ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L7
  mov    $0x0,%eax
  jmp    L3
L7:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L3:
  leave
  ret
```

assistant: ```c
char *strdup(const char *s) {
  size_t l = strlen(s);
  char *d = malloc(l + 1);
  if (!d) {
    return NULL;
  }
  return memcpy(d, s, l + 1);
}
```
user: What is the c source code of the assembly code below:

```asm
<strdup>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rbp
  mov    %rbp,%rdi
  call   <malloc@plt>
  test   %rax,%rax
  je     L3
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L3:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *strdup(const char *s) {
  size_t l = strlen(s);
  char *d = malloc(l + 1);
  if (!d) {
    return NULL;
  }
  return memcpy(d, s, l + 1);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%r12
  mov    %r12,%rdi
  call   <malloc@plt>
  test   %rax,%rax
  je     L0
  add    $0x8,%rsp
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <memcpy@plt>
L0:
  pop    %rdx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strdup(const char *s) {
  size_t l = strlen(s);
  char *d = malloc(l + 1);
  if (!d) {
    return NULL;
  }
  return memcpy(d, s, l + 1);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%r12
  mov    %r12,%rdi
  call   <malloc@plt>
  test   %rax,%rax
  je     L2
  add    $0x8,%rsp
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <memcpy@plt>
L2:
  pop    %rdx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strdup(const char *s) {
  size_t l = strlen(s);
  char *d = malloc(l + 1);
  if (!d) {
    return NULL;
  }
  return memcpy(d, s, l + 1);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x2a,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    $0x0,%eax
  call   <bar@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x2a,-0x4(%rbp)
  je     L0
  call   <abort@plt>
L0:
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int a = 42;
  asm("" : "+r"(a));
  bar();
  asm("" : "+r"(a));
  if (a != 42)
    __builtin_abort();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x2a,%ebx
  mov    $0x0,%eax
  call   <bar@plt>
  cmp    $0x2a,%ebx
  jne    L3
  mov    $0x0,%eax
  pop    %rbx
  ret
L3:
  call   <abort@plt>
```

assistant: ```c
int main() {
  int a = 42;
  asm("" : "+r"(a));
  bar();
  asm("" : "+r"(a));
  if (a != 42)
    __builtin_abort();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  mov    $0x2a,%ebx
  call   <bar@plt>
  cmp    $0x2a,%ebx
  jne    1080 <main.cold>
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int a = 42;
  asm("" : "+r"(a));
  bar();
  asm("" : "+r"(a));
  if (a != 42)
    __builtin_abort();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  mov    $0x2a,%ebx
  call   <bar@plt>
  cmp    $0x2a,%ebx
  jne    1080 <main.cold>
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int a = 42;
  asm("" : "+r"(a));
  bar();
  asm("" : "+r"(a));
  if (a != 42)
    __builtin_abort();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<itos>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %esi,-0x38(%rbp)
  mov    %rdx,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x3736353433323130,%rax
  movabs $0x4645444342413938,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  movb   $0x0,-0x10(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x24(%rbp)
L13:
  mov    -0x24(%rbp),%eax
  cltd
  idivl  -0x38(%rbp)
  mov    %eax,-0x24(%rbp)
  addq   $0x1,-0x40(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jg     L13
  mov    -0x40(%rbp),%rax
  movb   $0x0,(%rax)
L4:
  subq   $0x1,-0x40(%rbp)
  mov    -0x34(%rbp),%eax
  cltd
  idivl  -0x38(%rbp)
  mov    %edx,%eax
  cltq
  movzbl -0x20(%rbp,%rax,1),%edx
  mov    -0x40(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x34(%rbp),%eax
  cltd
  idivl  -0x38(%rbp)
  mov    %eax,-0x34(%rbp)
  cmpl   $0x0,-0x34(%rbp)
  jg     L4
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
void itos(int i, int base, char *to) {
  char const digits[] = "0123456789ABCDEF";
  int aux = i;
  do {
    aux /= base;
    to++;
  } while (aux > 0);
  *to = '\0';
  do {
    to--;
    *to = digits[i % base];
    i /= base;
  } while (i > 0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<itos>:
  endbr64
  sub    $0x28,%rsp
  mov    %rdx,%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movabs $0x3736353433323130,%rax
  movabs $0x4645444342413938,%rdx
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  movb   $0x0,0x10(%rsp)
  mov    %edi,%edx
L11:
  mov    %edx,%eax
  cltd
  idiv   %esi
  mov    %eax,%edx
  add    $0x1,%rcx
  test   %eax,%eax
  jg     L11
  movb   $0x0,(%rcx)
L14:
  sub    $0x1,%rcx
  mov    %edi,%eax
  cltd
  idiv   %esi
  movslq %edx,%rdx
  movzbl (%rsp,%rdx,1),%edx
  mov    %dl,(%rcx)
  mov    %eax,%edi
  test   %eax,%eax
  jg     L14
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x28,%rsp
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void itos(int i, int base, char *to) {
  char const digits[] = "0123456789ABCDEF";
  int aux = i;
  do {
    aux /= base;
    to++;
  } while (aux > 0);
  *to = '\0';
  do {
    to--;
    *to = digits[i % base];
    i /= base;
  } while (i > 0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<itos>:
  endbr64
  sub    $0x28,%rsp
  movdqa D7(%rip),%xmm0
  mov    %rdx,%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movb   $0x0,0x10(%rsp)
  mov    %edi,%eax
  movaps %xmm0,(%rsp)
  xchg   %ax,%ax
L26:
  cltd
  add    $0x1,%rcx
  idiv   %esi
  test   %eax,%eax
  jg     L26
  movb   $0x0,(%rcx)
  xchg   %ax,%ax
L6:
  mov    %edi,%eax
  sub    $0x1,%rcx
  cltd
  idiv   %esi
  movslq %edx,%rdx
  mov    %eax,%edi
  movzbl (%rsp,%rdx,1),%eax
  mov    %al,(%rcx)
  test   %edi,%edi
  jg     L6
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x28,%rsp
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D7
type_infered: byte[16]

tool: {"D7": [0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x41,0x42,0x43,0x44,0x45,0x46]}

assistant: ```c
void itos(int i, int base, char *to) {
  char const digits[] = "0123456789ABCDEF";
  int aux = i;
  do {
    aux /= base;
    to++;
  } while (aux > 0);
  *to = '\0';
  do {
    to--;
    *to = digits[i % base];
    i /= base;
  } while (i > 0);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<itos>:
  endbr64
  sub    $0x28,%rsp
  movdqa D9(%rip),%xmm0
  mov    %rdx,%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movb   $0x0,0x10(%rsp)
  mov    %edi,%eax
  movaps %xmm0,(%rsp)
  xchg   %ax,%ax
L16:
  cltd
  add    $0x1,%rcx
  idiv   %esi
  test   %eax,%eax
  jg     L16
  movb   $0x0,(%rcx)
  xchg   %ax,%ax
L17:
  mov    %edi,%eax
  sub    $0x1,%rcx
  cltd
  idiv   %esi
  movslq %edx,%rdx
  mov    %eax,%edi
  movzbl (%rsp,%rdx,1),%eax
  mov    %al,(%rcx)
  test   %edi,%edi
  jg     L17
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x28,%rsp
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D9
type_guessed: byte[16]

tool: {"D9": [0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x41,0x42,0x43,0x44,0x45,0x46]}

assistant: ```c
void itos(int i, int base, char *to) {
  char const digits[] = "0123456789ABCDEF";
  int aux = i;
  do {
    aux /= base;
    to++;
  } while (aux > 0);
  *to = '\0';
  do {
    to--;
    *to = digits[i % base];
    i /= base;
  } while (i > 0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ft_putendl_fd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <ft_strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x1c(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <ft_putchar_fd@plt>
  nop
  leave
  ret
```

assistant: ```c
void ft_putendl_fd(char const *s, int fd) {
  int l;
  l = ft_strlen(s);
  write(fd, s, l);
  ft_putchar_fd('\n', fd);
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ft_putendl_fd>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  call   <ft_strlen@plt>
  mov    %eax,%edx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <write@plt>
  mov    %ebx,%esi
  mov    $0xa,%edi
  call   <ft_putchar_fd@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void ft_putendl_fd(char const *s, int fd) {
  int l;
  l = ft_strlen(s);
  write(fd, s, l);
  ft_putchar_fd('\n', fd);
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ft_putendl_fd>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  sub    $0x8,%rsp
  call   <ft_strlen@plt>
  mov    %r12,%rsi
  mov    %ebp,%edi
  mov    %eax,%edx
  call   <write@plt>
  add    $0x8,%rsp
  mov    %ebp,%esi
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1090 <ft_putchar_fd@plt>
```

assistant: ```c
void ft_putendl_fd(char const *s, int fd) {
  int l;
  l = ft_strlen(s);
  write(fd, s, l);
  ft_putchar_fd('\n', fd);
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_putendl_fd>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  sub    $0x8,%rsp
  call   <ft_strlen@plt>
  mov    %r12,%rsi
  mov    %ebp,%edi
  mov    %eax,%edx
  call   <write@plt>
  add    $0x8,%rsp
  mov    %ebp,%esi
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1090 <ft_putchar_fd@plt>
```

assistant: ```c
void ft_putendl_fd(char const *s, int fd) {
  int l;
  l = ft_strlen(s);
  write(fd, s, l);
  ft_putchar_fd('\n', fd);
  return;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<test_recursive_call>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jle    L27
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%ecx
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    %ecx,%edi
  call   <test_recursive_call>
  mov    -0x4(%rbp),%edx
  add    %edx,%eax
  jmp    L4
L27:
  cmpq   $0x0,-0x10(%rbp)
  je     L33
  mov    -0x18(%rbp),%rax
  mov    -0x10(%rbp),%rdx
  mov    %rax,%rdi
  call   *%rdx
  mov    $0x0,%eax
  jmp    L4
L33:
  jmp    L33
L4:
  leave
  ret
```

assistant: ```c
int test_recursive_call(int level, void (*callback_func)(void *), void *data) {
  if (level > 0) {
    return test_recursive_call(level - 1, callback_func, data) + level;
  } else if (callback_func != NULL) {
    callback_func(data);
  } else {
    while (1) {
    }
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<test_recursive_call>:
  endbr64
  push   %rbx
  test   %edi,%edi
  jg     L19
  test   %rsi,%rsi
  je     L38
  mov    %rdx,%rdi
  call   *%rsi
  mov    $0x0,%eax
L3:
  pop    %rbx
  ret
L19:
  mov    %edi,%ebx
  lea    -0x1(%rdi),%edi
  call   <test_recursive_call>
  add    %ebx,%eax
  jmp    L3
L38:
  jmp    L38
```

assistant: ```c
int test_recursive_call(int level, void (*callback_func)(void *), void *data) {
  if (level > 0) {
    return test_recursive_call(level - 1, callback_func, data) + level;
  } else if (callback_func != NULL) {
    callback_func(data);
  } else {
    while (1) {
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<test_recursive_call>:
  endbr64
  mov    %edi,%eax
  push   %r12
  mov    %rdx,%rdi
  xor    %r12d,%r12d
  test   %eax,%eax
  jle    L28
  nopw   0x0(%rax,%rax,1)
L1:
  add    %eax,%r12d
  sub    $0x1,%eax
  jne    L1
L28:
  test   %rsi,%rsi
  je     L35
  call   *%rsi
  mov    %r12d,%eax
  pop    %r12
  ret
L35:
  jmp    L35
```

assistant: ```c
int test_recursive_call(int level, void (*callback_func)(void *), void *data) {
  if (level > 0) {
    return test_recursive_call(level - 1, callback_func, data) + level;
  } else if (callback_func != NULL) {
    callback_func(data);
  } else {
    while (1) {
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test_recursive_call>:
  endbr64
  mov    %edi,%eax
  push   %r12
  mov    %rdx,%rdi
  test   %eax,%eax
  jle    L156
  mov    %eax,%r8d
  cmp    $0xa,%eax
  jle    L118
  movd   %eax,%xmm4
  mov    %eax,%ecx
  movdqa D9(%rip),%xmm3
  xor    %edx,%edx
  pshufd $0x0,%xmm4,%xmm1
  paddd  D4(%rip),%xmm1
  shr    $0x2,%ecx
  pxor   %xmm0,%xmm0
  nopl   0x0(%rax,%rax,1)
L21:
  movdqa %xmm1,%xmm2
  add    $0x1,%edx
  paddd  %xmm3,%xmm1
  paddd  %xmm2,%xmm0
  cmp    %edx,%ecx
  jne    L21
  movdqa %xmm0,%xmm1
  mov    %eax,%edx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%edx
  paddd  %xmm1,%xmm0
  sub    %edx,%eax
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%r12d
  cmp    %edx,%r8d
  je     L33
L113:
  mov    %eax,%edx
  add    %eax,%r12d
  sub    $0x1,%edx
  je     L33
  add    %edx,%r12d
  mov    %eax,%edx
  sub    $0x2,%edx
  je     L33
  add    %edx,%r12d
  mov    %eax,%edx
  sub    $0x3,%edx
  je     L33
  add    %edx,%r12d
  mov    %eax,%edx
  sub    $0x4,%edx
  je     L33
  add    %edx,%r12d
  mov    %eax,%edx
  sub    $0x5,%edx
  je     L33
  add    %edx,%r12d
  mov    %eax,%edx
  sub    $0x6,%edx
  je     L33
  add    %edx,%r12d
  mov    %eax,%edx
  sub    $0x7,%edx
  je     L33
  add    %edx,%r12d
  mov    %eax,%edx
  sub    $0x8,%edx
  je     L33
  add    %r12d,%edx
  lea    -0x9(%rax,%rdx,1),%r12d
L33:
  test   %rsi,%rsi
  je     L72
  call   *%rsi
  mov    %r12d,%eax
  pop    %r12
  ret
L72:
  jmp    L72
  nopl   (%rax)
L156:
  xor    %r12d,%r12d
  jmp    L33
L118:
  xor    %r12d,%r12d
  jmp    L113
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
data_type: byte[16]
name: extract_symbolic_data
arguments: data_label: D4
data_type: byte[16]

tool: {"D9": [0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff]}

tool: {"D4": [0x00,0x00,0x00,0x00,0xff,0xff,0xff,0xff,0xfe,0xff,0xff,0xff,0xfd,0xff,0xff,0xff]}

assistant: ```c
int test_recursive_call(int level, void (*callback_func)(void *), void *data) {
  if (level > 0) {
    return test_recursive_call(level - 1, callback_func, data) + level;
  } else if (callback_func != NULL) {
    callback_func(data);
  } else {
    while (1) {
    }
  }
  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
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x2c(%rbp),%rax
  mov    %rax,%rsi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movss  -0x30(%rbp),%xmm0
  movss  -0x24(%rbp),%xmm1
  subss  %xmm1,%xmm0
  movd   %xmm0,%eax
  mov    $0x2,%edi
  movd   %eax,%xmm0
  call   <pow@plt>
  mov    %rax,%rbx
  movss  -0x2c(%rbp),%xmm0
  movss  -0x20(%rbp),%xmm1
  subss  %xmm1,%xmm0
  movd   %xmm0,%eax
  mov    $0x2,%edi
  movd   %eax,%xmm0
  call   <pow@plt>
  add    %rbx,%rax
  mov    %rax,%rdi
  call   <sqrt@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x1c(%rbp)
  movss  -0x28(%rbp),%xmm0
  comiss -0x1c(%rbp),%xmm0
  jbe    L33
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L33:
  movss  -0x28(%rbp),%xmm1
  movss  -0x1c(%rbp),%xmm0
  comiss %xmm1,%xmm0
  jbe    L34
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L26
L34:
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L26:
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D39
data_type: string
name: parse_data
arguments: tag: D36
data_type: string
name: parse_data
arguments: tag: D20
data_type: string
name: parse_data
arguments: tag: D40
data_type: string

tool: D39: %f

tool: D36: Point is inside the Circle.

tool: D20: Point is outside the Circle.

tool: D40: Point is on the Circle.

assistant: ```c
int main() {
  float x, y, r, x1, y1, d;
  scanf("%f", &x);
  scanf("%f", &y);
  scanf("%f", &r);
  scanf("%f", &x1);
  scanf("%f", &y1);
  d = sqrt(pow(x - x1, 2) + pow(y - y1, 2));
  if (d < r) {
    printf("Point is inside the Circle.");
  }
  if (d > r) {
    printf("Point is outside the Circle.");
  } else {
    printf("Point is on the Circle.");
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  lea    D5(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x18(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x1c(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x20(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x24(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movss  0x14(%rsp),%xmm0
  subss  0x20(%rsp),%xmm0
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %rax,%rbx
  movss  0x18(%rsp),%xmm0
  subss  0x24(%rsp),%xmm0
  mov    $0x2,%edi
  call   <pow@plt>
  lea    (%rbx,%rax,1),%rdi
  call   <sqrt@plt>
  movaps %xmm0,%xmm1
  movss  %xmm0,0xc(%rsp)
  movss  0x1c(%rsp),%xmm0
  comiss %xmm1,%xmm0
  ja     L35
L30:
  movss  0xc(%rsp),%xmm2
  comiss 0x1c(%rsp),%xmm2
  jbe    L31
  lea    D1(%rip),%rdi
  call   <printf@plt>
L34:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  ret
L35:
  lea    D24(%rip),%rdi
  call   <printf@plt>
  jmp    L30
L31:
  lea    D15(%rip),%rdi
  call   <printf@plt>
  jmp    L34
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D5
data_type: string
name: parse_data
arguments: tag: D1
data_type: string
name: parse_data
arguments: tag: D24
data_type: string
name: parse_data
arguments: tag: D15
data_type: string

tool: D5: %f

tool: D1: Point is outside the Circle.

tool: D24: Point is inside the Circle.

tool: D15: Point is on the Circle.

assistant: ```c
int main() {
  float x, y, r, x1, y1, d;
  scanf("%f", &x);
  scanf("%f", &y);
  scanf("%f", &r);
  scanf("%f", &x1);
  scanf("%f", &y1);
  d = sqrt(pow(x - x1, 2) + pow(y - y1, 2));
  if (d < r) {
    printf("Point is inside the Circle.");
  }
  if (d > r) {
    printf("Point is outside the Circle.");
  } else {
    printf("Point is on the Circle.");
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D10(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  call   <scanf@plt>
  lea    0x18(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x1c(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x20(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x24(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movss  0x14(%rsp),%xmm0
  mov    $0x2,%edi
  subss  0x20(%rsp),%xmm0
  call   <pow@plt>
  movss  0x18(%rsp),%xmm0
  mov    $0x2,%edi
  subss  0x24(%rsp),%xmm0
  mov    %rax,%rbx
  call   <pow@plt>
  lea    (%rbx,%rax,1),%rdi
  call   <sqrt@plt>
  movss  0x1c(%rsp),%xmm1
  comiss %xmm0,%xmm1
  ja     L39
L32:
  comiss %xmm1,%xmm0
  jbe    L46
  lea    D35(%rip),%rdi
  call   <printf@plt>
L13:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L46:
  lea    D8(%rip),%rdi
  call   <printf@plt>
  jmp    L13
L39:
  lea    D19(%rip),%rdi
  movss  %xmm0,0xc(%rsp)
  call   <printf@plt>
  movss  0x1c(%rsp),%xmm1
  movss  0xc(%rsp),%xmm0
  jmp    L32
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D10
datatype: string
name: retrieve_data
arguments: label: D35
datatype: string
name: retrieve_data
arguments: label: D8
datatype: string
name: retrieve_data
arguments: label: D19
datatype: string

tool: D10: %f

tool: D35: Point is outside the Circle.

tool: D8: Point is on the Circle.

tool: D19: Point is inside the Circle.

assistant: ```c
int main() {
  float x, y, r, x1, y1, d;
  scanf("%f", &x);
  scanf("%f", &y);
  scanf("%f", &r);
  scanf("%f", &x1);
  scanf("%f", &y1);
  d = sqrt(pow(x - x1, 2) + pow(y - y1, 2));
  if (d < r) {
    printf("Point is inside the Circle.");
  }
  if (d > r) {
    printf("Point is outside the Circle.");
  } else {
    printf("Point is on the Circle.");
  }
  return 0;
}
```