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
<append_sync_flush>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  and    $0x7,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jle    L16
  subq   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x1c(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  sub    $0x1,%eax
  and    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  jmp    L7
L16:
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
L7:
  mov    -0x20(%rbp),%eax
  cltq
  and    $0x1,%eax
  mov    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  shl    %cl,%rdx
  mov    %rdx,%rax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  or     %edx,%eax
  mov    %eax,-0x8(%rbp)
  addl   $0x3,-0x4(%rbp)
  cmpl   $0x8,-0x4(%rbp)
  jg     L51
  mov    $0x8,%eax
  jmp    L31
L51:
  mov    $0x10,%eax
L31:
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    $0xffff0000,%edx
  mov    %eax,%ecx
  shl    %cl,%rdx
  mov    %rdx,%rax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x8(%rbp)
  addl   $0x20,-0x4(%rbp)
  jmp    L84
L45:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
  sarl   $0x8,-0x8(%rbp)
  subl   $0x8,-0x4(%rbp)
L84:
  cmpl   $0x0,-0x4(%rbp)
  jg     L45
  cmpl   $0x5,-0x1c(%rbp)
  jg     L73
  cmpl   $0x0,-0x1c(%rbp)
  jne    L58
L73:
  mov    $0x5,%eax
  jmp    L66
L58:
  mov    $0x4,%eax
L66:
  pop    %rbp
  ret
```

assistant: ```c
int append_sync_flush(char *buf, int tebc, int final) {
  uint64_t flush;
  int shift = (tebc & 0x7);
  if (tebc > 0) {
    buf = buf - 1;
    *buf = *buf & (unsigned char)((1 << tebc) - 1);
  } else
    *buf = 0;
  flush = ((0x1ULL & final) << shift) | *buf;
  shift = shift + 3;
  shift = (shift <= 8) ? 8 : 16;
  flush |= (0xFFFF0000ULL) << shift;
  shift = shift + 32;
  while (shift > 0) {
    *buf++ = (unsigned char)(flush & 0xffULL);
    flush = flush >> 8;
    shift = shift - 8;
  }
  return (((tebc > 5) || (tebc == 0)) ? 5 : 4);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<append_sync_flush>:
  endbr64
  mov    %esi,%r8d
  mov    %esi,%r9d
  and    $0x7,%r9d
  test   %esi,%esi
  jle    L1
  lea    -0x1(%rdi),%rax
  mov    $0x1,%esi
  mov    %r8d,%ecx
  shl    %cl,%esi
  sub    $0x1,%esi
  and    %sil,-0x1(%rdi)
L12:
  and    $0x1,%edx
  mov    %r9d,%ecx
  shl    %cl,%rdx
  mov    %rdx,%rcx
  movsbl (%rax),%edx
  or     %ecx,%edx
  cmp    $0x6,%r9d
  setge  %cl
  movzbl %cl,%ecx
  lea    0x8(,%rcx,8),%ecx
  mov    $0xffff0000,%esi
  shl    %cl,%rsi
  or     %esi,%edx
  lea    0x1f(%rcx),%esi
  shr    $0x3,%esi
  add    $0x1,%esi
  mov    %esi,%esi
  add    %rax,%rsi
L21:
  add    $0x1,%rax
  mov    %dl,-0x1(%rax)
  sar    $0x8,%edx
  cmp    %rsi,%rax
  jne    L21
  cmp    $0x5,%r8d
  setg   %al
  test   %r8d,%r8d
  sete   %dl
  or     %edx,%eax
  movzbl %al,%eax
  add    $0x4,%eax
  ret
L1:
  movb   $0x0,(%rdi)
  mov    %rdi,%rax
  jmp    L12
```

assistant: ```c
int append_sync_flush(char *buf, int tebc, int final) {
  uint64_t flush;
  int shift = (tebc & 0x7);
  if (tebc > 0) {
    buf = buf - 1;
    *buf = *buf & (unsigned char)((1 << tebc) - 1);
  } else
    *buf = 0;
  flush = ((0x1ULL & final) << shift) | *buf;
  shift = shift + 3;
  shift = (shift <= 8) ? 8 : 16;
  flush |= (0xFFFF0000ULL) << shift;
  shift = shift + 32;
  while (shift > 0) {
    *buf++ = (unsigned char)(flush & 0xffULL);
    flush = flush >> 8;
    shift = shift - 8;
  }
  return (((tebc > 5) || (tebc == 0)) ? 5 : 4);
}
```
user: What is the c source code of the assembly code below:

```asm
<append_sync_flush>:
  endbr64
  mov    %esi,%r9d
  mov    %esi,%r8d
  and    $0x7,%r9d
  test   %esi,%esi
  jle    L38
  mov    $0x1,%eax
  mov    %r8d,%ecx
  lea    -0x1(%rdi),%rsi
  shl    %cl,%eax
  sub    $0x1,%eax
  and    -0x1(%rdi),%al
  mov    %al,-0x1(%rdi)
  movsbl %al,%eax
L22:
  and    $0x1,%edx
  mov    %r9d,%ecx
  shl    %cl,%rdx
  or     %edx,%eax
  mov    $0x30,%edx
  cmp    $0x5,%r9d
  jg     L19
  or     $0xff000000,%eax
  mov    $0x28,%edx
L19:
  sub    $0x1,%edx
  shr    $0x3,%edx
  add    $0x1,%edx
  add    %rsi,%rdx
  nopl   0x0(%rax,%rax,1)
L9:
  add    $0x1,%rsi
  mov    %al,-0x1(%rsi)
  sar    $0x8,%eax
  cmp    %rdx,%rsi
  jne    L9
  cmp    $0x5,%r8d
  setg   %al
  test   %r8d,%r8d
  sete   %dl
  or     %edx,%eax
  movzbl %al,%eax
  add    $0x4,%eax
  ret
  nopl   (%rax)
L38:
  movb   $0x0,(%rdi)
  mov    %rdi,%rsi
  xor    %eax,%eax
  jmp    L22
```

assistant: ```c
int append_sync_flush(char *buf, int tebc, int final) {
  uint64_t flush;
  int shift = (tebc & 0x7);
  if (tebc > 0) {
    buf = buf - 1;
    *buf = *buf & (unsigned char)((1 << tebc) - 1);
  } else
    *buf = 0;
  flush = ((0x1ULL & final) << shift) | *buf;
  shift = shift + 3;
  shift = (shift <= 8) ? 8 : 16;
  flush |= (0xFFFF0000ULL) << shift;
  shift = shift + 32;
  while (shift > 0) {
    *buf++ = (unsigned char)(flush & 0xffULL);
    flush = flush >> 8;
    shift = shift - 8;
  }
  return (((tebc > 5) || (tebc == 0)) ? 5 : 4);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<append_sync_flush>:
  endbr64
  mov    %esi,%r9d
  and    $0x7,%r9d
  test   %esi,%esi
  jle    L11
  mov    $0x1,%eax
  mov    %esi,%ecx
  lea    -0x1(%rdi),%r8
  shl    %cl,%eax
  sub    $0x1,%eax
  and    -0x1(%rdi),%al
  mov    %al,-0x1(%rdi)
  movsbl %al,%eax
L2:
  and    $0x1,%edx
  mov    %r9d,%ecx
  shl    %cl,%rdx
  or     %edx,%eax
  mov    $0x30,%edx
  cmp    $0x5,%r9d
  jg     L34
  or     $0xff000000,%eax
  mov    $0x28,%edx
L34:
  mov    %eax,(%r8)
  sar    $0x1f,%eax
  mov    %al,0x4(%r8)
  cmp    $0x28,%edx
  je     L3
  mov    %al,0x5(%r8)
L3:
  cmp    $0x5,%esi
  setg   %al
  test   %esi,%esi
  sete   %dl
  or     %edx,%eax
  movzbl %al,%eax
  add    $0x4,%eax
  ret
  nop
L11:
  movb   $0x0,(%rdi)
  mov    %rdi,%r8
  xor    %eax,%eax
  jmp    L2
```

assistant: ```c
int append_sync_flush(char *buf, int tebc, int final) {
  uint64_t flush;
  int shift = (tebc & 0x7);
  if (tebc > 0) {
    buf = buf - 1;
    *buf = *buf & (unsigned char)((1 << tebc) - 1);
  } else
    *buf = 0;
  flush = ((0x1ULL & final) << shift) | *buf;
  shift = shift + 3;
  shift = (shift <= 8) ? 8 : 16;
  flush |= (0xFFFF0000ULL) << shift;
  shift = shift + 32;
  while (shift > 0) {
    *buf++ = (unsigned char)(flush & 0xffULL);
    flush = flush >> 8;
    shift = shift - 8;
  }
  return (((tebc > 5) || (tebc == 0)) ? 5 : 4);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<aeron_number_of_leading_zeroes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L16
  mov    $0x20,%eax
  jmp    L1
L16:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <__builtin_clz@plt>
L1:
  leave
  ret
```

assistant: ```c
int aeron_number_of_leading_zeroes(int32_t value) {
  if (0 == value) {
    return 32;
  }
  return __builtin_clz(value);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<aeron_number_of_leading_zeroes>:
  endbr64
  mov    $0x20,%eax
  test   %rdi,%rdi
  je     L5
  sub    $0x8,%rsp
  call   <__builtin_clz@plt>
  add    $0x8,%rsp
  ret
L5:
  ret
```

assistant: ```c
int aeron_number_of_leading_zeroes(int32_t value) {
  if (0 == value) {
    return 32;
  }
  return __builtin_clz(value);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<aeron_number_of_leading_zeroes>:
  endbr64
  test   %rdi,%rdi
  je     L3
  jmp    1050 <__builtin_clz@plt>
  xchg   %ax,%ax
L3:
  mov    $0x20,%eax
  ret
```

assistant: ```c
int aeron_number_of_leading_zeroes(int32_t value) {
  if (0 == value) {
    return 32;
  }
  return __builtin_clz(value);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<aeron_number_of_leading_zeroes>:
  endbr64
  test   %rdi,%rdi
  je     L1
  jmp    1050 <__builtin_clz@plt>
  xchg   %ax,%ax
L1:
  mov    $0x20,%eax
  ret
```

assistant: ```c
int aeron_number_of_leading_zeroes(int32_t value) {
  if (0 == value) {
    return 32;
  }
  return __builtin_clz(value);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<sbpGethostname32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  cltq
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <gethostname@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <reduceLen32@plt>
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
int sbpGethostname32(char *name, uint32_t *namelen) {
  size_t i, len = (size_t)(*namelen);
  int res = gethostname(name, len);
  reduceLen32(name, namelen);
  return res;
}
```
user: What is the c source code of the assembly code below:

```asm
<sbpGethostname32>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  movslq (%rsi),%rsi
  call   <gethostname@plt>
  mov    %eax,%r12d
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <reduceLen32@plt>
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int sbpGethostname32(char *name, uint32_t *namelen) {
  size_t i, len = (size_t)(*namelen);
  int res = gethostname(name, len);
  reduceLen32(name, namelen);
  return res;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<sbpGethostname32>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq (%rsi),%rsi
  mov    %rdi,%rbp
  call   <gethostname@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    %eax,%r13d
  call   <reduceLen32@plt>
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int sbpGethostname32(char *name, uint32_t *namelen) {
  size_t i, len = (size_t)(*namelen);
  int res = gethostname(name, len);
  reduceLen32(name, namelen);
  return res;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<sbpGethostname32>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq (%rsi),%rsi
  mov    %rdi,%rbp
  call   <gethostname@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    %eax,%r13d
  call   <reduceLen32@plt>
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int sbpGethostname32(char *name, uint32_t *namelen) {
  size_t i, len = (size_t)(*namelen);
  int res = gethostname(name, len);
  reduceLen32(name, namelen);
  return res;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<sys$device_scan>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%r8
  mov    -0x30(%rbp),%rax
  mov    %rax,%r10
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x4d,%rax
  int    $0x81
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
int sys$device_scan(void *return_devnam, unsigned short int *retlen, void *search_devnam, void *itmlst, unsigned long long *contxt) {
  return ({
    unsigned long resultvar;
    register long int _a5 asm("r8") = (long)(contxt);
    register long int _a4 asm("r10") = (long)(itmlst);
    register long int _a3 asm("rdx") = (long)(search_devnam);
    register long int _a2 asm("rsi") = (long)(retlen);
    register long int _a1 asm("rdi") = (long)(return_devnam);
    asm volatile("movq %1, %%rax\n\t"
                 "int $0x81\n\t"
                 : "=a"(resultvar)
                 : "i"(77), "r"(_a1), "r"(_a2), "r"(_a3), "r"(_a4), "r"(_a5)
                 : "memory", "cc", "r11", "cx");
    (long)resultvar;
  });
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sys$device_scan>:
  endbr64
  mov    %rcx,%r10
  mov    $0x4d,%rax
  int    $0x81
  ret
```

assistant: ```c
int sys$device_scan(void *return_devnam, unsigned short int *retlen, void *search_devnam, void *itmlst, unsigned long long *contxt) {
  return ({
    unsigned long resultvar;
    register long int _a5 asm("r8") = (long)(contxt);
    register long int _a4 asm("r10") = (long)(itmlst);
    register long int _a3 asm("rdx") = (long)(search_devnam);
    register long int _a2 asm("rsi") = (long)(retlen);
    register long int _a1 asm("rdi") = (long)(return_devnam);
    asm volatile("movq %1, %%rax\n\t"
                 "int $0x81\n\t"
                 : "=a"(resultvar)
                 : "i"(77), "r"(_a1), "r"(_a2), "r"(_a3), "r"(_a4), "r"(_a5)
                 : "memory", "cc", "r11", "cx");
    (long)resultvar;
  });
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sys$device_scan>:
  endbr64
  mov    %rcx,%r10
  mov    $0x4d,%rax
  int    $0x81
  ret
```

assistant: ```c
int sys$device_scan(void *return_devnam, unsigned short int *retlen, void *search_devnam, void *itmlst, unsigned long long *contxt) {
  return ({
    unsigned long resultvar;
    register long int _a5 asm("r8") = (long)(contxt);
    register long int _a4 asm("r10") = (long)(itmlst);
    register long int _a3 asm("rdx") = (long)(search_devnam);
    register long int _a2 asm("rsi") = (long)(retlen);
    register long int _a1 asm("rdi") = (long)(return_devnam);
    asm volatile("movq %1, %%rax\n\t"
                 "int $0x81\n\t"
                 : "=a"(resultvar)
                 : "i"(77), "r"(_a1), "r"(_a2), "r"(_a3), "r"(_a4), "r"(_a5)
                 : "memory", "cc", "r11", "cx");
    (long)resultvar;
  });
}
```
user: Translate the below assembly code into C programming language:

```asm
<sys$device_scan>:
  endbr64
  mov    %rcx,%r10
  mov    $0x4d,%rax
  int    $0x81
  ret
```

assistant: ```c
int sys$device_scan(void *return_devnam, unsigned short int *retlen, void *search_devnam, void *itmlst, unsigned long long *contxt) {
  return ({
    unsigned long resultvar;
    register long int _a5 asm("r8") = (long)(contxt);
    register long int _a4 asm("r10") = (long)(itmlst);
    register long int _a3 asm("rdx") = (long)(search_devnam);
    register long int _a2 asm("rsi") = (long)(retlen);
    register long int _a1 asm("rdi") = (long)(return_devnam);
    asm volatile("movq %1, %%rax\n\t"
                 "int $0x81\n\t"
                 : "=a"(resultvar)
                 : "i"(77), "r"(_a1), "r"(_a2), "r"(_a3), "r"(_a4), "r"(_a5)
                 : "memory", "cc", "r11", "cx");
    (long)resultvar;
  });
}
```
user: Translate the below assembly code into C programming language:

```asm
<strndup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strnlen@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)
  je     L7
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L7:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *strndup(const char *str, size_t max) {
  size_t len = strnlen(str, max);
  char *res = malloc(len + 1);
  if (res) {
    memcpy(res, str, len);
    res[len] = '\0';
  }
  return res;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<strndup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  call   <strnlen@plt>
  mov    %rax,%rbp
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L0
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%rbx,%rbp,1)
L0:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strndup(const char *str, size_t max) {
  size_t len = strnlen(str, max);
  char *res = malloc(len + 1);
  if (res) {
    memcpy(res, str, len);
    res[len] = '\0';
  }
  return res;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strndup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strnlen@plt>
  lea    0x1(%rax),%rdi
  mov    %rax,%rbx
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L1
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
L1:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strndup(const char *str, size_t max) {
  size_t len = strnlen(str, max);
  char *res = malloc(len + 1);
  if (res) {
    memcpy(res, str, len);
    res[len] = '\0';
  }
  return res;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strndup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strnlen@plt>
  lea    0x1(%rax),%rdi
  mov    %rax,%rbx
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L6
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
L6:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strndup(const char *str, size_t max) {
  size_t len = strnlen(str, max);
  char *res = malloc(len + 1);
  if (res) {
    memcpy(res, str, len);
    res[len] = '\0';
  }
  return res;
}
```
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    $0x10,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L3
L10:
  lea    D12(%rip),%rax
  mov    %rax,%rsi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <benchmark@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <print_results@plt>
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L3:
  cmpl   $0x2,-0xc(%rbp)
  jle    L10
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D12
type_infered: string
name: read_data
arguments: label: D22
type_infered: string
name: read_data
arguments: label: D1
type_infered: string

tool: D12: is-positive-zerof

tool: D22: # c::%s


tool: D1: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::%s\n", "is-positive-zerof");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  mov    $0x0,%ebx
  lea    D21(%rip),%r13
  lea    D15(%rip),%r12
  lea    D29(%rip),%rbp
L4:
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <benchmark@plt>
  call   <print_results@plt>
  add    $0x1,%ebx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L4
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D21
data_type: string
name: retrieve_data
arguments: tag: D15
data_type: string
name: retrieve_data
arguments: tag: D29
data_type: string

tool: D21: is-positive-zerof

tool: D15: # c::%s


tool: D29: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::%s\n", "is-positive-zerof");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  xor    %edi,%edi
  lea    D11(%rip),%r13
  push   %r12
  lea    D28(%rip),%r12
  push   %rbp
  lea    D22(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
L6:
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  xor    %eax,%eax
  call   <benchmark@plt>
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L6
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D11
data_type: string
name: extract_symbolic_data
arguments: label: D28
data_type: string
name: extract_symbolic_data
arguments: label: D22
data_type: string

tool: D11: is-positive-zerof

tool: D28: # c::%s


tool: D22: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::%s\n", "is-positive-zerof");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  xor    %edi,%edi
  lea    D23(%rip),%r13
  push   %r12
  lea    D19(%rip),%r12
  push   %rbp
  lea    D9(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
L3:
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  xor    %eax,%eax
  call   <benchmark@plt>
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L3
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: read_data
arguments: label: D23
data_type: string
name: read_data
arguments: label: D19
data_type: string
name: read_data
arguments: label: D9
data_type: string

tool: D23: is-positive-zerof

tool: D19: # c::%s


tool: D9: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::%s\n", "is-positive-zerof");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```