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
<itoaBuffer>:
  endbr64
  sub    $0x28,%rsp
  mov    %edi,%edx
  mov    %rsi,%r9
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  js     L14
  je     L79
L20:
  mov    %rsp,%rdi
  mov    $0x0,%esi
L58:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%r8d
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %r8d,%edx
  mov    %esi,%r10d
  add    $0x1,%esi
  add    $0x30,%eax
  mov    %al,(%rdi)
  add    $0x1,%rdi
  test   %r8d,%r8d
  jne    L58
  test   %esi,%esi
  je     L79
  mov    %r9,%rax
  test   %r10d,%r10d
  jns    L34
  jmp    L42
L14:
  movb   $0x2d,(%rsi)
  neg    %edx
  lea    0x1(%rsi),%r9
  jmp    L20
L79:
  movb   $0x30,(%rsp)
  mov    $0x0,%r10d
L34:
  movslq %r10d,%rax
  mov    %r9,%rdx
L12:
  movzbl (%rsp,%rax,1),%ecx
  mov    %cl,(%rdx)
  add    $0x1,%rdx
  sub    $0x1,%rax
  test   %eax,%eax
  jns    L12
  test   %r10d,%r10d
  mov    $0x0,%eax
  cmovs  %eax,%r10d
  lea    0x1(%r10),%eax
  add    %r9,%rax
L42:
  movb   $0x0,(%rax)
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x28,%rsp
  ret
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *itoaBuffer(int value, char *buf) {
  char num[20];
  int pos;
  pos = 0;
  if (value < 0) {
    *buf++ = '-';
    value = -value;
  }
  while (value != 0) {
    char c = value % 10;
    value = value / 10;
    num[pos++] = c + '0';
  }
  if (pos == 0)
    num[pos++] = '0';
  while (--pos >= 0) {
    *buf = num[pos];
    buf++;
  }
  *buf = 0;
  return buf;
}
```
user: Translate the below assembly code into C programming language:

```asm
<itoaBuffer>:
  endbr64
  sub    $0x28,%rsp
  mov    %edi,%edx
  mov    %rsi,%r8
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  js     L41
  je     L36
L21:
  xor    %esi,%esi
  mov    %rsp,%rcx
  mov    $0xcccccccd,%r10d
  cs nopw 0x0(%rax,%rax,1)
L31:
  mov    %edx,%eax
  mov    %esi,%r11d
  imul   %r10,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%r9d
  mov    %eax,%edi
  mov    %edx,%eax
  add    %r9d,%r9d
  mov    %edi,%edx
  sub    %r9d,%eax
  add    $0x30,%eax
  mov    %al,(%rcx,%rsi,1)
  add    $0x1,%rsi
  test   %edi,%edi
  jne    L31
  movzbl (%rsp),%eax
L35:
  lea    0x1(%r11),%edx
  mov    %al,(%rsp)
  movslq %r11d,%rax
  movslq %edx,%rdx
  add    %rcx,%rax
  add    %r8,%rdx
  nopw   0x0(%rax,%rax,1)
L45:
  movzbl (%rax),%ecx
  add    $0x1,%r8
  sub    $0x1,%rax
  mov    %cl,-0x1(%r8)
  cmp    %rdx,%r8
  jne    L45
  movb   $0x0,(%r8)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    %r8,%rax
  add    $0x28,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L41:
  movb   $0x2d,(%rsi)
  neg    %edx
  add    $0x1,%r8
  jmp    L21
  xchg   %ax,%ax
L36:
  xor    %r11d,%r11d
  mov    $0x30,%eax
  mov    %rsp,%rcx
  jmp    L35
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *itoaBuffer(int value, char *buf) {
  char num[20];
  int pos;
  pos = 0;
  if (value < 0) {
    *buf++ = '-';
    value = -value;
  }
  while (value != 0) {
    char c = value % 10;
    value = value / 10;
    num[pos++] = c + '0';
  }
  if (pos == 0)
    num[pos++] = '0';
  while (--pos >= 0) {
    *buf = num[pos];
    buf++;
  }
  *buf = 0;
  return buf;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<itoaBuffer>:
  endbr64
  sub    $0x28,%rsp
  mov    %edi,%edx
  mov    %rsi,%r8
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  js     L20
  je     L23
L21:
  xor    %esi,%esi
  mov    %rsp,%rcx
  mov    $0xcccccccd,%r10d
  cs nopw 0x0(%rax,%rax,1)
L42:
  mov    %edx,%eax
  mov    %esi,%r11d
  imul   %r10,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%r9d
  mov    %eax,%edi
  mov    %edx,%eax
  add    %r9d,%r9d
  mov    %edi,%edx
  sub    %r9d,%eax
  add    $0x30,%eax
  mov    %al,(%rcx,%rsi,1)
  add    $0x1,%rsi
  test   %edi,%edi
  jne    L42
  movzbl (%rsp),%eax
L40:
  lea    0x1(%r11),%edx
  mov    %al,(%rsp)
  movslq %r11d,%rax
  movslq %edx,%rdx
  add    %rcx,%rax
  add    %r8,%rdx
  nopw   0x0(%rax,%rax,1)
L58:
  movzbl (%rax),%ecx
  add    $0x1,%r8
  sub    $0x1,%rax
  mov    %cl,-0x1(%r8)
  cmp    %rdx,%r8
  jne    L58
  movb   $0x0,(%r8)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    %r8,%rax
  add    $0x28,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L20:
  movb   $0x2d,(%rsi)
  neg    %edx
  add    $0x1,%r8
  jmp    L21
  xchg   %ax,%ax
L23:
  xor    %r11d,%r11d
  mov    $0x30,%eax
  mov    %rsp,%rcx
  jmp    L40
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *itoaBuffer(int value, char *buf) {
  char num[20];
  int pos;
  pos = 0;
  if (value < 0) {
    *buf++ = '-';
    value = -value;
  }
  while (value != 0) {
    char c = value % 10;
    value = value / 10;
    num[pos++] = c + '0';
  }
  if (pos == 0)
    num[pos++] = '0';
  while (--pos >= 0) {
    *buf = num[pos];
    buf++;
  }
  *buf = 0;
  return buf;
}
```
user: 请从给定的汇编指令推导出C源代码:

```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,-0x10(%rbp)
  movb   $0x2a,-0x1a(%rbp)
  movb   $0x23,-0x19(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  lea    -0x2(%rax),%edx
  mov    -0x18(%rbp),%eax
  sub    $0x2,%eax
  imul   %edx,%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L91
L72:
  movl   $0x0,-0xc(%rbp)
  jmp    L53
L0:
  mov    $0x20,%edi
  call   <putchar@plt>
  addl   $0x1,-0xc(%rbp)
L53:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jl     L0
  cmpl   $0x1,-0x10(%rbp)
  jne    L46
  movsbl -0x1a(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
  jmp    L113
L46:
  mov    -0x18(%rbp),%eax
  cmp    $0x1,%eax
  jne    L111
  movl   $0x0,-0xc(%rbp)
  jmp    L96
L77:
  movsbl -0x1a(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0xc(%rbp)
L96:
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jle    L77
  jmp    L113
L111:
  movsbl -0x1a(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L39
L24:
  movsbl -0x19(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0xc(%rbp)
L39:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L24
  movsbl -0x1a(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x2,-0x10(%rbp)
  addl   $0x2,-0x14(%rbp)
L113:
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x18(%rbp)
L91:
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  jg     L72
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L49
  call   <__stack_chk_fail@plt>
L49:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D22
datatype: string
name: retrieve_data
arguments: data_label: D4
datatype: string
name: retrieve_data
arguments: data_label: D8
datatype: string

tool: D22: %d

tool: D4: %d


tool: D8: %c


assistant: ```c
int main() {
  int n, i = 0, j = 1, k;
  char c1 = '*', c2 = '#';
  scanf("%d", &n);
  printf("%d\n", (n - 2) * (n - 2));
  while (n > 0) {
    for (k = 0; k < n; k++)
      printf(" ");
    if (j == 1) {
      printf("%c\n", c1);
      j++;
      i++;
    } else if (n == 1) {
      for (k = 0; k <= j; k++)
        printf("%c", c1);
    } else {
      printf("%c", c1);
      for (k = 0; k < i; k++)
        printf("%c", c2);
      printf("%c\n", c1);
      j = j + 2;
      i = i + 2;
    }
    n--;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```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    D24(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x2(%rax),%esi
  imul   %esi,%esi
  lea    D15(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%ebp
  mov    $0x0,%r12d
  lea    D7(%rip),%r13
  cmpl   $0x0,0x4(%rsp)
  jg     L70
L115:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L36:
  mov    $0x2a,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%r12d
  mov    $0x2,%ebp
  jmp    L56
L54:
  mov    $0x2a,%edi
  call   <putchar@plt>
  test   %r12d,%r12d
  jle    L8
  mov    $0x0,%ebx
L116:
  mov    $0x23,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  cmp    %r12d,%ebx
  jne    L116
L8:
  mov    $0x2a,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x2,%ebp
  add    $0x2,%r12d
L56:
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  test   %eax,%eax
  jle    L115
L70:
  mov    0x4(%rsp),%eax
  mov    $0x0,%ebx
  test   %eax,%eax
  jle    L21
L11:
  mov    $0x20,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  mov    0x4(%rsp),%eax
  cmp    %ebx,%eax
  jg     L11
L21:
  cmp    $0x1,%ebp
  je     L36
  cmp    $0x1,%eax
  jne    L54
  test   %ebp,%ebp
  js     L56
  mov    $0x0,%ebx
L4:
  mov    $0x2a,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  cmp    %ebp,%ebx
  jle    L4
  jmp    L56
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D24
type_infered: string
name: parse_data
arguments: data_tag: D15
type_infered: string
name: parse_data
arguments: data_tag: D7
type_infered: string

tool: D24: %d

tool: D15: %d


tool: D7: %c


assistant: ```c
int main() {
  int n, i = 0, j = 1, k;
  char c1 = '*', c2 = '#';
  scanf("%d", &n);
  printf("%d\n", (n - 2) * (n - 2));
  while (n > 0) {
    for (k = 0; k < n; k++)
      printf(" ");
    if (j == 1) {
      printf("%c\n", c1);
      j++;
      i++;
    } else if (n == 1) {
      for (k = 0; k <= j; k++)
        printf("%c", c1);
    } else {
      printf("%c", c1);
      for (k = 0; k < i; k++)
        printf("%c", c2);
      printf("%c\n", c1);
      j = j + 2;
      i = i + 2;
    }
    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   %r13
  lea    D16(%rip),%rdi
  lea    D12(%rip),%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    $0x1,%ebp
  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),%eax
  lea    D24(%rip),%rdi
  lea    -0x2(%rax),%esi
  xor    %eax,%eax
  imul   %esi,%esi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L79
  xchg   %ax,%ax
L94:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L98:
  mov    $0x20,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  cmp    %ebx,%eax
  jg     L98
  cmp    $0x1,%ebp
  je     L7
  cmp    $0x1,%eax
  jne    L112
  xor    %ebx,%ebx
  nopl   (%rax)
L13:
  mov    $0x2a,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  cmp    %ebp,%ebx
  jle    L13
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  test   %eax,%eax
  jg     L94
L79:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L103
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L112:
  mov    $0x2a,%edi
  xor    %ebx,%ebx
  call   <putchar@plt>
  test   %r12d,%r12d
  je     L123
  nopl   0x0(%rax)
L52:
  mov    $0x23,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  cmp    %r12d,%ebx
  jne    L52
L123:
  mov    $0x2a,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x2,%ebp
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  add    $0x2,%r12d
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  test   %eax,%eax
  jg     L94
  jmp    L79
  nopl   (%rax)
L7:
  mov    $0x2a,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%r12d
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  mov    $0x2,%ebp
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  test   %eax,%eax
  jg     L94
  jmp    L79
L103:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D16
type_infered: string
name: read_data
arguments: tag: D12
type_infered: string
name: read_data
arguments: tag: D24
type_infered: string

tool: D16: %d

tool: D12: %c


tool: D24: %d


assistant: ```c
int main() {
  int n, i = 0, j = 1, k;
  char c1 = '*', c2 = '#';
  scanf("%d", &n);
  printf("%d\n", (n - 2) * (n - 2));
  while (n > 0) {
    for (k = 0; k < n; k++)
      printf(" ");
    if (j == 1) {
      printf("%c\n", c1);
      j++;
      i++;
    } else if (n == 1) {
      for (k = 0; k <= j; k++)
        printf("%c", c1);
    } else {
      printf("%c", c1);
      for (k = 0; k < i; k++)
        printf("%c", c2);
      printf("%c\n", c1);
      j = j + 2;
      i = i + 2;
    }
    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   %r13
  lea    D10(%rip),%rdi
  lea    D18(%rip),%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    $0x1,%ebp
  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),%eax
  lea    D17(%rip),%rdi
  lea    -0x2(%rax),%esi
  xor    %eax,%eax
  imul   %esi,%esi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L4
  xchg   %ax,%ax
L87:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L59:
  mov    $0x20,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  cmp    %ebx,%eax
  jg     L59
  cmp    $0x1,%ebp
  je     L16
  cmp    $0x1,%eax
  jne    L129
  xor    %ebx,%ebx
  nopl   (%rax)
L91:
  mov    $0x2a,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  cmp    %ebp,%ebx
  jle    L91
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  test   %eax,%eax
  jg     L87
L4:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L129:
  mov    $0x2a,%edi
  xor    %ebx,%ebx
  call   <putchar@plt>
  test   %r12d,%r12d
  je     L9
  nopl   0x0(%rax)
L98:
  mov    $0x23,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  cmp    %r12d,%ebx
  jne    L98
L9:
  mov    $0x2a,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x2,%ebp
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  add    $0x2,%r12d
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  test   %eax,%eax
  jg     L87
  jmp    L4
  nopl   (%rax)
L16:
  mov    $0x2a,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%r12d
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  mov    $0x2,%ebp
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  test   %eax,%eax
  jg     L87
  jmp    L4
L48:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D10
type: string
name: unpack_data
arguments: label: D18
type: string
name: unpack_data
arguments: label: D17
type: string

tool: D10: %d

tool: D18: %c


tool: D17: %d


assistant: ```c
int main() {
  int n, i = 0, j = 1, k;
  char c1 = '*', c2 = '#';
  scanf("%d", &n);
  printf("%d\n", (n - 2) * (n - 2));
  while (n > 0) {
    for (k = 0; k < n; k++)
      printf(" ");
    if (j == 1) {
      printf("%c\n", c1);
      j++;
      i++;
    } else if (n == 1) {
      for (k = 0; k <= j; k++)
        printf("%c", c1);
    } else {
      printf("%c", c1);
      for (k = 0; k < i; k++)
        printf("%c", c2);
      printf("%c\n", c1);
      j = j + 2;
      i = i + 2;
    }
    n--;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<print_progr_ps>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %ecx,%eax
  mov    %r8d,%esi
  mov    %r9d,%ecx
  mov    %edi,%edx
  mov    %dl,-0x4(%rbp)
  mov    %al,-0x8(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  mov    %ecx,%eax
  mov    %al,-0x20(%rbp)
  movsbl -0x8(%rbp),%esi
  movsbl -0x1c(%rbp),%r8d
  movsbl -0x20(%rbp),%edi
  movsbl -0x4(%rbp),%eax
  mov    -0x18(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  sub    $0x8,%rsp
  push   %rsi
  mov    %r8d,%r9d
  mov    %edi,%r8d
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x10,%rsp
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D3
type_guessed: string

tool: D3:  %3d%% of %s @ %s/s ETA %02d:%02d:%02d   

assistant: ```c
void print_progr_ps(char perc, char *size, char *bps, char eta_s, char eta_m, char eta_h) {
  printf("\r %3d%% of %s @ %s/s ETA %02d:%02d:%02d   ", perc, size, bps, eta_h, eta_m, eta_s);
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<print_progr_ps>:
  endbr64
  sub    $0x10,%rsp
  mov    %rsi,%rax
  mov    %r9d,%r10d
  movsbl %dil,%esi
  movsbl %cl,%ecx
  push   %rcx
  movsbl %r8b,%r9d
  movsbl %r10b,%r8d
  mov    %rdx,%rcx
  mov    %rax,%rdx
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x18,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D9
data_type: string

tool: D9:  %3d%% of %s @ %s/s ETA %02d:%02d:%02d   

assistant: ```c
void print_progr_ps(char perc, char *size, char *bps, char eta_s, char eta_m, char eta_h) {
  printf("\r %3d%% of %s @ %s/s ETA %02d:%02d:%02d   ", perc, size, bps, eta_h, eta_m, eta_s);
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print_progr_ps>:
  endbr64
  sub    $0x10,%rsp
  movsbl %cl,%ecx
  mov    %rsi,%r10
  mov    %r9d,%eax
  push   %rcx
  movsbl %dil,%esi
  movsbl %r8b,%r9d
  mov    %rdx,%rcx
  movsbl %al,%r8d
  mov    %r10,%rdx
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x18,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
type_infered: string

tool: D9:  %3d%% of %s @ %s/s ETA %02d:%02d:%02d   

assistant: ```c
void print_progr_ps(char perc, char *size, char *bps, char eta_s, char eta_m, char eta_h) {
  printf("\r %3d%% of %s @ %s/s ETA %02d:%02d:%02d   ", perc, size, bps, eta_h, eta_m, eta_s);
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<print_progr_ps>:
  endbr64
  sub    $0x10,%rsp
  movsbl %cl,%ecx
  mov    %rsi,%r10
  mov    %r9d,%eax
  push   %rcx
  movsbl %dil,%esi
  movsbl %r8b,%r9d
  mov    %rdx,%rcx
  movsbl %al,%r8d
  mov    %r10,%rdx
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x18,%rsp
  ret
```

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

tool: D7:  %3d%% of %s @ %s/s ETA %02d:%02d:%02d   

assistant: ```c
void print_progr_ps(char perc, char *size, char *bps, char eta_s, char eta_m, char eta_h) {
  printf("\r %3d%% of %s @ %s/s ETA %02d:%02d:%02d   ", perc, size, bps, eta_h, eta_m, eta_s);
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<dns_comp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jbe    L64
  mov    $0x0,%eax
  jmp    L57
L143:
  movl   $0x0,-0x4(%rbp)
  jmp    L113
L60:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  je     L0
  addl   $0x1,-0x4(%rbp)
L113:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L60
  jmp    L119
L0:
  nop
L119:
  cmpl   $0x3f,-0x4(%rbp)
  jle    L1
  mov    $0x0,%eax
  jmp    L57
L1:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x28(%rbp)
  jae    L89
  mov    $0x0,%eax
  jmp    L57
L89:
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  jmp    L79
L103:
  mov    -0x18(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x20(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L79:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  test   %eax,%eax
  jne    L103
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  jne    L64
  addq   $0x1,-0x18(%rbp)
L64:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L143
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  cmp    %rax,-0x28(%rbp)
  jae    L114
  mov    $0x0,%eax
  jmp    L57
L114:
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
L57:
  pop    %rbp
  ret
```

assistant: ```c
char *dns_comp(const char *host, char *comp, char *end) {
  int x;
  if (comp > end)
    return NULL;
  while (*host) {
    for (x = 0; host[x]; ++x)
      if (host[x] == '.')
        break;
    if (x > 63)
      return NULL;
    if (comp + x > end)
      return NULL;
    *(comp++) = x;
    while (x--)
      *(comp++) = *(host++);
    if (*host == '.')
      ++host;
  }
  if (comp + 1 > end)
    return NULL;
  *(comp++) = '\0';
  return comp;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dns_comp>:
  endbr64
  mov    %rdi,%rcx
  mov    %rdx,%r10
  cmp    %rdx,%rsi
  ja     L71
  cmpb   $0x0,(%rdi)
  jne    L72
L45:
  lea    0x1(%rsi),%rax
  cmp    %r10,%rax
  ja     L81
  movb   $0x0,(%rsi)
  ret
L148:
  cmp    $0x3f,%edi
  jg     L78
  lea    (%rsi,%r8,1),%rax
  cmp    %rax,%r10
  jb     L30
  lea    0x1(%rsi),%r11
  mov    %r8b,(%rsi)
  lea    -0x1(%rdi),%r9d
  test   %edi,%edi
  je     L8
  lea    -0x1(%r8),%edi
  mov    $0x0,%eax
L106:
  movzbl (%rcx,%rax,1),%edx
  mov    %dl,0x1(%rsi,%rax,1)
  mov    %rax,%rdx
  add    $0x1,%rax
  cmp    %rdi,%rdx
  jne    L106
  mov    %r9d,%eax
  add    $0x1,%rax
  add    %rax,%rcx
  lea    (%r11,%rax,1),%rsi
L69:
  cmpb   $0x2e,(%rcx)
  sete   %al
  movzbl %al,%eax
  add    %rax,%rcx
  cmpb   $0x0,(%rcx)
  je     L45
L72:
  movzbl (%rcx),%eax
  cmp    $0x2e,%al
  je     L7
  test   %al,%al
  je     L7
  mov    $0x1,%eax
L57:
  mov    %eax,%edi
  movzbl (%rcx,%rax,1),%edx
  mov    %rax,%r8
  add    $0x1,%rax
  cmp    $0x2e,%dl
  je     L148
  test   %dl,%dl
  jne    L57
  jmp    L148
L8:
  mov    %r11,%rsi
  jmp    L69
L71:
  mov    $0x0,%eax
  ret
L78:
  mov    $0x0,%eax
  ret
L30:
  mov    $0x0,%eax
  ret
L81:
  mov    $0x0,%eax
  ret
L127:
  mov    $0x0,%eax
  ret
L7:
  cmp    %rsi,%r10
  jb     L127
  movb   $0x0,(%rsi)
  lea    0x1(%rsi),%rsi
  jmp    L69
```

assistant: ```c
char *dns_comp(const char *host, char *comp, char *end) {
  int x;
  if (comp > end)
    return NULL;
  while (*host) {
    for (x = 0; host[x]; ++x)
      if (host[x] == '.')
        break;
    if (x > 63)
      return NULL;
    if (comp + x > end)
      return NULL;
    *(comp++) = x;
    while (x--)
      *(comp++) = *(host++);
    if (*host == '.')
      ++host;
  }
  if (comp + 1 > end)
    return NULL;
  *(comp++) = '\0';
  return comp;
}
```