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

```asm
<AddElementAndCount>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdx,(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    $0x0,%r12d
  test   %esi,%esi
  jle    L74
  mov    %rdi,%r13
  mov    %esi,%ebx
  mov    %ecx,%r14d
  mov    %r8,%rbp
  mov    (%r8),%r12d
  test   %r12d,%r12d
  je     L87
  mov    $0x0,%r12d
L74:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  mov    %r12d,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L87:
  call   <strlen@plt>
  mov    %eax,%r15d
  test   %eax,%eax
  jle    L74
  cmp    $0x1,%ebx
  jg     L23
  movb   $0x0,0x10(%rsp)
  mov    %r12d,0xc(%rsp)
L84:
  mov    0xc(%rsp),%eax
  lea    (%rax,%r15,1),%ebx
  cmp    %r14d,%ebx
  jl     L72
  addl   $0x1,0x0(%rbp)
  jmp    L74
L23:
  lea    0x10(%rsp),%rdi
  mov    %ebx,%edx
  lea    D8(%rip),%rsi
  call   <sprintf@plt>
  mov    %eax,0xc(%rsp)
  jmp    L84
L72:
  mov    %r15d,%edx
  mov    %r13,%rsi
  mov    (%rsp),%r14
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    0xc(%rsp),%edx
  add    $0x1,%edx
  lea    0x10(%rsp),%rsi
  movslq %r15d,%r15
  mov    %r14,%rdi
  add    %r15,%rdi
  call   <memcpy@plt>
  mov    %ebx,%r12d
  jmp    L74
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D8
data_type: string

tool: D8: %d

assistant: ```c
int AddElementAndCount(const char *szElement, int mult, char *szLinearCT, int nLenLinearCT, int *bOverflow) {
  char szMult[16];
  int len1, len2;
  if (mult > 0 && !*bOverflow && 0 < (len1 = strlen(szElement))) {
    if (mult > 1) {
      len2 = sprintf(szMult, "%d", mult);
    } else {
      len2 = 0;
      szMult[0] = '\0';
    }
    if (len1 + len2 < nLenLinearCT) {
      memcpy(szLinearCT, szElement, len1);
      memcpy(szLinearCT + len1, szMult, len2 + 1);
      return len1 + len2;
    } else {
      (*bOverflow)++;
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<AddElementAndCount>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L70
  mov    (%r8),%r13d
  mov    %r8,%rbx
  test   %r13d,%r13d
  je     L20
L70:
  xor    %r13d,%r13d
L40:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x38,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L20:
  mov    %ecx,%r14d
  mov    %rdi,%rbp
  mov    %esi,%r12d
  mov    %rdx,%r15
  call   <strlen@plt>
  mov    %eax,%ecx
  test   %eax,%eax
  jle    L70
  cmp    $0x1,%r12d
  je     L52
  mov    %r12d,%edx
  lea    0x10(%rsp),%rdi
  lea    D1(%rip),%rsi
  mov    %eax,0x8(%rsp)
  call   <sprintf@plt>
  mov    0x8(%rsp),%ecx
  mov    %eax,%r12d
  lea    (%rcx,%rax,1),%r8d
L80:
  cmp    %r8d,%r14d
  jg     L33
  addl   $0x1,(%rbx)
  jmp    L40
  nopw   0x0(%rax,%rax,1)
L52:
  movb   $0x0,0x10(%rsp)
  mov    %eax,%r8d
  xor    %r12d,%r12d
  jmp    L80
  nopl   (%rax)
L33:
  mov    %ecx,%edx
  mov    %rbp,%rsi
  mov    %r15,%rdi
  mov    %r8d,0xc(%rsp)
  mov    %ecx,0x8(%rsp)
  call   <memcpy@plt>
  movslq 0x8(%rsp),%rcx
  lea    0x1(%r12),%edx
  lea    0x10(%rsp),%rsi
  lea    (%r15,%rcx,1),%rdi
  call   <memcpy@plt>
  mov    0xc(%rsp),%r13d
  jmp    L40
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D1
data_type: string

tool: D1: %d

assistant: ```c
int AddElementAndCount(const char *szElement, int mult, char *szLinearCT, int nLenLinearCT, int *bOverflow) {
  char szMult[16];
  int len1, len2;
  if (mult > 0 && !*bOverflow && 0 < (len1 = strlen(szElement))) {
    if (mult > 1) {
      len2 = sprintf(szMult, "%d", mult);
    } else {
      len2 = 0;
      szMult[0] = '\0';
    }
    if (len1 + len2 < nLenLinearCT) {
      memcpy(szLinearCT, szElement, len1);
      memcpy(szLinearCT + len1, szMult, len2 + 1);
      return len1 + len2;
    } else {
      (*bOverflow)++;
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<AddElementAndCount>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L34
  mov    (%r8),%r13d
  mov    %r8,%rbx
  test   %r13d,%r13d
  je     L7
L34:
  xor    %r13d,%r13d
L4:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L89
  add    $0x38,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  mov    %ecx,%r14d
  mov    %rdi,%rbp
  mov    %esi,%r12d
  mov    %rdx,%r15
  call   <strlen@plt>
  mov    %eax,%ecx
  test   %eax,%eax
  jle    L34
  cmp    $0x1,%r12d
  je     L18
  mov    %r12d,%edx
  lea    0x10(%rsp),%rdi
  lea    D4(%rip),%rsi
  mov    %eax,0x8(%rsp)
  call   <sprintf@plt>
  mov    0x8(%rsp),%ecx
  mov    %eax,%r12d
  lea    (%rcx,%rax,1),%r8d
L86:
  cmp    %r8d,%r14d
  jg     L54
  addl   $0x1,(%rbx)
  jmp    L4
  nopw   0x0(%rax,%rax,1)
L18:
  movb   $0x0,0x10(%rsp)
  mov    %eax,%r8d
  xor    %r12d,%r12d
  jmp    L86
  nopl   (%rax)
L54:
  mov    %ecx,%edx
  mov    %rbp,%rsi
  mov    %r15,%rdi
  mov    %r8d,0xc(%rsp)
  mov    %ecx,0x8(%rsp)
  call   <memcpy@plt>
  movslq 0x8(%rsp),%rcx
  lea    0x1(%r12),%edx
  lea    0x10(%rsp),%rsi
  lea    (%r15,%rcx,1),%rdi
  call   <memcpy@plt>
  mov    0xc(%rsp),%r13d
  jmp    L4
L89:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D4
type_guessed: string

tool: D4: %d

assistant: ```c
int AddElementAndCount(const char *szElement, int mult, char *szLinearCT, int nLenLinearCT, int *bOverflow) {
  char szMult[16];
  int len1, len2;
  if (mult > 0 && !*bOverflow && 0 < (len1 = strlen(szElement))) {
    if (mult > 1) {
      len2 = sprintf(szMult, "%d", mult);
    } else {
      len2 = 0;
      szMult[0] = '\0';
    }
    if (len1 + len2 < nLenLinearCT) {
      memcpy(szLinearCT, szElement, len1);
      memcpy(szLinearCT + len1, szMult, len2 + 1);
      return len1 + len2;
    } else {
      (*bOverflow)++;
    }
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movb   $0x41,-0x1(%rbp)
  jmp    L17
L18:
  movsbl -0x1(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  movzbl -0x1(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x1(%rbp)
L17:
  cmpb   $0x5a,-0x1(%rbp)
  jle    L18
  mov    $0x0,%eax
  leave
  ret
```

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

tool: D9: %d 

assistant: ```c
int main(void) {
  char i;
  for (i = 'A'; i <= 'Z'; i++) {
    printf("%d ", (int)i);
    printf("\n");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x41,%ebx
  lea    D7(%rip),%rbp
L4:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  cmp    $0x5b,%ebx
  jne    L4
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D7
datatype: string

tool: D7: %d 

assistant: ```c
int main(void) {
  char i;
  for (i = 'A'; i <= 'Z'; i++) {
    printf("%d ", (int)i);
    printf("\n");
  }
  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   %rbp
  lea    D0(%rip),%rbp
  push   %rbx
  mov    $0x41,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L9:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  cmp    $0x5b,%ebx
  jne    L9
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D0
datatype: string

tool: D0: %d 

assistant: ```c
int main(void) {
  char i;
  for (i = 'A'; i <= 'Z'; i++) {
    printf("%d ", (int)i);
    printf("\n");
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D6(%rip),%rbp
  push   %rbx
  mov    $0x41,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L7:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  cmp    $0x5b,%ebx
  jne    L7
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D6
data_type: string

tool: D6: %d 

assistant: ```c
int main(void) {
  char i;
  for (i = 'A'; i <= 'Z'; i++) {
    printf("%d ", (int)i);
    printf("\n");
  }
  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
<BM_WriteLongInt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  and    $0xff000000,%eax
  lea    0xffffff(%rax),%rdx
  test   %rax,%rax
  cmovs  %rdx,%rax
  sar    $0x18,%rax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x28(%rbp),%rax
  and    $0xff0000,%eax
  lea    0xffff(%rax),%rdx
  test   %rax,%rax
  cmovs  %rdx,%rax
  sar    $0x10,%rax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  mov    %dl,(%rax)
  mov    -0x28(%rbp),%rax
  and    $0xff00,%eax
  lea    0xff(%rax),%rdx
  test   %rax,%rax
  cmovs  %rdx,%rax
  sar    $0x8,%rax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  add    $0x2,%rax
  mov    %dl,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x3,%rax
  mov    -0x28(%rbp),%rdx
  mov    %dl,(%rax)
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L3:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <putc@plt>
  addl   $0x1,-0x4(%rbp)
L17:
  cmpl   $0x3,-0x4(%rbp)
  jle    L3
  nop
  nop
  leave
  ret
```

assistant: ```c
void BM_WriteLongInt(FILE *fptr, char *s, long n) {
  int i;
  s[0] = (n & 0xff000000) / 16777216;
  s[1] = (n & 0x00ff0000) / 65536;
  s[2] = (n & 0x0000ff00) / 256;
  s[3] = (n & 0x000000ff);
  for (i = 0; i < 4; i++)
    putc(s[i], fptr);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<BM_WriteLongInt>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rdx,%rax
  and    $0xff000000,%eax
  sar    $0x18,%rax
  mov    %al,(%rsi)
  mov    %rdx,%rax
  and    $0xff0000,%eax
  sar    $0x10,%rax
  mov    %al,0x1(%rsi)
  movzbl %dh,%eax
  mov    %al,0x2(%rsi)
  mov    %dl,0x3(%rsi)
  mov    %rsi,%rbx
  lea    0x4(%rsi),%rbp
L0:
  movsbl (%rbx),%edi
  mov    %r12,%rsi
  call   <putc@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L0
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void BM_WriteLongInt(FILE *fptr, char *s, long n) {
  int i;
  s[0] = (n & 0xff000000) / 16777216;
  s[1] = (n & 0x00ff0000) / 65536;
  s[2] = (n & 0x0000ff00) / 256;
  s[3] = (n & 0x000000ff);
  for (i = 0; i < 4; i++)
    putc(s[i], fptr);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<BM_WriteLongInt>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    %rdx,%rdi
  bswap  %edx
  push   %rbp
  sar    $0x18,%rdi
  lea    0x4(%rsi),%rbp
  push   %rbx
  lea    0x1(%rsi),%rbx
  mov    %edx,(%rsi)
L10:
  movsbl %dil,%edi
  mov    %r12,%rsi
  call   <putc@plt>
  cmp    %rbp,%rbx
  je     L18
  movzbl (%rbx),%edi
  add    $0x1,%rbx
  jmp    L10
  nopl   0x0(%rax,%rax,1)
L18:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void BM_WriteLongInt(FILE *fptr, char *s, long n) {
  int i;
  s[0] = (n & 0xff000000) / 16777216;
  s[1] = (n & 0x00ff0000) / 65536;
  s[2] = (n & 0x0000ff00) / 256;
  s[3] = (n & 0x000000ff);
  for (i = 0; i < 4; i++)
    putc(s[i], fptr);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<BM_WriteLongInt>:
  endbr64
  push   %r12
  bswap  %edx
  lea    0x4(%rsi),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    %edx,(%rsi)
L6:
  movsbl (%rbx),%edi
  mov    %rbp,%rsi
  add    $0x1,%rbx
  call   <putc@plt>
  cmp    %rbx,%r12
  jne    L6
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void BM_WriteLongInt(FILE *fptr, char *s, long n) {
  int i;
  s[0] = (n & 0xff000000) / 16777216;
  s[1] = (n & 0x00ff0000) / 65536;
  s[2] = (n & 0x0000ff00) / 256;
  s[3] = (n & 0x000000ff);
  for (i = 0; i < 4; i++)
    putc(s[i], fptr);
}
```
user: What is the c source code of the assembly code below:

```asm
<funD>:
  endbr64
  cmp    $0x8000,%edi
  ja     L135
  cmp    $0x4000,%edi
  ja     L40
  cmp    $0x100,%edi
  ja     L28
  cmp    $0x80,%edi
  ja     L158
  cmp    $0x40,%edi
  ja     L117
  cmp    $0x2,%edi
  jbe    L61
  lea    -0x3(%rdi),%eax
  cmp    $0x3d,%eax
  ja     L59
  mov    %eax,%eax
  lea    D3(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x8,%eax
  ret
L59:
  mov    %edi,%eax
  ret
L28:
  cmp    $0x1000,%edi
  ja     L44
  cmp    $0x800,%edi
  ja     L31
  mov    $0x800,%eax
  cmp    $0x400,%edi
  ja     L104
  cmp    $0x201,%edi
  sbb    %eax,%eax
  and    $0xfffffe00,%eax
  add    $0x400,%eax
  ret
L44:
  cmp    $0x2001,%edi
  sbb    %eax,%eax
  and    $0xffffe000,%eax
  add    $0x4000,%eax
  ret
L135:
  cmp    $0x400000,%edi
  ja     L57
  cmp    $0x200000,%edi
  ja     L47
  cmp    $0x80000,%edi
  ja     L25
  cmp    $0x40000,%edi
  ja     L179
  mov    $0x40000,%eax
  cmp    $0x20000,%edi
  ja     L104
  cmp    $0x10001,%edi
  sbb    %eax,%eax
  mov    $0x0,%ax
  add    $0x20000,%eax
  ret
L25:
  cmp    $0x100001,%edi
  sbb    %eax,%eax
  and    $0xfff00000,%eax
  add    $0x200000,%eax
  ret
L57:
  cmp    $0x2000000,%edi
  ja     L92
  cmp    $0x1000000,%edi
  ja     L13
  cmp    $0x800001,%edi
  sbb    %eax,%eax
  and    $0xff800000,%eax
  add    $0x1000000,%eax
  ret
L92:
  mov    $0x4000000,%eax
  cmp    $0x4000000,%edi
  jbe    L104
  lea    -0x4000001(%rdi),%eax
  cmp    $0x4000000,%eax
  mov    $0x8000000,%eax
  cmovae %edi,%eax
  ret
  mov    $0x10,%eax
  ret
  mov    $0x20,%eax
  ret
  mov    $0x40,%eax
  ret
L40:
  mov    $0x8000,%eax
  ret
L158:
  mov    $0x100,%eax
  ret
L117:
  mov    $0x80,%eax
  ret
L61:
  mov    %edi,%eax
  ret
  mov    $0x4,%eax
  ret
L31:
  mov    $0x1000,%eax
  ret
L47:
  mov    $0x400000,%eax
  ret
L179:
  mov    $0x80000,%eax
  ret
L13:
  mov    $0x2000000,%eax
L104:
  ret
```

assistant: ```c
unsigned int funD(unsigned int x) {
  switch (x) {
  case 3 ... 4:
    return 0x4;
  case 5 ... 8:
    return 0x8;
  case 9 ... 16:
    return 0x10;
  case 17 ... 32:
    return 0x20;
  case 33 ... 64:
    return 0x40;
  case 65 ... 128:
    return 0x80;
  case 129 ... 256:
    return 0x100;
  case 257 ... 512:
    return 0x200;
  case 513 ... 1024:
    return 0x400;
  case 1025 ... 2048:
    return 0x800;
  case 2049 ... 4096:
    return 0x1000;
  case 4097 ... 8192:
    return 0x2000;
  case 8193 ... 16384:
    return 0x4000;
  case 16385 ... 32768:
    return 0x8000;
  case 32769 ... 65536:
    return 0x10000;
  case 0x10001 ... 0x20000:
    return 0x20000;
  case 0x20001 ... 0x40000:
    return 0x40000;
  case 0x40001 ... 0x80000:
    return 0x80000;
  case 0x80001 ... 0x100000:
    return 0x100000;
  case 0x100001 ... 0x200000:
    return 0x200000;
  case 0x200001 ... 0x400000:
    return 0x400000;
  case 0x400001 ... 0x800000:
    return 0x800000;
  case 0x800001 ... 0x1000000:
    return 0x1000000;
  case 0x1000001 ... 0x2000000:
    return 0x2000000;
  case 0x2000001 ... 0x4000000:
    return 0x4000000;
  case 0x4000001 ... 0x8000000:
    return 0x8000000;
  }
  return x;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  movl   $0x1,-0xc(%rbp)
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x1c(%rbp),%eax
  lea    -0x2(%rax),%edx
  mov    -0x1c(%rbp),%eax
  sub    $0x2,%eax
  imul   %edx,%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L50
L72:
  cmpl   $0x1,-0xc(%rbp)
  jne    L114
  movl   $0x0,-0x10(%rbp)
  jmp    L110
L113:
  mov    $0x20,%edi
  call   <putchar@plt>
  addl   $0x1,-0x10(%rbp)
L110:
  mov    -0x10(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L113
  mov    $0x2a,%edi
  call   <putchar@plt>
  subl   $0x1,-0x14(%rbp)
  jmp    L10
L114:
  movl   $0x0,-0x10(%rbp)
  jmp    L116
L12:
  mov    $0x20,%edi
  call   <putchar@plt>
  addl   $0x1,-0x10(%rbp)
L116:
  mov    -0x10(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L12
  mov    $0x2a,%edi
  call   <putchar@plt>
  movl   $0x0,-0x10(%rbp)
  jmp    L71
L4:
  mov    $0x23,%edi
  call   <putchar@plt>
  addl   $0x1,-0x10(%rbp)
L71:
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L4
  mov    $0x2a,%edi
  call   <putchar@plt>
L10:
  subl   $0x1,-0x18(%rbp)
  addl   $0x2,-0x14(%rbp)
  addl   $0x1,-0xc(%rbp)
  mov    $0xa,%edi
  call   <putchar@plt>
L50:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jne    L72
  movl   $0x0,-0x10(%rbp)
  jmp    L44
L108:
  mov    $0x2a,%edi
  call   <putchar@plt>
  addl   $0x1,-0x10(%rbp)
L44:
  mov    -0x1c(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  sub    $0x1,%eax
  cmp    %eax,-0x10(%rbp)
  jl     L108
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L90
  call   <__stack_chk_fail@plt>
L90:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D6
data_type: string
name: parse_data
arguments: data_tag: D16
data_type: string

tool: D6: %d

tool: D16: %d


assistant: ```c
int main() {
  int n, space, hashtag = 0, i, current = 1;
  scanf("%d", &n);
  space = n - 1;
  printf("%d\n", (n - 2) * (n - 2));
  while (current != n) {
    if (current == 1) {
      for (i = 0; i < space; i++)
        printf(" ");
      printf("*");
      hashtag--;
    } else {
      for (i = 0; i < space; i++)
        printf(" ");
      printf("*");
      for (i = 0; i < hashtag; i++)
        printf("#");
      printf("*");
    }
    space--;
    hashtag += 2;
    current++;
    printf("\n");
  }
  for (i = 0; i < current + n - 1; i++)
    printf("*");
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  lea    -0x1(%rsi),%ebp
  sub    $0x2,%esi
  imul   %esi,%esi
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%r13d
  cmp    $0x1,%r13d
  je     L20
  mov    $0x1,%r13d
  mov    $0x0,%r12d
  jmp    L149
L58:
  test   %ebp,%ebp
  jle    L101
  mov    $0x0,%ebx
L133:
  mov    $0x20,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  cmp    %ebp,%ebx
  jne    L133
L101:
  mov    $0x2a,%edi
  call   <putchar@plt>
  sub    $0x1,%r12d
  jmp    L138
L94:
  mov    $0x2a,%edi
  call   <putchar@plt>
L138:
  sub    $0x1,%ebp
  add    $0x2,%r12d
  add    $0x1,%r13d
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  cmp    %r13d,%eax
  je     L29
L149:
  cmp    $0x1,%r13d
  je     L58
  mov    $0x0,%ebx
  test   %ebp,%ebp
  jle    L98
L146:
  mov    $0x20,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  cmp    %ebp,%ebx
  jne    L146
L98:
  mov    $0x2a,%edi
  call   <putchar@plt>
  test   %r12d,%r12d
  jle    L94
  mov    $0x0,%ebx
L124:
  mov    $0x23,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  cmp    %r12d,%ebx
  jne    L124
  jmp    L94
L29:
  add    %r13d,%eax
  cmp    $0x1,%eax
  jle    L87
L20:
  mov    $0x0,%ebx
L145:
  mov    $0x2a,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  mov    %r13d,%eax
  add    0x4(%rsp),%eax
  sub    $0x1,%eax
  cmp    %ebx,%eax
  jg     L145
L87:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L99
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L99:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D9
datatype: string
name: read_data
arguments: data_tag: D2
datatype: string

tool: D9: %d

tool: D2: %d


assistant: ```c
int main() {
  int n, space, hashtag = 0, i, current = 1;
  scanf("%d", &n);
  space = n - 1;
  printf("%d\n", (n - 2) * (n - 2));
  while (current != n) {
    if (current == 1) {
      for (i = 0; i < space; i++)
        printf(" ");
      printf("*");
      hashtag--;
    } else {
      for (i = 0; i < space; i++)
        printf(" ");
      printf("*");
      for (i = 0; i < hashtag; i++)
        printf("#");
      printf("*");
    }
    space--;
    hashtag += 2;
    current++;
    printf("\n");
  }
  for (i = 0; i < current + n - 1; i++)
    printf("*");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  lea    D14(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  lea    -0x1(%rsi),%ebp
  sub    $0x2,%esi
  imul   %esi,%esi
  call   <printf@plt>
  mov    0x4(%rsp),%r13d
  cmp    $0x1,%r13d
  je     L105
  mov    $0x1,%r13d
  xor    %r12d,%r12d
  xchg   %ax,%ax
L119:
  xor    %ebx,%ebx
  cmp    $0x1,%r13d
  je     L107
  test   %ebp,%ebp
  jle    L67
L19:
  mov    $0x20,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  cmp    %ebp,%ebx
  jne    L19
L67:
  mov    $0x2a,%edi
  xor    %ebx,%ebx
  call   <putchar@plt>
  test   %r12d,%r12d
  je     L3
  nopw   0x0(%rax,%rax,1)
L77:
  mov    $0x23,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  cmp    %r12d,%ebx
  jne    L77
L3:
  mov    $0x2a,%edi
  call   <putchar@plt>
L86:
  mov    $0xa,%edi
  sub    $0x1,%ebp
  add    $0x2,%r12d
  add    $0x1,%r13d
  call   <putchar@plt>
  cmp    %r13d,0x4(%rsp)
  jne    L119
L105:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L21:
  mov    $0x2a,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  add    %r13d,%eax
  sub    $0x1,%eax
  cmp    %ebx,%eax
  jg     L21
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L107:
  test   %ebp,%ebp
  jle    L7
  nopl   0x0(%rax)
L44:
  mov    $0x20,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  cmp    %ebp,%ebx
  jne    L44
L7:
  mov    $0x2a,%edi
  sub    $0x1,%r12d
  call   <putchar@plt>
  jmp    L86
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D14
data_type: string
name: retrieve_data
arguments: label: D13
data_type: string

tool: D14: %d

tool: D13: %d


assistant: ```c
int main() {
  int n, space, hashtag = 0, i, current = 1;
  scanf("%d", &n);
  space = n - 1;
  printf("%d\n", (n - 2) * (n - 2));
  while (current != n) {
    if (current == 1) {
      for (i = 0; i < space; i++)
        printf(" ");
      printf("*");
      hashtag--;
    } else {
      for (i = 0; i < space; i++)
        printf(" ");
      printf("*");
      for (i = 0; i < hashtag; i++)
        printf("#");
      printf("*");
    }
    space--;
    hashtag += 2;
    current++;
    printf("\n");
  }
  for (i = 0; i < current + n - 1; i++)
    printf("*");
}
```