ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movq   $0x0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movb   $0xa,-0x1a(%rbp)
  jmp    L53
L18:
  addq   $0x1,-0x8(%rbp)
  movsbl -0x19(%rbp),%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L44
  cmpl   $0x0,-0x10(%rbp)
  jne    L44
  movl   $0x1,-0x10(%rbp)
  addl   $0x1,-0x18(%rbp)
L44:
  movsbl -0x19(%rbp),%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L32
  cmpl   $0x0,-0x10(%rbp)
  je     L32
  movl   $0x0,-0x10(%rbp)
L32:
  movzbl -0x19(%rbp),%eax
  mov    %al,-0x1a(%rbp)
L53:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x19(%rbp)
  cmpb   $0x7c,-0x19(%rbp)
  jne    L18
  cmpb   $0xa,-0x1a(%rbp)
  je     L41
  movl   $0x1,-0x14(%rbp)
L41:
  mov    -0xc(%rbp),%ecx
  mov    -0x18(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D22
type_infered: string
name: parse_data
arguments: data_tag: D5
type_infered: string
name: parse_data
arguments: data_tag: D25
type_infered: string

tool: D22: Enter text to analyze (| to terminate):

tool: D5: characters = %ld, words = %d, lines = %d, 

tool: D25: partial lines = %d


assistant: ```c
int main(void) {
  char c;
  char prev;
  long n_chars = 0L;
  int n_lines = 0;
  int n_words = 0;
  int p_lines = 0;
  bool inword = false;
  printf("Enter text to analyze (| to terminate):\n");
  prev = '\n';
  while ((c = getchar()) != '|') {
    n_chars++;
    if (!isspace(c) && !inword) {
      inword = true;
      n_words++;
    }
    if (isspace(c) && inword)
      inword = false;
    prev = c;
  }
  if (prev != '\n')
    p_lines = 1;
  printf("characters = %ld, words = %d, lines = %d, ", n_chars, n_words, n_lines);
  printf("partial lines = %d\n", p_lines);
  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
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%ebp
  mov    $0x0,%r14d
  mov    $0x0,%r12d
  mov    $0xa,%r15d
  mov    $0x0,%r13d
  jmp    L7
L32:
  mov    %r15d,%edi
  call   <isspace@plt>
  test   %ebp,%ebp
  je     L14
  test   %rax,%rax
  cmovne %r13d,%ebp
L14:
  mov    %ebx,%r15d
L7:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  cmp    $0x7c,%al
  je     L24
  add    $0x1,%r12
  movsbl %bl,%r15d
  mov    %r15d,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L32
  test   %ebp,%ebp
  jne    L32
  add    $0x1,%r14d
  mov    $0x1,%ebp
  jmp    L32
L24:
  mov    $0x0,%ecx
  mov    %r14d,%edx
  mov    %r12,%rsi
  lea    D23(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0xa,%r15b
  setne  %sil
  movzbl %sil,%esi
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D4
type_infered: string
name: retrieve_data
arguments: tag: D23
type_infered: string
name: retrieve_data
arguments: tag: D2
type_infered: string

tool: D4: Enter text to analyze (| to terminate):

tool: D23: characters = %ld, words = %d, lines = %d, 

tool: D2: partial lines = %d


assistant: ```c
int main(void) {
  char c;
  char prev;
  long n_chars = 0L;
  int n_lines = 0;
  int n_words = 0;
  int p_lines = 0;
  bool inword = false;
  printf("Enter text to analyze (| to terminate):\n");
  prev = '\n';
  while ((c = getchar()) != '|') {
    n_chars++;
    if (!isspace(c) && !inword) {
      inword = true;
      n_words++;
    }
    if (isspace(c) && inword)
      inword = false;
    prev = c;
  }
  if (prev != '\n')
    p_lines = 1;
  printf("characters = %ld, words = %d, lines = %d, ", n_chars, n_words, n_lines);
  printf("partial lines = %d\n", p_lines);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r15
  lea    D21(%rip),%rdi
  xor    %r15d,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    $0xa,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <puts@plt>
  jmp    L25
  nopl   0x0(%rax)
L14:
  movsbl %bl,%ebp
  add    $0x1,%r13
  mov    %ebp,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L10
  test   %r12d,%r12d
  jne    L10
  add    $0x1,%r15d
  mov    $0x1,%r12d
L10:
  mov    %ebp,%edi
  mov    %ebx,%ebp
  call   <isspace@plt>
  test   %rax,%rax
  cmovne %r14d,%r12d
L25:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  cmp    $0x7c,%al
  jne    L14
  xor    %ecx,%ecx
  mov    %r15d,%edx
  mov    %r13,%rsi
  xor    %eax,%eax
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %esi,%esi
  cmp    $0xa,%bpl
  lea    D15(%rip),%rdi
  setne  %sil
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D21
type_guessed: string
name: disassemble_data
arguments: data_tag: D1
type_guessed: string
name: disassemble_data
arguments: data_tag: D15
type_guessed: string

tool: D21: Enter text to analyze (| to terminate):

tool: D1: characters = %ld, words = %d, lines = %d, 

tool: D15: partial lines = %d


assistant: ```c
int main(void) {
  char c;
  char prev;
  long n_chars = 0L;
  int n_lines = 0;
  int n_words = 0;
  int p_lines = 0;
  bool inword = false;
  printf("Enter text to analyze (| to terminate):\n");
  prev = '\n';
  while ((c = getchar()) != '|') {
    n_chars++;
    if (!isspace(c) && !inword) {
      inword = true;
      n_words++;
    }
    if (isspace(c) && inword)
      inword = false;
    prev = c;
  }
  if (prev != '\n')
    p_lines = 1;
  printf("characters = %ld, words = %d, lines = %d, ", n_chars, n_words, n_lines);
  printf("partial lines = %d\n", p_lines);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D21(%rip),%rdi
  xor    %r15d,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    $0xa,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <puts@plt>
  jmp    L14
  nopl   0x0(%rax)
L5:
  movsbl %bl,%ebp
  add    $0x1,%r13
  mov    %ebp,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L37
  test   %r12d,%r12d
  jne    L37
  add    $0x1,%r15d
  mov    $0x1,%r12d
L37:
  mov    %ebp,%edi
  mov    %ebx,%ebp
  call   <isspace@plt>
  test   %rax,%rax
  cmovne %r14d,%r12d
L14:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  cmp    $0x7c,%al
  jne    L5
  xor    %ecx,%ecx
  mov    %r15d,%edx
  mov    %r13,%rsi
  xor    %eax,%eax
  lea    D29(%rip),%rdi
  call   <printf@plt>
  xor    %esi,%esi
  cmp    $0xa,%bpl
  lea    D24(%rip),%rdi
  setne  %sil
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: tag: D21
datatype: string
name: parse_data
arguments: tag: D29
datatype: string
name: parse_data
arguments: tag: D24
datatype: string

tool: D21: Enter text to analyze (| to terminate):

tool: D29: characters = %ld, words = %d, lines = %d, 

tool: D24: partial lines = %d


assistant: ```c
int main(void) {
  char c;
  char prev;
  long n_chars = 0L;
  int n_lines = 0;
  int n_words = 0;
  int p_lines = 0;
  bool inword = false;
  printf("Enter text to analyze (| to terminate):\n");
  prev = '\n';
  while ((c = getchar()) != '|') {
    n_chars++;
    if (!isspace(c) && !inword) {
      inword = true;
      n_words++;
    }
    if (isspace(c) && inword)
      inword = false;
    prev = c;
  }
  if (prev != '\n')
    p_lines = 1;
  printf("characters = %ld, words = %d, lines = %d, ", n_chars, n_words, n_lines);
  printf("partial lines = %d\n", p_lines);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<copy_crimp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  jmp    L39
L35:
  mov    -0x10(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L39:
  mov    -0x14(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x14(%rbp)
  test   %eax,%eax
  jg     L35
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  subq   $0x1,-0x8(%rbp)
  jmp    L15
L17:
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  subq   $0x1,-0x8(%rbp)
L15:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L17
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void copy_crimp(char *dp, char *sp, int len) {
  while (len-- > 0) {
    *dp++ = *sp++;
  }
  *dp = 0;
  for (dp--; *dp == ' '; dp--) {
    *dp = 0;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<copy_crimp>:
  endbr64
  test   %edx,%edx
  jle    L37
  mov    %edx,%r8d
  mov    $0x0,%eax
L25:
  movzbl (%rsi,%rax,1),%ecx
  mov    %cl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %r8,%rax
  jne    L25
  mov    %edx,%edx
  add    %rdx,%rdi
L37:
  movb   $0x0,(%rdi)
  lea    -0x1(%rdi),%rax
  cmpb   $0x20,-0x1(%rdi)
  jne    L35
L5:
  movb   $0x0,(%rax)
  sub    $0x1,%rax
  cmpb   $0x20,(%rax)
  je     L5
L35:
  ret
```

assistant: ```c
void copy_crimp(char *dp, char *sp, int len) {
  while (len-- > 0) {
    *dp++ = *sp++;
  }
  *dp = 0;
  for (dp--; *dp == ' '; dp--) {
    *dp = 0;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<copy_crimp>:
  endbr64
  test   %edx,%edx
  jle    L34
  lea    -0x1(%rdx),%ecx
  xor    %eax,%eax
  mov    %edx,%edx
  nop
L33:
  movzbl (%rsi,%rax,1),%r8d
  mov    %r8b,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L33
  add    %rdi,%rax
  lea    (%rdi,%rcx,1),%rdx
L15:
  movb   $0x0,(%rax)
  cmp    $0x20,%r8b
  jne    L45
  nopw   0x0(%rax,%rax,1)
L40:
  movb   $0x0,(%rdx)
  sub    $0x1,%rdx
  cmpb   $0x20,(%rdx)
  je     L40
L45:
  ret
  nopl   (%rax)
L34:
  movzbl -0x1(%rdi),%r8d
  lea    -0x1(%rdi),%rdx
  mov    %rdi,%rax
  jmp    L15
```

assistant: ```c
void copy_crimp(char *dp, char *sp, int len) {
  while (len-- > 0) {
    *dp++ = *sp++;
  }
  *dp = 0;
  for (dp--; *dp == ' '; dp--) {
    *dp = 0;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<copy_crimp>:
  endbr64
  mov    %rsi,%r9
  movslq %edx,%rsi
  mov    %rdi,%rcx
  lea    -0x1(%rsi),%r10d
  test   %esi,%esi
  jle    L60
  lea    0x1(%r9),%rdi
  mov    %rcx,%rdx
  mov    %r10d,%eax
  sub    %rdi,%rdx
  cmp    $0xe,%rdx
  jbe    L79
  cmp    $0x6,%r10d
  jbe    L79
  cmp    $0xe,%r10d
  jbe    L104
  mov    %esi,%edi
  xor    %edx,%edx
  shr    $0x4,%edi
  shl    $0x4,%rdi
  nop
L129:
  movdqu (%r9,%rdx,1),%xmm0
  movups %xmm0,(%rcx,%rdx,1)
  add    $0x10,%rdx
  cmp    %rdx,%rdi
  jne    L129
  mov    %esi,%edi
  movaps %xmm0,-0x18(%rsp)
  movzbl -0x9(%rsp),%edx
  and    $0xfffffff0,%edi
  mov    %edi,%r8d
  sub    %edi,%r10d
  lea    (%rcx,%r8,1),%r11
  add    %r9,%r8
  cmp    %edi,%esi
  je     L139
L22:
  sub    %edi,%esi
  lea    -0x1(%rsi),%edx
  cmp    $0x6,%edx
  jbe    L1
  mov    (%r9,%rdi,1),%rdx
  mov    %rdx,(%rcx,%rdi,1)
  mov    %esi,%edi
  shr    $0x38,%rdx
  and    $0xfffffff8,%edi
  mov    %edi,%r9d
  sub    %edi,%r10d
  add    %r9,%r11
  add    %r9,%r8
  cmp    %edi,%esi
  je     L139
L1:
  movzbl (%r8),%edx
  mov    %dl,(%r11)
  test   %r10d,%r10d
  jle    L139
  movzbl 0x1(%r8),%edx
  mov    %dl,0x1(%r11)
  cmp    $0x1,%r10d
  jle    L139
  movzbl 0x2(%r8),%edx
  mov    %dl,0x2(%r11)
  cmp    $0x2,%r10d
  jle    L139
  movzbl 0x3(%r8),%edx
  mov    %dl,0x3(%r11)
  cmp    $0x3,%r10d
  jle    L139
  movzbl 0x4(%r8),%edx
  mov    %dl,0x4(%r11)
  cmp    $0x4,%r10d
  jle    L139
  movzbl 0x5(%r8),%edx
  mov    %dl,0x5(%r11)
  cmp    $0x5,%r10d
  jle    L139
  movzbl 0x6(%r8),%edx
  mov    %dl,0x6(%r11)
L139:
  mov    %eax,%eax
  lea    0x1(%rcx,%rax,1),%rsi
  add    %rcx,%rax
L6:
  movb   $0x0,(%rsi)
  cmp    $0x20,%dl
  jne    L54
  nopl   0x0(%rax)
L9:
  movb   $0x0,(%rax)
  sub    $0x1,%rax
  cmpb   $0x20,(%rax)
  je     L9
L54:
  ret
  nopl   (%rax)
L79:
  xor    %edi,%edi
  nopw   0x0(%rax,%rax,1)
L125:
  movzbl (%r9,%rdi,1),%edx
  mov    %dl,(%rcx,%rdi,1)
  add    $0x1,%rdi
  cmp    %rsi,%rdi
  jne    L125
  jmp    L139
  nopl   0x0(%rax,%rax,1)
L60:
  movzbl -0x1(%rdi),%edx
  lea    -0x1(%rdi),%rax
  mov    %rdi,%rsi
  jmp    L6
L104:
  mov    %r9,%r8
  mov    %rcx,%r11
  xor    %edi,%edi
  jmp    L22
```

assistant: ```c
void copy_crimp(char *dp, char *sp, int len) {
  while (len-- > 0) {
    *dp++ = *sp++;
  }
  *dp = 0;
  for (dp--; *dp == ' '; dp--) {
    *dp = 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<timegm>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x46,-0x4(%rbp)
  jmp    L35
L59:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <is_leap@plt>
  test   %rax,%rax
  je     L25
  mov    $0x16e,%eax
  jmp    L51
L25:
  mov    $0x16d,%eax
L51:
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L35:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jl     L59
  movl   $0x0,-0x4(%rbp)
  jmp    L20
L10:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <is_leap@plt>
  mov    %rax,%rdx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  lea    0x0(,%rax,4),%rdx
  lea    D7(%rip),%rax
  mov    (%rdx,%rax,1),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L20:
  mov    -0x18(%rbp),%rax
  mov    0x4(%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jl     L10
  mov    -0x18(%rbp),%rax
  mov    0x8(%rax),%eax
  sub    $0x1,%eax
  add    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  shl    $0x3,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    0x20(%rax),%rax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  imul   $0x3c,%eax,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    0x18(%rax),%rax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  imul   $0x3c,%eax,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    0x10(%rax),%rax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D7
type: string

tool: D7: 

assistant: ```c
time_t timegm(struct tm *tm) {
  static const unsigned ndays[2][12] = {{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}, {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}};
  time_t res = 0;
  int i;
  for (i = 70; i < tm->tm_year; ++i)
    res += is_leap(i) ? 366 : 365;
  for (i = 0; i < tm->tm_mon; ++i)
    res += ndays[is_leap(tm->tm_year)][i];
  res += tm->tm_mday - 1;
  res *= 24;
  res += tm->tm_hour;
  res *= 60;
  res += tm->tm_min;
  res *= 60;
  res += tm->tm_sec;
  return res;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<timegm>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  cmpl   $0x46,(%rdi)
  jle    L41
  mov    $0x46,%r12d
  mov    $0x0,%ebx
L17:
  mov    %r12d,%edi
  call   <is_leap@plt>
  cmp    $0x1,%rax
  sbb    $0xfffffe92,%ebx
  add    $0x1,%r12d
  cmp    %r12d,0x0(%rbp)
  jg     L17
L23:
  cmpl   $0x0,0x4(%rbp)
  jle    L22
  mov    $0x0,%r12d
  lea    D6(%rip),%r13
L7:
  mov    0x0(%rbp),%edi
  call   <is_leap@plt>
  movslq %r12d,%rdx
  lea    (%rax,%rax,2),%rax
  lea    (%rdx,%rax,4),%rax
  add    0x0(%r13,%rax,4),%ebx
  add    $0x1,%r12d
  cmp    %r12d,0x4(%rbp)
  jg     L7
L22:
  mov    0x8(%rbp),%eax
  lea    -0x1(%rbx,%rax,1),%eax
  lea    (%rax,%rax,2),%eax
  shl    $0x3,%eax
  add    0x20(%rbp),%eax
  imul   $0x3c,%eax,%eax
  add    0x18(%rbp),%eax
  imul   $0x3c,%eax,%eax
  add    0x10(%rbp),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L41:
  mov    $0x0,%ebx
  jmp    L23
```

tool_calls: name: unpack_data
arguments: data_label: D6
type_infered: string

tool: D6: 

assistant: ```c
time_t timegm(struct tm *tm) {
  static const unsigned ndays[2][12] = {{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}, {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}};
  time_t res = 0;
  int i;
  for (i = 70; i < tm->tm_year; ++i)
    res += is_leap(i) ? 366 : 365;
  for (i = 0; i < tm->tm_mon; ++i)
    res += ndays[is_leap(tm->tm_year)][i];
  res += tm->tm_mday - 1;
  res *= 24;
  res += tm->tm_hour;
  res *= 60;
  res += tm->tm_min;
  res *= 60;
  res += tm->tm_sec;
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<timegm>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  cmpl   $0x46,(%rdi)
  jle    L43
  mov    $0x46,%r12d
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L49:
  mov    %r12d,%edi
  call   <is_leap@plt>
  cmp    $0x1,%rax
  sbb    $0xfffffe92,%ebx
  add    $0x1,%r12d
  cmp    %r12d,0x0(%rbp)
  jg     L49
L12:
  mov    0x4(%rbp),%ecx
  test   %ecx,%ecx
  jle    L4
  xor    %r12d,%r12d
  lea    D9(%rip),%r13
  nopl   (%rax)
L48:
  mov    0x0(%rbp),%edi
  call   <is_leap@plt>
  movslq %r12d,%rdx
  add    $0x1,%r12d
  lea    (%rax,%rax,2),%rax
  lea    (%rdx,%rax,4),%rax
  add    0x0(%r13,%rax,4),%ebx
  cmp    %r12d,0x4(%rbp)
  jg     L48
L4:
  mov    0x8(%rbp),%eax
  mov    0x20(%rbp),%edx
  lea    -0x1(%rbx,%rax,1),%eax
  lea    (%rax,%rax,2),%eax
  lea    (%rdx,%rax,8),%eax
  imul   $0x3c,%eax,%eax
  add    0x18(%rbp),%eax
  imul   $0x3c,%eax,%eax
  add    0x10(%rbp),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L43:
  xor    %ebx,%ebx
  jmp    L12
```

tool_calls: name: unpack_data
arguments: data_label: D9
datatype: string

tool: D9: 

assistant: ```c
time_t timegm(struct tm *tm) {
  static const unsigned ndays[2][12] = {{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}, {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}};
  time_t res = 0;
  int i;
  for (i = 70; i < tm->tm_year; ++i)
    res += is_leap(i) ? 366 : 365;
  for (i = 0; i < tm->tm_mon; ++i)
    res += ndays[is_leap(tm->tm_year)][i];
  res += tm->tm_mday - 1;
  res *= 24;
  res += tm->tm_hour;
  res *= 60;
  res += tm->tm_min;
  res *= 60;
  res += tm->tm_sec;
  return res;
}
```
user: What is the c source code of the assembly code below?

```asm
<timegm>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%edi
  cmp    $0x46,%edi
  jle    L37
  mov    $0x46,%ebp
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L69:
  mov    %ebp,%edi
  call   <is_leap@plt>
  test   %rax,%rax
  je     L84
  mov    (%r12),%edi
  add    $0x1,%ebp
  add    $0x16e,%ebx
  cmp    %ebp,%edi
  jg     L69
L41:
  mov    0x4(%r12),%ecx
  test   %ecx,%ecx
  jle    L74
  xor    %ebp,%ebp
  lea    D9(%rip),%r13
  jmp    L65
  nopl   (%rax)
L76:
  mov    (%r12),%edi
L65:
  call   <is_leap@plt>
  movslq %ebp,%rdx
  add    $0x1,%ebp
  lea    (%rax,%rax,2),%rax
  lea    (%rdx,%rax,4),%rax
  add    0x0(%r13,%rax,4),%ebx
  cmp    %ebp,0x4(%r12)
  jg     L76
L74:
  mov    0x8(%r12),%eax
  mov    0x20(%r12),%edx
  lea    -0x1(%rbx,%rax,1),%eax
  lea    (%rax,%rax,2),%eax
  lea    (%rdx,%rax,8),%eax
  imul   $0x3c,%eax,%eax
  add    0x18(%r12),%eax
  imul   $0x3c,%eax,%eax
  add    0x10(%r12),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L84:
  mov    (%r12),%edi
  add    $0x1,%ebp
  add    $0x16d,%ebx
  cmp    %ebp,%edi
  jg     L69
  jmp    L41
  nopw   0x0(%rax,%rax,1)
L37:
  xor    %ebx,%ebx
  jmp    L41
```

tool_calls: name: unpack_data
arguments: data_tag: D9
type_guessed: string

tool: D9: 

assistant: ```c
time_t timegm(struct tm *tm) {
  static const unsigned ndays[2][12] = {{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}, {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}};
  time_t res = 0;
  int i;
  for (i = 70; i < tm->tm_year; ++i)
    res += is_leap(i) ? 366 : 365;
  for (i = 0; i < tm->tm_mon; ++i)
    res += ndays[is_leap(tm->tm_year)][i];
  res += tm->tm_mday - 1;
  res *= 24;
  res += tm->tm_hour;
  res *= 60;
  res += tm->tm_min;
  res *= 60;
  res += tm->tm_sec;
  return res;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<get_EH_name>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  lea    -0x10(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <dwarf_get_EH_name@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x28(%rbp),%ecx
  mov    -0x24(%rbp),%esi
  mov    -0x14(%rbp),%eax
  mov    %ecx,%r8d
  lea    D5(%rip),%rcx
  mov    %eax,%edi
  call   <ellipname@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D5
data_type: string

tool: D5: EH

assistant: ```c
const char *get_EH_name(unsigned int val_in, int printonerr) {
  const char *v = 0;
  int res = dwarf_get_EH_name(val_in, &v);
  return ellipname(res, val_in, v, "EH", printonerr);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<get_EH_name>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsp,%rsi
  call   <dwarf_get_EH_name@plt>
  mov    %eax,%edi
  mov    %ebp,%r8d
  lea    D0(%rip),%rcx
  mov    (%rsp),%rdx
  mov    %ebx,%esi
  call   <ellipname@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
type_infered: string

tool: D0: EH

assistant: ```c
const char *get_EH_name(unsigned int val_in, int printonerr) {
  const char *v = 0;
  int res = dwarf_get_EH_name(val_in, &v);
  return ellipname(res, val_in, v, "EH", printonerr);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_EH_name>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movq   $0x0,(%rsp)
  call   <dwarf_get_EH_name@plt>
  mov    (%rsp),%rdx
  mov    %ebx,%r8d
  mov    %ebp,%esi
  mov    %eax,%edi
  lea    D6(%rip),%rcx
  call   <ellipname@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D6
datatype: string

tool: D6: EH

assistant: ```c
const char *get_EH_name(unsigned int val_in, int printonerr) {
  const char *v = 0;
  int res = dwarf_get_EH_name(val_in, &v);
  return ellipname(res, val_in, v, "EH", printonerr);
}
```