ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<md5>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  mov    $0x21,%edi
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rbp
  lea    0x10(%rsp),%rbx
  call   <malloc@plt>
  mov    %rbp,%rdi
  mov    %rax,%r13
  call   <MD5_Init@plt>
  mov    %r14,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <MD5_Update@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    %r13,%rbp
  call   <MD5_Final@plt>
  lea    0x20(%rsp),%r14
  lea    D5(%rip),%r12
  nopl   (%rax)
L19:
  movzbl (%rbx),%ecx
  mov    %rbp,%rdi
  mov    %r12,%rdx
  mov    $0x3,%esi
  add    $0x1,%rbx
  add    $0x2,%rbp
  call   <snprintf@plt>
  cmp    %r14,%rbx
  jne    L19
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x30,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D5
type_infered: string

tool: D5: %02x

assistant: ```c
char *md5(const char *str, size_t len) {
  MD5_CTX ctx;
  unsigned char digest[16];
  char *out = (char *)malloc(33);
  MD5_Init(&ctx);
  MD5_Update(&ctx, str, len);
  MD5_Final(digest, &ctx);
  for (int i = 0; i < 16; i++) {
    snprintf(&(out[i * 2]), 3, "%02x", (unsigned int)digest[i]);
  }
  return out;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<md5>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  mov    $0x21,%edi
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rbp
  lea    0x10(%rsp),%rbx
  call   <malloc@plt>
  mov    %rbp,%rdi
  mov    %rax,%r13
  call   <MD5_Init@plt>
  mov    %r14,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <MD5_Update@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    %r13,%rbp
  call   <MD5_Final@plt>
  lea    0x20(%rsp),%r14
  lea    D7(%rip),%r12
  nopl   (%rax)
L7:
  movzbl (%rbx),%ecx
  mov    %rbp,%rdi
  mov    %r12,%rdx
  mov    $0x3,%esi
  add    $0x1,%rbx
  add    $0x2,%rbp
  call   <snprintf@plt>
  cmp    %r14,%rbx
  jne    L7
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x30,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
type_infered: string

tool: D7: %02x

assistant: ```c
char *md5(const char *str, size_t len) {
  MD5_CTX ctx;
  unsigned char digest[16];
  char *out = (char *)malloc(33);
  MD5_Init(&ctx);
  MD5_Update(&ctx, str, len);
  MD5_Final(digest, &ctx);
  for (int i = 0; i < 16; i++) {
    snprintf(&(out[i * 2]), 3, "%02x", (unsigned int)digest[i]);
  }
  return out;
}
```
user: What is the c source code of the assembly code below:

```asm
<isfirst>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    $0x50,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x1c(%rbp)
  mov    -0x28(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x10(%rbp)
  jmp    L25
L38:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L47
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L34
  cmpl   $0x0,-0x1c(%rbp)
  jne    L47
L34:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x1c(%rbp)
L47:
  addq   $0x1,-0x28(%rbp)
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x10(%rbp)
L25:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L38
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *isfirst(char *str) {
  char *ret = malloc(80);
  int toknum = 0;
  char *next = str + 1;
  char *prev = str - 1;
  while (*str != '\0') {
    if (!isspace(*str) && (isspace(*prev) || toknum == 0)) {
      ret[toknum] = *str;
      toknum++;
    }
    str++;
    next++;
    prev++;
  }
  ret[toknum] = '\0';
  return ret;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<isfirst>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x50,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L35
  mov    $0x0,%ebp
  jmp    L25
L14:
  movzbl (%rbx),%edx
  movslq %ebp,%rax
  mov    %dl,(%r12,%rax,1)
  add    $0x1,%ebp
L31:
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L9
L25:
  movsbl %dil,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L31
  movsbl -0x1(%rbx),%edi
  call   <isspace@plt>
  test   %ebp,%ebp
  je     L14
  test   %rax,%rax
  jne    L14
  jmp    L31
L35:
  mov    $0x0,%ebp
L9:
  movslq %ebp,%rbp
  movb   $0x0,(%r12,%rbp,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *isfirst(char *str) {
  char *ret = malloc(80);
  int toknum = 0;
  char *next = str + 1;
  char *prev = str - 1;
  while (*str != '\0') {
    if (!isspace(*str) && (isspace(*prev) || toknum == 0)) {
      ret[toknum] = *str;
      toknum++;
    }
    str++;
    next++;
    prev++;
  }
  ret[toknum] = '\0';
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<isfirst>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x50,%edi
  call   <malloc@plt>
  movsbl (%rbx),%edi
  mov    %rax,%r12
  test   %dil,%dil
  je     L30
  xor    %ebp,%ebp
  jmp    L14
  nopl   0x0(%rax)
L64:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L24
L14:
  call   <isspace@plt>
  test   %rax,%rax
  jne    L64
  movsbl -0x1(%rbx),%edi
  call   <isspace@plt>
  test   %ebp,%ebp
  je     L35
  test   %rax,%rax
  je     L64
L35:
  movzbl (%rbx),%edx
  movslq %ebp,%rax
  add    $0x1,%rbx
  add    $0x1,%ebp
  mov    %dl,(%r12,%rax,1)
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L14
L24:
  movslq %ebp,%rbp
  mov    %r12,%rax
  add    %r12,%rbp
  movb   $0x0,0x0(%rbp)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L30:
  mov    %rax,%rbp
  mov    %r12,%rax
  movb   $0x0,0x0(%rbp)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *isfirst(char *str) {
  char *ret = malloc(80);
  int toknum = 0;
  char *next = str + 1;
  char *prev = str - 1;
  while (*str != '\0') {
    if (!isspace(*str) && (isspace(*prev) || toknum == 0)) {
      ret[toknum] = *str;
      toknum++;
    }
    str++;
    next++;
    prev++;
  }
  ret[toknum] = '\0';
  return ret;
}
```
user: What is the c source code of the assembly code below:

```asm
<isfirst>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x50,%edi
  call   <malloc@plt>
  movsbl (%rbx),%edi
  mov    %rax,%r12
  test   %dil,%dil
  je     L37
  xor    %ebp,%ebp
  jmp    L60
  nopl   0x0(%rax)
L8:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L11
L60:
  call   <isspace@plt>
  test   %rax,%rax
  jne    L8
  movsbl -0x1(%rbx),%edi
  call   <isspace@plt>
  test   %ebp,%ebp
  je     L18
  test   %rax,%rax
  je     L8
L18:
  movzbl (%rbx),%edx
  movslq %ebp,%rax
  add    $0x1,%rbx
  add    $0x1,%ebp
  mov    %dl,(%r12,%rax,1)
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L60
L11:
  movslq %ebp,%rbp
  mov    %r12,%rax
  add    %r12,%rbp
  movb   $0x0,0x0(%rbp)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L37:
  mov    %rax,%rbp
  mov    %r12,%rax
  movb   $0x0,0x0(%rbp)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *isfirst(char *str) {
  char *ret = malloc(80);
  int toknum = 0;
  char *next = str + 1;
  char *prev = str - 1;
  while (*str != '\0') {
    if (!isspace(*str) && (isspace(*prev) || toknum == 0)) {
      ret[toknum] = *str;
      toknum++;
    }
    str++;
    next++;
    prev++;
  }
  ret[toknum] = '\0';
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<print_localpage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  lea    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <get_file_content@plt>
  test   %eax,%eax
  jne    L15
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <parse_tldrpage@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  jmp    L2
L15:
  mov    $0x0,%eax
L2:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
int print_localpage(char const *path) {
  char *output = NULL;
  if (!get_file_content(path, &output, 0)) {
    parse_tldrpage(output);
    free(output);
    return 0;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<print_localpage>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsp,%rsi
  mov    $0x0,%edx
  call   <get_file_content@plt>
  test   %eax,%eax
  je     L10
L2:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L10:
  mov    (%rsp),%rdi
  call   <parse_tldrpage@plt>
  mov    (%rsp),%rdi
  call   <free@plt>
  jmp    L2
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int print_localpage(char const *path) {
  char *output = NULL;
  if (!get_file_content(path, &output, 0)) {
    parse_tldrpage(output);
    free(output);
    return 0;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<print_localpage>:
  endbr64
  sub    $0x18,%rsp
  xor    %edx,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movq   $0x0,(%rsp)
  call   <get_file_content@plt>
  test   %eax,%eax
  je     L7
L28:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
  nopl   (%rax)
L7:
  mov    (%rsp),%rdi
  call   <parse_tldrpage@plt>
  mov    (%rsp),%rdi
  call   <free@plt>
  jmp    L28
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int print_localpage(char const *path) {
  char *output = NULL;
  if (!get_file_content(path, &output, 0)) {
    parse_tldrpage(output);
    free(output);
    return 0;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<print_localpage>:
  endbr64
  sub    $0x18,%rsp
  xor    %edx,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movq   $0x0,(%rsp)
  call   <get_file_content@plt>
  test   %eax,%eax
  je     L22
L14:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
  nopl   (%rax)
L22:
  mov    (%rsp),%rdi
  call   <parse_tldrpage@plt>
  mov    (%rsp),%rdi
  call   <free@plt>
  jmp    L14
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int print_localpage(char const *path) {
  char *output = NULL;
  if (!get_file_content(path, &output, 0)) {
    parse_tldrpage(output);
    free(output);
    return 0;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<numDecodings>:
  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
  test   %rdi,%rdi
  je     L75
  mov    %rdi,%r15
  movzbl (%rdi),%eax
  cmp    $0x30,%al
  je     L125
  test   %al,%al
  je     L125
  movl   $0x1,(%rsp)
  movl   $0x1,0x4(%rsp)
  movzbl 0x1(%rdi),%eax
  mov    $0x0,%ebp
  test   %al,%al
  jne    L19
  mov    $0x1,%r14d
L114:
  sub    $0x1,%r14d
  mov    %r14d,%eax
  shr    $0x1f,%eax
  add    %eax,%r14d
  and    $0x1,%r14d
  sub    %eax,%r14d
  movslq %r14d,%rax
  mov    (%rsp,%rax,4),%r13d
  jmp    L44
L49:
  mov    %ebp,%edx
  shr    $0x1f,%edx
  lea    (%rdx,%rbp,1),%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cltq
  mov    (%rsp,%rax,4),%edx
  mov    %ebx,%eax
  shr    $0x1f,%eax
  add    %eax,%ebx
  and    $0x1,%ebx
  sub    %eax,%ebx
  movslq %ebx,%rbx
  mov    %edx,(%rsp,%rbx,4)
  jmp    L112
L76:
  mov    %ebx,%eax
  shr    $0x1f,%eax
  add    %eax,%ebx
  and    $0x1,%ebx
  sub    %eax,%ebx
  movslq %ebx,%rbx
  mov    %ebp,%edx
  shr    $0x1f,%edx
  lea    (%rdx,%rbp,1),%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cltq
  mov    (%rsp,%rax,4),%eax
  add    %eax,(%rsp,%rbx,4)
L112:
  add    $0x2,%r14d
  add    $0x1,%rbp
  movzbl 0x1(%r15,%rbp,1),%eax
  test   %al,%al
  je     L114
L19:
  mov    %ebp,%r14d
  lea    0x1(%rbp),%ebx
  movsbl (%r15,%rbp,1),%edx
  lea    -0xf0(%rdx,%rdx,4),%edx
  movsbl %al,%eax
  lea    (%rax,%rdx,2),%r12d
  lea    -0x30(%r12),%r13d
  mov    %r13d,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  test   %r13d,%r13d
  je     L44
  sub    $0x3a,%r12d
  cmp    $0x10,%r12d
  ja     L49
  cmp    $0xa,%r13d
  je     L112
  cmp    $0x14,%r13d
  jne    L76
  jmp    L112
L75:
  mov    $0x0,%r13d
  jmp    L44
L125:
  mov    $0x0,%r13d
L44:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  mov    %r13d,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D0
datatype: string

tool: D0: %d


assistant: ```c
int numDecodings(char *s) {
  if (!s || !*s)
    return 0;
  if (*s == '0')
    return 0;
  int result[2] = {1, 1};
  for (int i = 1;; ++i) {
    if (!s[i])
      return result[(i - 1) % 2];
    int tmp = (s[i - 1] - '0') * 10 + s[i] - '0';
    printf("%d\n", tmp);
    if (tmp == 0)
      return 0;
    if (tmp < 10 || tmp > 26) {
      result[i % 2] = result[(i - 1) % 2];
      continue;
    }
    if (tmp != 10 && tmp != 20)
      result[i % 2] += result[(i - 1) % 2];
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<numDecodings>:
  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
  test   %rdi,%rdi
  je     L17
  movsbl (%rdi),%eax
  mov    %rdi,%r15
  cmp    $0x30,%al
  je     L17
  test   %al,%al
  je     L17
  movsbl 0x1(%rdi),%edx
  xor    %ebx,%ebx
  lea    D5(%rip),%r12
  movabs $0x100000001,%rcx
  mov    %rcx,(%rsp)
  test   %dl,%dl
  jne    L138
  jmp    L95
  nopl   (%rax)
L123:
  cmp    $0xa,%r14d
  je     L92
  cmp    $0x14,%r14d
  je     L92
  mov    %ebp,%eax
  mov    %rbx,%rdx
  and    $0x1,%eax
  and    $0x1,%edx
  mov    (%rsp,%rdx,4),%edx
  add    %edx,(%rsp,%rax,4)
L92:
  add    $0x1,%rbx
  movsbl 0x1(%r15,%rbx,1),%edx
  test   %dl,%dl
  je     L118
L14:
  movsbl (%r15,%rbx,1),%eax
L138:
  lea    -0xf0(%rax,%rax,4),%eax
  mov    %r12,%rdi
  lea    0x1(%rbx),%ebp
  lea    (%rdx,%rax,2),%r13d
  lea    -0x30(%r13),%r14d
  mov    %r14d,%esi
  call   <printf@plt>
  test   %r14d,%r14d
  je     L17
  sub    $0x3a,%r13d
  cmp    $0x10,%r13d
  jbe    L123
  mov    %rbx,%rax
  add    $0x1,%rbx
  and    $0x1,%eax
  mov    (%rsp,%rax,4),%edx
  mov    %ebp,%eax
  and    $0x1,%eax
  mov    %edx,(%rsp,%rax,4)
  movsbl 0x1(%r15,%rbx,1),%edx
  test   %dl,%dl
  jne    L14
L118:
  and    $0x1,%ebp
  mov    (%rsp,%rbp,4),%eax
  jmp    L35
  nopl   0x0(%rax,%rax,1)
L17:
  xor    %eax,%eax
L35:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L82
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L95:
  mov    $0x1,%eax
  jmp    L35
L82:
  call   <__stack_chk_fail@plt>
```

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

tool: D5: %d


assistant: ```c
int numDecodings(char *s) {
  if (!s || !*s)
    return 0;
  if (*s == '0')
    return 0;
  int result[2] = {1, 1};
  for (int i = 1;; ++i) {
    if (!s[i])
      return result[(i - 1) % 2];
    int tmp = (s[i - 1] - '0') * 10 + s[i] - '0';
    printf("%d\n", tmp);
    if (tmp == 0)
      return 0;
    if (tmp < 10 || tmp > 26) {
      result[i % 2] = result[(i - 1) % 2];
      continue;
    }
    if (tmp != 10 && tmp != 20)
      result[i % 2] += result[(i - 1) % 2];
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<numDecodings>:
  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
  test   %rdi,%rdi
  je     L2
  movzbl (%rdi),%eax
  mov    %rdi,%r15
  cmp    $0x30,%al
  je     L2
  test   %al,%al
  je     L2
  mov    D7(%rip),%rax
  xor    %ebx,%ebx
  lea    D17(%rip),%r13
  mov    %rax,(%rsp)
  movsbl 0x1(%rdi),%eax
  test   %al,%al
  jne    L121
  jmp    L107
  nopw   0x0(%rax,%rax,1)
L77:
  cmp    $0xa,%r14d
  je     L52
  cmp    $0x14,%r14d
  je     L52
  mov    %r12d,%eax
  mov    %rbx,%rdx
  and    $0x1,%eax
  and    $0x1,%edx
  mov    (%rsp,%rdx,4),%edx
  add    %edx,(%rsp,%rax,4)
L52:
  add    $0x1,%rbx
  movsbl 0x1(%r15,%rbx,1),%eax
  test   %al,%al
  je     L73
L121:
  movsbl (%r15,%rbx,1),%edx
  mov    %r13,%rdi
  lea    0x1(%rbx),%r12d
  lea    -0xf0(%rdx,%rdx,4),%edx
  lea    (%rax,%rdx,2),%ebp
  lea    -0x30(%rbp),%r14d
  mov    %r14d,%esi
  call   <printf@plt>
  test   %r14d,%r14d
  je     L2
  sub    $0x3a,%ebp
  cmp    $0x10,%ebp
  jbe    L77
  mov    %rbx,%rax
  add    $0x1,%rbx
  and    $0x1,%eax
  mov    (%rsp,%rax,4),%edx
  mov    %r12d,%eax
  and    $0x1,%eax
  mov    %edx,(%rsp,%rax,4)
  movsbl 0x1(%r15,%rbx,1),%eax
  test   %al,%al
  jne    L121
L73:
  and    $0x1,%r12d
  mov    (%rsp,%r12,4),%eax
  jmp    L101
  nopl   (%rax)
L2:
  xor    %eax,%eax
L101:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L97
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L107:
  mov    $0x1,%eax
  jmp    L101
L97:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D7
data_type: byte[8]
name: read_data
arguments: data_label: D17
data_type: string

tool: {"D7": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: D17: %d


assistant: ```c
int numDecodings(char *s) {
  if (!s || !*s)
    return 0;
  if (*s == '0')
    return 0;
  int result[2] = {1, 1};
  for (int i = 1;; ++i) {
    if (!s[i])
      return result[(i - 1) % 2];
    int tmp = (s[i - 1] - '0') * 10 + s[i] - '0';
    printf("%d\n", tmp);
    if (tmp == 0)
      return 0;
    if (tmp < 10 || tmp > 26) {
      result[i % 2] = result[(i - 1) % 2];
      continue;
    }
    if (tmp != 10 && tmp != 20)
      result[i % 2] += result[(i - 1) % 2];
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x340,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x338(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x334(%rbp)
  jmp    L15
L45:
  lea    -0x330(%rbp),%rdx
  mov    -0x334(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x334(%rbp)
L15:
  mov    -0x338(%rbp),%eax
  cmp    %eax,-0x334(%rbp)
  jl     L45
  mov    -0x338(%rbp),%edx
  lea    -0x1a0(%rbp),%rcx
  lea    -0x330(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <derivative@plt>
  movl   $0x0,-0x334(%rbp)
  jmp    L23
L47:
  mov    -0x334(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x334(%rbp)
L23:
  mov    -0x338(%rbp),%eax
  cmp    %eax,-0x334(%rbp)
  jl     L47
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D26
type_guessed: string
name: retrieve_data
arguments: tag: D0
type_guessed: string

tool: D26: %d

tool: D0: %d 

assistant: ```c
int main() {
  int arr[100], result[100], i, N;
  scanf("%d", &N);
  for (i = 0; i < N; i++) {
    scanf("%d", &arr[i]);
  }
  derivative(arr, result, N);
  for (i = 0; i < N; i++) {
    printf("%d ", result[i]);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x340,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x338(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D29(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%edx
  test   %edx,%edx
  jle    L38
  lea    0x10(%rsp),%rbp
  mov    $0x0,%ebx
  lea    D29(%rip),%r12
L30:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  mov    0xc(%rsp),%edx
  add    $0x4,%rbp
  cmp    %ebx,%edx
  jg     L30
L38:
  lea    0x1a0(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  call   <derivative@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L45
  mov    $0x0,%ebx
  lea    D6(%rip),%rbp
L43:
  mov    0x1a0(%rsp,%rbx,4),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %ebx,0xc(%rsp)
  jg     L43
L45:
  mov    0x338(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  mov    $0x0,%eax
  add    $0x340,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D29
type: string
name: parse_data
arguments: tag: D6
type: string

tool: D29: %d

tool: D6: %d 

assistant: ```c
int main() {
  int arr[100], result[100], i, N;
  scanf("%d", &N);
  for (i = 0; i < N; i++) {
    scanf("%d", &arr[i]);
  }
  derivative(arr, result, N);
  for (i = 0; i < N; i++) {
    printf("%d ", result[i]);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  lea    D13(%rip),%r12
  push   %rbp
  mov    %r12,%rdi
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x348,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x338(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    0x10(%rsp),%r13
  call   <scanf@plt>
  mov    0xc(%rsp),%edx
  mov    %r13,%rbp
  test   %edx,%edx
  jle    L12
  nopw   0x0(%rax,%rax,1)
L20:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <scanf@plt>
  mov    0xc(%rsp),%edx
  cmp    %ebx,%edx
  jg     L20
L12:
  lea    0x1a0(%rsp),%rbp
  mov    %r13,%rdi
  mov    %rbp,%rsi
  call   <derivative@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L10
  xor    %ebx,%ebx
  lea    D2(%rip),%r12
  xchg   %ax,%ax
L6:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jg     L6
L10:
  mov    0x338(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x348,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D13
type_infered: string
name: parse_data
arguments: data_label: D2
type_infered: string

tool: D13: %d

tool: D2: %d 

assistant: ```c
int main() {
  int arr[100], result[100], i, N;
  scanf("%d", &N);
  for (i = 0; i < N; i++) {
    scanf("%d", &arr[i]);
  }
  derivative(arr, result, N);
  for (i = 0; i < N; i++) {
    printf("%d ", result[i]);
  }
  return 0;
}
```