ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    $0x2f,%esi
  mov    (%rbx),%rdi
  call   <strrchr@plt>
  lea    0x1(%rax),%r12
  test   %rax,%rax
  je     L25
L62:
  mov    $0x2e,%esi
  mov    %r12,%rdi
  call   <strchr@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jne    L99
  mov    %r12,%rdi
  call   <strlen@plt>
  movslq %eax,%rbx
  add    %r12,%rbx
L99:
  sub    %r12,%rbx
  lea    0x1(%rbx),%edi
  call   <OPENSSL_malloc@plt>
  mov    %ebx,%edx
  mov    %r12,%rsi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <strncpy@plt>
  movb   $0x0,0x0(%rbp,%rbx,1)
L5:
  movzbl 0x0(%rbp),%edi
  mov    %rbp,%rbx
  test   %dil,%dil
  je     L46
  nopl   0x0(%rax)
L33:
  call   <islower@plt>
  test   %rax,%rax
  jne    L28
  movzbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L33
L46:
  lea    D14(%rip),%rsi
  mov    %rbp,%rdi
  call   <strstr@plt>
  cmp    %rbx,%rax
  jbe    L92
  cmpb   $0x5f,-0x1(%rax)
  je     L93
L92:
  movb   $0x0,(%rax)
  mov    %rbp,%rsi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <OPENSSL_free@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L28:
  movzbl (%rbx),%edi
  add    $0x1,%rbx
  call   <toupper@plt>
  mov    %al,-0x1(%rbx)
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L33
  jmp    L46
L93:
  sub    $0x1,%rax
  jmp    L92
L25:
  mov    (%rbx),%rdi
  mov    $0x5c,%esi
  call   <strrchr@plt>
  test   %rax,%rax
  jne    L23
  mov    (%rbx),%r12
  test   %r12,%r12
  jne    L62
  lea    D10(%rip),%rdi
  call   <BUF_strdup@plt>
  mov    %rax,%rbp
  jmp    L5
L23:
  lea    0x1(%rax),%r12
  jmp    L62
```

tool_calls: name: retrieve_data
arguments: data_label: D14
type_guessed: string
name: retrieve_data
arguments: data_label: D4
type_guessed: string
name: retrieve_data
arguments: data_label: D10
type_guessed: string

tool: D14: TEST

tool: D4: No %s support


tool: D10: (unknown)

assistant: ```c
int main(int argc, char *argv[]) {
  char *p, *q = 0, *program;
  p = strrchr(argv[0], '/');
  if (!p)
    p = strrchr(argv[0], '\\');
  if (p)
    p++;
  if (!p)
    p = argv[0];
  if (p)
    q = strchr(p, '.');
  if (p && !q)
    q = p + strlen(p);
  if (!p)
    program = BUF_strdup("(unknown)");
  else {
    program = OPENSSL_malloc((q - p) + 1);
    strncpy(program, p, q - p);
    program[q - p] = '\0';
  }
  for (p = program; *p; p++)
    if (islower((unsigned char)(*p)))
      *p = toupper((unsigned char)(*p));
  q = strstr(program, "TEST");
  if (q > p && q[-1] == '_')
    q--;
  *q = '\0';
  printf("No %s support\n", program);
  OPENSSL_free(program);
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<bin2dec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L33
L48:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jg     L61
  cmpl   $0x0,-0x4(%rbp)
  jns    L25
L61:
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0xffffffff,%eax
  jmp    L41
L25:
  movl   $0x1,-0x18(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L6
L49:
  shll   -0x18(%rbp)
  subl   $0x1,-0x10(%rbp)
L6:
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jg     L49
  mov    -0x4(%rbp),%eax
  imul   -0x18(%rbp),%eax
  add    %eax,-0xc(%rbp)
  addl   $0x1,-0x14(%rbp)
L33:
  mov    -0x14(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L48
  mov    -0xc(%rbp),%eax
L41:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D7
datatype: string

tool: D7: 

 ERROR! BINARY has only 1 and 0!


assistant: ```c
int bin2dec(char *bin) {
  int b, k, m, n;
  int len, sum = 0;
  len = strlen(bin) - 1;
  for (k = 0; k <= len; k++) {
    n = (bin[k] - '0');
    if ((n > 1) || (n < 0)) {
      puts("\n\n ERROR! BINARY has only 1 and 0!\n");
      return (-1);
    }
    for (b = 1, m = len; m > k; m--) {
      b *= 2;
    }
    sum = sum + n * b;
  }
  return (sum);
}
```
user: What is the c source code of the assembly code below:

```asm
<bin2dec>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%r9d
  sub    $0x1,%r9d
  js     L74
  lea    -0x1(%rax),%r10d
  mov    $0x0,%edi
  mov    $0x0,%r8d
  jmp    L37
L50:
  lea    D3(%rip),%rdi
  call   <puts@plt>
  mov    $0xffffffff,%r8d
L38:
  mov    %r8d,%eax
  pop    %rbx
  ret
L55:
  imul   %edx,%esi
  add    %esi,%r8d
  lea    0x1(%rdi),%rax
  cmp    %r10,%rdi
  je     L38
  mov    %rax,%rdi
L37:
  movsbl (%rbx,%rdi,1),%esi
  sub    $0x30,%esi
  cmp    $0x1,%esi
  ja     L50
  mov    %edi,%ecx
  mov    %r9d,%eax
  mov    $0x1,%edx
  cmp    %edi,%r9d
  jle    L55
L77:
  add    %edx,%edx
  sub    $0x1,%eax
  cmp    %ecx,%eax
  jne    L77
  jmp    L55
L74:
  mov    $0x0,%r8d
  jmp    L38
```

tool_calls: name: unpack_data
arguments: data_label: D3
type_guessed: string

tool: D3: 

 ERROR! BINARY has only 1 and 0!


assistant: ```c
int bin2dec(char *bin) {
  int b, k, m, n;
  int len, sum = 0;
  len = strlen(bin) - 1;
  for (k = 0; k <= len; k++) {
    n = (bin[k] - '0');
    if ((n > 1) || (n < 0)) {
      puts("\n\n ERROR! BINARY has only 1 and 0!\n");
      return (-1);
    }
    for (b = 1, m = len; m > k; m--) {
      b *= 2;
    }
    sum = sum + n * b;
  }
  return (sum);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bin2dec>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  sub    $0x1,%eax
  js     L60
  xor    %r9d,%r9d
  mov    %eax,%edi
  xor    %r10d,%r10d
  movsbl (%rbx,%r9,1),%r8d
  sub    $0x30,%r8d
  cmp    $0x1,%r8d
  ja     L47
  nopl   0x0(%rax)
L3:
  mov    %r9d,%esi
  mov    %eax,%edx
  mov    $0x1,%ecx
  cmp    %r9d,%eax
  jle    L7
  nop
L23:
  sub    $0x1,%edx
  add    %ecx,%ecx
  cmp    %esi,%edx
  jne    L23
  imul   %ecx,%r8d
L7:
  add    %r8d,%r10d
  lea    0x1(%r9),%rdx
  cmp    %r9,%rdi
  je     L43
  mov    %rdx,%r9
  movsbl (%rbx,%r9,1),%r8d
  sub    $0x30,%r8d
  cmp    $0x1,%r8d
  jbe    L3
L47:
  lea    D0(%rip),%rdi
  call   <puts@plt>
  mov    $0xffffffff,%r10d
L43:
  mov    %r10d,%eax
  pop    %rbx
  ret
L60:
  xor    %r10d,%r10d
  jmp    L43
```

tool_calls: name: disassemble_data
arguments: tag: D0
type_guessed: string

tool: D0: 

 ERROR! BINARY has only 1 and 0!


assistant: ```c
int bin2dec(char *bin) {
  int b, k, m, n;
  int len, sum = 0;
  len = strlen(bin) - 1;
  for (k = 0; k <= len; k++) {
    n = (bin[k] - '0');
    if ((n > 1) || (n < 0)) {
      puts("\n\n ERROR! BINARY has only 1 and 0!\n");
      return (-1);
    }
    for (b = 1, m = len; m > k; m--) {
      b *= 2;
    }
    sum = sum + n * b;
  }
  return (sum);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bin2dec>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%r10d
  sub    $0x1,%r10d
  js     L15
  xor    %esi,%esi
  mov    %eax,%r9d
  mov    %r10d,%edi
  mov    %r10d,%r11d
  movsbl (%rbx,%rsi,1),%ecx
  xor    %r8d,%r8d
  sub    $0x30,%ecx
  cmp    $0x1,%ecx
  ja     L29
  nopl   0x0(%rax)
L24:
  cmp    %esi,%r10d
  jle    L53
  mov    %r9d,%eax
  sub    %esi,%eax
  cmp    $0x2,%eax
  je     L31
  movq   D2(%rip),%xmm0
  mov    %edi,%edx
  xor    %eax,%eax
  shr    %edx
  nopl   (%rax)
L17:
  add    $0x1,%eax
  pslld  $0x1,%xmm0
  cmp    %edx,%eax
  jne    L17
  movd   %xmm0,%eax
  pshufd $0xe5,%xmm0,%xmm1
  movd   %xmm1,%edx
  imul   %edx,%eax
  test   $0x1,%dil
  je     L47
L20:
  add    %eax,%eax
L47:
  imul   %eax,%ecx
L53:
  add    %ecx,%r8d
  lea    0x1(%rsi),%rax
  sub    $0x1,%edi
  cmp    %rsi,%r11
  je     L82
  mov    %rax,%rsi
  movsbl (%rbx,%rsi,1),%ecx
  sub    $0x30,%ecx
  cmp    $0x1,%ecx
  jbe    L24
L29:
  lea    D3(%rip),%rdi
  call   <puts@plt>
  mov    $0xffffffff,%r8d
L82:
  mov    %r8d,%eax
  pop    %rbx
  ret
L31:
  mov    $0x1,%eax
  jmp    L20
L15:
  xor    %r8d,%r8d
  jmp    L82
```

tool_calls: name: unpack_data
arguments: label: D2
datatype: byte[8]
name: unpack_data
arguments: label: D3
datatype: string

tool: {"D2": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: D3: 

 ERROR! BINARY has only 1 and 0!


assistant: ```c
int bin2dec(char *bin) {
  int b, k, m, n;
  int len, sum = 0;
  len = strlen(bin) - 1;
  for (k = 0; k <= len; k++) {
    n = (bin[k] - '0');
    if ((n > 1) || (n < 0)) {
      puts("\n\n ERROR! BINARY has only 1 and 0!\n");
      return (-1);
    }
    for (b = 1, m = len; m > k; m--) {
      b *= 2;
    }
    sum = sum + n * b;
  }
  return (sum);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  movq   $0x2a,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  jmp    L5
L9:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x10(%rbp),%edx
  add    $0x1,%edx
  imul   %edx,%eax
  cltq
  add    %rax,-0x8(%rbp)
  addl   $0x1,-0x10(%rbp)
L5:
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jg     L9
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
long hash(char *word) {
  int counter = 0, lenght;
  long res = 42;
  lenght = strlen(word);
  while (counter <= lenght - 1) {
    res = res + word[counter] * (counter + 1);
    counter++;
  }
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<hash>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L13
  lea    0x1(%rax),%esi
  mov    $0x1,%edx
  mov    $0x2a,%ecx
L20:
  movsbl -0x1(%rbx,%rdx,1),%eax
  imul   %edx,%eax
  cltq
  add    %rax,%rcx
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L20
L17:
  mov    %rcx,%rax
  pop    %rbx
  ret
L13:
  mov    $0x2a,%ecx
  jmp    L17
```

assistant: ```c
long hash(char *word) {
  int counter = 0, lenght;
  long res = 42;
  lenght = strlen(word);
  while (counter <= lenght - 1) {
    res = res + word[counter] * (counter + 1);
    counter++;
  }
  return res;
}
```
user: Translate the below assembly code into C programming language:

```asm
<hash>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L7
  lea    0x1(%rax),%ecx
  mov    $0x1,%edx
  mov    $0x2a,%r8d
  nop
L1:
  movsbl -0x1(%rbx,%rdx,1),%eax
  imul   %edx,%eax
  add    $0x1,%rdx
  cltq
  add    %rax,%r8
  cmp    %rdx,%rcx
  jne    L1
  mov    %r8,%rax
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  mov    $0x2a,%r8d
  pop    %rbx
  mov    %r8,%rax
  ret
```

assistant: ```c
long hash(char *word) {
  int counter = 0, lenght;
  long res = 42;
  lenght = strlen(word);
  while (counter <= lenght - 1) {
    res = res + word[counter] * (counter + 1);
    counter++;
  }
  return res;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    -0x8(%rbp),%eax
  imul   $0x191,%eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int a, b;
  a = b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b +
      b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b;
  return a;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  int a, b;
  a = b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b +
      b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b;
  return a;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int a, b;
  a = b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b +
      b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b;
  return a;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int a, b;
  a = b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b +
      b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b + b;
  return a;
}
```
user: Translate the below assembly code into C programming language:

```asm
<nand_write>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x28(%rbp)
  movq   $0x0,-0x10(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %rax,-0x8(%rbp)
  mov    -0x24(%rbp),%eax
  and    $0xfff,%eax
  mov    %eax,-0x14(%rbp)
  mov    $0x0,%eax
  call   <nand_select@plt>
  jmp    L2
L27:
  mov    $0x80,%edi
  call   <nand_cmd@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <nand_send_addr@plt>
  jmp    L37
L11:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x30(%rbp),%rdx
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <nand_send_data@plt>
  addq   $0x1,-0x8(%rbp)
  addl   $0x1,-0x14(%rbp)
L37:
  cmpl   $0xfff,-0x14(%rbp)
  jg     L33
  mov    -0x28(%rbp),%eax
  cmp    %rax,-0x10(%rbp)
  jb     L11
L33:
  mov    $0x10,%edi
  call   <nand_cmd@plt>
  mov    $0x0,%eax
  call   <wait_ready@plt>
  movl   $0x0,-0x14(%rbp)
L2:
  mov    -0x28(%rbp),%eax
  cmp    %rax,-0x10(%rbp)
  jb     L27
  mov    $0x0,%eax
  call   <nand_deselect@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int nand_write(unsigned int nand_start, unsigned char *buf, unsigned int len) {
  unsigned long count = 0;
  unsigned long addr = nand_start;
  int i = nand_start % 4096;
  nand_select();
  while (count < len) {
    nand_cmd(0x80);
    nand_send_addr(addr);
    for (; i < 4096 && count < len; i++) {
      nand_send_data(buf[count++]);
      addr++;
    }
    nand_cmd(0x10);
    wait_ready();
    i = 0;
  }
  nand_deselect();
  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
<nand_write>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r14d
  mov    %rsi,%r15
  mov    %edx,%r12d
  mov    %edi,%r13d
  and    $0xfff,%r14d
  mov    $0x0,%eax
  call   <nand_select@plt>
  mov    %r12d,%r12d
  test   %r12,%r12
  je     L19
  mov    $0x0,%ebp
  jmp    L22
L24:
  mov    %rdx,%r13
  mov    %rbx,%rbp
L56:
  mov    $0x10,%edi
  call   <nand_cmd@plt>
  mov    $0x0,%eax
  call   <wait_ready@plt>
  mov    $0x0,%r14d
  cmp    %rbp,%r12
  jbe    L19
L22:
  mov    $0x80,%edi
  call   <nand_cmd@plt>
  mov    %r13,%rdi
  call   <nand_send_addr@plt>
  cmp    %rbp,%r12
  jbe    L56
  mov    %rbp,%rbx
L60:
  add    $0x1,%rbx
  movzbl -0x1(%r15,%rbx,1),%edi
  call   <nand_send_data@plt>
  lea    (%rbx,%r13,1),%rax
  sub    %rbp,%rax
  mov    %rax,%rdx
  lea    (%r14,%rbx,1),%eax
  sub    %ebp,%eax
  cmp    %r12,%rbx
  jae    L24
  cmp    $0xfff,%eax
  jle    L60
  jmp    L24
L19:
  mov    $0x0,%eax
  call   <nand_deselect@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int nand_write(unsigned int nand_start, unsigned char *buf, unsigned int len) {
  unsigned long count = 0;
  unsigned long addr = nand_start;
  int i = nand_start % 4096;
  nand_select();
  while (count < len) {
    nand_cmd(0x80);
    nand_send_addr(addr);
    for (; i < 4096 && count < len; i++) {
      nand_send_data(buf[count++]);
      addr++;
    }
    nand_cmd(0x10);
    wait_ready();
    i = 0;
  }
  nand_deselect();
  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
<nand_write>:
  endbr64
  push   %r15
  xor    %eax,%eax
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %rbx,%r14
  and    $0xfff,%r14d
  sub    $0x8,%rsp
  call   <nand_select@plt>
  test   %r12,%r12
  je     L37
  xor    %ebp,%ebp
  nopw   0x0(%rax,%rax,1)
L20:
  mov    $0x80,%edi
  mov    %rbp,%r15
  call   <nand_cmd@plt>
  mov    %rbx,%rdi
  call   <nand_send_addr@plt>
  mov    $0xfff,%eax
  sub    %r14d,%eax
  lea    0x1(%rbp,%rax,1),%rbp
  nopw   0x0(%rax,%rax,1)
L18:
  cmp    %r15,%r12
  jbe    L28
  add    $0x1,%r15
  add    $0x1,%rbx
  movzbl -0x1(%r13,%r15,1),%edi
  call   <nand_send_data@plt>
  cmp    %r15,%rbp
  jne    L18
  mov    $0x10,%edi
  xor    %r14d,%r14d
  call   <nand_cmd@plt>
  xor    %eax,%eax
  call   <wait_ready@plt>
  cmp    %r12,%rbp
  jb     L20
L37:
  xor    %eax,%eax
  call   <nand_deselect@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L28:
  mov    $0x10,%edi
  call   <nand_cmd@plt>
  xor    %eax,%eax
  call   <wait_ready@plt>
  jmp    L37
```

assistant: ```c
int nand_write(unsigned int nand_start, unsigned char *buf, unsigned int len) {
  unsigned long count = 0;
  unsigned long addr = nand_start;
  int i = nand_start % 4096;
  nand_select();
  while (count < len) {
    nand_cmd(0x80);
    nand_send_addr(addr);
    for (; i < 4096 && count < len; i++) {
      nand_send_data(buf[count++]);
      addr++;
    }
    nand_cmd(0x10);
    wait_ready();
    i = 0;
  }
  nand_deselect();
  return 0;
}
```