ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<parse_section>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  lea    0x1(%rdi),%rbp
  push   %rbx
  mov    %rbp,%rdi
  mov    %rsi,%rbx
  mov    $0x5d,%esi
  call   <strchr@plt>
  test   %rax,%rax
  je     L5
  movb   $0x0,(%rax)
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    *%rax
  cs nopw 0x0(%rax,%rax,1)
L5:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
bool parse_section(char *buf, bool (*sfunc)(const char *section, void *private_data), void *private_data) {
  char *p, *q;
  p = buf + 1;
  q = strchr(p, ']');
  if (q == NULL) {
    return false;
  }
  *q = '\0';
  return sfunc(p, private_data);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<bitScan_fun_ulong>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x28,-0x4(%rbp)
  jmp    L12
L1:
  subl   $0x1,-0x4(%rbp)
  shrq   -0x18(%rbp)
L12:
  cmpq   $0x0,-0x18(%rbp)
  jne    L1
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned bitScan_fun_ulong(unsigned long x) {
  unsigned r = 40;
  while (x) {
    --r;
    x >>= 1;
  }
  return r;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<bitScan_fun_ulong>:
  endbr64
  test   %rdi,%rdi
  je     L4
  mov    $0x28,%eax
L1:
  sub    $0x1,%eax
  shr    %rdi
  jne    L1
  ret
L4:
  mov    $0x28,%eax
  ret
```

assistant: ```c
unsigned bitScan_fun_ulong(unsigned long x) {
  unsigned r = 40;
  while (x) {
    --r;
    x >>= 1;
  }
  return r;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bitScan_fun_ulong>:
  endbr64
  mov    $0x28,%eax
  test   %rdi,%rdi
  je     L12
  xchg   %ax,%ax
L17:
  sub    $0x1,%eax
  shr    %rdi
  jne    L17
  ret
  nopl   0x0(%rax)
L12:
  ret
```

assistant: ```c
unsigned bitScan_fun_ulong(unsigned long x) {
  unsigned r = 40;
  while (x) {
    --r;
    x >>= 1;
  }
  return r;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<bitScan_fun_ulong>:
  endbr64
  mov    $0x28,%eax
  test   %rdi,%rdi
  je     L5
  xchg   %ax,%ax
L17:
  sub    $0x1,%eax
  shr    %rdi
  jne    L17
  ret
  nopl   0x0(%rax)
L5:
  ret
```

assistant: ```c
unsigned bitScan_fun_ulong(unsigned long x) {
  unsigned r = 40;
  while (x) {
    --r;
    x >>= 1;
  }
  return r;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<has_wwn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    0x1b0(%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0xf000,%eax
  cmp    $0x5000,%eax
  sete   %al
  movzbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
int has_wwn(const uint8_t *const identify) {
  uint16_t word;
  word = *((uint16_t *)identify + 108);
  return ((word & 0xf000) == 0x5000) ? 1 : 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<has_wwn>:
  endbr64
  mov    0x1b0(%rdi),%eax
  and    $0xf000,%eax
  cmp    $0x5000,%eax
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int has_wwn(const uint8_t *const identify) {
  uint16_t word;
  word = *((uint16_t *)identify + 108);
  return ((word & 0xf000) == 0x5000) ? 1 : 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<has_wwn>:
  endbr64
  mov    0x1b0(%rdi),%eax
  and    $0xf000,%eax
  cmp    $0x5000,%eax
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int has_wwn(const uint8_t *const identify) {
  uint16_t word;
  word = *((uint16_t *)identify + 108);
  return ((word & 0xf000) == 0x5000) ? 1 : 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<has_wwn>:
  endbr64
  mov    0x1b0(%rdi),%eax
  and    $0xf000,%eax
  cmp    $0x5000,%eax
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int has_wwn(const uint8_t *const identify) {
  uint16_t word;
  word = *((uint16_t *)identify + 108);
  return ((word & 0xf000) == 0x5000) ? 1 : 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<stringutils_is_empty_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L71
L88:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L46
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  je     L46
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  je     L46
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x23,%al
  je     L46
  mov    $0x0,%eax
  jmp    L54
L46:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x23,%al
  jne    L62
  mov    $0x1,%eax
  jmp    L54
L62:
  addl   $0x1,-0x4(%rbp)
L71:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L88
  mov    $0x1,%eax
L54:
  pop    %rbp
  ret
```

assistant: ```c
int stringutils_is_empty_line(const char *line) {
  int index = 0;
  while (line[index] != '\0') {
    if (line[index] != ' ' && line[index] != '\n' && line[index] != '\t' && line[index] != '#') {
      return 0;
    }
    if (line[index] == '#') {
      return 1;
    }
    index++;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<stringutils_is_empty_line>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L0
  add    $0x1,%rdi
  movabs $0x900000600,%rdx
  jmp    L53
L28:
  cmp    $0x23,%al
  je     L56
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%eax
  test   %al,%al
  je     L18
L53:
  cmp    $0x23,%al
  ja     L12
  bt     %rax,%rdx
  jb     L28
  mov    $0x0,%eax
  ret
L18:
  mov    $0x1,%eax
  ret
L0:
  mov    $0x1,%eax
  ret
L12:
  mov    $0x0,%eax
  ret
L56:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int stringutils_is_empty_line(const char *line) {
  int index = 0;
  while (line[index] != '\0') {
    if (line[index] != ' ' && line[index] != '\n' && line[index] != '\t' && line[index] != '#') {
      return 0;
    }
    if (line[index] == '#') {
      return 1;
    }
    index++;
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<stringutils_is_empty_line>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L34
  movabs $0xfffffff6fffff9ff,%rdx
  add    $0x1,%rdi
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L31:
  bt     %rax,%rdx
  jb     L57
  cmp    $0x23,%al
  je     L34
  movzbl (%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  je     L34
L18:
  cmp    $0x23,%al
  jbe    L31
L57:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L34:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int stringutils_is_empty_line(const char *line) {
  int index = 0;
  while (line[index] != '\0') {
    if (line[index] != ' ' && line[index] != '\n' && line[index] != '\t' && line[index] != '#') {
      return 0;
    }
    if (line[index] == '#') {
      return 1;
    }
    index++;
  }
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<stringutils_is_empty_line>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L45
  movabs $0xfffffff6fffff9ff,%rdx
  add    $0x1,%rdi
  jmp    L7
  nopl   0x0(%rax,%rax,1)
L21:
  bt     %rax,%rdx
  jb     L17
  cmp    $0x23,%al
  je     L45
  movzbl (%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  je     L45
L7:
  cmp    $0x23,%al
  jbe    L21
L17:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L45:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int stringutils_is_empty_line(const char *line) {
  int index = 0;
  while (line[index] != '\0') {
    if (line[index] != ' ' && line[index] != '\n' && line[index] != '\t' && line[index] != '#') {
      return 0;
    }
    if (line[index] == '#') {
      return 1;
    }
    index++;
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<reverse_char_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  sub    $0x1,%rax
  add    %rax,-0x18(%rbp)
  jmp    L12
L1:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  subq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L12:
  mov    -0xc(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L1
  mov    -0x24(%rbp),%eax
  cltq
  neg    %rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void reverse_char_array(char *input, char **output, int num_chars) {
  char *temp;
  int i;
  i = 0;
  temp = *output;
  input = input + (num_chars - 1);
  while (i < num_chars) {
    *temp = *input;
    input--;
    temp++;
    i++;
  }
  *output = temp - num_chars;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<reverse_char_array>:
  endbr64
  mov    (%rsi),%r9
  movslq %edx,%r10
  lea    -0x1(%r10,%rdi,1),%rax
  test   %edx,%edx
  jle    L1
  lea    -0x1(%rdx),%r8d
  not    %r8
  add    %rax,%r8
  mov    %r9,%rcx
L7:
  movzbl (%rax),%edi
  mov    %dil,(%rcx)
  sub    $0x1,%rax
  add    $0x1,%rcx
  cmp    %r8,%rax
  jne    L7
  mov    %edx,%edx
  add    %rdx,%r9
L1:
  sub    %r10,%r9
  mov    %r9,(%rsi)
  ret
```

assistant: ```c
void reverse_char_array(char *input, char **output, int num_chars) {
  char *temp;
  int i;
  i = 0;
  temp = *output;
  input = input + (num_chars - 1);
  while (i < num_chars) {
    *temp = *input;
    input--;
    temp++;
    i++;
  }
  *output = temp - num_chars;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse_char_array>:
  endbr64
  movslq %edx,%r10
  mov    (%rsi),%r9
  lea    -0x1(%r10,%rdi,1),%rax
  test   %edx,%edx
  jle    L4
  lea    -0x1(%rdx),%r8d
  mov    %r9,%rcx
  not    %r8
  add    %rax,%r8
L7:
  movzbl (%rax),%edi
  sub    $0x1,%rax
  add    $0x1,%rcx
  mov    %dil,-0x1(%rcx)
  cmp    %r8,%rax
  jne    L7
  mov    %edx,%edx
  add    %rdx,%r9
L4:
  sub    %r10,%r9
  mov    %r9,(%rsi)
  ret
```

assistant: ```c
void reverse_char_array(char *input, char **output, int num_chars) {
  char *temp;
  int i;
  i = 0;
  temp = *output;
  input = input + (num_chars - 1);
  while (i < num_chars) {
    *temp = *input;
    input--;
    temp++;
    i++;
  }
  *output = temp - num_chars;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<reverse_char_array>:
  endbr64
  movslq %edx,%r10
  mov    (%rsi),%r9
  lea    -0x1(%r10,%rdi,1),%rax
  test   %edx,%edx
  jle    L5
  lea    -0x1(%rdx),%r8d
  mov    %r9,%rcx
  not    %r8
  add    %rax,%r8
L12:
  movzbl (%rax),%edi
  sub    $0x1,%rax
  add    $0x1,%rcx
  mov    %dil,-0x1(%rcx)
  cmp    %r8,%rax
  jne    L12
  mov    %edx,%edx
  add    %rdx,%r9
L5:
  sub    %r10,%r9
  mov    %r9,(%rsi)
  ret
```

assistant: ```c
void reverse_char_array(char *input, char **output, int num_chars) {
  char *temp;
  int i;
  i = 0;
  temp = *output;
  input = input + (num_chars - 1);
  while (i < num_chars) {
    *temp = *input;
    input--;
    temp++;
    i++;
  }
  *output = temp - num_chars;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<cube_copy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L0:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x4(%rbp)
L13:
  cmpl   $0x17,-0x4(%rbp)
  jle    L0
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void cube_copy(int *a, int *b) {
  int i;
  for (i = 0; i < 24; i++)
    b[i] = a[i];
}
```
user: What is the c source code of the assembly code below?

```asm
<cube_copy>:
  endbr64
  mov    $0x0,%eax
L0:
  mov    (%rdi,%rax,1),%edx
  mov    %edx,(%rsi,%rax,1)
  add    $0x4,%rax
  cmp    $0x60,%rax
  jne    L0
  ret
```

assistant: ```c
void cube_copy(int *a, int *b) {
  int i;
  for (i = 0; i < 24; i++)
    b[i] = a[i];
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<cube_copy>:
  endbr64
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L3:
  mov    (%rdi,%rax,1),%edx
  mov    %edx,(%rsi,%rax,1)
  add    $0x4,%rax
  cmp    $0x60,%rax
  jne    L3
  ret
```

assistant: ```c
void cube_copy(int *a, int *b) {
  int i;
  for (i = 0; i < 24; i++)
    b[i] = a[i];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cube_copy>:
  endbr64
  lea    0x4(%rdi),%rdx
  mov    %rsi,%rax
  sub    %rdx,%rax
  cmp    $0x8,%rax
  jbe    L13
  movdqu (%rdi),%xmm0
  movups %xmm0,(%rsi)
  movdqu 0x10(%rdi),%xmm1
  movups %xmm1,0x10(%rsi)
  movdqu 0x20(%rdi),%xmm2
  movups %xmm2,0x20(%rsi)
  movdqu 0x30(%rdi),%xmm3
  movups %xmm3,0x30(%rsi)
  movdqu 0x40(%rdi),%xmm4
  movups %xmm4,0x40(%rsi)
  movdqu 0x50(%rdi),%xmm5
  movups %xmm5,0x50(%rsi)
  ret
L13:
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L19:
  mov    (%rdi,%rax,1),%edx
  mov    %edx,(%rsi,%rax,1)
  add    $0x4,%rax
  cmp    $0x60,%rax
  jne    L19
  ret
```

assistant: ```c
void cube_copy(int *a, int *b) {
  int i;
  for (i = 0; i < 24; i++)
    b[i] = a[i];
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<xmodem_read_block>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%eax
  mov    %rsi,-0x20(%rbp)
  mov    %al,-0x14(%rbp)
  mov    $0x1,%edi
  call   <serial_recv_byte@plt>
  mov    %al,-0x6(%rbp)
  movzbl -0x6(%rbp),%eax
  cmp    -0x14(%rbp),%al
  je     L70
  mov    $0xffffffff,%eax
  jmp    L32
L70:
  mov    $0x1,%edi
  call   <serial_recv_byte@plt>
  xor    %al,-0x6(%rbp)
  cmpb   $0xff,-0x6(%rbp)
  je     L6
  mov    $0xffffffff,%eax
  jmp    L32
L6:
  movb   $0x0,-0x7(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L71
L24:
  mov    $0x1,%edi
  call   <serial_recv_byte@plt>
  mov    %al,-0x5(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movzbl -0x5(%rbp),%edx
  mov    %dl,(%rax)
  movzbl -0x5(%rbp),%eax
  add    %al,-0x7(%rbp)
  addl   $0x1,-0x4(%rbp)
L71:
  cmpl   $0x7f,-0x4(%rbp)
  jle    L24
  mov    $0x1,%edi
  call   <serial_recv_byte@plt>
  xor    %al,-0x7(%rbp)
  cmpb   $0x0,-0x7(%rbp)
  je     L44
  mov    $0xffffffff,%eax
  jmp    L32
L44:
  mov    -0x4(%rbp),%eax
L32:
  leave
  ret
```

assistant: ```c
int xmodem_read_block(unsigned char block_number, char *buf) {
  unsigned char c, block_num, check_sum;
  int i;
  block_num = serial_recv_byte(1);
  if (block_num != block_number)
    return -1;
  block_num ^= serial_recv_byte(1);
  if (block_num != 0xff)
    return -1;
  check_sum = 0;
  for (i = 0; i < 128; i++) {
    c = serial_recv_byte(1);
    *(buf++) = c;
    check_sum += c;
  }
  check_sum ^= serial_recv_byte(1);
  if (check_sum)
    return -1;
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<xmodem_read_block>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %rsi,%rbx
  mov    $0x1,%edi
  call   <serial_recv_byte@plt>
  cmp    %r12b,%al
  jne    L38
  mov    %eax,%ebp
  mov    $0x1,%edi
  call   <serial_recv_byte@plt>
  xor    %eax,%ebp
  cmp    $0xff,%bpl
  jne    L5
  lea    0x80(%rbx),%r12
  mov    $0x0,%ebp
L32:
  mov    $0x1,%edi
  call   <serial_recv_byte@plt>
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  add    %eax,%ebp
  cmp    %r12,%rbx
  jne    L32
  mov    $0x1,%edi
  call   <serial_recv_byte@plt>
  cmp    %bpl,%al
  mov    $0x80,%eax
  mov    $0xffffffff,%edx
  cmovne %edx,%eax
L15:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L38:
  mov    $0xffffffff,%eax
  jmp    L15
L5:
  mov    $0xffffffff,%eax
  jmp    L15
```

assistant: ```c
int xmodem_read_block(unsigned char block_number, char *buf) {
  unsigned char c, block_num, check_sum;
  int i;
  block_num = serial_recv_byte(1);
  if (block_num != block_number)
    return -1;
  block_num ^= serial_recv_byte(1);
  if (block_num != 0xff)
    return -1;
  check_sum = 0;
  for (i = 0; i < 128; i++) {
    c = serial_recv_byte(1);
    *(buf++) = c;
    check_sum += c;
  }
  check_sum ^= serial_recv_byte(1);
  if (check_sum)
    return -1;
  return i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<xmodem_read_block>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  mov    $0x1,%edi
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <serial_recv_byte@plt>
  cmp    %r12b,%al
  jne    L31
  mov    $0x1,%edi
  mov    %eax,%ebp
  call   <serial_recv_byte@plt>
  xor    %eax,%ebp
  cmp    $0xff,%bpl
  jne    L31
  lea    0x80(%rbx),%r12
  xor    %ebp,%ebp
  nopw   0x0(%rax,%rax,1)
L45:
  mov    $0x1,%edi
  add    $0x1,%rbx
  call   <serial_recv_byte@plt>
  mov    %al,-0x1(%rbx)
  add    %eax,%ebp
  cmp    %r12,%rbx
  jne    L45
  mov    $0x1,%edi
  call   <serial_recv_byte@plt>
  mov    %eax,%r8d
  mov    $0x80,%eax
  cmp    %bpl,%r8b
  jne    L31
L26:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L31:
  mov    $0xffffffff,%eax
  jmp    L26
```

assistant: ```c
int xmodem_read_block(unsigned char block_number, char *buf) {
  unsigned char c, block_num, check_sum;
  int i;
  block_num = serial_recv_byte(1);
  if (block_num != block_number)
    return -1;
  block_num ^= serial_recv_byte(1);
  if (block_num != 0xff)
    return -1;
  check_sum = 0;
  for (i = 0; i < 128; i++) {
    c = serial_recv_byte(1);
    *(buf++) = c;
    check_sum += c;
  }
  check_sum ^= serial_recv_byte(1);
  if (check_sum)
    return -1;
  return i;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<xmodem_read_block>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  mov    $0x1,%edi
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <serial_recv_byte@plt>
  cmp    %r12b,%al
  jne    L18
  mov    $0x1,%edi
  mov    %eax,%ebp
  call   <serial_recv_byte@plt>
  xor    %eax,%ebp
  cmp    $0xff,%bpl
  jne    L18
  lea    0x80(%rbx),%r12
  xor    %ebp,%ebp
  nopw   0x0(%rax,%rax,1)
L0:
  mov    $0x1,%edi
  add    $0x1,%rbx
  call   <serial_recv_byte@plt>
  mov    %al,-0x1(%rbx)
  add    %eax,%ebp
  cmp    %r12,%rbx
  jne    L0
  mov    $0x1,%edi
  call   <serial_recv_byte@plt>
  mov    %eax,%r8d
  mov    $0x80,%eax
  cmp    %bpl,%r8b
  jne    L18
L3:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  mov    $0xffffffff,%eax
  jmp    L3
```

assistant: ```c
int xmodem_read_block(unsigned char block_number, char *buf) {
  unsigned char c, block_num, check_sum;
  int i;
  block_num = serial_recv_byte(1);
  if (block_num != block_number)
    return -1;
  block_num ^= serial_recv_byte(1);
  if (block_num != 0xff)
    return -1;
  check_sum = 0;
  for (i = 0; i < 128; i++) {
    c = serial_recv_byte(1);
    *(buf++) = c;
    check_sum += c;
  }
  check_sum ^= serial_recv_byte(1);
  if (check_sum)
    return -1;
  return i;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x6262626161626162,%rax
  movabs $0x6261616161616262,%rdx
  mov    %rax,-0x40(%rbp)
  mov    %rdx,-0x38(%rbp)
  movabs $0x6162626162616162,%rax
  movabs $0x6262616161616161,%rdx
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
  movabs $0x6262626161626161,%rax
  movabs $0x6262626161626261,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  movw   $0x6262,-0x10(%rbp)
  movb   $0x0,-0xe(%rbp)
  movabs $0x6262622a2a61622a,%rax
  mov    %rax,-0x4a(%rbp)
  movw   $0x62,-0x42(%rbp)
  lea    -0x4a(%rbp),%rdx
  lea    -0x40(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <isMatch@plt>
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

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

tool: D4: %d


assistant: ```c
int main(int argc, char *argv[]) {
  char s[] = {"babaabbbbbaaaaabbaababbaaaaaaabbaabaabbbabbaabbbbb"};
  char p[] = {"*ba**bbbb"};
  printf("%d\n", isMatch(s, p));
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movabs $0x6262626161626162,%rax
  movabs $0x6261616161616262,%rdx
  mov    %rax,0x10(%rsp)
  mov    %rdx,0x18(%rsp)
  movabs $0x6162626162616162,%rax
  movabs $0x6262616161616161,%rdx
  mov    %rax,0x20(%rsp)
  mov    %rdx,0x28(%rsp)
  movabs $0x6262626161626161,%rax
  movabs $0x6262626161626261,%rdx
  mov    %rax,0x30(%rsp)
  mov    %rdx,0x38(%rsp)
  movw   $0x6262,0x40(%rsp)
  movb   $0x0,0x42(%rsp)
  movabs $0x6262622a2a61622a,%rax
  mov    %rax,0x6(%rsp)
  movw   $0x62,0xe(%rsp)
  lea    0x6(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  mov    $0x0,%eax
  call   <isMatch@plt>
  mov    %eax,%esi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x58,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D6
type_infered: string

tool: D6: %d


assistant: ```c
int main(int argc, char *argv[]) {
  char s[] = {"babaabbbbbaaaaabbaababbaaaaaaabbaabaabbbabbaabbbbb"};
  char p[] = {"*ba**bbbb"};
  printf("%d\n", isMatch(s, p));
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  movdqa D29(%rip),%xmm0
  mov    $0x62,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    $0x6262,%eax
  lea    0x6(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  movaps %xmm0,0x10(%rsp)
  movdqa D23(%rip),%xmm0
  mov    %ax,0x40(%rsp)
  movabs $0x6262622a2a61622a,%rax
  movaps %xmm0,0x20(%rsp)
  movdqa D13(%rip),%xmm0
  mov    %rax,0x6(%rsp)
  xor    %eax,%eax
  movb   $0x0,0x42(%rsp)
  mov    %dx,0xe(%rsp)
  movaps %xmm0,0x30(%rsp)
  call   <isMatch@plt>
  lea    D37(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x58,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D29
type_infered: byte[16]
name: disassemble_data
arguments: tag: D23
type_infered: byte[16]
name: disassemble_data
arguments: tag: D13
type_infered: byte[16]
name: disassemble_data
arguments: tag: D37
type_infered: string

tool: {"D29": [0x62,0x61,0x62,0x61,0x61,0x62,0x62,0x62,0x62,0x62,0x61,0x61,0x61,0x61,0x61,0x62]}

tool: {"D23": [0x62,0x61,0x61,0x62,0x61,0x62,0x62,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x62,0x62]}

tool: {"D13": [0x61,0x61,0x62,0x61,0x61,0x62,0x62,0x62,0x61,0x62,0x62,0x61,0x61,0x62,0x62,0x62]}

tool: D37: %d


assistant: ```c
int main(int argc, char *argv[]) {
  char s[] = {"babaabbbbbaaaaabbaababbaaaaaaabbaabaabbbabbaabbbbb"};
  char p[] = {"*ba**bbbb"};
  printf("%d\n", isMatch(s, p));
}
```