ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<randUpperChar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movb   $0x30,-0x1(%rbp)
  jmp    L16
L3:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x4ec4ec4f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x3,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  add    $0x41,%eax
  mov    %al,-0x1(%rbp)
L16:
  movsbl -0x1(%rbp),%eax
  mov    %eax,%edi
  call   <isUpperAlpha@plt>
  test   %eax,%eax
  je     L3
  movzbl -0x1(%rbp),%eax
  leave
  ret
```

assistant: ```c
char randUpperChar() {
  char alpha = '0';
  while (!isUpperAlpha(alpha)) {
    alpha = (rand() % 26) + 65;
  }
  return alpha;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<randUpperChar>:
  endbr64
  push   %rbx
  mov    $0x30,%ebx
  jmp    L8
L1:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x23,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  lea    0x41(%rax),%ebx
L8:
  movsbl %bl,%edi
  call   <isUpperAlpha@plt>
  test   %eax,%eax
  je     L1
  mov    %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
char randUpperChar() {
  char alpha = '0';
  while (!isUpperAlpha(alpha)) {
    alpha = (rand() % 26) + 65;
  }
  return alpha;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<randUpperChar>:
  endbr64
  push   %r12
  mov    $0x30,%r12d
  jmp    L4
  xchg   %ax,%ax
L15:
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x23,%rdx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  lea    0x41(%rax),%r12d
L4:
  movsbl %r12b,%edi
  call   <isUpperAlpha@plt>
  test   %eax,%eax
  je     L15
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
char randUpperChar() {
  char alpha = '0';
  while (!isUpperAlpha(alpha)) {
    alpha = (rand() % 26) + 65;
  }
  return alpha;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<randUpperChar>:
  endbr64
  push   %r12
  mov    $0x30,%r12d
  jmp    L0
  xchg   %ax,%ax
L1:
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x23,%rdx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  lea    0x41(%rax),%r12d
L0:
  movsbl %r12b,%edi
  call   <isUpperAlpha@plt>
  test   %eax,%eax
  je     L1
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
char randUpperChar() {
  char alpha = '0';
  while (!isUpperAlpha(alpha)) {
    alpha = (rand() % 26) + 65;
  }
  return alpha;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<cdb_hashadd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,%eax
  mov    %al,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  shl    $0x5,%eax
  add    %eax,-0x4(%rbp)
  movzbl -0x8(%rbp),%eax
  xor    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int cdb_hashadd(unsigned int h, unsigned char c) {
  h += (h << 5);
  return h ^ c;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<cdb_hashadd>:
  endbr64
  mov    %edi,%eax
  shl    $0x5,%eax
  lea    (%rax,%rdi,1),%eax
  movzbl %sil,%esi
  xor    %esi,%eax
  ret
```

assistant: ```c
unsigned int cdb_hashadd(unsigned int h, unsigned char c) {
  h += (h << 5);
  return h ^ c;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cdb_hashadd>:
  endbr64
  mov    %edi,%eax
  movzbl %sil,%esi
  shl    $0x5,%eax
  add    %edi,%eax
  xor    %esi,%eax
  ret
```

assistant: ```c
unsigned int cdb_hashadd(unsigned int h, unsigned char c) {
  h += (h << 5);
  return h ^ c;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<cdb_hashadd>:
  endbr64
  mov    %edi,%eax
  movzbl %sil,%esi
  shl    $0x5,%eax
  add    %edi,%eax
  xor    %esi,%eax
  ret
```

assistant: ```c
unsigned int cdb_hashadd(unsigned int h, unsigned char c) {
  h += (h << 5);
  return h ^ c;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dump_pt_regs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %rdx,-0x30(%rbp)
  mov    %rcx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x18(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jns    L19
  mov    $0x0,%edi
  call   <malloc@plt>
  mov    -0x30(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x18(%rbp),%eax
  jmp    L12
L19:
  mov    $0x8,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  js     L23
  mov    -0x38(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <dump_long_unsigned_int_from_mem@plt>
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x18(%rbp)
L23:
  lea    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <copy_params@plt>
  mov    -0x30(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x10(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <free_dumped_params@plt>
  mov    -0x18(%rbp),%eax
L12:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

assistant: ```c
int dump_pt_regs(int depth, unsigned int addr, char **dumped_params, struct target *target) {
  char **dumped_type_params;
  unsigned int arm_tracing_uregs = addr;
  int len = 0;
  if (depth < 0) {
    *dumped_params = malloc(0);
    return len;
  }
  dumped_type_params = malloc(1 * (sizeof(char *)));
  if (depth >= 0) {
    len += dump_long_unsigned_int_from_mem(arm_tracing_uregs, &dumped_type_params[0], target);
  }
  *dumped_params = copy_params(dumped_type_params, 1, &len);
  free_dumped_params(dumped_type_params, 1);
  return len;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dump_pt_regs>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdx,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  test   %edi,%edi
  js     L8
  mov    %esi,%r12d
  mov    %rcx,%rbp
  mov    $0x8,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdx
  mov    %rax,%rsi
  mov    %r12d,%edi
  call   <dump_long_unsigned_int_from_mem@plt>
  add    %eax,0x4(%rsp)
  lea    0x4(%rsp),%rdx
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <copy_params@plt>
  mov    %rax,0x0(%r13)
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <free_dumped_params@plt>
  mov    0x4(%rsp),%eax
L16:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  mov    $0x0,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%r13)
  mov    0x4(%rsp),%eax
  jmp    L16
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dump_pt_regs(int depth, unsigned int addr, char **dumped_params, struct target *target) {
  char **dumped_type_params;
  unsigned int arm_tracing_uregs = addr;
  int len = 0;
  if (depth < 0) {
    *dumped_params = malloc(0);
    return len;
  }
  dumped_type_params = malloc(1 * (sizeof(char *)));
  if (depth >= 0) {
    len += dump_long_unsigned_int_from_mem(arm_tracing_uregs, &dumped_type_params[0], target);
  }
  *dumped_params = copy_params(dumped_type_params, 1, &len);
  free_dumped_params(dumped_type_params, 1);
  return len;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<dump_pt_regs>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  test   %edi,%edi
  js     L3
  mov    $0x8,%edi
  mov    %rcx,%r13
  mov    %esi,%r12d
  call   <malloc@plt>
  mov    %r13,%rdx
  mov    %r12d,%edi
  mov    %rax,%rbp
  mov    %rax,%rsi
  call   <dump_long_unsigned_int_from_mem@plt>
  lea    0x4(%rsp),%rdx
  mov    %rbp,%rdi
  add    %eax,0x4(%rsp)
  mov    $0x1,%esi
  call   <copy_params@plt>
  mov    $0x1,%esi
  mov    %rbp,%rdi
  mov    %rax,(%rbx)
  call   <free_dumped_params@plt>
  mov    0x4(%rsp),%eax
L9:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %edi,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    0x4(%rsp),%eax
  jmp    L9
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dump_pt_regs(int depth, unsigned int addr, char **dumped_params, struct target *target) {
  char **dumped_type_params;
  unsigned int arm_tracing_uregs = addr;
  int len = 0;
  if (depth < 0) {
    *dumped_params = malloc(0);
    return len;
  }
  dumped_type_params = malloc(1 * (sizeof(char *)));
  if (depth >= 0) {
    len += dump_long_unsigned_int_from_mem(arm_tracing_uregs, &dumped_type_params[0], target);
  }
  *dumped_params = copy_params(dumped_type_params, 1, &len);
  free_dumped_params(dumped_type_params, 1);
  return len;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<dump_pt_regs>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  test   %edi,%edi
  js     L11
  mov    $0x8,%edi
  mov    %rcx,%r13
  mov    %esi,%r12d
  call   <malloc@plt>
  mov    %r13,%rdx
  mov    %r12d,%edi
  mov    %rax,%rbp
  mov    %rax,%rsi
  call   <dump_long_unsigned_int_from_mem@plt>
  lea    0x4(%rsp),%rdx
  mov    %rbp,%rdi
  add    %eax,0x4(%rsp)
  mov    $0x1,%esi
  call   <copy_params@plt>
  mov    $0x1,%esi
  mov    %rbp,%rdi
  mov    %rax,(%rbx)
  call   <free_dumped_params@plt>
  mov    0x4(%rsp),%eax
L14:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L11:
  xor    %edi,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    0x4(%rsp),%eax
  jmp    L14
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dump_pt_regs(int depth, unsigned int addr, char **dumped_params, struct target *target) {
  char **dumped_type_params;
  unsigned int arm_tracing_uregs = addr;
  int len = 0;
  if (depth < 0) {
    *dumped_params = malloc(0);
    return len;
  }
  dumped_type_params = malloc(1 * (sizeof(char *)));
  if (depth >= 0) {
    len += dump_long_unsigned_int_from_mem(arm_tracing_uregs, &dumped_type_params[0], target);
  }
  *dumped_params = copy_params(dumped_type_params, 1, &len);
  free_dumped_params(dumped_type_params, 1);
  return len;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<calendar_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    %r8,-0x28(%rbp)
  mov    %r9,-0x30(%rbp)
  mov    -0x30(%rbp),%r8
  mov    -0x28(%rbp),%rdi
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  mov    %r8,%r9
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <calendar@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int calendar_(int *year, int *month, int *day, int *hour, int *minute, int *second) {
  calendar(year, month, day, hour, minute, second);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

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

assistant: ```c
int calendar_(int *year, int *month, int *day, int *hour, int *minute, int *second) {
  calendar(year, month, day, hour, minute, second);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<calendar_>:
  endbr64
  sub    $0x8,%rsp
  call   <calendar@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int calendar_(int *year, int *month, int *day, int *hour, int *minute, int *second) {
  calendar(year, month, day, hour, minute, second);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<calendar_>:
  endbr64
  sub    $0x8,%rsp
  call   <calendar@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int calendar_(int *year, int *month, int *day, int *hour, int *minute, int *second) {
  calendar(year, month, day, hour, minute, second);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rtrim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  nop
L9:
  subq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  je     L9
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *rtrim(char *string, char junk) {
  char *original = string + strlen(string);
  while (*--original == junk)
    ;
  *(original + 1) = '\0';
  return string;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<rtrim>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  call   <strlen@plt>
  cltq
  add    %rbp,%rax
L2:
  mov    %rax,%rdx
  sub    $0x1,%rax
  cmp    %bl,(%rax)
  je     L2
  movb   $0x0,(%rdx)
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *rtrim(char *string, char junk) {
  char *original = string + strlen(string);
  while (*--original == junk)
    ;
  *(original + 1) = '\0';
  return string;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<rtrim>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <strlen@plt>
  cltq
  add    %r12,%rax
  nopw   0x0(%rax,%rax,1)
L8:
  mov    %rax,%rdx
  sub    $0x1,%rax
  cmp    %bl,(%rax)
  je     L8
  movb   $0x0,(%rdx)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *rtrim(char *string, char junk) {
  char *original = string + strlen(string);
  while (*--original == junk)
    ;
  *(original + 1) = '\0';
  return string;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<rtrim>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <strlen@plt>
  cltq
  add    %r12,%rax
  nopw   0x0(%rax,%rax,1)
L4:
  mov    %rax,%rdx
  sub    $0x1,%rax
  cmp    %bl,(%rax)
  je     L4
  movb   $0x0,(%rdx)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *rtrim(char *string, char junk) {
  char *original = string + strlen(string);
  while (*--original == junk)
    ;
  *(original + 1) = '\0';
  return string;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<parse_ipv4>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%rbx
  mov    %rsi,0x8(%rsp)
  mov    %rdx,%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rdx)
  movl   $0x0,0x10(%rsp)
  lea    0x10(%rsp),%rdi
  mov    $0x10,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  lea    0x1(%rbx),%rbp
  movsbl (%rbx),%ebx
  test   %ebx,%ebx
  je     L106
  lea    0x10(%rsp),%r12
  mov    $0x0,%r13d
  lea    D9(%rip),%r14
  jmp    L152
L90:
  mov    %eax,(%r12)
  mov    $0x1,%r13d
L37:
  add    $0x1,%rbp
  movsbl -0x1(%rbp),%ebx
  test   %ebx,%ebx
  je     L106
L152:
  mov    %ebx,%esi
  mov    %r14,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L54
  mov    (%r12),%edx
  lea    (%rdx,%rdx,4),%edx
  sub    %r14,%rax
  lea    (%rax,%rdx,2),%eax
  cmp    $0xff,%eax
  ja     L85
  test   %r13d,%r13d
  jne    L90
  mov    (%r15),%ecx
  lea    0x1(%rcx),%edx
  mov    %edx,(%r15)
  cmp    $0x4,%edx
  jle    L90
  jmp    L13
L54:
  cmp    $0x2e,%ebx
  jne    L88
  test   %r13d,%r13d
  je     L88
  cmpl   $0x4,(%r15)
  je     L23
  lea    0x4(%r12),%rax
  movl   $0x0,0x4(%r12)
  cmpb   $0x0,0x0(%rbp)
  je     L46
  mov    %rax,%r12
  mov    $0x0,%r13d
  jmp    L37
L106:
  lea    0x10(%rsp),%rsi
  mov    $0x10,%edx
  mov    0x8(%rsp),%rdi
  call   <memcpy@plt>
  mov    $0x1,%r13d
L13:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L119
  mov    %r13d,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L85:
  mov    $0x0,%r13d
  jmp    L13
L88:
  mov    $0x0,%r13d
  jmp    L13
L23:
  mov    $0x0,%r13d
  jmp    L13
L46:
  mov    $0x0,%r13d
  jmp    L13
L119:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D9
type_infered: string

tool: D9: 0123456789

assistant: ```c
int parse_ipv4(const char *src, uint32_t *dst, int *bytes) {
  static const char digits[] = "0123456789";
  int saw_digit, ch;
  uint8_t tmp[4], *tp;
  saw_digit = 0;
  *bytes = 0;
  *(tp = tmp) = 0;
  memset(tmp, 0, sizeof(tmp));
  while ((ch = *src++) != '\0') {
    const char *pch;
    if ((pch = strchr(digits, ch)) != NULL) {
      unsigned int new = *tp * 10 + (pch - digits);
      if (new > 255)
        return (0);
      if (!saw_digit) {
        if (++(*bytes) > 4)
          return (0);
        saw_digit = 1;
      }
      *tp = new;
    } else if (ch == '.' && saw_digit) {
      if (*bytes == 4)
        return (0);
      *++tp = 0;
      saw_digit = 0;
      if (!(*src))
        return 0;
    } else
      return (0);
  }
  memcpy(dst, tmp, sizeof(tmp));
  return (1);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<parse_ipv4>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  lea    0x1(%r13),%r14
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x38,%rsp
  mov    %rsi,0x8(%rsp)
  lea    0x10(%rsp),%rdi
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rdx)
  mov    $0x10,%edx
  movl   $0x0,0x10(%rsp)
  mov    %rdi,(%rsp)
  call   <memset@plt>
  movsbl 0x0(%r13),%r13d
  test   %r13d,%r13d
  je     L0
  mov    (%rsp),%r15
  xor    %r12d,%r12d
  lea    D8(%rip),%rbp
  jmp    L83
  nopw   0x0(%rax,%rax,1)
L117:
  mov    (%r15),%edx
  sub    %rbp,%rax
  lea    (%rdx,%rdx,4),%edx
  lea    (%rax,%rdx,2),%eax
  cmp    $0xff,%eax
  ja     L3
  test   %r12d,%r12d
  jne    L99
  mov    (%rbx),%ecx
  lea    0x1(%rcx),%edx
  mov    %edx,(%rbx)
  cmp    $0x4,%edx
  jg     L3
L99:
  movsbl (%r14),%r13d
  mov    %eax,(%r15)
  mov    $0x1,%r12d
  add    $0x1,%r14
  test   %r13d,%r13d
  je     L0
L83:
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L117
  cmp    $0x2e,%r13d
  jne    L3
  test   %r12d,%r12d
  je     L3
  cmpl   $0x4,(%rbx)
  je     L3
  movsbl (%r14),%r13d
  movl   $0x0,0x4(%r15)
  lea    0x4(%r15),%rax
  test   %r13b,%r13b
  je     L3
  mov    %rax,%r15
  xor    %r12d,%r12d
  add    $0x1,%r14
  test   %r13d,%r13d
  jne    L83
L0:
  mov    (%rsp),%rsi
  mov    0x8(%rsp),%rdi
  mov    $0x10,%edx
  call   <memcpy@plt>
  mov    $0x1,%eax
  jmp    L18
  nopl   0x0(%rax)
L3:
  xor    %eax,%eax
L18:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D8
datatype: string

tool: D8: 0123456789

assistant: ```c
int parse_ipv4(const char *src, uint32_t *dst, int *bytes) {
  static const char digits[] = "0123456789";
  int saw_digit, ch;
  uint8_t tmp[4], *tp;
  saw_digit = 0;
  *bytes = 0;
  *(tp = tmp) = 0;
  memset(tmp, 0, sizeof(tmp));
  while ((ch = *src++) != '\0') {
    const char *pch;
    if ((pch = strchr(digits, ch)) != NULL) {
      unsigned int new = *tp * 10 + (pch - digits);
      if (new > 255)
        return (0);
      if (!saw_digit) {
        if (++(*bytes) > 4)
          return (0);
        saw_digit = 1;
      }
      *tp = new;
    } else if (ch == '.' && saw_digit) {
      if (*bytes == 4)
        return (0);
      *++tp = 0;
      saw_digit = 0;
      if (!(*src))
        return 0;
    } else
      return (0);
  }
  memcpy(dst, tmp, sizeof(tmp));
  return (1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<parse_ipv4>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  lea    0x1(%r13),%r14
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x38,%rsp
  mov    %rsi,0x8(%rsp)
  lea    0x10(%rsp),%rdi
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rdx)
  mov    $0x10,%edx
  movl   $0x0,0x10(%rsp)
  mov    %rdi,(%rsp)
  call   <memset@plt>
  movsbl 0x0(%r13),%r13d
  test   %r13d,%r13d
  je     L106
  mov    (%rsp),%r15
  xor    %r12d,%r12d
  lea    D1(%rip),%rbp
  jmp    L61
  nopw   0x0(%rax,%rax,1)
L62:
  mov    (%r15),%edx
  sub    %rbp,%rax
  lea    (%rdx,%rdx,4),%edx
  lea    (%rax,%rdx,2),%eax
  cmp    $0xff,%eax
  ja     L121
  test   %r12d,%r12d
  jne    L113
  mov    (%rbx),%ecx
  lea    0x1(%rcx),%edx
  mov    %edx,(%rbx)
  cmp    $0x4,%edx
  jg     L121
L113:
  movsbl (%r14),%r13d
  mov    %eax,(%r15)
  mov    $0x1,%r12d
  add    $0x1,%r14
  test   %r13d,%r13d
  je     L106
L61:
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L62
  cmp    $0x2e,%r13d
  jne    L121
  test   %r12d,%r12d
  je     L121
  cmpl   $0x4,(%rbx)
  je     L121
  movsbl (%r14),%r13d
  movl   $0x0,0x4(%r15)
  lea    0x4(%r15),%rax
  test   %r13b,%r13b
  je     L121
  mov    %rax,%r15
  xor    %r12d,%r12d
  add    $0x1,%r14
  test   %r13d,%r13d
  jne    L61
L106:
  mov    (%rsp),%rsi
  mov    0x8(%rsp),%rdi
  mov    $0x10,%edx
  call   <memcpy@plt>
  mov    $0x1,%eax
  jmp    L71
  nopl   0x0(%rax)
L121:
  xor    %eax,%eax
L71:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L78
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L78:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D1
type_infered: string

tool: D1: 0123456789

assistant: ```c
int parse_ipv4(const char *src, uint32_t *dst, int *bytes) {
  static const char digits[] = "0123456789";
  int saw_digit, ch;
  uint8_t tmp[4], *tp;
  saw_digit = 0;
  *bytes = 0;
  *(tp = tmp) = 0;
  memset(tmp, 0, sizeof(tmp));
  while ((ch = *src++) != '\0') {
    const char *pch;
    if ((pch = strchr(digits, ch)) != NULL) {
      unsigned int new = *tp * 10 + (pch - digits);
      if (new > 255)
        return (0);
      if (!saw_digit) {
        if (++(*bytes) > 4)
          return (0);
        saw_digit = 1;
      }
      *tp = new;
    } else if (ch == '.' && saw_digit) {
      if (*bytes == 4)
        return (0);
      *++tp = 0;
      saw_digit = 0;
      if (!(*src))
        return 0;
    } else
      return (0);
  }
  memcpy(dst, tmp, sizeof(tmp));
  return (1);
}
```