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
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xfc0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xfc0(%rbp)
  movl   $0x0,-0xfbc(%rbp)
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L76
L11:
  lea    -0xfb0(%rbp),%rdx
  mov    -0xfbc(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xfbc(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  cmp    $0x2a,%eax
  je     L51
  addl   $0x1,-0xfc0(%rbp)
  addl   $0x1,-0xfbc(%rbp)
L76:
  cmpl   $0x3e7,-0xfc0(%rbp)
  jle    L11
  jmp    L88
L51:
  nop
L88:
  movl   $0x0,-0xfbc(%rbp)
  jmp    L116
L35:
  mov    -0xfbc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xfb8(%rbp)
  jmp    L57
L6:
  mov    -0xfbc(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%edx
  mov    -0xfb8(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  jle    L104
  mov    -0xfbc(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  mov    %eax,-0xfb4(%rbp)
  mov    -0xfb8(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%edx
  mov    -0xfbc(%rbp),%eax
  cltq
  mov    %edx,-0xfb0(%rbp,%rax,4)
  mov    -0xfb8(%rbp),%eax
  cltq
  mov    -0xfb4(%rbp),%edx
  mov    %edx,-0xfb0(%rbp,%rax,4)
L104:
  addl   $0x1,-0xfb8(%rbp)
L57:
  mov    -0xfb8(%rbp),%eax
  cmp    -0xfc0(%rbp),%eax
  jl     L6
  addl   $0x1,-0xfbc(%rbp)
L116:
  mov    -0xfbc(%rbp),%eax
  cmp    -0xfc0(%rbp),%eax
  jl     L35
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0xfbc(%rbp)
  jmp    L37
L53:
  mov    -0xfbc(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xfbc(%rbp)
L37:
  mov    -0xfbc(%rbp),%eax
  cmp    -0xfc0(%rbp),%eax
  jl     L53
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L91
  call   <__stack_chk_fail@plt>
L91:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D10
type_infered: string
name: extract_symbolic_data
arguments: label: D39
type_infered: string
name: extract_symbolic_data
arguments: label: D26
type_infered: string
name: extract_symbolic_data
arguments: label: D16
type_infered: string

tool: D10: Enter numbers for sorting, stop with 42: 

tool: D39: %d

tool: D26: Sorted list: 

tool: D16: %d 


assistant: ```c
int main() {
  int a[1000];
  int s = 0, i = 0, j, temp;
  printf("Enter numbers for sorting, stop with 42: \n");
  while (s < 1000) {
    scanf("%d", &a[i]);
    if (a[i] == 42) {
      break;
    }
    s++;
    i++;
  }
  for (i = 0; i < s; i++) {
    for (j = i + 1; j < s; j++) {
      if (a[i] > a[j]) {
        temp = a[i];
        a[i] = a[j];
        a[j] = temp;
      }
    }
  }
  printf("Sorted list: \n");
  for (i = 0; i < s; i++) {
    printf("%d \n", a[i]);
  }
  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
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xfb8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  lea    D15(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rbp
  mov    $0x0,%r12d
  lea    D0(%rip),%r13
L50:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  cmpl   $0x2a,0x0(%rbp)
  je     L23
  add    $0x1,%r12d
  add    $0x4,%rbp
  cmp    $0x3e8,%r12d
  jne    L50
  jmp    L60
L19:
  add    $0x4,%rax
  cmp    %rdi,%rax
  je     L124
L7:
  mov    (%rsi),%edx
  mov    0x4(%rax),%ecx
  cmp    %ecx,%edx
  jle    L19
  mov    %ecx,(%rsi)
  mov    %edx,0x4(%rax)
  jmp    L19
L124:
  add    $0x1,%r8
  add    $0x4,%rsi
L41:
  cmp    %r12,%r8
  je     L88
  mov    %r9d,%eax
  sub    %r8d,%eax
  lea    -0x1(%r8,%rax,1),%rax
  lea    (%r10,%rax,4),%rdi
  mov    %rsi,%rax
  jmp    L7
L23:
  test   %r12d,%r12d
  jle    L33
L60:
  lea    -0x1(%r12),%r9d
  mov    %r9d,%ebp
  mov    %r12d,%r12d
  mov    %rbx,%rsi
  mov    $0x1,%r8d
  lea    0x4(%rbx),%r10
  jmp    L41
L33:
  lea    D1(%rip),%rdi
  call   <puts@plt>
  jmp    L40
L88:
  lea    D1(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rbx,%rbp,4),%r12
  lea    D38(%rip),%rbp
L96:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %rbx,%r12
  jne    L96
L40:
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  mov    $0x0,%eax
  add    $0xfb8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D15
type_guessed: string
name: extract_symbolic_data
arguments: label: D0
type_guessed: string
name: extract_symbolic_data
arguments: label: D1
type_guessed: string
name: extract_symbolic_data
arguments: label: D38
type_guessed: string

tool: D15: Enter numbers for sorting, stop with 42: 

tool: D0: %d

tool: D1: Sorted list: 

tool: D38: %d 


assistant: ```c
int main() {
  int a[1000];
  int s = 0, i = 0, j, temp;
  printf("Enter numbers for sorting, stop with 42: \n");
  while (s < 1000) {
    scanf("%d", &a[i]);
    if (a[i] == 42) {
      break;
    }
    s++;
    i++;
  }
  for (i = 0; i < s; i++) {
    for (j = i + 1; j < s; j++) {
      if (a[i] > a[j]) {
        temp = a[i];
        a[i] = a[j];
        a[j] = temp;
      }
    }
  }
  printf("Sorted list: \n");
  for (i = 0; i < s; i++) {
    printf("%d \n", a[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
  lea    D27(%rip),%rdi
  lea    D2(%rip),%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0xfb8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <puts@plt>
  mov    %rbx,%r12
  jmp    L45
  nopl   0x0(%rax)
L79:
  add    $0x1,%ebp
  add    $0x4,%r12
  cmp    $0x3e8,%ebp
  je     L20
L45:
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  cmpl   $0x2a,(%r12)
  jne    L79
  test   %ebp,%ebp
  je     L36
L20:
  mov    $0x1,%r8d
  lea    -0x1(%rbp),%r10d
  lea    0x4(%rbx),%r9
  mov    %rbx,%rsi
  cmp    %r8d,%ebp
  jle    L68
  nopw   0x0(%rax,%rax,1)
L55:
  mov    %r10d,%eax
  sub    %r8d,%eax
  lea    -0x1(%r8,%rax,1),%rax
  lea    (%r9,%rax,4),%rdi
  mov    %rsi,%rax
  nopw   0x0(%rax,%rax,1)
L42:
  mov    (%rsi),%edx
  mov    0x4(%rax),%ecx
  cmp    %ecx,%edx
  jle    L104
  mov    %ecx,(%rsi)
  mov    %edx,0x4(%rax)
L104:
  add    $0x4,%rax
  cmp    %rdi,%rax
  jne    L42
  add    $0x1,%r8
  add    $0x4,%rsi
  cmp    %r8d,%ebp
  jg     L55
L68:
  lea    D44(%rip),%rdi
  xor    %r12d,%r12d
  lea    D29(%rip),%r13
  call   <puts@plt>
  nopw   0x0(%rax,%rax,1)
L16:
  mov    (%rbx,%r12,4),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%r12
  call   <printf@plt>
  cmp    %r12d,%ebp
  jg     L16
L37:
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L83
  add    $0xfb8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L36:
  lea    D44(%rip),%rdi
  call   <puts@plt>
  jmp    L37
L83:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D27
data_type: string
name: unpack_data
arguments: tag: D2
data_type: string
name: unpack_data
arguments: tag: D44
data_type: string
name: unpack_data
arguments: tag: D29
data_type: string

tool: D27: Enter numbers for sorting, stop with 42: 

tool: D2: %d

tool: D44: Sorted list: 

tool: D29: %d 


assistant: ```c
int main() {
  int a[1000];
  int s = 0, i = 0, j, temp;
  printf("Enter numbers for sorting, stop with 42: \n");
  while (s < 1000) {
    scanf("%d", &a[i]);
    if (a[i] == 42) {
      break;
    }
    s++;
    i++;
  }
  for (i = 0; i < s; i++) {
    for (j = i + 1; j < s; j++) {
      if (a[i] > a[j]) {
        temp = a[i];
        a[i] = a[j];
        a[j] = temp;
      }
    }
  }
  printf("Sorted list: \n");
  for (i = 0; i < s; i++) {
    printf("%d \n", a[i]);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r13
  lea    D0(%rip),%rdi
  lea    D10(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0xfb8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <puts@plt>
  mov    %rbp,%r12
  jmp    L54
  nopl   0x0(%rax)
L77:
  add    $0x1,%ebx
  add    $0x4,%r12
  cmp    $0x3e8,%ebx
  je     L65
L54:
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  cmpl   $0x2a,(%r12)
  jne    L77
  test   %ebx,%ebx
  je     L47
  cmp    $0x1,%ebx
  je     L13
L65:
  lea    0x4(%rsp),%rdi
  mov    $0x1,%r8d
  nopl   0x0(%rax)
L102:
  mov    %rdi,%rax
  mov    %r8d,%edx
  xchg   %ax,%ax
L20:
  mov    -0x4(%rdi),%ecx
  mov    (%rax),%esi
  cmp    %esi,%ecx
  jle    L93
  mov    %esi,-0x4(%rdi)
  mov    %ecx,(%rax)
L93:
  add    $0x1,%edx
  add    $0x4,%rax
  cmp    %ebx,%edx
  jl     L20
  add    $0x1,%r8d
  add    $0x4,%rdi
  cmp    %r8d,%ebx
  jne    L102
L13:
  lea    D37(%rip),%rdi
  xor    %r12d,%r12d
  lea    D14(%rip),%r13
  call   <puts@plt>
  nopl   0x0(%rax)
L66:
  mov    0x0(%rbp,%r12,4),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%r12
  call   <printf@plt>
  cmp    %r12d,%ebx
  jg     L66
L73:
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  add    $0xfb8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L47:
  lea    D37(%rip),%rdi
  call   <puts@plt>
  jmp    L73
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D0
type: string
name: retrieve_data
arguments: tag: D10
type: string
name: retrieve_data
arguments: tag: D37
type: string
name: retrieve_data
arguments: tag: D14
type: string

tool: D0: Enter numbers for sorting, stop with 42: 

tool: D10: %d

tool: D37: Sorted list: 

tool: D14: %d 


assistant: ```c
int main() {
  int a[1000];
  int s = 0, i = 0, j, temp;
  printf("Enter numbers for sorting, stop with 42: \n");
  while (s < 1000) {
    scanf("%d", &a[i]);
    if (a[i] == 42) {
      break;
    }
    s++;
    i++;
  }
  for (i = 0; i < s; i++) {
    for (j = i + 1; j < s; j++) {
      if (a[i] > a[j]) {
        temp = a[i];
        a[i] = a[j];
        a[j] = temp;
      }
    }
  }
  printf("Sorted list: \n");
  for (i = 0; i < s; i++) {
    printf("%d \n", a[i]);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<capitalisepath>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    D5(%rip),%rax
  movq   $0x0,(%rax)
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,0x0(%r13)
  test   %rax,%rax
  je     L163
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  sub    $0x1,%ebx
  je     L84
  mov    %ebx,%eax
  add    %r12,%rax
L58:
  cmpb   $0x2f,(%rax)
  je     L69
  sub    $0x1,%rax
  sub    $0x1,%ebx
  jne    L58
  jmp    L84
L69:
  test   %ebx,%ebx
  je     L84
  mov    %ebx,%r14d
  mov    $0x0,%ebp
  movb   $0x0,0xf(%rsp)
  jmp    L83
L85:
  cmp    $0x2f,%dil
  jne    L127
  mov    0x0(%r13),%rax
  movb   $0x2f,(%rax)
  jmp    L64
L127:
  mov    0x0(%r13),%r15
  movzbl %dil,%edi
  call   <toupper@plt>
  mov    %al,(%r15)
  jmp    L64
L57:
  mov    %rbp,%r15
  add    0x0(%r13),%r15
  movzbl %dil,%edi
  call   <toupper@plt>
  mov    %al,(%r15)
L64:
  add    $0x1,%rbp
  cmp    %r14,%rbp
  je     L84
L83:
  movzbl (%r12,%rbp,1),%edi
  movzbl 0xf(%rsp),%eax
  mov    %dil,0xf(%rsp)
  test   %dil,%dil
  je     L84
  test   %ebp,%ebp
  je     L85
  cmp    $0x2f,%al
  je     L57
  cmp    $0x20,%al
  je     L57
  mov    0x0(%r13),%rax
  mov    %dil,(%rax,%rbp,1)
  jmp    L64
L8:
  movzbl (%r12,%rbp,1),%edx
  mov    0x0(%r13),%rax
  mov    %dl,(%rax,%rbp,1)
  add    $0x1,%ebx
L84:
  mov    %ebx,%ebp
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %rax,%rbp
  jb     L8
  mov    0x0(%r13),%rdx
  movb   $0x0,(%rdx,%rax,1)
  mov    0x0(%r13),%rdi
  call   <strlen@plt>
  mov    D0(%rip),%rdx
  cmp    %rax,(%rdx)
  mov    $0x0,%edx
  cmovb  %edx,%eax
L61:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L163:
  mov    $0xffffffff,%eax
  jmp    L61
```

assistant: ```c
int capitalisepath(const char *string, char **result) {
  unsigned char lastchar = 0;
  unsigned long len;
  errno = 0;
  if (!(*result = (char *)malloc(strlen(string) + 1))) {
    ;
    return -1;
  }
  unsigned int max;
  for (max = strlen(string) - 1; max > 0; --max) {
    unsigned char c = string[max];
    if (c == '/')
      break;
  }
  for (unsigned int n = 0; n < max; n++) {
    unsigned char c = string[n];
    if (0 == c)
      break;
    if (0 == n) {
      lastchar = c;
      if (lastchar != '/')
        (*result)[n] = toupper(c);
      else
        (*result)[n] = c;
    } else {
      if (lastchar == '/' || lastchar == ' ')
        (*result)[n] = toupper(c);
      else
        (*result)[n] = c;
      lastchar = c;
    }
  }
  for (; max < strlen(string); max++) {
    unsigned char c = string[max];
    (*result)[max] = c;
  }
  (*result)[strlen(string)] = '\0';
  len = strlen(*result);
  if (len > INT_MAX) {
    return 0;
  } else {
    return (int)len;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fls>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x20,-0x4(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L28
  mov    $0x0,%eax
  jmp    L37
L28:
  mov    -0x14(%rbp),%eax
  mov    $0x0,%ax
  test   %eax,%eax
  jne    L43
  shll   $0x10,-0x14(%rbp)
  subl   $0x10,-0x4(%rbp)
L43:
  mov    -0x14(%rbp),%eax
  and    $0xff000000,%eax
  test   %eax,%eax
  jne    L65
  shll   $0x8,-0x14(%rbp)
  subl   $0x8,-0x4(%rbp)
L65:
  mov    -0x14(%rbp),%eax
  and    $0xf0000000,%eax
  test   %eax,%eax
  jne    L47
  shll   $0x4,-0x14(%rbp)
  subl   $0x4,-0x4(%rbp)
L47:
  mov    -0x14(%rbp),%eax
  and    $0xc0000000,%eax
  test   %eax,%eax
  jne    L12
  shll   $0x2,-0x14(%rbp)
  subl   $0x2,-0x4(%rbp)
L12:
  cmpl   $0x0,-0x14(%rbp)
  js     L33
  shll   -0x14(%rbp)
  subl   $0x1,-0x4(%rbp)
L33:
  mov    -0x4(%rbp),%eax
L37:
  pop    %rbp
  ret
```

assistant: ```c
int fls(int x) {
  int r = 32;
  if (!x)
    return 0;
  if (!(x & 0xffff0000u)) {
    x <<= 16;
    r -= 16;
  }
  if (!(x & 0xff000000u)) {
    x <<= 8;
    r -= 8;
  }
  if (!(x & 0xf0000000u)) {
    x <<= 4;
    r -= 4;
  }
  if (!(x & 0xc0000000u)) {
    x <<= 2;
    r -= 2;
  }
  if (!(x & 0x80000000u)) {
    x <<= 1;
    r -= 1;
  }
  return r;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<fls>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  je     L38
  mov    $0x20,%eax
  test   $0xffff0000,%edi
  jne    L16
  shl    $0x10,%edi
  mov    $0x10,%eax
L16:
  test   $0xff000000,%edi
  jne    L39
  shl    $0x8,%edi
  sub    $0x8,%eax
L39:
  test   $0xf0000000,%edi
  jne    L22
  shl    $0x4,%edi
  sub    $0x4,%eax
L22:
  test   $0xc0000000,%edi
  jne    L25
  shl    $0x2,%edi
  sub    $0x2,%eax
L25:
  cmp    $0x80000000,%edi
  sbb    $0x0,%eax
L38:
  ret
```

assistant: ```c
int fls(int x) {
  int r = 32;
  if (!x)
    return 0;
  if (!(x & 0xffff0000u)) {
    x <<= 16;
    r -= 16;
  }
  if (!(x & 0xff000000u)) {
    x <<= 8;
    r -= 8;
  }
  if (!(x & 0xf0000000u)) {
    x <<= 4;
    r -= 4;
  }
  if (!(x & 0xc0000000u)) {
    x <<= 2;
    r -= 2;
  }
  if (!(x & 0x80000000u)) {
    x <<= 1;
    r -= 1;
  }
  return r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fls>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  je     L31
  mov    $0x20,%eax
  test   $0xffff0000,%edi
  jne    L19
  shl    $0x10,%edi
  mov    $0x10,%eax
L19:
  test   $0xff000000,%edi
  jne    L39
  shl    $0x8,%edi
  sub    $0x8,%eax
L39:
  test   $0xf0000000,%edi
  jne    L30
  shl    $0x4,%edi
  sub    $0x4,%eax
L30:
  test   $0xc0000000,%edi
  jne    L18
  shl    $0x2,%edi
  sub    $0x2,%eax
L18:
  cmp    $0x80000000,%edi
  sbb    $0x0,%eax
L31:
  ret
```

assistant: ```c
int fls(int x) {
  int r = 32;
  if (!x)
    return 0;
  if (!(x & 0xffff0000u)) {
    x <<= 16;
    r -= 16;
  }
  if (!(x & 0xff000000u)) {
    x <<= 8;
    r -= 8;
  }
  if (!(x & 0xf0000000u)) {
    x <<= 4;
    r -= 4;
  }
  if (!(x & 0xc0000000u)) {
    x <<= 2;
    r -= 2;
  }
  if (!(x & 0x80000000u)) {
    x <<= 1;
    r -= 1;
  }
  return r;
}
```
user: What is the c source code of the assembly code below:

```asm
<fls>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  je     L27
  test   $0xffff0000,%edi
  je     L10
  mov    $0x18,%edx
  mov    $0x20,%eax
L14:
  test   $0xff000000,%edi
  jne    L12
  shl    $0x8,%edi
  mov    %edx,%eax
L12:
  test   $0xf0000000,%edi
  jne    L2
  shl    $0x4,%edi
  sub    $0x4,%eax
L2:
  test   $0xc0000000,%edi
  jne    L17
  shl    $0x2,%edi
  sub    $0x2,%eax
L17:
  cmp    $0x80000000,%edi
  sbb    $0x0,%eax
L27:
  ret
  nop
L10:
  shl    $0x10,%edi
  mov    $0x8,%edx
  mov    $0x10,%eax
  jmp    L14
```

assistant: ```c
int fls(int x) {
  int r = 32;
  if (!x)
    return 0;
  if (!(x & 0xffff0000u)) {
    x <<= 16;
    r -= 16;
  }
  if (!(x & 0xff000000u)) {
    x <<= 8;
    r -= 8;
  }
  if (!(x & 0xf0000000u)) {
    x <<= 4;
    r -= 4;
  }
  if (!(x & 0xc0000000u)) {
    x <<= 2;
    r -= 2;
  }
  if (!(x & 0x80000000u)) {
    x <<= 1;
    r -= 1;
  }
  return r;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x14(%rbp),%rcx
  lea    -0x18(%rbp),%rdx
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x1,-0x10(%rbp)
  jmp    L50
L67:
  movl   $0x1,-0xc(%rbp)
  jmp    L26
L2:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  je     L69
  cmpl   $0x1,-0xc(%rbp)
  je     L69
  cmpl   $0x1,-0x10(%rbp)
  je     L69
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jne    L58
L69:
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L71
L58:
  mov    $0x20,%edi
  call   <putchar@plt>
L71:
  addl   $0x1,-0xc(%rbp)
L26:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jle    L2
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x10(%rbp)
L50:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jle    L67
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L47
  call   <__stack_chk_fail@plt>
L47:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D12
type: string
name: retrieve_data
arguments: data_label: D6
type: string

tool: D12: %d %d %d

tool: D6: %d

assistant: ```c
int main() {
  int N, width, height;
  scanf("%d %d %d", &N, &width, &height);
  int i, j;
  for (i = 1; i <= height; i++) {
    for (j = 1; j <= width; j++) {
      if (j == width || j == 1 || i == 1 || i == height)
        printf("%d", N);
      else
        printf(" ");
    }
    printf("\n");
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  mov    $0x1,%ebp
  lea    D12(%rip),%r12
  cmpl   $0x0,0x14(%rsp)
  jg     L98
L75:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L47
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L59:
  mov    0xc(%rsp),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L18:
  add    $0x1,%ebx
  mov    0x10(%rsp),%eax
  cmp    %ebx,%eax
  jl     L31
L13:
  cmp    %ebx,%eax
  je     L59
  cmp    $0x1,%ebx
  je     L59
  cmp    $0x1,%ebp
  je     L59
  cmp    %ebp,0x14(%rsp)
  je     L59
  mov    $0x20,%edi
  call   <putchar@plt>
  jmp    L18
L31:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%ebp
  cmp    %ebp,0x14(%rsp)
  jl     L75
L98:
  mov    0x10(%rsp),%eax
  mov    $0x1,%ebx
  test   %eax,%eax
  jg     L13
  jmp    L31
L47:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D13
type_guessed: string
name: read_data
arguments: data_tag: D12
type_guessed: string

tool: D13: %d %d %d

tool: D12: %d

assistant: ```c
int main() {
  int N, width, height;
  scanf("%d %d %d", &N, &width, &height);
  int i, j;
  for (i = 1; i <= height; i++) {
    for (j = 1; j <= width; j++) {
      if (j == width || j == 1 || i == 1 || i == height)
        printf("%d", N);
      else
        printf(" ");
    }
    printf("\n");
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D11(%rip),%rdi
  lea    D3(%rip),%r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jle    L103
  nopl   0x0(%rax,%rax,1)
L64:
  mov    0x10(%rsp),%eax
  mov    $0x1,%ebx
  test   %eax,%eax
  jg     L106
  jmp    L49
  nop
L92:
  mov    $0x20,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  mov    0x10(%rsp),%eax
  cmp    %ebx,%eax
  jl     L49
  nopl   (%rax)
L106:
  cmp    %ebx,%eax
  je     L24
  cmp    $0x1,%ebx
  je     L24
  cmp    $0x1,%ebp
  je     L24
  cmp    %ebp,0x14(%rsp)
  jne    L92
L24:
  mov    0xc(%rsp),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  mov    0x10(%rsp),%eax
  cmp    %ebx,%eax
  jge    L106
L49:
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  cmp    %ebp,0x14(%rsp)
  jge    L64
L103:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L69
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L69:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D11
type_infered: string
name: retrieve_data
arguments: data_tag: D3
type_infered: string

tool: D11: %d %d %d

tool: D3: %d

assistant: ```c
int main() {
  int N, width, height;
  scanf("%d %d %d", &N, &width, &height);
  int i, j;
  for (i = 1; i <= height; i++) {
    for (j = 1; j <= width; j++) {
      if (j == width || j == 1 || i == 1 || i == height)
        printf("%d", N);
      else
        printf(" ");
    }
    printf("\n");
  }
  return 0;
}
```