, uint8_t *addr) {
  uint8_t bit_place = nr % 8;
  if (nr > 7) {
    addr += (nr / 8);
  }
  int oldbit = *addr & (0x1 << bit_place);
  *addr &= ~(0x1 << bit_place);
  return oldbit;
}
```