ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<expand_zeros>:
  endbr64
  mov    %edx,%r9d
  mov    %rdi,%r8
  mov    %rsi,%r10
  mov    %rsi,%rdx
  xor    %eax,%eax
L50:
  lea    (%r9,%rdx,1),%edi
  lea    -0x1(%rdi),%r11d
  jmp    L16
  nopl   0x0(%rax)
L70:
  movzbl (%rdx),%esi
  lea    0x1(%rdx),%rcx
  add    $0x1,%r8
  mov    %sil,-0x1(%r8)
  test   %sil,%sil
  je     L47
  mov    %rcx,%rdx
L16:
  mov    %edi,%ecx
  mov    %r11d,%r9d
  sub    %edx,%ecx
  sub    %edx,%r9d
  test   %ecx,%ecx
  jg     L70
  mov    %edx,%eax
  sub    %r10d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L47:
  movzbl (%rcx),%esi
  add    $0x2,%rdx
  cmp    %r9d,%esi
  cmovg  %r9d,%esi
  movzbl %sil,%ecx
  sub    %ecx,%r9d
  test   %sil,%sil
  je     L50
  movzbl %sil,%esi
  cmp    $0x8,%esi
  jb     L40
  lea    0x8(%r8),%rdi
  mov    %r8,%rcx
  movq   $0x0,(%r8)
  movq   $0x0,-0x8(%r8,%rsi,1)
  and    $0xfffffffffffffff8,%rdi
  sub    %rdi,%rcx
  add    %esi,%ecx
  shr    $0x3,%ecx
  rep stos %rax,%es:(%rdi)
L33:
  add    %rsi,%r8
  jmp    L50
  nopl   0x0(%rax)
L40:
  test   $0x4,%sil
  jne    L38
  test   %esi,%esi
  je     L33
  movb   $0x0,(%r8)
  test   $0x2,%sil
  je     L33
  xor    %ecx,%ecx
  mov    %cx,-0x2(%r8,%rsi,1)
  jmp    L33
L38:
  movl   $0x0,(%r8)
  movl   $0x0,-0x4(%r8,%rsi,1)
  jmp    L33
```

assistant: ```c
int expand_zeros(char *dest, const char *src, int length) {
  const char *src_ptr = src;
  while (length-- > 0) {
    *dest++ = *src_ptr++;
    if (dest[-1] == 0) {
      unsigned char chr = *src_ptr++;
      chr = (chr > length ? length : chr);
      length -= chr;
      while (chr-- > 0)
        *dest++ = 0;
    }
  }
  return src_ptr - src;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<expand_zeros>:
  endbr64
  mov    %edx,%r9d
  mov    %rdi,%r8
  mov    %rsi,%r10
  mov    %rsi,%rdx
  xor    %eax,%eax
L69:
  lea    (%r9,%rdx,1),%edi
  lea    -0x1(%rdi),%r11d
  jmp    L48
  nopl   0x0(%rax)
L90:
  movzbl (%rdx),%esi
  lea    0x1(%rdx),%rcx
  add    $0x1,%r8
  mov    %sil,-0x1(%r8)
  test   %sil,%sil
  je     L83
  mov    %rcx,%rdx
L48:
  mov    %edi,%ecx
  mov    %r11d,%r9d
  sub    %edx,%ecx
  sub    %edx,%r9d
  test   %ecx,%ecx
  jg     L90
  mov    %edx,%eax
  sub    %r10d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L83:
  movzbl (%rcx),%esi
  add    $0x2,%rdx
  cmp    %r9d,%esi
  cmovg  %r9d,%esi
  movzbl %sil,%ecx
  sub    %ecx,%r9d
  test   %sil,%sil
  je     L69
  movzbl %sil,%esi
  cmp    $0x8,%esi
  jb     L88
  lea    0x8(%r8),%rdi
  mov    %r8,%rcx
  movq   $0x0,(%r8)
  movq   $0x0,-0x8(%r8,%rsi,1)
  and    $0xfffffffffffffff8,%rdi
  sub    %rdi,%rcx
  add    %esi,%ecx
  shr    $0x3,%ecx
  rep stos %rax,%es:(%rdi)
L94:
  add    %rsi,%r8
  jmp    L69
  nopl   0x0(%rax)
L88:
  test   $0x4,%sil
  jne    L6
  test   %esi,%esi
  je     L94
  movb   $0x0,(%r8)
  test   $0x2,%sil
  je     L94
  xor    %ecx,%ecx
  mov    %cx,-0x2(%r8,%rsi,1)
  jmp    L94
L6:
  movl   $0x0,(%r8)
  movl   $0x0,-0x4(%r8,%rsi,1)
  jmp    L94
```

assistant: ```c
int expand_zeros(char *dest, const char *src, int length) {
  const char *src_ptr = src;
  while (length-- > 0) {
    *dest++ = *src_ptr++;
    if (dest[-1] == 0) {
      unsigned char chr = *src_ptr++;
      chr = (chr > length ? length : chr);
      length -= chr;
      while (chr-- > 0)
        *dest++ = 0;
    }
  }
  return src_ptr - src;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mp_encode_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  cmpl   $0xf,-0xc(%rbp)
  jg     L32
  mov    -0xc(%rbp),%eax
  or     $0x90,%al
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <mp_store_u8@plt>
  jmp    L3
L32:
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0xc(%rbp)
  jg     L26
  mov    -0x8(%rbp),%rax
  mov    $0xdc,%esi
  mov    %rax,%rdi
  call   <mp_store_u8@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <mp_store_u16@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  jmp    L3
L26:
  mov    -0x8(%rbp),%rax
  mov    $0xdd,%esi
  mov    %rax,%rdi
  call   <mp_store_u8@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <mp_store_u32@plt>
L3:
  leave
  ret
```

assistant: ```c
char *mp_encode_array(char *data, uint32_t size) {
  if (size <= 15) {
    return mp_store_u8(data, 0x90 | size);
  } else if (size <= UINT16_MAX) {
    data = mp_store_u8(data, 0xdc);
    data = mp_store_u16(data, size);
    return data;
  } else {
    data = mp_store_u8(data, 0xdd);
    return mp_store_u32(data, size);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mp_encode_array>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  cmp    $0xf,%esi
  jle    L9
  mov    D9(%rip),%rax
  cmp    %esi,(%rax)
  jl     L34
  mov    $0xdc,%esi
  call   <mp_store_u8@plt>
  mov    %rax,%rdi
  mov    %ebx,%esi
  call   <mp_store_u16@plt>
L29:
  pop    %rbx
  ret
L9:
  or     $0x90,%bl
  mov    %ebx,%esi
  call   <mp_store_u8@plt>
  jmp    L29
L34:
  mov    $0xdd,%esi
  call   <mp_store_u8@plt>
  mov    %rax,%rdi
  mov    %ebx,%esi
  call   <mp_store_u32@plt>
  jmp    L29
```

assistant: ```c
char *mp_encode_array(char *data, uint32_t size) {
  if (size <= 15) {
    return mp_store_u8(data, 0x90 | size);
  } else if (size <= UINT16_MAX) {
    data = mp_store_u8(data, 0xdc);
    data = mp_store_u16(data, size);
    return data;
  } else {
    data = mp_store_u8(data, 0xdd);
    return mp_store_u32(data, size);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<mp_encode_array>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  cmp    $0xf,%esi
  jle    L8
  mov    D6(%rip),%rax
  cmp    %esi,(%rax)
  jl     L5
  mov    $0xdc,%esi
  call   <mp_store_u8@plt>
  mov    %ebp,%esi
  pop    %rbp
  mov    %rax,%rdi
  jmp    1070 <mp_store_u16@plt>
  nopl   0x0(%rax)
L5:
  mov    $0xdd,%esi
  call   <mp_store_u8@plt>
  mov    %ebp,%esi
  pop    %rbp
  mov    %rax,%rdi
  jmp    1080 <mp_store_u32@plt>
  nopl   (%rax)
L8:
  or     $0x90,%sil
  pop    %rbp
  jmp    1090 <mp_store_u8@plt>
```

assistant: ```c
char *mp_encode_array(char *data, uint32_t size) {
  if (size <= 15) {
    return mp_store_u8(data, 0x90 | size);
  } else if (size <= UINT16_MAX) {
    data = mp_store_u8(data, 0xdc);
    data = mp_store_u16(data, size);
    return data;
  } else {
    data = mp_store_u8(data, 0xdd);
    return mp_store_u32(data, size);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mp_encode_array>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  cmp    $0xf,%esi
  jle    L11
  mov    D5(%rip),%rax
  cmp    %esi,(%rax)
  jl     L18
  mov    $0xdc,%esi
  call   <mp_store_u8@plt>
  mov    %ebp,%esi
  pop    %rbp
  mov    %rax,%rdi
  jmp    1070 <mp_store_u16@plt>
  nopl   0x0(%rax)
L18:
  mov    $0xdd,%esi
  call   <mp_store_u8@plt>
  mov    %ebp,%esi
  pop    %rbp
  mov    %rax,%rdi
  jmp    1080 <mp_store_u32@plt>
  nopl   (%rax)
L11:
  or     $0x90,%sil
  pop    %rbp
  jmp    1090 <mp_store_u8@plt>
```

assistant: ```c
char *mp_encode_array(char *data, uint32_t size) {
  if (size <= 15) {
    return mp_store_u8(data, 0x90 | size);
  } else if (size <= UINT16_MAX) {
    data = mp_store_u8(data, 0xdc);
    data = mp_store_u16(data, size);
    return data;
  } else {
    data = mp_store_u8(data, 0xdd);
    return mp_store_u32(data, size);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cstrjoin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L18
L21:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  add    %eax,-0xc(%rbp)
  addl   $0x1,-0x14(%rbp)
L18:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L44
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L21
L44:
  addl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <memalloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L60
  mov    $0x0,%eax
  jmp    L9
L60:
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  cmpq   $0x0,-0x28(%rbp)
  jne    L67
  mov    -0x8(%rbp),%rax
  jmp    L9
L67:
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L7
L15:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rcx
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <cstrcat@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x20,(%rax)
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  addl   $0x1,-0x14(%rbp)
L7:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L107
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L15
L107:
  cmpl   $0x0,-0x10(%rbp)
  je     L25
  subl   $0x1,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L25:
  mov    -0x8(%rbp),%rax
L9:
  leave
  ret
```

assistant: ```c
char *cstrjoin(char **args) {
  char *buf;
  int i, l, buflen = 0;
  for (i = 0; args[i] != NULL && *args[i]; i++)
    buflen += strlen(args[i]) + 1;
  buflen++;
  if ((buf = (char *)memalloc(buflen)) == NULL)
    return NULL;
  *buf = 0;
  if (args == NULL)
    return buf;
  l = 0;
  for (i = 0; args[i] != NULL && *args[i]; i++) {
    cstrcat(buf, args[i], buflen);
    l = strlen(buf);
    buf[l++] = ' ';
    buf[l] = 0;
  }
  if (l)
    buf[--l] = 0;
  return buf;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<cstrjoin>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    (%rdi),%rdi
  test   %rdi,%rdi
  je     L65
  lea    0x8(%r12),%rbx
  mov    $0x0,%ebp
L98:
  cmpb   $0x0,(%rdi)
  je     L3
  call   <strlen@plt>
  lea    0x1(%rbp,%rax,1),%ebp
  add    $0x8,%rbx
  mov    -0x8(%rbx),%rdi
  test   %rdi,%rdi
  jne    L98
L3:
  add    $0x1,%ebp
  mov    %ebp,%edi
  call   <memalloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L68
  movb   $0x0,(%rax)
  mov    (%r12),%rsi
  test   %rsi,%rsi
  je     L12
  add    $0x8,%r12
  mov    $0x0,%edx
L2:
  cmpb   $0x0,(%rsi)
  je     L36
  mov    %ebp,%edx
  mov    %rbx,%rdi
  call   <cstrcat@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edx
  cltq
  movb   $0x20,(%rbx,%rax,1)
  movslq %edx,%rax
  movb   $0x0,(%rbx,%rax,1)
  add    $0x8,%r12
  mov    -0x8(%r12),%rsi
  test   %rsi,%rsi
  jne    L2
L36:
  test   %edx,%edx
  je     L12
  sub    $0x1,%edx
  movslq %edx,%rdx
  movb   $0x0,(%rbx,%rdx,1)
L12:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L65:
  mov    $0x0,%ebp
  jmp    L3
L68:
  mov    $0x0,%ebx
  jmp    L12
```

assistant: ```c
char *cstrjoin(char **args) {
  char *buf;
  int i, l, buflen = 0;
  for (i = 0; args[i] != NULL && *args[i]; i++)
    buflen += strlen(args[i]) + 1;
  buflen++;
  if ((buf = (char *)memalloc(buflen)) == NULL)
    return NULL;
  *buf = 0;
  if (args == NULL)
    return buf;
  l = 0;
  for (i = 0; args[i] != NULL && *args[i]; i++) {
    cstrcat(buf, args[i], buflen);
    l = strlen(buf);
    buf[l++] = ' ';
    buf[l] = 0;
  }
  if (l)
    buf[--l] = 0;
  return buf;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<cstrjoin>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    (%rdi),%rdi
  mov    $0x1,%ebx
  test   %rdi,%rdi
  je     L21
  lea    0x8(%rbp),%r12
  xor    %ebx,%ebx
  jmp    L94
L76:
  call   <strlen@plt>
  mov    (%r12),%rdi
  add    $0x8,%r12
  lea    0x1(%rbx,%rax,1),%ebx
  test   %rdi,%rdi
  je     L62
L94:
  cmpb   $0x0,(%rdi)
  jne    L76
L62:
  add    $0x1,%ebx
L21:
  mov    %ebx,%edi
  call   <memalloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L71
  movb   $0x0,(%rax)
  mov    0x0(%rbp),%rsi
  test   %rsi,%rsi
  je     L71
  add    $0x8,%rbp
  xor    %edx,%edx
  jmp    L28
  nopl   0x0(%rax)
L85:
  mov    %ebx,%edx
  mov    %r12,%rdi
  add    $0x8,%rbp
  call   <cstrcat@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edx
  cltq
  movb   $0x20,(%r12,%rax,1)
  movslq %edx,%rax
  movb   $0x0,(%r12,%rax,1)
  mov    -0x8(%rbp),%rsi
  test   %rsi,%rsi
  je     L61
L28:
  cmpb   $0x0,(%rsi)
  jne    L85
L61:
  test   %edx,%edx
  je     L71
  sub    $0x1,%edx
  movslq %edx,%rdx
  movb   $0x0,(%r12,%rdx,1)
L71:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *cstrjoin(char **args) {
  char *buf;
  int i, l, buflen = 0;
  for (i = 0; args[i] != NULL && *args[i]; i++)
    buflen += strlen(args[i]) + 1;
  buflen++;
  if ((buf = (char *)memalloc(buflen)) == NULL)
    return NULL;
  *buf = 0;
  if (args == NULL)
    return buf;
  l = 0;
  for (i = 0; args[i] != NULL && *args[i]; i++) {
    cstrcat(buf, args[i], buflen);
    l = strlen(buf);
    buf[l++] = ' ';
    buf[l] = 0;
  }
  if (l)
    buf[--l] = 0;
  return buf;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<cstrjoin>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    (%rdi),%rdi
  mov    $0x1,%ebx
  test   %rdi,%rdi
  je     L78
  lea    0x8(%rbp),%r12
  xor    %ebx,%ebx
  jmp    L4
L50:
  call   <strlen@plt>
  mov    (%r12),%rdi
  add    $0x8,%r12
  lea    0x1(%rbx,%rax,1),%ebx
  test   %rdi,%rdi
  je     L95
L4:
  cmpb   $0x0,(%rdi)
  jne    L50
L95:
  add    $0x1,%ebx
L78:
  mov    %ebx,%edi
  call   <memalloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L63
  movb   $0x0,(%rax)
  mov    0x0(%rbp),%rsi
  test   %rsi,%rsi
  je     L63
  add    $0x8,%rbp
  xor    %edx,%edx
  jmp    L75
  nopl   0x0(%rax)
L89:
  mov    %ebx,%edx
  mov    %r12,%rdi
  add    $0x8,%rbp
  call   <cstrcat@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edx
  cltq
  movb   $0x20,(%r12,%rax,1)
  movslq %edx,%rax
  movb   $0x0,(%r12,%rax,1)
  mov    -0x8(%rbp),%rsi
  test   %rsi,%rsi
  je     L14
L75:
  cmpb   $0x0,(%rsi)
  jne    L89
L14:
  test   %edx,%edx
  je     L63
  sub    $0x1,%edx
  movslq %edx,%rdx
  movb   $0x0,(%r12,%rdx,1)
L63:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *cstrjoin(char **args) {
  char *buf;
  int i, l, buflen = 0;
  for (i = 0; args[i] != NULL && *args[i]; i++)
    buflen += strlen(args[i]) + 1;
  buflen++;
  if ((buf = (char *)memalloc(buflen)) == NULL)
    return NULL;
  *buf = 0;
  if (args == NULL)
    return buf;
  l = 0;
  for (i = 0; args[i] != NULL && *args[i]; i++) {
    cstrcat(buf, args[i], buflen);
    l = strlen(buf);
    buf[l++] = ' ';
    buf[l] = 0;
  }
  if (l)
    buf[--l] = 0;
  return buf;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <mfunc@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  mfunc();
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

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

assistant: ```c
int main() {
  mfunc();
  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
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <mfunc@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  mfunc();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

assistant: ```c
int main() {
  mfunc();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x12c,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x14,-0x4(%rbp)
  jmp    L14
L13:
  mov    -0x10(%rbp),%eax
  movd   %eax,%xmm0
  call   <Temtrans@plt>
  movq   %xmm0,%rdx
  mov    -0x10(%rbp),%eax
  movq   %rdx,%xmm1
  movd   %eax,%xmm0
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0x4(%rbp),%xmm0
  movss  -0x10(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x10(%rbp)
L14:
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0xc(%rbp),%xmm0
  comiss -0x10(%rbp),%xmm0
  jae    L13
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D0
data_type: string

tool: D0: %3.1f	%3.1f


assistant: ```c
main() {
  float f;
  int max = 300, min = 0, step = 20;
  while (f <= max) {
    printf("%3.1f\t%3.1f\n", f, Temtrans(f));
    f = f + step;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  pxor   %xmm5,%xmm5
  movss  %xmm5,0xc(%rsp)
  lea    D12(%rip),%rbx
L2:
  movss  0xc(%rsp),%xmm0
  call   <Temtrans@plt>
  movapd %xmm0,%xmm1
  movss  0xc(%rsp),%xmm0
  mov    %rbx,%rdi
  call   <printf@plt>
  movss  D1(%rip),%xmm2
  addss  0xc(%rsp),%xmm2
  movss  %xmm2,0xc(%rsp)
  movss  D26(%rip),%xmm4
  comiss %xmm2,%xmm4
  jae    L2
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: label: D12
type_guessed: string
name: disassemble_data
arguments: label: D1
type_guessed: byte[4]
name: disassemble_data
arguments: label: D26
type_guessed: byte[4]

tool: D12: %3.1f	%3.1f


tool: {"D1": [0x00,0x00,0xa0,0x41]}

tool: {"D26": [0x00,0x00,0x96,0x43]}

assistant: ```c
main() {
  float f;
  int max = 300, min = 0, step = 20;
  while (f <= max) {
    printf("%3.1f\t%3.1f\n", f, Temtrans(f));
    f = f + step;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  movss  D3(%rip),%xmm4
  pxor   %xmm2,%xmm2
  comiss %xmm2,%xmm4
  jb     L19
  push   %rbx
  lea    D2(%rip),%rbx
  sub    $0x10,%rsp
  nopl   0x0(%rax)
L18:
  movaps %xmm2,%xmm0
  movss  %xmm2,0xc(%rsp)
  call   <Temtrans@plt>
  movss  0xc(%rsp),%xmm2
  mov    %rbx,%rdi
  movapd %xmm0,%xmm1
  movaps %xmm2,%xmm0
  call   <printf@plt>
  movss  0xc(%rsp),%xmm2
  addss  D5(%rip),%xmm2
  movss  D3(%rip),%xmm3
  comiss %xmm2,%xmm3
  jae    L18
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L19:
  xor    %eax,%eax
  ret
```

tool_calls: name: read_data
arguments: label: D3
data_type: byte[4]
name: read_data
arguments: label: D2
data_type: string
name: read_data
arguments: label: D5
data_type: byte[4]

tool: {"D3": [0x00,0x00,0x96,0x43]}

tool: D2: %3.1f	%3.1f


tool: {"D5": [0x00,0x00,0xa0,0x41]}

assistant: ```c
main() {
  float f;
  int max = 300, min = 0, step = 20;
  while (f <= max) {
    printf("%3.1f\t%3.1f\n", f, Temtrans(f));
    f = f + step;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  movss  D34(%rip),%xmm4
  pxor   %xmm2,%xmm2
  comiss %xmm2,%xmm4
  jb     L14
  push   %rbx
  lea    D8(%rip),%rbx
  sub    $0x10,%rsp
  nopl   0x0(%rax)
L2:
  movaps %xmm2,%xmm0
  movss  %xmm2,0xc(%rsp)
  call   <Temtrans@plt>
  movss  0xc(%rsp),%xmm2
  mov    %rbx,%rdi
  movapd %xmm0,%xmm1
  movaps %xmm2,%xmm0
  call   <printf@plt>
  movss  0xc(%rsp),%xmm2
  addss  D16(%rip),%xmm2
  movss  D34(%rip),%xmm3
  comiss %xmm2,%xmm3
  jae    L2
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L14:
  xor    %eax,%eax
  ret
```

tool_calls: name: parse_data
arguments: data_label: D34
type_guessed: byte[4]
name: parse_data
arguments: data_label: D8
type_guessed: string
name: parse_data
arguments: data_label: D16
type_guessed: byte[4]

tool: {"D34": [0x00,0x00,0x96,0x43]}

tool: D8: %3.1f	%3.1f


tool: {"D16": [0x00,0x00,0xa0,0x41]}

assistant: ```c
main() {
  float f;
  int max = 300, min = 0, step = 20;
  while (f <= max) {
    printf("%3.1f\t%3.1f\n", f, Temtrans(f));
    f = f + step;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x14(%rbp)
  jmp    L40
L52:
  movl   $0x0,-0x10(%rbp)
  jmp    L66
L10:
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
L66:
  mov    -0x18(%rbp),%eax
  sub    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jl     L10
  movl   $0x0,-0xc(%rbp)
  jmp    L68
L61:
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L68:
  mov    -0x14(%rbp),%eax
  add    %eax,%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L61
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L40:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jle    L52
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L62
  call   <__stack_chk_fail@plt>
L62:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D23
type_guessed: string
name: disassemble_data
arguments: tag: D32
type_guessed: string
name: disassemble_data
arguments: tag: D6
type_guessed: string
name: disassemble_data
arguments: tag: D22
type_guessed: string

tool: D23: %d

tool: D32:  

tool: D6: *

tool: D22: 


assistant: ```c
int main() {
  int count, i, a, b;
  scanf("%d", &b);
  for (count = 0; count <= b; ++count) {
    for (i = 0; i < b - count; ++i) {
      printf(" ");
    }
    for (a = 0; a < 2 * count - 1; ++a) {
      printf("*");
    }
    printf("\n");
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  mov    $0xffffffff,%r15d
  mov    $0x0,%ebp
  lea    D39(%rip),%r14
  lea    D5(%rip),%r13
  cmpl   $0x0,0x4(%rsp)
  jns    L43
L69:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L11:
  mov    %r14,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  mov    0x4(%rsp),%eax
  sub    %ebp,%eax
  cmp    %ebx,%eax
  jg     L11
L45:
  mov    %r15d,%r12d
  test   %r15d,%r15d
  jle    L31
  mov    $0x0,%ebx
L41:
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,%r12d
  jne    L41
L31:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  add    $0x1,%ebp
  add    $0x2,%r15d
  cmp    %ebp,0x4(%rsp)
  jl     L69
L43:
  mov    $0x0,%ebx
  cmp    %ebp,0x4(%rsp)
  jg     L11
  jmp    L45
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D14
datatype: string
name: retrieve_data
arguments: label: D39
datatype: string
name: retrieve_data
arguments: label: D5
datatype: string
name: retrieve_data
arguments: label: D6
datatype: string

tool: D14: %d

tool: D39:  

tool: D5: *

tool: D6: 


assistant: ```c
int main() {
  int count, i, a, b;
  scanf("%d", &b);
  for (count = 0; count <= b; ++count) {
    for (i = 0; i < b - count; ++i) {
      printf(" ");
    }
    for (a = 0; a < 2 * count - 1; ++a) {
      printf("*");
    }
    printf("\n");
  }
  return 0;
}
```