ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<__fill_table>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%r12d
  mov    %rdx,%rbp
  mov    %rdx,%rdi
  call   <strlen@plt>
  test   %r12d,%r12d
  jle    L31
  mov    %rbx,%rdx
  lea    -0x1(%r12),%ecx
  lea    0x4(%rbx,%rcx,4),%rcx
L25:
  mov    %eax,(%rdx)
  add    $0x4,%rdx
  cmp    %rcx,%rdx
  jne    L25
L31:
  cmp    $0x1,%eax
  jle    L12
  lea    -0x2(%rax),%edx
  lea    0x1(%rbp,%rdx,1),%rdi
  mov    %rbp,%rdx
  lea    -0x1(%rax,%rbp,1),%esi
L34:
  movsbq (%rdx),%rax
  mov    %esi,%ecx
  sub    %edx,%ecx
  mov    %ecx,(%rbx,%rax,4)
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  jne    L34
L12:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void __fill_table(int *shift_table, int size, char *pattern) {
  int i;
  int len = strlen(pattern);
  for (i = 0; i < size; ++i) {
    shift_table[i] = len;
  }
  for (i = 0; i < len - 1; ++i) {
    shift_table[pattern[i]] = len - i - 1;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<__fill_table>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  mov    %rdx,%rdi
  push   %rbx
  mov    %rdx,%rbx
  call   <strlen@plt>
  test   %r12d,%r12d
  jle    L14
  lea    -0x1(%r12),%ecx
  mov    %rbp,%rdx
  lea    0x4(%rbp,%rcx,4),%rcx
  nopl   0x0(%rax,%rax,1)
L0:
  mov    %eax,(%rdx)
  add    $0x4,%rdx
  cmp    %rcx,%rdx
  jne    L0
L14:
  cmp    $0x1,%eax
  jle    L5
  lea    -0x2(%rax),%edx
  lea    -0x1(%rax,%rbx,1),%esi
  lea    0x1(%rbx,%rdx,1),%rdi
  mov    %rbx,%rdx
  nop
L12:
  movsbq (%rdx),%rax
  mov    %esi,%ecx
  sub    %edx,%ecx
  add    $0x1,%rdx
  mov    %ecx,0x0(%rbp,%rax,4)
  cmp    %rdi,%rdx
  jne    L12
L5:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void __fill_table(int *shift_table, int size, char *pattern) {
  int i;
  int len = strlen(pattern);
  for (i = 0; i < size; ++i) {
    shift_table[i] = len;
  }
  for (i = 0; i < len - 1; ++i) {
    shift_table[pattern[i]] = len - i - 1;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<__fill_table>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  mov    %rdx,%rdi
  push   %rbx
  mov    %rdx,%rbx
  call   <strlen@plt>
  mov    %eax,%edx
  test   %r12d,%r12d
  jle    L0
  lea    -0x1(%r12),%eax
  cmp    $0x2,%eax
  jbe    L82
  mov    %r12d,%ecx
  movd   %edx,%xmm1
  mov    %rbp,%rax
  shr    $0x2,%ecx
  pshufd $0x0,%xmm1,%xmm0
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rbp,%rcx,1),%rcx
  nopl   0x0(%rax)
L58:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L58
  mov    %r12d,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%r12b
  je     L0
L50:
  movslq %eax,%rcx
  mov    %edx,0x0(%rbp,%rcx,4)
  lea    0x0(,%rcx,4),%rsi
  lea    0x1(%rax),%ecx
  cmp    %ecx,%r12d
  jle    L0
  add    $0x2,%eax
  mov    %edx,0x4(%rbp,%rsi,1)
  cmp    %eax,%r12d
  jle    L0
  mov    %edx,0x8(%rbp,%rsi,1)
L0:
  cmp    $0x1,%edx
  jle    L71
  lea    -0x2(%rdx),%eax
  lea    -0x1(%rdx,%rbx,1),%esi
  lea    0x1(%rbx,%rax,1),%rdi
  mov    %rbx,%rax
  nopl   0x0(%rax,%rax,1)
L87:
  movsbq (%rax),%rdx
  mov    %esi,%ecx
  sub    %eax,%ecx
  add    $0x1,%rax
  mov    %ecx,0x0(%rbp,%rdx,4)
  cmp    %rdi,%rax
  jne    L87
L71:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L82:
  xor    %eax,%eax
  jmp    L50
```

assistant: ```c
void __fill_table(int *shift_table, int size, char *pattern) {
  int i;
  int len = strlen(pattern);
  for (i = 0; i < size; ++i) {
    shift_table[i] = len;
  }
  for (i = 0; i < len - 1; ++i) {
    shift_table[pattern[i]] = len - i - 1;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<break_string>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r13
  mov    (%rsi),%ebx
  movzbl (%rdi),%eax
  test   %al,%al
  je     L148
  mov    %rdi,%rdx
  mov    $0x0,%r12d
  jmp    L205
L39:
  cmpb   $0x20,(%rdx)
  je     L155
L28:
  add    $0x1,%rdx
  movzbl (%rdx),%eax
  test   %al,%al
  je     L6
L205:
  sub    $0x9,%eax
  cmp    $0x1,%al
  ja     L39
  movb   $0x20,(%rdx)
L155:
  add    $0x1,%r12d
  jmp    L28
L148:
  mov    $0x0,%r12d
L6:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%edi
  cmp    %r12d,%ebx
  jg     L124
  mov    %eax,%r9d
  mov    $0x1,%r11d
  mov    $0x1,%r10d
  cmp    $0x1,%ebx
  jg     L173
  jmp    L203
L124:
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  jne    L63
L194:
  add    $0x1,%r12d
  mov    %r12d,0x0(%r13)
L203:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L191:
  add    $0x1,%rbp
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L194
L63:
  cmp    $0x20,%al
  jne    L191
  movb   $0xa,0x0(%rbp)
  jmp    L191
L135:
  movb   $0xa,(%rdx)
  add    $0x1,%r11d
L22:
  cmp    %ebx,%r11d
  jge    L203
  add    $0x1,%r10d
  add    %edi,%r9d
  cmp    %r10d,%ebx
  je     L203
L173:
  mov    %r9d,%eax
  cltd
  idiv   %ebx
  movslq %eax,%r8
  add    %rbp,%r8
  mov    %r8,%rdx
  mov    %eax,%ecx
  mov    %r8,%rsi
  sub    %r8d,%eax
  jmp    L181
L213:
  add    $0x1,%rsi
  add    $0x1,%ecx
  sub    $0x1,%rdx
L181:
  mov    %eax,%r8d
  add    %edx,%r8d
  js     L23
  cmpb   $0x20,(%rdx)
  je     L135
L193:
  cmp    %ecx,%edi
  jle    L213
  cmpb   $0x20,(%rsi)
  jne    L213
  movb   $0xa,(%rsi)
  add    $0x1,%r11d
  jmp    L22
L23:
  cmp    %ecx,%edi
  jg     L193
  jmp    L22
```

assistant: ```c
void break_string(char *s, int *nlines) {
  int i, j, j0, c, nl, len;
  nl = (*nlines);
  for (i = c = 0; s[i] != '\0'; i++) {
    if (s[i] == '\n' || s[i] == '\t')
      s[i] = ' ';
    if (s[i] == ' ')
      c++;
  }
  len = strlen(s);
  if (nl > c) {
    for (i = 0; s[i] != '\0'; i++)
      if (s[i] == ' ')
        s[i] = '\n';
    (*nlines) = c + 1;
    return;
  }
  for (i = c = 1; i < nl; i++) {
    j0 = len * i / nl;
    for (j = 0; j0 - j >= 0 || j0 + j < len; j++) {
      if (j0 - j >= 0 && s[j0 - j] == ' ') {
        s[j0 - j] = '\n';
        c++;
        break;
      }
      if (j0 + j < len && s[j0 + j] == ' ') {
        s[j0 + j] = '\n';
        c++;
        break;
      }
    }
    if (c >= nl)
      break;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<break_string>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%eax
  mov    (%rsi),%ebp
  test   %al,%al
  je     L100
  mov    %rdi,%rdx
  xor    %r12d,%r12d
  jmp    L188
  nopl   0x0(%rax)
L189:
  cmpb   $0x20,(%rdx)
  je     L4
  movzbl 0x1(%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L62
L188:
  sub    $0x9,%eax
  cmp    $0x1,%al
  ja     L189
  movb   $0x20,(%rdx)
L4:
  movzbl 0x1(%rdx),%eax
  add    $0x1,%rdx
  add    $0x1,%r12d
  test   %al,%al
  jne    L188
L62:
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%edi
  cmp    %r12d,%ebp
  jg     L48
  mov    %eax,%r8d
  mov    $0x1,%r10d
  mov    $0x1,%r9d
  cmp    $0x1,%ebp
  jle    L55
  nopw   0x0(%rax,%rax,1)
L64:
  mov    %r8d,%eax
  cltd
  idiv   %ebp
  movslq %eax,%rdx
  mov    %eax,%esi
  add    %rbx,%rdx
  sub    %edx,%esi
  jmp    L68
  nopw   0x0(%rax,%rax,1)
L14:
  cmpb   $0x20,(%rdx)
  je     L66
  cmp    %edi,%eax
  jl     L186
L32:
  add    $0x1,%eax
  sub    $0x1,%rdx
L68:
  mov    %esi,%ecx
  add    %edx,%ecx
  jns    L14
  cmp    %edi,%eax
  jge    L132
L186:
  movslq %eax,%rcx
  add    %rbx,%rcx
  cmpb   $0x20,(%rcx)
  jne    L32
  movb   $0xa,(%rcx)
  add    $0x1,%r10d
L132:
  cmp    %ebp,%r10d
  jge    L55
  add    $0x1,%r9d
  add    %edi,%r8d
  cmp    %r9d,%ebp
  jne    L64
L55:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L1:
  cmp    $0x20,%al
  jne    L61
  movb   $0xa,(%rbx)
L61:
  add    $0x1,%rbx
L48:
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L1
  add    $0x1,%r12d
  mov    %r12d,0x0(%r13)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L66:
  movb   $0xa,(%rdx)
  add    $0x1,%r10d
  jmp    L132
L100:
  xor    %r12d,%r12d
  jmp    L62
```

assistant: ```c
void break_string(char *s, int *nlines) {
  int i, j, j0, c, nl, len;
  nl = (*nlines);
  for (i = c = 0; s[i] != '\0'; i++) {
    if (s[i] == '\n' || s[i] == '\t')
      s[i] = ' ';
    if (s[i] == ' ')
      c++;
  }
  len = strlen(s);
  if (nl > c) {
    for (i = 0; s[i] != '\0'; i++)
      if (s[i] == ' ')
        s[i] = '\n';
    (*nlines) = c + 1;
    return;
  }
  for (i = c = 1; i < nl; i++) {
    j0 = len * i / nl;
    for (j = 0; j0 - j >= 0 || j0 + j < len; j++) {
      if (j0 - j >= 0 && s[j0 - j] == ' ') {
        s[j0 - j] = '\n';
        c++;
        break;
      }
      if (j0 + j < len && s[j0 + j] == ' ') {
        s[j0 + j] = '\n';
        c++;
        break;
      }
    }
    if (c >= nl)
      break;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<break_string>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%eax
  mov    (%rsi),%ebp
  test   %al,%al
  je     L79
  mov    %rdi,%rdx
  xor    %r12d,%r12d
  jmp    L104
  nopl   0x0(%rax)
L206:
  cmpb   $0x20,(%rdx)
  je     L133
  movzbl 0x1(%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L179
L104:
  sub    $0x9,%eax
  cmp    $0x1,%al
  ja     L206
  movb   $0x20,(%rdx)
L133:
  movzbl 0x1(%rdx),%eax
  add    $0x1,%rdx
  add    $0x1,%r12d
  test   %al,%al
  jne    L104
L179:
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%edi
  cmp    %r12d,%ebp
  jg     L202
  mov    %eax,%r8d
  mov    $0x1,%r10d
  mov    $0x1,%r9d
  cmp    $0x1,%ebp
  jle    L111
  xchg   %ax,%ax
L124:
  mov    %r8d,%eax
  cltd
  idiv   %ebp
  movslq %eax,%rdx
  mov    %eax,%esi
  add    %rbx,%rdx
  sub    %edx,%esi
  jmp    L120
  nopw   0x0(%rax,%rax,1)
L50:
  cmpb   $0x20,(%rdx)
  je     L49
  cmp    %edi,%eax
  jl     L84
L137:
  add    $0x1,%eax
  sub    $0x1,%rdx
L120:
  mov    %esi,%ecx
  add    %edx,%ecx
  jns    L50
  cmp    %edi,%eax
  jge    L151
L84:
  movslq %eax,%rcx
  add    %rbx,%rcx
  cmpb   $0x20,(%rcx)
  jne    L137
  movb   $0xa,(%rcx)
  add    $0x1,%r10d
L151:
  cmp    %ebp,%r10d
  jge    L111
  add    $0x1,%r9d
  add    %edi,%r8d
  cmp    %r9d,%ebp
  jne    L124
L111:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L52:
  add    $0x1,%rbx
L202:
  movzbl (%rbx),%eax
  test   %al,%al
  je     L29
L21:
  cmp    $0x20,%al
  jne    L52
  movb   $0xa,(%rbx)
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  test   %al,%al
  jne    L21
L29:
  add    $0x1,%r12d
  mov    %r12d,0x0(%r13)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L49:
  movb   $0xa,(%rdx)
  add    $0x1,%r10d
  jmp    L151
L79:
  xor    %r12d,%r12d
  jmp    L179
```

assistant: ```c
void break_string(char *s, int *nlines) {
  int i, j, j0, c, nl, len;
  nl = (*nlines);
  for (i = c = 0; s[i] != '\0'; i++) {
    if (s[i] == '\n' || s[i] == '\t')
      s[i] = ' ';
    if (s[i] == ' ')
      c++;
  }
  len = strlen(s);
  if (nl > c) {
    for (i = 0; s[i] != '\0'; i++)
      if (s[i] == ' ')
        s[i] = '\n';
    (*nlines) = c + 1;
    return;
  }
  for (i = c = 1; i < nl; i++) {
    j0 = len * i / nl;
    for (j = 0; j0 - j >= 0 || j0 + j < len; j++) {
      if (j0 - j >= 0 && s[j0 - j] == ' ') {
        s[j0 - j] = '\n';
        c++;
        break;
      }
      if (j0 + j < len && s[j0 + j] == ' ') {
        s[j0 + j] = '\n';
        c++;
        break;
      }
    }
    if (c >= nl)
      break;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmp    $0x1,%eax
  sete   %al
  mov    %al,-0x19(%rbp)
  jmp    L10
L0:
  mov    -0x18(%rbp),%rax
  add    %rax,-0x10(%rbp)
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmp    $0x1,%eax
  sete   %al
  mov    %al,-0x19(%rbp)
L10:
  cmpb   $0x0,-0x19(%rbp)
  jne    L0
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D35
type_guessed: string
name: extract_symbolic_data
arguments: label: D15
type_guessed: string
name: extract_symbolic_data
arguments: label: D10
type_guessed: string
name: extract_symbolic_data
arguments: label: D19
type_guessed: string
name: extract_symbolic_data
arguments: label: D51
type_guessed: string

tool: D35: Please enter an integer to be summed 

tool: D15: (q to quit): 

tool: D10: %ld

tool: D19: Please enter next integer (q to quit): 

tool: D51: Those integers sum to %ld.


assistant: ```c
int main(void) {
  long num;
  long sum = 0L;
  _Bool input_is_good;
  printf("Please enter an integer to be summed ");
  printf("(q to quit): ");
  input_is_good = (scanf("%ld", &num) == 1);
  while (input_is_good) {
    sum = sum + num;
    printf("Please enter next integer (q to quit): ");
    input_is_good = (scanf("%ld", &num) == 1);
  }
  printf("Those integers sum to %ld.\n", sum);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```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    D44(%rip),%rdi
  call   <printf@plt>
  lea    D58(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D50(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L8
  mov    $0x0,%ebx
  lea    D27(%rip),%r13
  mov    %rsp,%r12
  lea    D50(%rip),%rbp
L5:
  add    (%rsp),%rbx
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L5
L36:
  mov    %rbx,%rsi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  mov    $0x0,%ebx
  jmp    L36
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D44
data_type: string
name: read_data
arguments: data_tag: D58
data_type: string
name: read_data
arguments: data_tag: D50
data_type: string
name: read_data
arguments: data_tag: D27
data_type: string
name: read_data
arguments: data_tag: D3
data_type: string

tool: D44: Please enter an integer to be summed 

tool: D58: (q to quit): 

tool: D50: %ld

tool: D27: Please enter next integer (q to quit): 

tool: D3: Those integers sum to %ld.


assistant: ```c
int main(void) {
  long num;
  long sum = 0L;
  _Bool input_is_good;
  printf("Please enter an integer to be summed ");
  printf("(q to quit): ");
  input_is_good = (scanf("%ld", &num) == 1);
  while (input_is_good) {
    sum = sum + num;
    printf("Please enter next integer (q to quit): ");
    input_is_good = (scanf("%ld", &num) == 1);
  }
  printf("Those integers sum to %ld.\n", sum);
  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    D12(%rip),%rdi
  push   %r12
  lea    D11(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <printf@plt>
  lea    D2(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L21
  lea    D27(%rip),%r13
  nopl   (%rax)
L10:
  mov    %r13,%rdi
  xor    %eax,%eax
  add    (%rsp),%rbp
  call   <printf@plt>
  xor    %eax,%eax
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L10
L21:
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D25(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D12
type_infered: string
name: parse_data
arguments: tag: D11
type_infered: string
name: parse_data
arguments: tag: D2
type_infered: string
name: parse_data
arguments: tag: D27
type_infered: string
name: parse_data
arguments: tag: D25
type_infered: string

tool: D12: Please enter an integer to be summed 

tool: D11: %ld

tool: D2: (q to quit): 

tool: D27: Please enter next integer (q to quit): 

tool: D25: Those integers sum to %ld.


assistant: ```c
int main(void) {
  long num;
  long sum = 0L;
  _Bool input_is_good;
  printf("Please enter an integer to be summed ");
  printf("(q to quit): ");
  input_is_good = (scanf("%ld", &num) == 1);
  while (input_is_good) {
    sum = sum + num;
    printf("Please enter next integer (q to quit): ");
    input_is_good = (scanf("%ld", &num) == 1);
  }
  printf("Those integers sum to %ld.\n", sum);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  lea    D33(%rip),%rdi
  push   %r12
  lea    D36(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <printf@plt>
  lea    D41(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L13
  lea    D21(%rip),%r13
  nopl   (%rax)
L8:
  mov    %r13,%rdi
  xor    %eax,%eax
  add    (%rsp),%rbp
  call   <printf@plt>
  xor    %eax,%eax
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L8
L13:
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D46(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D33
type_infered: string
name: retrieve_data
arguments: data_label: D36
type_infered: string
name: retrieve_data
arguments: data_label: D41
type_infered: string
name: retrieve_data
arguments: data_label: D21
type_infered: string
name: retrieve_data
arguments: data_label: D46
type_infered: string

tool: D33: Please enter an integer to be summed 

tool: D36: %ld

tool: D41: (q to quit): 

tool: D21: Please enter next integer (q to quit): 

tool: D46: Those integers sum to %ld.


assistant: ```c
int main(void) {
  long num;
  long sum = 0L;
  _Bool input_is_good;
  printf("Please enter an integer to be summed ");
  printf("(q to quit): ");
  input_is_good = (scanf("%ld", &num) == 1);
  while (input_is_good) {
    sum = sum + num;
    printf("Please enter next integer (q to quit): ");
    input_is_good = (scanf("%ld", &num) == 1);
  }
  printf("Those integers sum to %ld.\n", sum);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<REMOVE_WS>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  jmp    L12
L21:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
L12:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L21
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  je     L21
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void REMOVE_WS(char **LINE) {
  while (((**LINE) == ' ') || ((**LINE) == '	'))
    (*LINE)++;
}
```
user: Translate the below assembly code into C programming language:

```asm
<REMOVE_WS>:
  endbr64
  mov    (%rdi),%rax
  movzbl (%rax),%edx
  cmp    $0x20,%dl
  je     L8
  cmp    $0x9,%dl
  jne    L10
L8:
  add    $0x1,%rax
L34:
  mov    %rax,(%rdi)
  movzbl (%rax),%edx
  add    $0x1,%rax
  cmp    $0x20,%dl
  je     L34
  cmp    $0x9,%dl
  je     L34
L10:
  ret
```

assistant: ```c
void REMOVE_WS(char **LINE) {
  while (((**LINE) == ' ') || ((**LINE) == '	'))
    (*LINE)++;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<REMOVE_WS>:
  endbr64
  mov    (%rdi),%rax
  movzbl (%rax),%edx
  cmp    $0x20,%dl
  je     L37
  cmp    $0x9,%dl
  jne    L1
L37:
  add    $0x1,%rax
  nopl   0x0(%rax,%rax,1)
L7:
  mov    %rax,(%rdi)
  movzbl (%rax),%edx
  add    $0x1,%rax
  cmp    $0x20,%dl
  je     L7
  cmp    $0x9,%dl
  je     L7
L1:
  ret
```

assistant: ```c
void REMOVE_WS(char **LINE) {
  while (((**LINE) == ' ') || ((**LINE) == '	'))
    (*LINE)++;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<REMOVE_WS>:
  endbr64
  mov    (%rdi),%rax
  movzbl (%rax),%edx
  cmp    $0x20,%dl
  je     L13
  cmp    $0x9,%dl
  jne    L24
L13:
  add    $0x1,%rax
  nopl   0x0(%rax,%rax,1)
L35:
  mov    %rax,(%rdi)
  movzbl (%rax),%edx
  add    $0x1,%rax
  cmp    $0x20,%dl
  je     L35
  cmp    $0x9,%dl
  je     L35
L24:
  ret
```

assistant: ```c
void REMOVE_WS(char **LINE) {
  while (((**LINE) == ' ') || ((**LINE) == '	'))
    (*LINE)++;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<wpa_cipher_bits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x10,-0x4(%rbp)
  je     L64
  cmpl   $0x10,-0x4(%rbp)
  jg     L94
  cmpl   $0x8,-0x4(%rbp)
  je     L96
  cmpl   $0x8,-0x4(%rbp)
  jg     L94
  cmpl   $0x2,-0x4(%rbp)
  je     L12
  cmpl   $0x4,-0x4(%rbp)
  je     L9
  jmp    L94
L64:
  mov    $0x80,%eax
  jmp    L8
L96:
  mov    $0x100,%eax
  jmp    L8
L9:
  mov    $0x68,%eax
  jmp    L8
L12:
  mov    $0x28,%eax
  jmp    L8
L94:
  mov    $0x0,%eax
L8:
  pop    %rbp
  ret
```

assistant: ```c
int wpa_cipher_bits(int cipher) {
  switch (cipher) {
  case (1 << (4)):
    return 128;
  case (1 << (3)):
    return 256;
  case (1 << (2)):
    return 104;
  case (1 << (1)):
    return 40;
  default:
    return 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<wpa_cipher_bits>:
  endbr64
  cmp    $0x8,%edi
  je     L20
  jg     L13
  mov    $0x28,%eax
  cmp    $0x2,%edi
  je     L15
  cmp    $0x4,%edi
  mov    $0x0,%eax
  mov    $0x68,%edx
  cmove  %edx,%eax
  ret
L13:
  cmp    $0x10,%edi
  sete   %al
  movzbl %al,%eax
  shl    $0x7,%eax
  ret
L20:
  mov    $0x100,%eax
L15:
  ret
```

assistant: ```c
int wpa_cipher_bits(int cipher) {
  switch (cipher) {
  case (1 << (4)):
    return 128;
  case (1 << (3)):
    return 256;
  case (1 << (2)):
    return 104;
  case (1 << (1)):
    return 40;
  default:
    return 0;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<wpa_cipher_bits>:
  endbr64
  sub    $0x2,%edi
  xor    %eax,%eax
  cmp    $0xe,%edi
  ja     L6
  lea    D1(%rip),%rax
  mov    (%rax,%rdi,4),%eax
L6:
  ret
```

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

tool: D1: (

assistant: ```c
int wpa_cipher_bits(int cipher) {
  switch (cipher) {
  case (1 << (4)):
    return 128;
  case (1 << (3)):
    return 256;
  case (1 << (2)):
    return 104;
  case (1 << (1)):
    return 40;
  default:
    return 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<wpa_cipher_bits>:
  endbr64
  sub    $0x2,%edi
  xor    %eax,%eax
  cmp    $0xe,%edi
  ja     L1
  lea    D6(%rip),%rax
  mov    (%rax,%rdi,4),%eax
L1:
  ret
```

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

tool: D6: (

assistant: ```c
int wpa_cipher_bits(int cipher) {
  switch (cipher) {
  case (1 << (4)):
    return 128;
  case (1 << (3)):
    return 256;
  case (1 << (2)):
    return 104;
  case (1 << (1)):
    return 40;
  default:
    return 0;
  }
}
```