ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<clearstring>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L12
L17:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  addl   $0x1,-0x8(%rbp)
L12:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L17
  nop
  nop
  leave
  ret
```

assistant: ```c
void clearstring(char *pStr) {
  int len = strlen(pStr);
  int i;
  for (i = 0; i < len; i++) {
    pStr[i] = '\0';
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<clearstring>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L16
  mov    %rbx,%rdx
  lea    -0x1(%rax),%eax
  lea    0x1(%rbx,%rax,1),%rax
L18:
  movb   $0x0,(%rdx)
  add    $0x1,%rdx
  cmp    %rax,%rdx
  jne    L18
L16:
  pop    %rbx
  ret
```

assistant: ```c
void clearstring(char *pStr) {
  int len = strlen(pStr);
  int i;
  for (i = 0; i < len; i++) {
    pStr[i] = '\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
<clearstring>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  test   %eax,%eax
  jle    L9
  mov    %rbp,%rdi
  movslq %eax,%rdx
  xor    %esi,%esi
  pop    %rbp
  jmp    1070 <memset@plt>
  nop
L9:
  pop    %rbp
  ret
```

assistant: ```c
void clearstring(char *pStr) {
  int len = strlen(pStr);
  int i;
  for (i = 0; i < len; i++) {
    pStr[i] = '\0';
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<clearstring>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  test   %eax,%eax
  jle    L7
  mov    %rbp,%rdi
  movslq %eax,%rdx
  xor    %esi,%esi
  pop    %rbp
  jmp    1070 <memset@plt>
  nop
L7:
  pop    %rbp
  ret
```

assistant: ```c
void clearstring(char *pStr) {
  int len = strlen(pStr);
  int i;
  for (i = 0; i < len; i++) {
    pStr[i] = '\0';
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<count_spaces>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L23
L3:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L23
  addl   $0x1,-0x4(%rbp)
L23:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L3
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int count_spaces(const char *str) {
  int count = 0;
  while (*str != '\0') {
    if (*str++ == ' ')
      count++;
  }
  return count;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<count_spaces>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L27
  mov    $0x0,%edx
L16:
  add    $0x1,%rdi
  cmp    $0x20,%al
  sete   %al
  movzbl %al,%eax
  add    %eax,%edx
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L16
L18:
  mov    %edx,%eax
  ret
L27:
  mov    $0x0,%edx
  jmp    L18
```

assistant: ```c
int count_spaces(const char *str) {
  int count = 0;
  while (*str != '\0') {
    if (*str++ == ' ')
      count++;
  }
  return count;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<count_spaces>:
  endbr64
  movzbl (%rdi),%eax
  xor    %r8d,%r8d
  test   %al,%al
  je     L18
  xchg   %ax,%ax
L4:
  add    $0x1,%rdi
  cmp    $0x20,%al
  sete   %al
  movzbl %al,%eax
  add    %eax,%r8d
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L4
L18:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int count_spaces(const char *str) {
  int count = 0;
  while (*str != '\0') {
    if (*str++ == ' ')
      count++;
  }
  return count;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<count_spaces>:
  endbr64
  movzbl (%rdi),%eax
  xor    %r8d,%r8d
  test   %al,%al
  je     L7
  xchg   %ax,%ax
L1:
  add    $0x1,%rdi
  cmp    $0x20,%al
  sete   %al
  movzbl %al,%eax
  add    %eax,%r8d
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L1
L7:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int count_spaces(const char *str) {
  int count = 0;
  while (*str != '\0') {
    if (*str++ == ' ')
      count++;
  }
  return count;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<test19>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    %r8d,-0x24(%rbp)
  mov    %r9d,-0x28(%rbp)
  mov    -0x14(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  setne  %al
  movzbl %al,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  je     L17
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jne    L19
L17:
  mov    $0x1,%eax
  jmp    L15
L19:
  mov    $0x0,%eax
L15:
  pop    %rbp
  ret
```

assistant: ```c
int test19(int a, int b, int c, int d, int e, int f) {
  int x = a != b;
  int y = c != 0;
  int z = d == e;
  return x == y || x == z;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<test19>:
  endbr64
  mov    %edx,%eax
  cmp    %esi,%edi
  setne  %dl
  movzbl %dl,%edx
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  cmp    %edx,%eax
  sete   %al
  cmp    %r8d,%ecx
  sete   %cl
  movzbl %cl,%ecx
  cmp    %edx,%ecx
  sete   %dl
  or     %edx,%eax
  movzbl %al,%eax
  ret
```

assistant: ```c
int test19(int a, int b, int c, int d, int e, int f) {
  int x = a != b;
  int y = c != 0;
  int z = d == e;
  return x == y || x == z;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<test19>:
  endbr64
  mov    %edx,%r9d
  xor    %edx,%edx
  cmp    %esi,%edi
  setne  %dl
  xor    %eax,%eax
  test   %r9d,%r9d
  setne  %al
  cmp    %edx,%eax
  sete   %al
  cmp    %r8d,%ecx
  sete   %cl
  movzbl %cl,%ecx
  cmp    %edx,%ecx
  sete   %dl
  or     %edx,%eax
  movzbl %al,%eax
  ret
```

assistant: ```c
int test19(int a, int b, int c, int d, int e, int f) {
  int x = a != b;
  int y = c != 0;
  int z = d == e;
  return x == y || x == z;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ar_auc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %rdi,-0x78(%rbp)
  mov    %esi,-0x7c(%rbp)
  mov    %edx,-0x80(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  fldz
  fstpt  -0x50(%rbp)
  fldz
  fstpt  -0x30(%rbp)
  fldt   0xe92(%rip)        # 2000 <_fini+0xdec>
  fstpt  -0x40(%rbp)
  jmp    L46
L27:
  lea    -0x60(%rbp),%rdi
  lea    -0x70(%rbp),%rcx
  mov    -0x80(%rbp),%edx
  mov    -0x7c(%rbp),%esi
  mov    -0x78(%rbp),%rax
  push   -0x38(%rbp)
  push   -0x40(%rbp)
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <ar_test@plt>
  add    $0x10,%rsp
  fldt   -0x60(%rbp)
  fldt   -0x30(%rbp)
  fsubp  %st,%st(1)
  fldz
  fcomip %st(1),%st
  fstp   %st(0)
  jbe    L24
  fldt   -0x60(%rbp)
  fldt   -0x30(%rbp)
  fsubp  %st,%st(1)
  fchs
  jmp    L36
L24:
  fldt   -0x60(%rbp)
  fldt   -0x30(%rbp)
  fsubp  %st,%st(1)
L36:
  fldt   -0x70(%rbp)
  fmulp  %st,%st(1)
  fstpt  -0x20(%rbp)
  fldt   -0x50(%rbp)
  fldt   -0x20(%rbp)
  faddp  %st,%st(1)
  fstpt  -0x50(%rbp)
  fldt   -0x60(%rbp)
  fstpt  -0x30(%rbp)
  fldt   -0x40(%rbp)
  fldt   0xe30(%rip)        # 2010 <_fini+0xdfc>
  fsubrp %st,%st(1)
  fstpt  -0x40(%rbp)
L46:
  fldt   0xe35(%rip)        # 2020 <_fini+0xe0c>
  fldt   -0x40(%rbp)
  fcomip %st(1),%st
  fstp   %st(0)
  jae    L27
  fldt   -0x50(%rbp)
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L19
  fstp   %st(0)
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

assistant: ```c
long double ar_auc(struct probans *ans, unsigned int num_probs, int tclass) {
  long double auc = 0;
  long double t;
  long double fpr_last = 0;
  for (t = 1.01; t >= -0.01; t -= 0.001) {
    long double tpr, fpr;
    ar_test(ans, num_probs, tclass, t, &tpr, &fpr);
    long double auc_column = ((fpr_last - fpr) < 0 ? ((fpr_last - fpr) * -1) : (fpr_last - fpr)) * tpr;
    auc += auc_column;
    fpr_last = fpr;
  }
  return auc;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ar_auc>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    %edx,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  fldz
  fstpt  0x20(%rsp)
  fldt   0x20(%rsp)
  fldl   0xe92(%rip)        # 2000 <_fini+0xdf8>
  fstpt  (%rsp)
  fstpt  0x10(%rsp)
  lea    0x40(%rsp),%r13
  jmp    L40
L35:
  fldt   0x30(%rsp)
  fmulp  %st,%st(1)
  fldt   0x10(%rsp)
  faddp  %st,%st(1)
  fstpt  0x10(%rsp)
  fldt   (%rsp)
  fsubl  0xe73(%rip)        # 2008 <_fini+0xe00>
  fld    %st(0)
  fstpt  (%rsp)
  fldl   0xe70(%rip)        # 2010 <_fini+0xe08>
  fxch   %st(1)
  fcomip %st(1),%st
  fstp   %st(0)
  jb     L5
L40:
  lea    0x30(%rsp),%rcx
  push   0x8(%rsp)
  push   0x8(%rsp)
  mov    %r13,%r8
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <ar_test@plt>
  fldt   0x30(%rsp)
  fldt   0x50(%rsp)
  fld    %st(0)
  fstpt  0x30(%rsp)
  fsubrp %st,%st(1)
  add    $0x10,%rsp
  fldz
  fcomip %st(1),%st
  jbe    L35
  fchs
  jmp    L35
L5:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  fldt   0x10(%rsp)
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long double ar_auc(struct probans *ans, unsigned int num_probs, int tclass) {
  long double auc = 0;
  long double t;
  long double fpr_last = 0;
  for (t = 1.01; t >= -0.01; t -= 0.001) {
    long double tpr, fpr;
    ar_test(ans, num_probs, tclass, t, &tpr, &fpr);
    long double auc_column = ((fpr_last - fpr) < 0 ? ((fpr_last - fpr) * -1) : (fpr_last - fpr)) * tpr;
    auc += auc_column;
    fpr_last = fpr;
  }
  return auc;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ar_auc>:
  endbr64
  push   %r14
  fldz
  push   %r13
  fld    %st(0)
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x60,%rsp
  fldl   0xe9f(%rip)        # 2000 <_fini+0xdf4>
  fxch   %st(2)
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    0x40(%rsp),%r14
  lea    0x30(%rsp),%r13
  jmp    L18
  nop
L1:
  fxch   %st(2)
L18:
  fstpt  0x20(%rsp)
  sub    $0x10,%rsp
  mov    %ebx,%edx
  mov    %r14,%r8
  mov    %r13,%rcx
  mov    %ebp,%esi
  mov    %r12,%rdi
  fstpt  0x20(%rsp)
  fld    %st(0)
  fstpt  (%rsp)
  fstpt  0x10(%rsp)
  call   <ar_test@plt>
  fldt   0x20(%rsp)
  fldt   0x50(%rsp)
  pop    %rax
  pop    %rdx
  fsubr  %st,%st(1)
  fldz
  fcomip %st(2),%st
  fldt   (%rsp)
  fldt   0x20(%rsp)
  jbe    L30
  fxch   %st(3)
  fchs
  fxch   %st(3)
L30:
  fldt   0x30(%rsp)
  fmulp  %st,%st(4)
  faddp  %st,%st(3)
  fsubl  0xe32(%rip)        # 2008 <_fini+0xdfc>
  fldl   0xe34(%rip)        # 2010 <_fini+0xe04>
  fxch   %st(1)
  fcomi  %st(1),%st
  fstp   %st(1)
  jae    L1
  fstp   %st(0)
  fstp   %st(0)
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L15:
  fstp   %st(0)
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long double ar_auc(struct probans *ans, unsigned int num_probs, int tclass) {
  long double auc = 0;
  long double t;
  long double fpr_last = 0;
  for (t = 1.01; t >= -0.01; t -= 0.001) {
    long double tpr, fpr;
    ar_test(ans, num_probs, tclass, t, &tpr, &fpr);
    long double auc_column = ((fpr_last - fpr) < 0 ? ((fpr_last - fpr) * -1) : (fpr_last - fpr)) * tpr;
    auc += auc_column;
    fpr_last = fpr;
  }
  return auc;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ar_auc>:
  endbr64
  push   %r14
  fldz
  push   %r13
  fld    %st(0)
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x60,%rsp
  fldl   0xe9f(%rip)        # 2000 <_fini+0xdf4>
  fxch   %st(2)
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    0x40(%rsp),%r14
  lea    0x30(%rsp),%r13
  jmp    L28
  nop
L14:
  fxch   %st(2)
L28:
  fstpt  0x20(%rsp)
  sub    $0x10,%rsp
  mov    %ebx,%edx
  mov    %r14,%r8
  mov    %r13,%rcx
  mov    %ebp,%esi
  mov    %r12,%rdi
  fstpt  0x20(%rsp)
  fld    %st(0)
  fstpt  (%rsp)
  fstpt  0x10(%rsp)
  call   <ar_test@plt>
  fldt   0x20(%rsp)
  fldt   0x50(%rsp)
  pop    %rax
  pop    %rdx
  fsubr  %st,%st(1)
  fldz
  fcomip %st(2),%st
  fldt   (%rsp)
  fldt   0x20(%rsp)
  jbe    L26
  fxch   %st(3)
  fchs
  fxch   %st(3)
L26:
  fldt   0x30(%rsp)
  fmulp  %st,%st(4)
  faddp  %st,%st(3)
  fsubl  0xe32(%rip)        # 2008 <_fini+0xdfc>
  fldl   0xe34(%rip)        # 2010 <_fini+0xe04>
  fxch   %st(1)
  fcomi  %st(1),%st
  fstp   %st(1)
  jae    L14
  fstp   %st(0)
  fstp   %st(0)
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L36:
  fstp   %st(0)
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long double ar_auc(struct probans *ans, unsigned int num_probs, int tclass) {
  long double auc = 0;
  long double t;
  long double fpr_last = 0;
  for (t = 1.01; t >= -0.01; t -= 0.001) {
    long double tpr, fpr;
    ar_test(ans, num_probs, tclass, t, &tpr, &fpr);
    long double auc_column = ((fpr_last - fpr) < 0 ? ((fpr_last - fpr) * -1) : (fpr_last - fpr)) * tpr;
    auc += auc_column;
    fpr_last = fpr;
  }
  return auc;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<luaI_addquoted>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <luaL_check_lstr@plt>
  mov    %rax,-0x10(%rbp)
  mov    $0x22,%edi
  call   <luaL_addchar@plt>
  jmp    L5
L72:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    $0x5c,%eax
  je     L32
  cmp    $0x5c,%eax
  jg     L39
  cmp    $0x22,%eax
  je     L32
  cmp    $0x22,%eax
  jg     L39
  test   %eax,%eax
  je     L60
  cmp    $0xa,%eax
  jne    L39
L32:
  mov    $0x5c,%edi
  call   <luaL_addchar@plt>
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <luaL_addchar@plt>
  jmp    L17
L60:
  mov    $0x4,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <addnchar@plt>
  jmp    L17
L39:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <luaL_addchar@plt>
L17:
  addq   $0x1,-0x10(%rbp)
L5:
  mov    -0x18(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  test   %rax,%rax
  jne    L72
  mov    $0x22,%edi
  call   <luaL_addchar@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D8
type: string

tool: D8: \000

assistant: ```c
void luaI_addquoted(int arg) {
  long l;
  char *s = luaL_check_lstr(arg, &l);
  luaL_addchar('"');
  while (l--) {
    switch (*s) {
    case '"':
    case '\\':
    case '\n':
      luaL_addchar('\\');
      luaL_addchar(*s);
      break;
    case '\0':
      addnchar("\\000", 4);
      break;
    default:
      luaL_addchar(*s);
    }
    s++;
  }
  luaL_addchar('"');
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<luaI_addquoted>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <luaL_check_lstr@plt>
  mov    %rax,%rbx
  mov    $0x22,%edi
  call   <luaL_addchar@plt>
  mov    (%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,(%rsp)
  test   %rax,%rax
  je     L15
  lea    D6(%rip),%rbp
  jmp    L98
L97:
  cmp    $0x22,%dil
  je     L57
  cmp    $0x5c,%dil
  jne    L72
L57:
  mov    $0x5c,%edi
  call   <luaL_addchar@plt>
  movsbl (%rbx),%edi
  call   <luaL_addchar@plt>
L23:
  add    $0x1,%rbx
  mov    (%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,(%rsp)
  test   %rax,%rax
  je     L15
L98:
  movzbl (%rbx),%edi
  cmp    $0xa,%dil
  je     L57
  jg     L97
  test   %dil,%dil
  jne    L72
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <addnchar@plt>
  jmp    L23
L72:
  movsbl %dil,%edi
  call   <luaL_addchar@plt>
  jmp    L23
L15:
  mov    $0x22,%edi
  call   <luaL_addchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D6
type_guessed: string

tool: D6: \000

assistant: ```c
void luaI_addquoted(int arg) {
  long l;
  char *s = luaL_check_lstr(arg, &l);
  luaL_addchar('"');
  while (l--) {
    switch (*s) {
    case '"':
    case '\\':
    case '\n':
      luaL_addchar('\\');
      luaL_addchar(*s);
      break;
    case '\0':
      addnchar("\\000", 4);
      break;
    default:
      luaL_addchar(*s);
    }
    s++;
  }
  luaL_addchar('"');
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<luaI_addquoted>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <luaL_check_lstr@plt>
  mov    $0x22,%edi
  mov    %rax,%rbx
  call   <luaL_addchar@plt>
  mov    (%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,(%rsp)
  test   %rax,%rax
  je     L41
  lea    D2(%rip),%rbp
  jmp    L83
  nopl   0x0(%rax)
L68:
  test   %dil,%dil
  jne    L9
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <addnchar@plt>
L98:
  mov    (%rsp),%rax
  add    $0x1,%rbx
  lea    -0x1(%rax),%rdx
  mov    %rdx,(%rsp)
  test   %rax,%rax
  je     L41
L83:
  movsbl (%rbx),%edi
  cmp    $0xa,%dil
  je     L52
  jle    L68
  cmp    $0x22,%dil
  je     L52
  cmp    $0x5c,%dil
  jne    L9
L52:
  mov    $0x5c,%edi
  call   <luaL_addchar@plt>
  movsbl (%rbx),%edi
  call   <luaL_addchar@plt>
  jmp    L98
  nopw   0x0(%rax,%rax,1)
L9:
  call   <luaL_addchar@plt>
  jmp    L98
  nop
L41:
  mov    $0x22,%edi
  call   <luaL_addchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L47
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L47:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D2
type_guessed: string

tool: D2: \000

assistant: ```c
void luaI_addquoted(int arg) {
  long l;
  char *s = luaL_check_lstr(arg, &l);
  luaL_addchar('"');
  while (l--) {
    switch (*s) {
    case '"':
    case '\\':
    case '\n':
      luaL_addchar('\\');
      luaL_addchar(*s);
      break;
    case '\0':
      addnchar("\\000", 4);
      break;
    default:
      luaL_addchar(*s);
    }
    s++;
  }
  luaL_addchar('"');
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<luaI_addquoted>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <luaL_check_lstr@plt>
  mov    $0x22,%edi
  mov    %rax,%rbx
  call   <luaL_addchar@plt>
  mov    (%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,(%rsp)
  test   %rax,%rax
  je     L100
  lea    D8(%rip),%rbp
  jmp    L7
  nopl   0x0(%rax)
L70:
  test   %dil,%dil
  jne    L6
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <addnchar@plt>
L18:
  mov    (%rsp),%rax
  add    $0x1,%rbx
  lea    -0x1(%rax),%rdx
  mov    %rdx,(%rsp)
  test   %rax,%rax
  je     L100
L7:
  movsbl (%rbx),%edi
  cmp    $0xa,%dil
  je     L77
  jle    L70
  cmp    $0x22,%dil
  je     L77
  cmp    $0x5c,%dil
  jne    L6
L77:
  mov    $0x5c,%edi
  call   <luaL_addchar@plt>
  movsbl (%rbx),%edi
  call   <luaL_addchar@plt>
  jmp    L18
  nopw   0x0(%rax,%rax,1)
L6:
  call   <luaL_addchar@plt>
  jmp    L18
  nop
L100:
  mov    $0x22,%edi
  call   <luaL_addchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D8
type: string

tool: D8: \000

assistant: ```c
void luaI_addquoted(int arg) {
  long l;
  char *s = luaL_check_lstr(arg, &l);
  luaL_addchar('"');
  while (l--) {
    switch (*s) {
    case '"':
    case '\\':
    case '\n':
      luaL_addchar('\\');
      luaL_addchar(*s);
      break;
    case '\0':
      addnchar("\\000", 4);
      break;
    default:
      luaL_addchar(*s);
    }
    s++;
  }
  luaL_addchar('"');
}
```
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
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movss  -0x14(%rbp),%xmm1
  movss  -0x10(%rbp),%xmm0
  addss  %xmm1,%xmm0
  movss  D60(%rip),%xmm1
  divss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  movss  -0xc(%rbp),%xmm0
  comiss D40(%rip),%xmm0
  jb     L28
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L3
L28:
  lea    D66(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L3:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D13
data_type: string
name: read_data
arguments: tag: D43
data_type: string
name: read_data
arguments: tag: D54
data_type: string
name: read_data
arguments: tag: D60
data_type: byte[4]
name: read_data
arguments: tag: D40
data_type: f32
name: read_data
arguments: tag: D50
data_type: string
name: read_data
arguments: tag: D66
data_type: string

tool: D13: Digite a nota da primeira prova: 

tool: D43: %f

tool: D54: Digite a nota da segunda prova: 

tool: {"D60": [0x00,0x00,0x00,0x40]}

tool: D40: 6

tool: D50: 
APROVADO!!!



tool: D66: 
REPROVADO!!!



assistant: ```c
int main() {
  float x, y, z;
  printf("Digite a nota da primeira prova: ");
  scanf("%f", &x);
  printf("Digite a nota da segunda prova: ");
  scanf("%f", &y);
  z = ((x + y) / 2);
  if (z >= 6.0) {
    printf("\nAPROVADO!!!\n\n");
  } else {
    printf("\nREPROVADO!!!\n\n");
  }
  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   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D15(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D11(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movss  (%rsp),%xmm0
  addss  0x4(%rsp),%xmm0
  mulss  D9(%rip),%xmm0
  comiss D23(%rip),%xmm0
  jb     L22
  lea    D14(%rip),%rdi
  call   <printf@plt>
L8:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L22:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  jmp    L8
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D28
data_type: string
name: extract_symbolic_data
arguments: data_label: D15
data_type: string
name: extract_symbolic_data
arguments: data_label: D11
data_type: string
name: extract_symbolic_data
arguments: data_label: D9
data_type: byte[4]
name: extract_symbolic_data
arguments: data_label: D23
data_type: f32
name: extract_symbolic_data
arguments: data_label: D14
data_type: string
name: extract_symbolic_data
arguments: data_label: D7
data_type: string

tool: D28: Digite a nota da primeira prova: 

tool: D15: %f

tool: D11: Digite a nota da segunda prova: 

tool: {"D9": [0x00,0x00,0x00,0x3f]}

tool: D23: 6

tool: D14: 
APROVADO!!!



tool: D7: 
REPROVADO!!!



assistant: ```c
int main() {
  float x, y, z;
  printf("Digite a nota da primeira prova: ");
  scanf("%f", &x);
  printf("Digite a nota da segunda prova: ");
  scanf("%f", &y);
  z = ((x + y) / 2);
  if (z >= 6.0) {
    printf("\nAPROVADO!!!\n\n");
  } else {
    printf("\nREPROVADO!!!\n\n");
  }
  return 0;
}
```