ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D12(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D35(%rip),%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <log10@plt>
  cmp    $0x3,%eax
  jne    L7
  mov    0x4(%rsp),%r8d
  xor    %ecx,%ecx
  mov    %r8d,%edx
  test   %r8d,%r8d
  je     L16
  nopw   0x0(%rax,%rax,1)
L62:
  movslq %edx,%rax
  lea    (%rcx,%rcx,4),%esi
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rsi,2),%ecx
  mov    %eax,%edx
  test   %eax,%eax
  jne    L62
  movl   $0x0,0x4(%rsp)
  cmp    %ecx,%r8d
  je     L16
  mov    %r8d,%esi
  lea    D40(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L31:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L51
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L7:
  lea    D29(%rip),%rdi
  call   <puts@plt>
  jmp    L31
L16:
  mov    %r8d,%esi
  lea    D41(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L31
L51:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D12
data_type: string
name: parse_data
arguments: data_tag: D35
data_type: string
name: parse_data
arguments: data_tag: D40
data_type: string
name: parse_data
arguments: data_tag: D29
data_type: string
name: parse_data
arguments: data_tag: D41
data_type: string

tool: D12: Enter 4 digit number

tool: D35: %d

tool: D40: %d is not a palindrome 


tool: D29: Not 4 digit 

tool: D41: %d is a palindrome 


assistant: ```c
main() {
  int num, rev = 0, n;
  printf("Enter 4 digit number\n");
  scanf("%d", &num);
  int l = (int)log10(num) + 1;
  if (l == 4) {
    n = num;
    while (num != 0) {
      rev = rev * 10 + num % 10;
      num /= 10;
    }
    if (n == rev)
      printf("%d is a palindrome \n", n);
    else
      printf("%d is not a palindrome \n", n);
  } else
    printf("Not 4 digit \n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<mem_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L0
  mov    $0x0,%eax
  jmp    L1
L0:
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L22
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0x0,%eax
  call   <abort@plt>
L22:
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
L1:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D4
data_type: string

tool: D4: malloc failure

assistant: ```c
char *mem_str(const char *str, int len) {
  char *str_new;
  if (!str)
    return 0;
  {
    if (!((str_new) = (char *)calloc((len + 1), sizeof(char)))) {
      perror("malloc failure");
      abort();
    }
  };
  memcpy(str_new, str, len);
  *(str_new + len) = 0;
  return str_new;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<mem_str>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  test   %rdi,%rdi
  je     L28
  mov    %esi,%ebx
  lea    0x1(%rsi),%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L9
L5:
  mov    %ebx,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  movslq %ebx,%rbx
  movb   $0x0,0x0(%rbp,%rbx,1)
L27:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  lea    D2(%rip),%rdi
  call   <perror@plt>
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L5
L28:
  mov    %rdi,%rbp
  jmp    L27
```

tool_calls: name: parse_data
arguments: data_tag: D2
data_type: string

tool: D2: malloc failure

assistant: ```c
char *mem_str(const char *str, int len) {
  char *str_new;
  if (!str)
    return 0;
  {
    if (!((str_new) = (char *)calloc((len + 1), sizeof(char)))) {
      perror("malloc failure");
      abort();
    }
  };
  memcpy(str_new, str, len);
  *(str_new + len) = 0;
  return str_new;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<mem_str>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L3
  movslq %esi,%rbx
  mov    %rdi,%rbp
  mov    $0x1,%esi
  lea    0x1(%rbx),%edi
  call   <calloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L10
L24:
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L10:
  lea    D8(%rip),%rdi
  call   <perror@plt>
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L24
  nopl   (%rax)
L3:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: label: D8
data_type: string

tool: D8: malloc failure

assistant: ```c
char *mem_str(const char *str, int len) {
  char *str_new;
  if (!str)
    return 0;
  {
    if (!((str_new) = (char *)calloc((len + 1), sizeof(char)))) {
      perror("malloc failure");
      abort();
    }
  };
  memcpy(str_new, str, len);
  *(str_new + len) = 0;
  return str_new;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<mem_str>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L5
  movslq %esi,%rbx
  mov    %rdi,%rbp
  mov    $0x1,%esi
  lea    0x1(%rbx),%edi
  call   <calloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L14
L19:
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  lea    D4(%rip),%rdi
  call   <perror@plt>
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L19
  nopl   (%rax)
L5:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
type: string

tool: D4: malloc failure

assistant: ```c
char *mem_str(const char *str, int len) {
  char *str_new;
  if (!str)
    return 0;
  {
    if (!((str_new) = (char *)calloc((len + 1), sizeof(char)))) {
      perror("malloc failure");
      abort();
    }
  };
  memcpy(str_new, str, len);
  *(str_new + len) = 0;
  return str_new;
}
```
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
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x64,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
type: string

tool: D3: You have travelled %d kms

assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  printf("You have travelled %d kms", distance);
  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
  sub    $0x8,%rsp
  mov    $0x64,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D5
datatype: string

tool: D5: You have travelled %d kms

assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  printf("You have travelled %d kms", distance);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type_guessed: string

tool: D1: You have travelled %d kms

assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  printf("You have travelled %d kms", distance);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D5
type_infered: string

tool: D5: You have travelled %d kms

assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  printf("You have travelled %d kms", distance);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<MY_LOG10>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jg     L0
  mov    $0x0,%eax
  jmp    L13
L0:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <Flog10@plt>
  lea    0x4d104(%rax),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sar    $0x10,%eax
L13:
  leave
  ret
```

assistant: ```c
int MY_LOG10(int x) {
  if (x <= 0)
    return 0;
  else
    return (10 * (Flog10((int)x) + 0x4d104)) >> 16;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<MY_LOG10>:
  endbr64
  mov    $0x0,%eax
  test   %edi,%edi
  jg     L2
  ret
L2:
  sub    $0x8,%rsp
  call   <Flog10@plt>
  lea    0x181514(%rax,%rax,4),%eax
  add    %eax,%eax
  sar    $0x10,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int MY_LOG10(int x) {
  if (x <= 0)
    return 0;
  else
    return (10 * (Flog10((int)x) + 0x4d104)) >> 16;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<MY_LOG10>:
  endbr64
  test   %edi,%edi
  jg     L0
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  sub    $0x8,%rsp
  call   <Flog10@plt>
  add    $0x8,%rsp
  lea    0x181514(%rax,%rax,4),%eax
  add    %eax,%eax
  sar    $0x10,%eax
  ret
```

assistant: ```c
int MY_LOG10(int x) {
  if (x <= 0)
    return 0;
  else
    return (10 * (Flog10((int)x) + 0x4d104)) >> 16;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<MY_LOG10>:
  endbr64
  test   %edi,%edi
  jg     L3
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  sub    $0x8,%rsp
  call   <Flog10@plt>
  add    $0x8,%rsp
  lea    0x181514(%rax,%rax,4),%eax
  add    %eax,%eax
  sar    $0x10,%eax
  ret
```

assistant: ```c
int MY_LOG10(int x) {
  if (x <= 0)
    return 0;
  else
    return (10 * (Flog10((int)x) + 0x4d104)) >> 16;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<uint128_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%rax
  mov    %rsi,%rcx
  mov    %rcx,%rdx
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  or     -0x30(%rbp),%rax
  test   %rax,%rax
  jne    L5
  lea    D6(%rip),%rax
  jmp    L6
L5:
  mov    $0x28,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x27,%rax
  mov    %rax,-0x20(%rbp)
  jmp    L29
L11:
  mov    -0x30(%rbp),%rax
  mov    -0x28(%rbp),%rdx
  mov    %rax,%rcx
  mov    %rdx,%rbx
  mov    %rcx,%rsi
  mov    %rdx,%rax
  mov    $0xa,%edx
  mov    $0x0,%ecx
  mov    %rsi,%rdi
  mov    %rax,%rsi
  call   <__umodti3>
  mov    %rax,%rcx
  mov    %rdx,%rbx
  subq   $0x1,-0x20(%rbp)
  lea    D7(%rip),%rax
  add    %rcx,%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x30(%rbp),%rax
  mov    -0x28(%rbp),%rdx
  mov    %rax,%rcx
  mov    %rdx,%rbx
  mov    %rcx,%rsi
  mov    %rdx,%rax
  mov    $0xa,%edx
  mov    $0x0,%ecx
  mov    %rsi,%rdi
  mov    %rax,%rsi
  call   <__udivti3>
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
L29:
  mov    -0x28(%rbp),%rax
  or     -0x30(%rbp),%rax
  test   %rax,%rax
  jne    L11
  mov    -0x18(%rbp),%rax
L6:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
type: string
name: retrieve_data
arguments: data_tag: D7
type: string

tool: D6: 0

tool: D7: 0123456789

assistant: ```c
char *uint128_str(unsigned __int128 n) {
  if (n == 0)
    return "0";
  char *str = (char *)malloc(40 * sizeof(char));
  char *s = str + (40 * sizeof(char)) - 1;
  while (n != 0) {
    *--s = "0123456789"[n % 10];
    n /= 10;
  }
  return str;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<uint128_str>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  lea    D12(%rip),%r14
  mov    %rsi,%rax
  or     %rdi,%rax
  jne    L18
L8:
  mov    %r14,%rax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L18:
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    $0x28,%edi
  call   <malloc@plt>
  lea    0x27(%rax),%r8
  lea    D7(%rip),%rbp
  movabs $0xcccccccccccccccd,%rcx
  movabs $0xcccccccccccccccc,%r9
  mov    $0xa,%ebx
  mov    $0x9,%r11d
  mov    $0x0,%r10d
  mov    %rax,%r15
L4:
  sub    $0x1,%r8
  mov    %r12,%rsi
  add    %r13,%rsi
  adc    $0x0,%rsi
  mov    %rsi,%rax
  mul    %rcx
  mov    %rdx,%rax
  shr    $0x2,%rax
  and    $0xfffffffffffffffc,%rdx
  add    %rax,%rdx
  sub    %rdx,%rsi
  mov    %rsi,(%rsp)
  movq   $0x0,0x8(%rsp)
  mov    %r12,%rax
  mov    %r13,%rdx
  sub    (%rsp),%rax
  sbb    0x8(%rsp),%rdx
  mov    %rdx,%rsi
  imul   %rcx,%rsi
  mov    %rax,0x20(%rsp)
  mov    %rdx,0x28(%rsp)
  mov    %rax,%rdx
  imul   %r9,%rdx
  add    %rdx,%rsi
  mov    0x20(%rsp),%rax
  mul    %rcx
  add    %rsi,%rdx
  shrd   $0x1,%rdx,%rax
  shr    %rdx
  mul    %rbx
  mov    %r12,%rsi
  sub    %rax,%rsi
  movzbl 0x0(%rbp,%rsi,1),%eax
  mov    %al,(%r8)
  mov    %r12,%rdi
  mov    %r13,%rsi
  mov    %r12,%r14
  add    %r13,%r14
  adc    $0x0,%r14
  mov    %r14,%rax
  mul    %rcx
  mov    %rdx,%rax
  shr    $0x2,%rax
  and    $0xfffffffffffffffc,%rdx
  add    %rax,%rdx
  sub    %rdx,%r14
  mov    %r14,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  sub    0x10(%rsp),%r12
  sbb    0x18(%rsp),%r13
  mov    %r13,%r14
  imul   %rcx,%r14
  mov    %r12,%rax
  imul   %r9,%rax
  add    %rax,%r14
  mov    %r12,%rax
  mul    %rcx
  mov    %rax,%r12
  mov    %rdx,%r13
  add    %r14,%r13
  shrd   $0x1,%r13,%r12
  shr    %r13
  cmp    %rdi,%r11
  mov    %r10,%rax
  sbb    %rsi,%rax
  jb     L4
  mov    %r15,%r14
  jmp    L8
```

tool_calls: name: unpack_data
arguments: data_tag: D12
data_type: string
name: unpack_data
arguments: data_tag: D7
data_type: string

tool: D12: 0

tool: D7: 0123456789

assistant: ```c
char *uint128_str(unsigned __int128 n) {
  if (n == 0)
    return "0";
  char *str = (char *)malloc(40 * sizeof(char));
  char *s = str + (40 * sizeof(char)) - 1;
  while (n != 0) {
    *--s = "0123456789"[n % 10];
    n /= 10;
  }
  return str;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<uint128_str>:
  endbr64
  push   %r15
  lea    D1(%rip),%rax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rax,0x18(%rsp)
  mov    %rsi,%rax
  or     %rdi,%rax
  jne    L22
L26:
  mov    0x18(%rsp),%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L22:
  mov    %rdi,%r12
  mov    $0x28,%edi
  mov    %rsi,%r13
  mov    $0x9,%ebx
  call   <malloc@plt>
  lea    D5(%rip),%r11
  xor    %r9d,%r9d
  movabs $0xcccccccccccccccd,%rsi
  mov    %rax,0x18(%rsp)
  lea    0x27(%rax),%rdi
  mov    $0x5,%r8d
  movabs $0xcccccccccccccccc,%r10
  nopw   0x0(%rax,%rax,1)
L7:
  mov    %r12,%rcx
  mov    %r12,%rbp
  add    %r13,%rcx
  adc    $0x0,%rcx
  xor    %r15d,%r15d
  mov    %rcx,%rax
  mul    %rsi
  mov    %rdx,%rax
  and    $0xfffffffffffffffc,%rdx
  shr    $0x2,%rax
  add    %rax,%rdx
  mov    %r12,%rax
  sub    %rdx,%rcx
  mov    %r13,%rdx
  sub    %rcx,%rax
  mov    %rcx,%r14
  sbb    %r15,%rdx
  mov    %rax,(%rsp)
  sub    $0x1,%rdi
  mov    %rdx,%rcx
  mov    %rdx,0x8(%rsp)
  mov    %rax,%rdx
  mov    (%rsp),%rax
  imul   %r10,%rdx
  imul   %rsi,%rcx
  add    %rdx,%rcx
  mul    %rsi
  add    %rcx,%rdx
  mov    %rax,%rcx
  and    $0x1,%ecx
  shrd   $0x1,%rdx,%rax
  imul   %r8,%rcx
  shr    %rdx
  mov    %rax,%r12
  mov    %r9,%rax
  add    %r14,%rcx
  cmp    %rbp,%rbx
  movzbl (%r11,%rcx,1),%ecx
  mov    %cl,(%rdi)
  mov    %r13,%rcx
  mov    %rdx,%r13
  sbb    %rcx,%rax
  jb     L7
  jmp    L26
```

tool_calls: name: read_data
arguments: label: D1
data_type: string
name: read_data
arguments: label: D5
data_type: string

tool: D1: 0

tool: D5: 0123456789

assistant: ```c
char *uint128_str(unsigned __int128 n) {
  if (n == 0)
    return "0";
  char *str = (char *)malloc(40 * sizeof(char));
  char *s = str + (40 * sizeof(char)) - 1;
  while (n != 0) {
    *--s = "0123456789"[n % 10];
    n /= 10;
  }
  return str;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<uint128_str>:
  endbr64
  push   %r15
  lea    D16(%rip),%rax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rax,0x18(%rsp)
  mov    %rsi,%rax
  or     %rdi,%rax
  jne    L17
L23:
  mov    0x18(%rsp),%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  mov    %rdi,%r12
  mov    $0x28,%edi
  mov    %rsi,%r13
  mov    $0x9,%ebx
  call   <malloc@plt>
  lea    D0(%rip),%r11
  xor    %r9d,%r9d
  movabs $0xcccccccccccccccd,%rsi
  mov    %rax,0x18(%rsp)
  lea    0x27(%rax),%rdi
  mov    $0x5,%r8d
  movabs $0xcccccccccccccccc,%r10
  nopw   0x0(%rax,%rax,1)
L22:
  mov    %r12,%rcx
  mov    %r12,%rbp
  add    %r13,%rcx
  adc    $0x0,%rcx
  xor    %r15d,%r15d
  mov    %rcx,%rax
  mul    %rsi
  mov    %rdx,%rax
  and    $0xfffffffffffffffc,%rdx
  shr    $0x2,%rax
  add    %rax,%rdx
  mov    %r12,%rax
  sub    %rdx,%rcx
  mov    %r13,%rdx
  sub    %rcx,%rax
  mov    %rcx,%r14
  sbb    %r15,%rdx
  mov    %rax,(%rsp)
  sub    $0x1,%rdi
  mov    %rdx,%rcx
  mov    %rdx,0x8(%rsp)
  mov    %rax,%rdx
  mov    (%rsp),%rax
  imul   %r10,%rdx
  imul   %rsi,%rcx
  add    %rdx,%rcx
  mul    %rsi
  add    %rcx,%rdx
  mov    %rax,%rcx
  and    $0x1,%ecx
  shrd   $0x1,%rdx,%rax
  imul   %r8,%rcx
  shr    %rdx
  mov    %rax,%r12
  mov    %r9,%rax
  add    %r14,%rcx
  cmp    %rbp,%rbx
  movzbl (%r11,%rcx,1),%ecx
  mov    %cl,(%rdi)
  mov    %r13,%rcx
  mov    %rdx,%r13
  sbb    %rcx,%rax
  jb     L22
  jmp    L23
```

tool_calls: name: retrieve_data
arguments: tag: D16
datatype: string
name: retrieve_data
arguments: tag: D0
datatype: string

tool: D16: 0

tool: D0: 0123456789

assistant: ```c
char *uint128_str(unsigned __int128 n) {
  if (n == 0)
    return "0";
  char *str = (char *)malloc(40 * sizeof(char));
  char *s = str + (40 * sizeof(char)) - 1;
  while (n != 0) {
    *--s = "0123456789"[n % 10];
    n /= 10;
  }
  return str;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strchrs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%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)
  jmp    L25
L45:
  movq   $0x0,-0x10(%rbp)
  jmp    L18
L48:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L51
  mov    -0x18(%rbp),%rax
  jmp    L36
L51:
  addq   $0x1,-0x10(%rbp)
L18:
  mov    -0x10(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L48
  addq   $0x1,-0x18(%rbp)
L25:
  mov    -0x18(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L45
  mov    $0x0,%eax
L36:
  leave
  ret
```

assistant: ```c
const char *strchrs(const char *haystack, const char *chars, size_t nchars) {
  const char *end = haystack + strlen(haystack);
  while (haystack < end) {
    for (size_t i = 0; i < nchars; i++) {
      if (*haystack == chars[i]) {
        return haystack;
      }
    }
    haystack++;
  }
  return NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strchrs>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %rdx,%rbp
  call   <strlen@plt>
  movslq %eax,%rsi
  add    %rbx,%rsi
  lea    0x0(%rbp,%r12,1),%rcx
  cmp    %rsi,%rbx
  jae    L25
L16:
  test   %rbp,%rbp
  je     L2
  movzbl (%rbx),%edx
  mov    %r12,%rax
L5:
  cmp    (%rax),%dl
  je     L27
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L5
L2:
  add    $0x1,%rbx
  cmp    %rbx,%rsi
  jne    L16
  mov    $0x0,%ebx
  jmp    L27
L25:
  mov    $0x0,%ebx
L27:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
const char *strchrs(const char *haystack, const char *chars, size_t nchars) {
  const char *end = haystack + strlen(haystack);
  while (haystack < end) {
    for (size_t i = 0; i < nchars; i++) {
      if (*haystack == chars[i]) {
        return haystack;
      }
    }
    haystack++;
  }
  return NULL;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strchrs>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  call   <strlen@plt>
  lea    (%rbx,%rbp,1),%r8
  movslq %eax,%rdi
  add    %r12,%rdi
  cmp    %rdi,%r12
  jae    L4
L15:
  test   %rbx,%rbx
  je     L43
  movzbl (%r12),%ecx
  mov    %rbp,%rax
  jmp    L33
  nopl   0x0(%rax)
L39:
  add    $0x1,%rax
  cmp    %r8,%rax
  je     L43
L33:
  cmp    (%rax),%cl
  jne    L39
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L43:
  add    $0x1,%r12
  cmp    %r12,%rdi
  jne    L15
L4:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
const char *strchrs(const char *haystack, const char *chars, size_t nchars) {
  const char *end = haystack + strlen(haystack);
  while (haystack < end) {
    for (size_t i = 0; i < nchars; i++) {
      if (*haystack == chars[i]) {
        return haystack;
      }
    }
    haystack++;
  }
  return NULL;
}
```
user: What is the c source code of the assembly code below:

```asm
<strchrs>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  call   <strlen@plt>
  lea    (%rbx,%rbp,1),%r8
  movslq %eax,%rdi
  add    %r12,%rdi
  cmp    %rdi,%r12
  jae    L54
L45:
  test   %rbx,%rbx
  je     L36
  movzbl (%r12),%ecx
  mov    %rbp,%rax
  jmp    L56
  nopl   0x0(%rax)
L11:
  add    $0x1,%rax
  cmp    %r8,%rax
  je     L36
L56:
  cmp    (%rax),%cl
  jne    L11
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L36:
  add    $0x1,%r12
  cmp    %r12,%rdi
  jne    L45
L54:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
const char *strchrs(const char *haystack, const char *chars, size_t nchars) {
  const char *end = haystack + strlen(haystack);
  while (haystack < end) {
    for (size_t i = 0; i < nchars; i++) {
      if (*haystack == chars[i]) {
        return haystack;
      }
    }
    haystack++;
  }
  return NULL;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<fd_max>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  lea    -0xc(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <getrlimit@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
int fd_max(void) {
  struct rlimit rlim;
  getrlimit(RLIMIT_NOFILE, &rlim);
  return rlim.rlim_max;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fd_max>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  mov    D8(%rip),%rax
  mov    (%rax),%edi
  call   <getrlimit@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int fd_max(void) {
  struct rlimit rlim;
  getrlimit(RLIMIT_NOFILE, &rlim);
  return rlim.rlim_max;
}
```