ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<Check>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    $0x1,%r12d
L14:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  lea    -0x1(%r12),%eax
  cmp    %eax,%edx
  jl     L21
  mov    %r12,%rbx
L4:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jl     L35
  add    $0x1,%rbx
  movzbl -0x1(%rbp,%rbx,1),%eax
  cmp    %al,-0x1(%rbp,%r12,1)
  jne    L4
  mov    $0x1,%eax
L37:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L35:
  add    $0x1,%r12
  jmp    L14
L21:
  mov    $0x0,%eax
  jmp    L37
```

assistant: ```c
int Check(char *string) {
  int i, j;
  for (i = 0; i <= strlen(string); i++) {
    for (j = i + 1; j <= strlen(string); j++)
      if (string[i] == string[j])
        return 1;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<Check>:
  endbr64
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
L12:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%r8d
  lea    -0x1(%r12),%eax
  cmp    %eax,%r8d
  jl     L17
  mov    %r12,%rbx
  jmp    L30
  nopl   0x0(%rax,%rax,1)
L46:
  add    $0x1,%rbx
  movzbl -0x1(%rbp,%rbx,1),%eax
  cmp    %al,-0x1(%rbp,%r12,1)
  je     L40
L30:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jge    L46
  add    $0x1,%r12
  jmp    L12
  nopw   0x0(%rax,%rax,1)
L40:
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
L17:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int Check(char *string) {
  int i, j;
  for (i = 0; i <= strlen(string); i++) {
    for (j = i + 1; j <= strlen(string); j++)
      if (string[i] == string[j])
        return 1;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<Check>:
  endbr64
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
L10:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%r8d
  lea    -0x1(%r12),%eax
  cmp    %eax,%r8d
  jl     L42
  mov    %r12,%rbx
  jmp    L39
  nopl   0x0(%rax,%rax,1)
L15:
  add    $0x1,%rbx
  movzbl -0x1(%rbp,%rbx,1),%eax
  cmp    %al,-0x1(%rbp,%r12,1)
  je     L14
L39:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jge    L15
  add    $0x1,%r12
  jmp    L10
  nopw   0x0(%rax,%rax,1)
L14:
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
L42:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int Check(char *string) {
  int i, j;
  for (i = 0; i <= strlen(string); i++) {
    for (j = i + 1; j <= strlen(string); j++)
      if (string[i] == string[j])
        return 1;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x34(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L26:
  cmp    %rdx,%rsp
  je     L31
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L26
L31:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L5
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L5:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <g@plt>
  mov    -0x20(%rbp),%rax
  mov    0x8(%rax),%eax
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int f(int x, int y[x]) {
  int buf[x];
  g(buf);
  return buf[2];
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movslq %edi,%rdi
  lea    0xf(,%rdi,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L17:
  cmp    %rdx,%rsp
  je     L35
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L17
L35:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L27
  orq    $0x0,-0x8(%rsp,%rax,1)
L27:
  lea    0x3(%rsp),%rdi
  mov    %rdi,%rbx
  shr    $0x2,%rbx
  and    $0xfffffffffffffffc,%rdi
  call   <g@plt>
  mov    0x8(,%rbx,4),%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  mov    -0x8(%rbp),%rbx
  leave
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int f(int x, int y[x]) {
  int buf[x];
  g(buf);
  return buf[2];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<f>:
  endbr64
  push   %rbp
  movslq %edi,%rdi
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    0xf(,%rdi,4),%rax
  mov    %rsp,%rcx
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L7
L29:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L29
L7:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L47
L14:
  lea    0x3(%rsp),%rdi
  mov    %rdi,%rbx
  and    $0xfffffffffffffffc,%rdi
  call   <g@plt>
  shr    $0x2,%rbx
  mov    0x8(,%rbx,4),%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  mov    -0x8(%rbp),%rbx
  leave
  ret
  nopl   0x0(%rax,%rax,1)
L47:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L14
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int f(int x, int y[x]) {
  int buf[x];
  g(buf);
  return buf[2];
}
```
user: What is the c source code of the assembly code below:

```asm
<f>:
  endbr64
  push   %rbp
  movslq %edi,%rdi
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    0xf(,%rdi,4),%rax
  mov    %rsp,%rcx
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L42
L15:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L15
L42:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L20
L36:
  lea    0x3(%rsp),%rdi
  mov    %rdi,%rbx
  and    $0xfffffffffffffffc,%rdi
  call   <g@plt>
  shr    $0x2,%rbx
  mov    0x8(,%rbx,4),%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L32
  mov    -0x8(%rbp),%rbx
  leave
  ret
  nopl   0x0(%rax,%rax,1)
L20:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L36
L32:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int f(int x, int y[x]) {
  int buf[x];
  g(buf);
  return buf[2];
}
```
user: Translate the below assembly code into C programming language:

```asm
<next_huffman_value>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  lea    -0x14(%rbp),%rdx
  mov    -0x10(%rbp),%rsi
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x0,%edx
  mov    %rax,%rdi
  call   <rec_parcours_abr@plt>
  mov    -0x14(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
int8_t next_huffman_value(struct abr *table, char *code) {
  char *direction = code;
  uint8_t symb;
  rec_parcours_abr(table, direction, 0, &symb);
  return symb;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<next_huffman_value>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rcx
  mov    $0x0,%edx
  call   <rec_parcours_abr@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int8_t next_huffman_value(struct abr *table, char *code) {
  char *direction = code;
  uint8_t symb;
  rec_parcours_abr(table, direction, 0, &symb);
  return symb;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<next_huffman_value>:
  endbr64
  sub    $0x18,%rsp
  xor    %edx,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rcx
  call   <rec_parcours_abr@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int8_t next_huffman_value(struct abr *table, char *code) {
  char *direction = code;
  uint8_t symb;
  rec_parcours_abr(table, direction, 0, &symb);
  return symb;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<next_huffman_value>:
  endbr64
  sub    $0x18,%rsp
  xor    %edx,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rcx
  call   <rec_parcours_abr@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int8_t next_huffman_value(struct abr *table, char *code) {
  char *direction = code;
  uint8_t symb;
  rec_parcours_abr(table, direction, 0, &symb);
  return symb;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<trimStr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L8
  mov    D30(%rip),%rax
  mov    (%rax),%edx
  mov    D13(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  jmp    L98
L8:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L39
  mov    D22(%rip),%rax
  mov    (%rax),%edx
  mov    D13(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  jmp    L98
L39:
  mov    -0x28(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L78
L37:
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  addq   $0x1,-0x18(%rbp)
L78:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L6
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L37
L6:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  jmp    L34
L68:
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  subq   $0x1,-0x10(%rbp)
L34:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L50
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L68
L50:
  mov    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
L98:
  leave
  ret
```

assistant: ```c
int trimStr(char *s) {
  if (!s) {
    errno = EINVAL;
    return -1;
  }
  char *copy = (char *)malloc(strlen(s) + 1);
  if (!copy) {
    errno = ENOMEM;
    return -1;
  }
  strcpy(copy, s);
  char *p = NULL;
  char *e = NULL;
  for (p = copy; *p && isspace(*p); *p = 0, p++)
    ;
  for (e = p + strlen(p) - 1; *e && isspace(*e); *e = 0, e--)
    ;
  strcpy(s, p);
  free(copy);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<trimStr>:
  endbr64
  test   %rdi,%rdi
  je     L8
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L61
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  movzbl 0x0(%rbp),%edi
  mov    %rbp,%rbx
  test   %dil,%dil
  je     L75
L7:
  movsbl %dil,%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L75
  movb   $0x0,(%rbx)
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L7
L75:
  mov    %rbx,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbx,%rax,1),%r13
  movzbl 0x0(%r13),%edi
  test   %dil,%dil
  je     L86
L26:
  movsbl %dil,%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L86
  movb   $0x0,0x0(%r13)
  sub    $0x1,%r13
  movzbl 0x0(%r13),%edi
  test   %dil,%dil
  jne    L26
L86:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    $0x0,%eax
L9:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  mov    D18(%rip),%rax
  mov    (%rax),%edx
  mov    D19(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
L61:
  mov    D33(%rip),%rax
  mov    (%rax),%edx
  mov    D19(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  jmp    L9
```

assistant: ```c
int trimStr(char *s) {
  if (!s) {
    errno = EINVAL;
    return -1;
  }
  char *copy = (char *)malloc(strlen(s) + 1);
  if (!copy) {
    errno = ENOMEM;
    return -1;
  }
  strcpy(copy, s);
  char *p = NULL;
  char *e = NULL;
  for (p = copy; *p && isspace(*p); *p = 0, p++)
    ;
  for (e = p + strlen(p) - 1; *e && isspace(*e); *e = 0, e--)
    ;
  strcpy(s, p);
  free(copy);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<trimStr>:
  endbr64
  test   %rdi,%rdi
  je     L47
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L88
  mov    %rax,%rdi
  mov    %r13,%rsi
  mov    %r12,%rbp
  call   <strcpy@plt>
  movsbl (%r12),%edi
  test   %dil,%dil
  jne    L105
  jmp    L56
  nopl   (%rax)
L60:
  movb   $0x0,0x0(%rbp)
  movsbl 0x1(%rbp),%edi
  add    $0x1,%rbp
  test   %dil,%dil
  je     L56
L105:
  call   <isspace@plt>
  test   %rax,%rax
  jne    L60
L56:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbp,%rax,1),%rbx
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L103
  jmp    L75
  nopl   0x0(%rax)
L50:
  movb   $0x0,(%rbx)
  movsbl -0x1(%rbx),%edi
  sub    $0x1,%rbx
  test   %dil,%dil
  je     L75
L103:
  call   <isspace@plt>
  test   %rax,%rax
  jne    L50
L75:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <strcpy@plt>
  mov    %r12,%rdi
  call   <free@plt>
  xor    %eax,%eax
L71:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L88:
  mov    D38(%rip),%rax
  mov    (%rax),%edx
  mov    D5(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  jmp    L71
L47:
  mov    D21(%rip),%rax
  mov    (%rax),%edx
  mov    D5(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int trimStr(char *s) {
  if (!s) {
    errno = EINVAL;
    return -1;
  }
  char *copy = (char *)malloc(strlen(s) + 1);
  if (!copy) {
    errno = ENOMEM;
    return -1;
  }
  strcpy(copy, s);
  char *p = NULL;
  char *e = NULL;
  for (p = copy; *p && isspace(*p); *p = 0, p++)
    ;
  for (e = p + strlen(p) - 1; *e && isspace(*e); *e = 0, e--)
    ;
  strcpy(s, p);
  free(copy);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<trimStr>:
  endbr64
  test   %rdi,%rdi
  je     L95
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L106
  mov    %rax,%rdi
  mov    %r13,%rsi
  mov    %r12,%rbp
  call   <strcpy@plt>
  movsbl (%r12),%edi
  test   %dil,%dil
  jne    L50
  jmp    L43
  nopl   (%rax)
L58:
  movb   $0x0,0x0(%rbp)
  movsbl 0x1(%rbp),%edi
  add    $0x1,%rbp
  test   %dil,%dil
  je     L43
L50:
  call   <isspace@plt>
  test   %rax,%rax
  jne    L58
L43:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbp,%rax,1),%rbx
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L0
  jmp    L87
  nopl   0x0(%rax)
L84:
  movb   $0x0,(%rbx)
  movsbl -0x1(%rbx),%edi
  sub    $0x1,%rbx
  test   %dil,%dil
  je     L87
L0:
  call   <isspace@plt>
  test   %rax,%rax
  jne    L84
L87:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <strcpy@plt>
  mov    %r12,%rdi
  call   <free@plt>
  xor    %eax,%eax
L103:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L106:
  mov    D22(%rip),%rax
  mov    (%rax),%edx
  mov    D5(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  jmp    L103
L95:
  mov    D11(%rip),%rax
  mov    (%rax),%edx
  mov    D5(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int trimStr(char *s) {
  if (!s) {
    errno = EINVAL;
    return -1;
  }
  char *copy = (char *)malloc(strlen(s) + 1);
  if (!copy) {
    errno = ENOMEM;
    return -1;
  }
  strcpy(copy, s);
  char *p = NULL;
  char *e = NULL;
  for (p = copy; *p && isspace(*p); *p = 0, p++)
    ;
  for (e = p + strlen(p) - 1; *e && isspace(*e); *e = 0, e--)
    ;
  strcpy(s, p);
  free(copy);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<warnx>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L9
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L9:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb4(%rbp),%edx
  mov    -0xc8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <vwarnx@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  nop
  leave
  ret
```

assistant: ```c
void warnx(const char *fmt, ...) {
  va_list ap;
  va_start(ap, fmt);
  vwarnx(fmt, ap);
  va_end(ap);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<warnx>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <vwarnx@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void warnx(const char *fmt, ...) {
  va_list ap;
  va_start(ap, fmt);
  vwarnx(fmt, ap);
  va_end(ap);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<warnx>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <vwarnx@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void warnx(const char *fmt, ...) {
  va_list ap;
  va_start(ap, fmt);
  vwarnx(fmt, ap);
  va_end(ap);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<warnx>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <vwarnx@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void warnx(const char *fmt, ...) {
  va_list ap;
  va_start(ap, fmt);
  vwarnx(fmt, ap);
  va_end(ap);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  test   %eax,%eax
  je     L1
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L1:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D4
datatype: string

tool: D4: %d

assistant: ```c
int main() {
  int a = 1;
  int b;
  if (a--)
    printf("%d", a);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
datatype: string

tool: D0: %d

assistant: ```c
int main() {
  int a = 1;
  int b;
  if (a--)
    printf("%d", a);
  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
  sub    $0x8,%rsp
  xor    %esi,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D9: %d

assistant: ```c
int main() {
  int a = 1;
  int b;
  if (a--)
    printf("%d", a);
  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
  sub    $0x8,%rsp
  xor    %esi,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
type: string

tool: D9: %d

assistant: ```c
int main() {
  int a = 1;
  int b;
  if (a--)
    printf("%d", a);
  return 0;
}
```