ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<print_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  subl   $0x1,-0xc(%rbp)
  jmp    L7
L18:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  subl   $0x1,-0xc(%rbp)
L7:
  cmpl   $0x0,-0xc(%rbp)
  jns    L18
  nop
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D9
data_type: string

tool: D9: %d

assistant: ```c
void print_int(int result[], int m) {
  for (m--; m >= 0; m--)
    printf("%d", result[m]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print_int>:
  endbr64
  sub    $0x1,%esi
  js     L5
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  movslq %esi,%rbx
  lea    D6(%rip),%r12
L2:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  call   <printf@plt>
  sub    $0x1,%rbx
  test   %ebx,%ebx
  jns    L2
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
type: string

tool: D6: %d

assistant: ```c
void print_int(int result[], int m) {
  for (m--; m >= 0; m--)
    printf("%d", result[m]);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<print_int>:
  endbr64
  sub    $0x1,%esi
  js     L16
  push   %r12
  lea    D2(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %esi,%rbx
  nopw   0x0(%rax,%rax,1)
L6:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  sub    $0x1,%rbx
  call   <printf@plt>
  test   %ebx,%ebx
  jns    L6
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L16:
  ret
```

tool_calls: name: read_data
arguments: data_label: D2
type_infered: string

tool: D2: %d

assistant: ```c
void print_int(int result[], int m) {
  for (m--; m >= 0; m--)
    printf("%d", result[m]);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<print_int>:
  endbr64
  sub    $0x1,%esi
  js     L19
  push   %r12
  lea    D6(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %esi,%rbx
  nopw   0x0(%rax,%rax,1)
L13:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  sub    $0x1,%rbx
  call   <printf@plt>
  test   %ebx,%ebx
  jns    L13
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L19:
  ret
```

tool_calls: name: parse_data
arguments: data_label: D6
type_guessed: string

tool: D6: %d

assistant: ```c
void print_int(int result[], int m) {
  for (m--; m >= 0; m--)
    printf("%d", result[m]);
}
```
user: What is the c source code of the assembly code below?

```asm
<negate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  negl   -0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int negate(int number) {
  number = -number;
  return number;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<negate>:
  endbr64
  mov    %edi,%eax
  neg    %eax
  ret
```

assistant: ```c
int negate(int number) {
  number = -number;
  return number;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<negate>:
  endbr64
  mov    %edi,%eax
  neg    %eax
  ret
```

assistant: ```c
int negate(int number) {
  number = -number;
  return number;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<negate>:
  endbr64
  mov    %edi,%eax
  neg    %eax
  ret
```

assistant: ```c
int negate(int number) {
  number = -number;
  return number;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<M11_f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <debug@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <debug@plt>
  mov    $0x1,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D9
datatype: string
name: disassemble_data
arguments: data_tag: D0
datatype: string

tool: D9: M11_f(const char *s) is not yet implemented!


tool: D0: %s


assistant: ```c
int M11_f(char *s) {
  debug("M11_f(const char *s) is not yet implemented!\n");
  debug("%s\n", s);
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<M11_f>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  lea    D5(%rip),%rdi
  mov    $0x0,%eax
  call   <debug@plt>
  mov    %rbx,%rsi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <debug@plt>
  mov    $0x1,%eax
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: data_tag: D5
data_type: string
name: read_data
arguments: data_tag: D16
data_type: string

tool: D5: M11_f(const char *s) is not yet implemented!


tool: D16: %s


assistant: ```c
int M11_f(char *s) {
  debug("M11_f(const char *s) is not yet implemented!\n");
  debug("%s\n", s);
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<M11_f>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  lea    D4(%rip),%rdi
  call   <debug@plt>
  mov    %rbp,%rsi
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  call   <debug@plt>
  mov    $0x1,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D4
type_guessed: string
name: unpack_data
arguments: tag: D11
type_guessed: string

tool: D4: M11_f(const char *s) is not yet implemented!


tool: D11: %s


assistant: ```c
int M11_f(char *s) {
  debug("M11_f(const char *s) is not yet implemented!\n");
  debug("%s\n", s);
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<M11_f>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  lea    D11(%rip),%rdi
  call   <debug@plt>
  mov    %rbp,%rsi
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  call   <debug@plt>
  mov    $0x1,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D11
type: string
name: unpack_data
arguments: data_tag: D7
type: string

tool: D11: M11_f(const char *s) is not yet implemented!


tool: D7: %s


assistant: ```c
int M11_f(char *s) {
  debug("M11_f(const char *s) is not yet implemented!\n");
  debug("%s\n", s);
  return 1;
}
```
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    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x6c(%rbp)
  jmp    L31
L18:
  lea    -0x60(%rbp),%rdx
  mov    -0x6c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x6c(%rbp)
L31:
  mov    -0x70(%rbp),%eax
  cmp    %eax,-0x6c(%rbp)
  jl     L18
  lea    -0x60(%rbp),%rax
  mov    %rax,-0x68(%rbp)
  mov    -0x70(%rbp),%edx
  mov    -0x68(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <sort@plt>
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x6c(%rbp)
  jmp    L43
L4:
  mov    -0x6c(%rbp),%eax
  cltq
  mov    -0x60(%rbp,%rax,4),%edx
  mov    -0x6c(%rbp),%eax
  mov    %eax,%esi
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x6c(%rbp)
L43:
  mov    -0x70(%rbp),%eax
  cmp    %eax,-0x6c(%rbp)
  jl     L4
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D43
type_guessed: string
name: parse_data
arguments: label: D26
type_guessed: string
name: parse_data
arguments: label: D25
type_guessed: string
name: parse_data
arguments: label: D48
type_guessed: string
name: parse_data
arguments: label: D50
type_guessed: string

tool: D43: input n:

tool: D26: %d

tool: D25: Please input these numbers:

tool: D48: Now,the sequence is :

tool: D50: %d,%d


assistant: ```c
int main() {
  void sort(int *p, int m);
  int i, n;
  int *p, num[20];
  printf("input n:");
  scanf("%d", &n);
  printf("Please input these numbers:\n");
  for (i = 0; i < n; i++) {
    scanf("%d", &num[i]);
  }
  p = num;
  sort(p, n);
  printf("Now,the sequence is :\n");
  for (i = 0; i < n; i++) {
    printf("%d,%d\n", i, num[i]);
  }
  printf("\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    D26(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  lea    D47(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%esi
  test   %esi,%esi
  jle    L6
  lea    0x10(%rsp),%rbp
  mov    $0x0,%ebx
  lea    D9(%rip),%r12
L0:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  mov    0xc(%rsp),%esi
  add    $0x4,%rbp
  cmp    %ebx,%esi
  jg     L0
L6:
  lea    0x10(%rsp),%rdi
  call   <sort@plt>
  lea    D21(%rip),%rdi
  call   <puts@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L39
  mov    $0x0,%ebx
  lea    D57(%rip),%rbp
L33:
  mov    0x10(%rsp,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %ebx,0xc(%rsp)
  jg     L33
L39:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  mov    $0x0,%eax
  add    $0x70,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D26
datatype: string
name: extract_symbolic_data
arguments: data_tag: D9
datatype: string
name: extract_symbolic_data
arguments: data_tag: D47
datatype: string
name: extract_symbolic_data
arguments: data_tag: D21
datatype: string
name: extract_symbolic_data
arguments: data_tag: D57
datatype: string

tool: D26: input n:

tool: D9: %d

tool: D47: Please input these numbers:

tool: D21: Now,the sequence is :

tool: D57: %d,%d


assistant: ```c
int main() {
  void sort(int *p, int m);
  int i, n;
  int *p, num[20];
  printf("input n:");
  scanf("%d", &n);
  printf("Please input these numbers:\n");
  for (i = 0; i < n; i++) {
    scanf("%d", &num[i]);
  }
  p = num;
  sort(p, n);
  printf("Now,the sequence is :\n");
  for (i = 0; i < n; i++) {
    printf("%d,%d\n", i, num[i]);
  }
  printf("\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
  lea    D28(%rip),%rdi
  lea    D7(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  mov    %r12,%rbp
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  lea    D32(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%esi
  test   %esi,%esi
  jle    L11
  nopl   0x0(%rax)
L27:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  cmp    %ebx,%esi
  jg     L27
L11:
  mov    %r12,%rdi
  call   <sort@plt>
  lea    D5(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L1
  xor    %ebx,%ebx
  lea    D16(%rip),%rbp
  nop
L4:
  mov    (%r12,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jg     L4
L1:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D28
data_type: string
name: retrieve_data
arguments: data_tag: D7
data_type: string
name: retrieve_data
arguments: data_tag: D32
data_type: string
name: retrieve_data
arguments: data_tag: D5
data_type: string
name: retrieve_data
arguments: data_tag: D16
data_type: string

tool: D28: input n:

tool: D7: %d

tool: D32: Please input these numbers:

tool: D5: Now,the sequence is :

tool: D16: %d,%d


assistant: ```c
int main() {
  void sort(int *p, int m);
  int i, n;
  int *p, num[20];
  printf("input n:");
  scanf("%d", &n);
  printf("Please input these numbers:\n");
  for (i = 0; i < n; i++) {
    scanf("%d", &num[i]);
  }
  p = num;
  sort(p, n);
  printf("Now,the sequence is :\n");
  for (i = 0; i < n; i++) {
    printf("%d,%d\n", i, num[i]);
  }
  printf("\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r13
  lea    D49(%rip),%rdi
  lea    D8(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  mov    %r12,%rbp
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  lea    D34(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%esi
  test   %esi,%esi
  jle    L19
  nopl   0x0(%rax)
L27:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  cmp    %ebx,%esi
  jg     L27
L19:
  mov    %r12,%rdi
  call   <sort@plt>
  lea    D16(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L33
  xor    %ebx,%ebx
  lea    D14(%rip),%rbp
  nop
L5:
  mov    (%r12,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jg     L5
L33:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L42
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L42:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D49
type_guessed: string
name: extract_symbolic_data
arguments: label: D8
type_guessed: string
name: extract_symbolic_data
arguments: label: D34
type_guessed: string
name: extract_symbolic_data
arguments: label: D16
type_guessed: string
name: extract_symbolic_data
arguments: label: D14
type_guessed: string

tool: D49: input n:

tool: D8: %d

tool: D34: Please input these numbers:

tool: D16: Now,the sequence is :

tool: D14: %d,%d


assistant: ```c
int main() {
  void sort(int *p, int m);
  int i, n;
  int *p, num[20];
  printf("input n:");
  scanf("%d", &n);
  printf("Please input these numbers:\n");
  for (i = 0; i < n; i++) {
    scanf("%d", &num[i]);
  }
  p = num;
  sort(p, n);
  printf("Now,the sequence is :\n");
  for (i = 0; i < n; i++) {
    printf("%d,%d\n", i, num[i]);
  }
  printf("\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<distrifunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L166
L15:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jle    L53
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5a,%al
  jg     L53
  mov    -0x18(%rbp),%rax
  sub    $0x1,%rax
  cmp    $0x39,%rax
  jbe    L53
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x20(%rbp)
  jmp    L58
L53:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jle    L109
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5a,%al
  jg     L109
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x30(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x30(%rbp)
  jmp    L58
L109:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L165
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jg     L165
  cmpl   $0x3,-0x4(%rbp)
  je     L165
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x28(%rbp)
  addl   $0x1,-0x4(%rbp)
  jmp    L58
L165:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L34
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jg     L34
  cmpl   $0x3,-0x4(%rbp)
  jne    L34
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x30(%rbp),%rax
  mov    %dl,(%rax)
  addl   $0x1,-0x8(%rbp)
  jmp    L58
L34:
  mov    -0x38(%rbp),%rax
  movl   $0x1,(%rax)
L58:
  movl   $0xa,-0xc(%rbp)
  addl   $0x1,-0xc(%rbp)
L166:
  cmpl   $0x9,-0xc(%rbp)
  jg     L45
  cmpl   $0x1,-0x8(%rbp)
  jle    L15
L45:
  nop
  pop    %rbp
  ret
```

assistant: ```c
char distrifunc(char *input, char *pware, char *ppro, char *pquali, int *a) {
  int count, c, b = 0;
  for (count = 0, c = 0; count < 10 && c < 2; ++count) {
    if (*input >= 'A' && *input <= 'Z' && input - 1 > '9') {
      *pware = *input;
      ++input;
      ++pware;
    } else if (*input >= 'A' && *input <= 'Z') {
      *pquali = *input;
      ++input;
      ++pquali;
    } else if (*input >= '0' && *input <= '9' && b != 3) {
      *ppro = *input;
      ++input;
      ++ppro;
      ++b;
    } else if (*input >= '0' && *input <= '9' && b == 3) {
      *pquali = *input;
      ++c;
    } else
      *a = 1;
    count = 10;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<distrifunc>:
  endbr64
  movzbl (%rdi),%eax
  lea    -0x41(%rax),%r9d
  cmp    $0x19,%r9b
  ja     L3
  sub    $0x1,%rdi
  cmp    $0x39,%rdi
  jbe    L12
  mov    %al,(%rsi)
  ret
L12:
  mov    %al,(%rcx)
  ret
L3:
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  jbe    L27
  movl   $0x1,(%r8)
  ret
L27:
  mov    %al,(%rdx)
  ret
```

assistant: ```c
char distrifunc(char *input, char *pware, char *ppro, char *pquali, int *a) {
  int count, c, b = 0;
  for (count = 0, c = 0; count < 10 && c < 2; ++count) {
    if (*input >= 'A' && *input <= 'Z' && input - 1 > '9') {
      *pware = *input;
      ++input;
      ++pware;
    } else if (*input >= 'A' && *input <= 'Z') {
      *pquali = *input;
      ++input;
      ++pquali;
    } else if (*input >= '0' && *input <= '9' && b != 3) {
      *ppro = *input;
      ++input;
      ++ppro;
      ++b;
    } else if (*input >= '0' && *input <= '9' && b == 3) {
      *pquali = *input;
      ++c;
    } else
      *a = 1;
    count = 10;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<distrifunc>:
  endbr64
  movzbl (%rdi),%eax
  lea    -0x41(%rax),%r9d
  cmp    $0x19,%r9b
  jbe    L2
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  ja     L0
  mov    %al,(%rdx)
  ret
  nopl   0x0(%rax)
L2:
  sub    $0x1,%rdi
  cmp    $0x39,%rdi
  jbe    L13
  mov    %al,(%rsi)
  ret
  nopl   (%rax)
L13:
  mov    %al,(%rcx)
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  movl   $0x1,(%r8)
  ret
```

assistant: ```c
char distrifunc(char *input, char *pware, char *ppro, char *pquali, int *a) {
  int count, c, b = 0;
  for (count = 0, c = 0; count < 10 && c < 2; ++count) {
    if (*input >= 'A' && *input <= 'Z' && input - 1 > '9') {
      *pware = *input;
      ++input;
      ++pware;
    } else if (*input >= 'A' && *input <= 'Z') {
      *pquali = *input;
      ++input;
      ++pquali;
    } else if (*input >= '0' && *input <= '9' && b != 3) {
      *ppro = *input;
      ++input;
      ++ppro;
      ++b;
    } else if (*input >= '0' && *input <= '9' && b == 3) {
      *pquali = *input;
      ++c;
    } else
      *a = 1;
    count = 10;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<distrifunc>:
  endbr64
  movzbl (%rdi),%eax
  lea    -0x41(%rax),%r9d
  cmp    $0x19,%r9b
  jbe    L10
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  ja     L13
  mov    %al,(%rdx)
  ret
  nopl   0x0(%rax)
L10:
  sub    $0x1,%rdi
  cmp    $0x39,%rdi
  jbe    L1
  mov    %al,(%rsi)
  ret
  nopl   (%rax)
L1:
  mov    %al,(%rcx)
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  movl   $0x1,(%r8)
  ret
```

assistant: ```c
char distrifunc(char *input, char *pware, char *ppro, char *pquali, int *a) {
  int count, c, b = 0;
  for (count = 0, c = 0; count < 10 && c < 2; ++count) {
    if (*input >= 'A' && *input <= 'Z' && input - 1 > '9') {
      *pware = *input;
      ++input;
      ++pware;
    } else if (*input >= 'A' && *input <= 'Z') {
      *pquali = *input;
      ++input;
      ++pquali;
    } else if (*input >= '0' && *input <= '9' && b != 3) {
      *ppro = *input;
      ++input;
      ++ppro;
      ++b;
    } else if (*input >= '0' && *input <= '9' && b == 3) {
      *pquali = *input;
      ++c;
    } else
      *a = 1;
    count = 10;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<start_irssi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %rdi,-0x78(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x70(%rbp)
  movq   $0x0,-0x68(%rbp)
  movq   $0x0,-0x60(%rbp)
  movq   $0x0,-0x58(%rbp)
  movq   $0x0,-0x50(%rbp)
  movq   $0x0,-0x48(%rbp)
  movq   $0x0,-0x40(%rbp)
  movq   $0x0,-0x38(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    -0x78(%rbp),%rdx
  lea    -0x70(%rbp),%rax
  mov    %rdx,%rcx
  lea    D16(%rip),%rdx
  mov    $0x64,%esi
  mov    %rax,%rdi
  call   <snprintf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <system@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D16
type_guessed: string
name: disassemble_data
arguments: label: D6
type_guessed: string

tool: D16: su --command='screen -dmS irc_logger irssi' %s


tool: D6: %s

assistant: ```c
void start_irssi(char nickname[]) {
  char chaine[100] = {0};
  snprintf(chaine, 100, "su --command='screen -dmS irc_logger irssi' %s\n", nickname);
  printf("%s", chaine);
  system(chaine);
}
```
user: Translate the below assembly code into C programming language:

```asm
<start_irssi>:
  endbr64
  push   %rbx
  sub    $0x70,%rsp
  mov    %rdi,%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x38(%rsp)
  movq   $0x0,0x40(%rsp)
  movq   $0x0,0x48(%rsp)
  movq   $0x0,0x50(%rsp)
  movq   $0x0,0x58(%rsp)
  movl   $0x0,0x60(%rsp)
  mov    %rsp,%rbx
  lea    D13(%rip),%rdx
  mov    $0x64,%esi
  mov    %rbx,%rdi
  call   <snprintf@plt>
  mov    %rbx,%rsi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <system@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x70,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D13
type_guessed: string
name: retrieve_data
arguments: data_tag: D18
type_guessed: string

tool: D13: su --command='screen -dmS irc_logger irssi' %s


tool: D18: %s

assistant: ```c
void start_irssi(char nickname[]) {
  char chaine[100] = {0};
  snprintf(chaine, 100, "su --command='screen -dmS irc_logger irssi' %s\n", nickname);
  printf("%s", chaine);
  system(chaine);
}
```