ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r14
  lea    D23(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L18
  mov    $0x1,%r13d
  lea    D20(%rip),%rbp
  lea    D19(%rip),%r14
  nopl   (%rax)
L55:
  mov    $0x1,%ebx
  nopl   (%rax)
L8:
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%r13d
  jge    L8
  mov    %r14,%rdi
  add    $0x1,%r13d
  call   <printf@plt>
  mov    0x4(%rsp),%r12d
  cmp    %r13d,%r12d
  jge    L55
  test   %r12d,%r12d
  jle    L18
  nopl   0x0(%rax,%rax,1)
L14:
  mov    $0x1,%ebx
  nopl   (%rax)
L51:
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%r12d
  jge    L51
  mov    %r14,%rdi
  call   <printf@plt>
  sub    $0x1,%r12d
  jne    L14
L18:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L52
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L52:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D23
type_guessed: string
name: read_data
arguments: tag: D20
type_guessed: string
name: read_data
arguments: tag: D19
type_guessed: string

tool: D23: %d

tool: D20: *

tool: D19: 


assistant: ```c
int main(void) {
  int j, n = 1, number;
  scanf("%d", &number);
  for (; n <= number; n++) {
    for (j = 1; j <= n; j++) {
      printf("*");
    }
    printf("\n");
  }
  for (n = number; n >= 1; n--) {
    for (j = 1; j <= n; j++)
      printf("*");
    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   %r14
  lea    D14(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L15
  mov    $0x1,%r13d
  lea    D17(%rip),%rbp
  lea    D25(%rip),%r14
  nopl   (%rax)
L40:
  mov    $0x1,%ebx
  nopl   (%rax)
L42:
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%r13d
  jge    L42
  mov    %r14,%rdi
  add    $0x1,%r13d
  call   <printf@plt>
  mov    0x4(%rsp),%r12d
  cmp    %r13d,%r12d
  jge    L40
  test   %r12d,%r12d
  jle    L15
  nopl   0x0(%rax,%rax,1)
L65:
  mov    $0x1,%ebx
  nopl   (%rax)
L53:
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%r12d
  jge    L53
  mov    %r14,%rdi
  call   <printf@plt>
  sub    $0x1,%r12d
  jne    L65
L15:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D14
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D17
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D25
type_guessed: string

tool: D14: %d

tool: D17: *

tool: D25: 


assistant: ```c
int main(void) {
  int j, n = 1, number;
  scanf("%d", &number);
  for (; n <= number; n++) {
    for (j = 1; j <= n; j++) {
      printf("*");
    }
    printf("\n");
  }
  for (n = number; n >= 1; n--) {
    for (j = 1; j <= n; j++)
      printf("*");
    printf("\n");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<kdb_buf_feed_cmdline>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <kbd_buf_parse_string@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int kdb_buf_feed_cmdline(const char *param, void *extra_param) {
  kbd_buf_parse_string(param);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<kdb_buf_feed_cmdline>:
  endbr64
  sub    $0x8,%rsp
  call   <kbd_buf_parse_string@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int kdb_buf_feed_cmdline(const char *param, void *extra_param) {
  kbd_buf_parse_string(param);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<kdb_buf_feed_cmdline>:
  endbr64
  sub    $0x8,%rsp
  call   <kbd_buf_parse_string@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int kdb_buf_feed_cmdline(const char *param, void *extra_param) {
  kbd_buf_parse_string(param);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<kdb_buf_feed_cmdline>:
  endbr64
  sub    $0x8,%rsp
  call   <kbd_buf_parse_string@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int kdb_buf_feed_cmdline(const char *param, void *extra_param) {
  kbd_buf_parse_string(param);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D98(%rip),%rbx
  mov    %rbx,0x8(%rsp)
  lea    0x8(%rsp),%rsi
  mov    %rbx,%rdx
  lea    D45(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  movsbl (%rax),%esi
  mov    %esi,%edx
  lea    D89(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  movsbl 0x5(%rax),%esi
  lea    D76(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,0x10(%rsp)
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rdx
  lea    D35(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x6,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  lea    D24(%rip),%rdi
  call   <puts@plt>
  mov    %rbx,%rsi
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  mov    %rbx,%rsi
  lea    D55(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movb   $0x58,(%rbx)
  mov    %rbx,%rsi
  lea    D44(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  lea    D30(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D98
type: string
name: unpack_data
arguments: data_tag: D45
type: string
name: unpack_data
arguments: data_tag: D89
type: string
name: unpack_data
arguments: data_tag: D76
type: string
name: unpack_data
arguments: data_tag: D35
type: string
name: unpack_data
arguments: data_tag: D24
type: string
name: unpack_data
arguments: data_tag: D25
type: string
name: unpack_data
arguments: data_tag: D55
type: string
name: unpack_data
arguments: data_tag: D44
type: string
name: unpack_data
arguments: data_tag: D30
type: string

tool: D98: Hello

tool: D45: s is at %p; it points to a string at %p


tool: D89: The ASCII value for %c is %d


tool: D76: The ASCII value for '\0' is %d



tool: D35: s2 is at %p; it points to a string at %p


tool: D24: Enter a string of AT MOST 5 LETTERS!!!

tool: D25: %s

tool: D55: You entered %s


tool: D44: I mutated it to %s


tool: D30: The length of what you entered is %d


assistant: ```c
int main() {
  const char *s = "Hello";
  printf("s is at %p; it points to a string at %p\n", &s, s);
  printf("The ASCII value for %c is %d\n", s[0], (int)s[0]);
  printf("The ASCII value for '\\0' is %d\n\n", (int)s[5]);
  const char *s2 = "Hello";
  printf("s2 is at %p; it points to a string at %p\n", &s2, s2);
  char *m = (char *)malloc(6 * sizeof(char));
  printf("Enter a string of AT MOST 5 LETTERS!!!\n");
  scanf("%s", m);
  printf("You entered %s\n", m);
  m[0] = 'X';
  printf("I mutated it to %s\n", m);
  printf("The length of what you entered is %d\n", strlen(m));
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r12
  lea    D30(%rip),%r12
  lea    D45(%rip),%rdi
  push   %rbp
  mov    %r12,%rdx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  mov    %r12,0x8(%rsp)
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  lea    D91(%rip),%rdi
  movsbl (%rax),%esi
  xor    %eax,%eax
  mov    %esi,%edx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  lea    D27(%rip),%rdi
  movsbl 0x5(%rax),%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rdx
  lea    0x10(%rsp),%rsi
  xor    %eax,%eax
  lea    D66(%rip),%rdi
  mov    %r12,0x10(%rsp)
  call   <printf@plt>
  mov    $0x6,%edi
  call   <malloc@plt>
  lea    D38(%rip),%rdi
  mov    %rax,%rbp
  call   <puts@plt>
  mov    %rbp,%rsi
  lea    D85(%rip),%rdi
  call   <scanf@plt>
  mov    %rbp,%rsi
  lea    D10(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  movb   $0x58,0x0(%rbp)
  mov    %rbp,%rsi
  xor    %eax,%eax
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    D48(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D30
data_type: string
name: disassemble_data
arguments: label: D45
data_type: string
name: disassemble_data
arguments: label: D91
data_type: string
name: disassemble_data
arguments: label: D27
data_type: string
name: disassemble_data
arguments: label: D66
data_type: string
name: disassemble_data
arguments: label: D38
data_type: string
name: disassemble_data
arguments: label: D85
data_type: string
name: disassemble_data
arguments: label: D10
data_type: string
name: disassemble_data
arguments: label: D4
data_type: string
name: disassemble_data
arguments: label: D48
data_type: string

tool: D30: Hello

tool: D45: s is at %p; it points to a string at %p


tool: D91: The ASCII value for %c is %d


tool: D27: The ASCII value for '\0' is %d



tool: D66: s2 is at %p; it points to a string at %p


tool: D38: Enter a string of AT MOST 5 LETTERS!!!

tool: D85: %s

tool: D10: You entered %s


tool: D4: I mutated it to %s


tool: D48: The length of what you entered is %d


assistant: ```c
int main() {
  const char *s = "Hello";
  printf("s is at %p; it points to a string at %p\n", &s, s);
  printf("The ASCII value for %c is %d\n", s[0], (int)s[0]);
  printf("The ASCII value for '\\0' is %d\n\n", (int)s[5]);
  const char *s2 = "Hello";
  printf("s2 is at %p; it points to a string at %p\n", &s2, s2);
  char *m = (char *)malloc(6 * sizeof(char));
  printf("Enter a string of AT MOST 5 LETTERS!!!\n");
  scanf("%s", m);
  printf("You entered %s\n", m);
  m[0] = 'X';
  printf("I mutated it to %s\n", m);
  printf("The length of what you entered is %d\n", strlen(m));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r12
  lea    D99(%rip),%r12
  lea    D74(%rip),%rdi
  push   %rbp
  mov    %r12,%rdx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  mov    %r12,0x8(%rsp)
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  lea    D67(%rip),%rdi
  movsbl (%rax),%esi
  xor    %eax,%eax
  mov    %esi,%edx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  lea    D6(%rip),%rdi
  movsbl 0x5(%rax),%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rdx
  lea    0x10(%rsp),%rsi
  xor    %eax,%eax
  lea    D96(%rip),%rdi
  mov    %r12,0x10(%rsp)
  call   <printf@plt>
  mov    $0x6,%edi
  call   <malloc@plt>
  lea    D43(%rip),%rdi
  mov    %rax,%rbp
  call   <puts@plt>
  mov    %rbp,%rsi
  lea    D94(%rip),%rdi
  call   <scanf@plt>
  mov    %rbp,%rsi
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  movb   $0x58,0x0(%rbp)
  mov    %rbp,%rsi
  xor    %eax,%eax
  lea    D47(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    D93(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D99
datatype: string
name: unpack_data
arguments: data_label: D74
datatype: string
name: unpack_data
arguments: data_label: D67
datatype: string
name: unpack_data
arguments: data_label: D6
datatype: string
name: unpack_data
arguments: data_label: D96
datatype: string
name: unpack_data
arguments: data_label: D43
datatype: string
name: unpack_data
arguments: data_label: D94
datatype: string
name: unpack_data
arguments: data_label: D38
datatype: string
name: unpack_data
arguments: data_label: D47
datatype: string
name: unpack_data
arguments: data_label: D93
datatype: string

tool: D99: Hello

tool: D74: s is at %p; it points to a string at %p


tool: D67: The ASCII value for %c is %d


tool: D6: The ASCII value for '\0' is %d



tool: D96: s2 is at %p; it points to a string at %p


tool: D43: Enter a string of AT MOST 5 LETTERS!!!

tool: D94: %s

tool: D38: You entered %s


tool: D47: I mutated it to %s


tool: D93: The length of what you entered is %d


assistant: ```c
int main() {
  const char *s = "Hello";
  printf("s is at %p; it points to a string at %p\n", &s, s);
  printf("The ASCII value for %c is %d\n", s[0], (int)s[0]);
  printf("The ASCII value for '\\0' is %d\n\n", (int)s[5]);
  const char *s2 = "Hello";
  printf("s2 is at %p; it points to a string at %p\n", &s2, s2);
  char *m = (char *)malloc(6 * sizeof(char));
  printf("Enter a string of AT MOST 5 LETTERS!!!\n");
  scanf("%s", m);
  printf("You entered %s\n", m);
  m[0] = 'X';
  printf("I mutated it to %s\n", m);
  printf("The length of what you entered is %d\n", strlen(m));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<s2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0xb,-0x4(%rbp)
  jg     L33
  cmpl   $0xa,-0x4(%rbp)
  jge    L40
  jmp    L51
L33:
  cmpl   $0xc,-0x4(%rbp)
  je     L38
  jmp    L51
L40:
  mov    $0x2,%eax
  jmp    L55
L38:
  mov    $0x7b,%eax
  jmp    L55
L51:
  mov    -0x4(%rbp),%eax
  add    $0x2,%eax
L55:
  pop    %rbp
  ret
```

assistant: ```c
__attribute__((no)) int s2(int x) {
  switch (x) {
  case 10:
  case 11:
    return 2;
  case 12:
    return 123;
  default:
    return x + 2;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<s2>:
  endbr64
  cmp    $0xb,%edi
  jle    L27
  mov    $0x7b,%eax
  cmp    $0xc,%edi
  jne    L22
L25:
  ret
L27:
  mov    $0x2,%eax
  cmp    $0x9,%edi
  jg     L20
L22:
  lea    0x2(%rdi),%eax
  jmp    L25
L20:
  ret
```

assistant: ```c
__attribute__((no)) int s2(int x) {
  switch (x) {
  case 10:
  case 11:
    return 2;
  case 12:
    return 123;
  default:
    return x + 2;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<s2>:
  endbr64
  cmp    $0xb,%edi
  jle    L16
  mov    $0x7b,%eax
  cmp    $0xc,%edi
  jne    L27
  ret
  nopl   0x0(%rax)
L16:
  mov    $0x2,%eax
  cmp    $0x9,%edi
  jg     L25
L27:
  lea    0x2(%rdi),%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L25:
  ret
```

assistant: ```c
__attribute__((no)) int s2(int x) {
  switch (x) {
  case 10:
  case 11:
    return 2;
  case 12:
    return 123;
  default:
    return x + 2;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<s2>:
  endbr64
  cmp    $0xb,%edi
  jle    L2
  mov    $0x7b,%eax
  cmp    $0xc,%edi
  jne    L8
  ret
  nopl   0x0(%rax)
L2:
  mov    $0x2,%eax
  cmp    $0x9,%edi
  jg     L20
L8:
  lea    0x2(%rdi),%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L20:
  ret
```

assistant: ```c
__attribute__((no)) int s2(int x) {
  switch (x) {
  case 10:
  case 11:
    return 2;
  case 12:
    return 123;
  default:
    return x + 2;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<soma>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L14
  mov    $0x0,%eax
  jmp    L17
L14:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <soma>
  mov    -0x4(%rbp),%edx
  add    %edx,%eax
L17:
  leave
  ret
```

assistant: ```c
int soma(int n) {
  if (n == 0)
    return 0;
  return n + soma(n - 1);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<soma>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  test   %edi,%edi
  jne    L18
L9:
  mov    %ebx,%eax
  pop    %rbx
  ret
L18:
  lea    -0x1(%rdi),%edi
  call   <soma>
  add    %eax,%ebx
  jmp    L9
```

assistant: ```c
int soma(int n) {
  if (n == 0)
    return 0;
  return n + soma(n - 1);
}
```
user: What is the c source code of the assembly code below?

```asm
<soma>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  je     L3
  nopw   0x0(%rax,%rax,1)
L5:
  add    %edi,%eax
  sub    $0x1,%edi
  jne    L5
L3:
  ret
```

assistant: ```c
int soma(int n) {
  if (n == 0)
    return 0;
  return n + soma(n - 1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<soma>:
  endbr64
  test   %edi,%edi
  je     L50
  lea    -0x1(%rdi),%edx
  mov    %edi,%ecx
  cmp    $0x9,%edx
  jbe    L101
  movd   %edi,%xmm4
  mov    %edi,%edx
  movdqa D6(%rip),%xmm3
  xor    %eax,%eax
  pshufd $0x0,%xmm4,%xmm1
  paddd  D13(%rip),%xmm1
  shr    $0x2,%edx
  pxor   %xmm0,%xmm0
  xchg   %ax,%ax
L122:
  movdqa %xmm1,%xmm2
  add    $0x1,%eax
  paddd  %xmm3,%xmm1
  paddd  %xmm2,%xmm0
  cmp    %eax,%edx
  jne    L122
  movdqa %xmm0,%xmm1
  mov    %edi,%edx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%edx
  paddd  %xmm1,%xmm0
  sub    %edx,%edi
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  cmp    %edx,%ecx
  je     L78
  lea    -0x1(%rdi),%edx
L125:
  add    %edi,%eax
  test   %edx,%edx
  je     L78
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x2,%edx
  je     L78
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x3,%edx
  je     L78
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x4,%edx
  je     L78
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x5,%edx
  je     L78
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x6,%edx
  je     L78
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x7,%edx
  je     L78
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x8,%edx
  je     L78
  add    %eax,%edx
  lea    -0x9(%rdi,%rdx,1),%eax
L78:
  ret
  nopl   0x0(%rax,%rax,1)
L50:
  xor    %eax,%eax
  ret
L101:
  xor    %eax,%eax
  jmp    L125
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D6
datatype: byte[16]
name: extract_symbolic_data
arguments: data_label: D13
datatype: byte[16]

tool: {"D6": [0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff]}

tool: {"D13": [0x00,0x00,0x00,0x00,0xff,0xff,0xff,0xff,0xfe,0xff,0xff,0xff,0xfd,0xff,0xff,0xff]}

assistant: ```c
int soma(int n) {
  if (n == 0)
    return 0;
  return n + soma(n - 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<char2int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L101
L43:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L102
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jle    L107
L102:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L124
  mov    $0xffffffff,%eax
  jmp    L39
L107:
  addl   $0x1,-0x10(%rbp)
L101:
  cmpl   $0x1f,-0x10(%rbp)
  jle    L43
  jmp    L25
L124:
  nop
L25:
  cmpl   $0x20,-0x10(%rbp)
  jne    L109
  mov    $0xfffffffd,%eax
  jmp    L39
L109:
  cmpl   $0x0,-0x10(%rbp)
  jne    L31
  mov    $0xfffffffe,%eax
  jmp    L39
L31:
  movl   $0x0,-0xc(%rbp)
  jmp    L0
L68:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x4(%rbp)
  subl   $0x30,-0x4(%rbp)
  mov    -0x10(%rbp),%eax
  sub    $0x1,%eax
  sub    -0xc(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <mypow@plt>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L0:
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jl     L68
  mov    -0x8(%rbp),%eax
L39:
  leave
  ret
```

assistant: ```c
int char2int(char *s) {
  int i = 0, j, r = 0, aux;
  for (i = 0; i < 32; i++) {
    if (!((s[i] >= '0') && (s[i] <= '9'))) {
      if (s[i] == '\0')
        break;
      else
        return -1;
    }
  }
  if (i == 32)
    return -3;
  if (i == 0)
    return -2;
  for (j = 0; j < i; j++) {
    aux = (int)s[j];
    aux -= 48;
    aux *= (int)mypow(10, i - 1 - j);
    r += aux;
  }
  return r;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<char2int>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rdi,%rdx
  mov    $0x0,%eax
L74:
  movzbl (%rdx),%ecx
  lea    -0x30(%rcx),%esi
  cmp    $0x9,%sil
  ja     L54
  add    $0x1,%eax
  add    $0x1,%rdx
  cmp    $0x20,%eax
  jne    L74
  mov    $0xfffffffd,%r12d
  jmp    L119
L54:
  test   %cl,%cl
  jne    L36
  cmp    $0x20,%eax
  je     L27
  test   %eax,%eax
  je     L39
  jle    L7
  mov    %eax,%r14d
  lea    0x1(%rax),%r15d
  mov    $0x1,%ebp
  mov    $0x0,%r12d
L33:
  movsbl -0x1(%r13,%rbp,1),%ebx
  sub    $0x30,%ebx
  mov    %r14d,%esi
  sub    %ebp,%esi
  mov    $0xa,%edi
  call   <mypow@plt>
  imul   %eax,%ebx
  add    %ebx,%r12d
  add    $0x1,%rbp
  cmp    %r15,%rbp
  jne    L33
L119:
  mov    %r12d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L36:
  mov    $0xffffffff,%r12d
  jmp    L119
L27:
  mov    $0xfffffffd,%r12d
  jmp    L119
L39:
  mov    $0xfffffffe,%r12d
  jmp    L119
L7:
  mov    $0x0,%r12d
  jmp    L119
```

assistant: ```c
int char2int(char *s) {
  int i = 0, j, r = 0, aux;
  for (i = 0; i < 32; i++) {
    if (!((s[i] >= '0') && (s[i] <= '9'))) {
      if (s[i] == '\0')
        break;
      else
        return -1;
    }
  }
  if (i == 32)
    return -3;
  if (i == 0)
    return -2;
  for (j = 0; j < i; j++) {
    aux = (int)s[j];
    aux -= 48;
    aux *= (int)mypow(10, i - 1 - j);
    r += aux;
  }
  return r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<char2int>:
  endbr64
  push   %r13
  xor    %eax,%eax
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  jmp    L62
  nopl   (%rax)
L34:
  add    $0x1,%rax
  cmp    $0x20,%rax
  je     L40
L62:
  movzbl (%r12,%rax,1),%edx
  mov    %eax,%ebp
  lea    -0x30(%rdx),%ecx
  cmp    $0x9,%cl
  jbe    L34
  test   %dl,%dl
  jne    L4
  test   %eax,%eax
  je     L36
  sub    $0x1,%ebp
  xor    %r13d,%r13d
  nop
L59:
  movsbl (%r12),%ebx
  mov    %ebp,%esi
  mov    $0xa,%edi
  add    $0x1,%r12
  call   <mypow@plt>
  sub    $0x30,%ebx
  imul   %eax,%ebx
  add    %ebx,%r13d
  sub    $0x1,%ebp
  jae    L59
L54:
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L40:
  add    $0x8,%rsp
  mov    $0xfffffffd,%r13d
  pop    %rbx
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L36:
  mov    $0xfffffffe,%r13d
  jmp    L54
L4:
  mov    $0xffffffff,%r13d
  jmp    L54
```

assistant: ```c
int char2int(char *s) {
  int i = 0, j, r = 0, aux;
  for (i = 0; i < 32; i++) {
    if (!((s[i] >= '0') && (s[i] <= '9'))) {
      if (s[i] == '\0')
        break;
      else
        return -1;
    }
  }
  if (i == 32)
    return -3;
  if (i == 0)
    return -2;
  for (j = 0; j < i; j++) {
    aux = (int)s[j];
    aux -= 48;
    aux *= (int)mypow(10, i - 1 - j);
    r += aux;
  }
  return r;
}
```