ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<version>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x6f69736976655224,%rax
  movabs $0x3833312e31203a6e,%rdx
  mov    %rax,-0x90(%rbp)
  mov    %rdx,-0x88(%rbp)
  movq   $0x2420,-0x80(%rbp)
  movq   $0x0,-0x78(%rbp)
  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   $0xb,-0x94(%rbp)
  jmp    L15
L36:
  addl   $0x1,-0x94(%rbp)
L15:
  mov    -0x94(%rbp),%eax
  cltq
  movzbl -0x90(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L36
  mov    -0x94(%rbp),%eax
  cltq
  movzbl -0x90(%rbp,%rax,1),%eax
  cmp    $0x2e,%al
  je     L36
  mov    -0x94(%rbp),%eax
  cltq
  movb   $0x0,-0x90(%rbp,%rax,1)
  lea    -0x90(%rbp),%rax
  add    $0xb,%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L39
  call   <__stack_chk_fail@plt>
L39:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D2
datatype: string

tool: D2: CAP - C Auxilary Preprocessor - version %s


assistant: ```c
void version() {
  char ver[128] = "$Revision: 1.138 $";
  int i = 11;
  while (isdigit(ver[i]) || (ver[i] == '.'))
    i++;
  ver[i] = 0;
  printf("CAP - C Auxilary Preprocessor - version %s\n", ver + 11);
}
```
user: What is the c source code of the assembly code below:

```asm
<version>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  movabs $0x6f69736976655224,%rax
  movabs $0x3833312e31203a6e,%rdx
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  movq   $0x2420,0x10(%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)
  movq   $0x0,0x60(%rsp)
  movq   $0x0,0x68(%rsp)
  movq   $0x0,0x70(%rsp)
  movq   $0x0,0x78(%rsp)
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    $0xb,%r12d
  sub    %eax,%r12d
  jmp    L26
L5:
  add    $0x1,%rbx
L26:
  lea    (%r12,%rbx,1),%ebp
  movsbl 0xb(%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L5
  cmpb   $0x2e,0xb(%rbx)
  je     L5
  movslq %ebp,%rbp
  movb   $0x0,(%rsp,%rbp,1)
  lea    0xb(%rsp),%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x90,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D2
type_infered: string

tool: D2: CAP - C Auxilary Preprocessor - version %s


assistant: ```c
void version() {
  char ver[128] = "$Revision: 1.138 $";
  int i = 11;
  while (isdigit(ver[i]) || (ver[i] == '.'))
    i++;
  ver[i] = 0;
  printf("CAP - C Auxilary Preprocessor - version %s\n", ver + 11);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<version>:
  endbr64
  push   %r13
  mov    $0x31,%edi
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0xb,%ebx
  sub    $0x98,%rsp
  movdqa D5(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  lea    0x1(%rsp),%r12
  movq   $0x2420,0x10(%rsp)
  movaps %xmm0,(%rsp)
  pxor   %xmm0,%xmm0
  movq   $0x0,0x18(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  movaps %xmm0,0x60(%rsp)
  movaps %xmm0,0x70(%rsp)
  jmp    L34
  nopl   0x0(%rax,%rax,1)
L26:
  movsbl (%r12,%rbx,1),%edi
  add    $0x1,%rbx
L34:
  movslq %ebx,%rbp
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L26
  cmpb   $0x2e,0x0(%r13,%rbx,1)
  je     L26
  lea    0xb(%rsp),%rsi
  lea    D8(%rip),%rdi
  movb   $0x0,(%rsp,%rbp,1)
  call   <printf@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x98,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D5
type: byte[16]
name: parse_data
arguments: tag: D8
type: string

tool: {"D5": [0x24,0x52,0x65,0x76,0x69,0x73,0x69,0x6f,0x6e,0x3a,0x20,0x31,0x2e,0x31,0x33,0x38]}

tool: D8: CAP - C Auxilary Preprocessor - version %s


assistant: ```c
void version() {
  char ver[128] = "$Revision: 1.138 $";
  int i = 11;
  while (isdigit(ver[i]) || (ver[i] == '.'))
    i++;
  ver[i] = 0;
  printf("CAP - C Auxilary Preprocessor - version %s\n", ver + 11);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<version>:
  endbr64
  push   %r12
  mov    $0xb,%r12d
  push   %rbp
  push   %rbx
  sub    $0x90,%rsp
  movdqa D1(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  movq   $0x2420,0x10(%rsp)
  movaps %xmm0,(%rsp)
  pxor   %xmm0,%xmm0
  sub    %ebx,%r12d
  movq   $0x0,0x18(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  movaps %xmm0,0x60(%rsp)
  movaps %xmm0,0x70(%rsp)
  jmp    L22
L2:
  add    $0x1,%rbx
L22:
  movsbl 0xb(%rbx),%edi
  lea    (%r12,%rbx,1),%ebp
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L2
  cmpb   $0x2e,0xb(%rbx)
  je     L2
  movslq %ebp,%rbp
  lea    0xb(%rsp),%rsi
  lea    D2(%rip),%rdi
  movb   $0x0,(%rsp,%rbp,1)
  call   <printf@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x90,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
type_infered: byte[16]
name: retrieve_data
arguments: data_tag: D2
type_infered: string

tool: {"D1": [0x24,0x52,0x65,0x76,0x69,0x73,0x69,0x6f,0x6e,0x3a,0x20,0x31,0x2e,0x31,0x33,0x38]}

tool: D2: CAP - C Auxilary Preprocessor - version %s


assistant: ```c
void version() {
  char ver[128] = "$Revision: 1.138 $";
  int i = 11;
  while (isdigit(ver[i]) || (ver[i] == '.'))
    i++;
  ver[i] = 0;
  printf("CAP - C Auxilary Preprocessor - version %s\n", ver + 11);
}
```
user: What is the c source code of the assembly code below:

```asm
<name_in_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L14
L20:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x28(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L1
  mov    $0x1,%eax
  jmp    L27
L1:
  addl   $0x1,-0x4(%rbp)
L14:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L20
  mov    $0x0,%eax
L27:
  leave
  ret
```

assistant: ```c
int name_in_array(const int num, const char **arr, const char *name) {
  int i;
  for (i = 0; i < num; i++) {
    if (!strcmp(arr[i], name))
      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
<name_in_array>:
  endbr64
  test   %edi,%edi
  jle    L28
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbp
  mov    %rsi,%rbx
  lea    -0x1(%rdi),%eax
  lea    0x8(%rsi,%rax,8),%r12
L0:
  mov    %rbp,%rsi
  mov    (%rbx),%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L14
  add    $0x8,%rbx
  cmp    %r12,%rbx
  jne    L0
  mov    $0x0,%eax
  jmp    L24
L28:
  mov    $0x0,%eax
  ret
L14:
  mov    $0x1,%eax
L24:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int name_in_array(const int num, const char **arr, const char *name) {
  int i;
  for (i = 0; i < num; i++) {
    if (!strcmp(arr[i], name))
      return 1;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<name_in_array>:
  endbr64
  test   %edi,%edi
  jle    L33
  lea    -0x1(%rdi),%eax
  push   %r12
  lea    0x8(%rsi,%rax,8),%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L6
  nopl   0x0(%rax)
L5:
  add    $0x8,%rbx
  cmp    %r12,%rbx
  je     L11
L6:
  mov    (%rbx),%rdi
  mov    %rbp,%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L5
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L33:
  xor    %eax,%eax
  ret
```

assistant: ```c
int name_in_array(const int num, const char **arr, const char *name) {
  int i;
  for (i = 0; i < num; i++) {
    if (!strcmp(arr[i], name))
      return 1;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<name_in_array>:
  endbr64
  test   %edi,%edi
  jle    L2
  lea    -0x1(%rdi),%eax
  push   %r12
  lea    0x8(%rsi,%rax,8),%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L1
  nopl   0x0(%rax)
L37:
  add    $0x8,%rbx
  cmp    %r12,%rbx
  je     L3
L1:
  mov    (%rbx),%rdi
  mov    %rbp,%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L37
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L2:
  xor    %eax,%eax
  ret
```

assistant: ```c
int name_in_array(const int num, const char **arr, const char *name) {
  int i;
  for (i = 0; i < num; i++) {
    if (!strcmp(arr[i], name))
      return 1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<wrout>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L88
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L75
L42:
  addq   $0x1,-0x10(%rbp)
L75:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L42
  mov    -0x10(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  je     L72
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  jmp    L89
L116:
  mov    -0x10(%rbp),%rax
  sub    -0x28(%rbp),%rax
  mov    %rax,%rdx
  mov    D17(%rip),%rax
  mov    (%rax),%eax
  mov    -0x28(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jle    L38
  mov    -0x8(%rbp),%rax
  add    %rax,-0x28(%rbp)
  jmp    L89
L38:
  cmpq   $0xffffffffffffffff,-0x8(%rbp)
  jne    L90
  mov    D6(%rip),%rax
  mov    (%rax),%edx
  mov    D11(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jne    L90
L89:
  mov    -0x28(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jb     L116
L90:
  mov    D6(%rip),%rax
  mov    -0x14(%rbp),%edx
  mov    %edx,(%rax)
  jmp    L26
L88:
  nop
  jmp    L26
L72:
  nop
L26:
  leave
  ret
```

assistant: ```c
void wrout(const char *p) {
  const char *q;
  int saved_errno;
  if (!p)
    return;
  q = p;
  while (*q)
    q++;
  if (q == p)
    return;
  saved_errno = errno;
  while (p < q) {
    ssize_t n = write(STDOUT_FILENO, p, (size_t)(q - p));
    if (n > 0)
      p += n;
    else if (n != (ssize_t)-1 || errno != EINTR)
      break;
  }
  errno = saved_errno;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<wrout>:
  endbr64
  test   %rdi,%rdi
  je     L80
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  cmpb   $0x0,(%rdi)
  je     L48
  mov    %rdi,%rbp
L70:
  add    $0x1,%rbp
  cmpb   $0x0,0x0(%rbp)
  jne    L70
  cmp    %rbp,%rbx
  je     L48
  mov    D39(%rip),%rax
  mov    (%rax),%r13d
  jae    L7
  mov    D12(%rip),%r12
  jmp    L53
L45:
  add    %rax,%rbx
L4:
  cmp    %rbp,%rbx
  jae    L7
L53:
  mov    %rbp,%rdx
  sub    %rbx,%rdx
  mov    %rbx,%rsi
  mov    (%r12),%edi
  call   <write@plt>
  test   %rax,%rax
  jg     L45
  cmp    $0xffffffffffffffff,%rax
  jne    L7
  mov    D39(%rip),%rdx
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,(%rdx)
  je     L4
L7:
  mov    D39(%rip),%rax
  mov    %r13d,(%rax)
L48:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L80:
  ret
```

assistant: ```c
void wrout(const char *p) {
  const char *q;
  int saved_errno;
  if (!p)
    return;
  q = p;
  while (*q)
    q++;
  if (q == p)
    return;
  saved_errno = errno;
  while (p < q) {
    ssize_t n = write(STDOUT_FILENO, p, (size_t)(q - p));
    if (n > 0)
      p += n;
    else if (n != (ssize_t)-1 || errno != EINTR)
      break;
  }
  errno = saved_errno;
}
```
user: Translate the below assembly code into C programming language:

```asm
<wrout>:
  endbr64
  test   %rdi,%rdi
  je     L28
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  cmpb   $0x0,(%rdi)
  mov    %rdi,%rbx
  je     L16
  mov    %rdi,%r14
L26:
  add    $0x1,%r14
  cmpb   $0x0,(%r14)
  jne    L26
  cmp    %r14,%rbx
  je     L16
  mov    D8(%rip),%r12
  mov    (%r12),%r13d
  jae    L9
  mov    D24(%rip),%rbp
  jmp    L86
  nopl   (%rax)
L89:
  add    %rax,%rbx
L80:
  cmp    %rbx,%r14
  jbe    L9
L86:
  mov    %r14,%rdx
  mov    0x0(%rbp),%edi
  mov    %rbx,%rsi
  sub    %rbx,%rdx
  call   <write@plt>
  test   %rax,%rax
  jg     L89
  cmp    $0xffffffffffffffff,%rax
  jne    L9
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,(%r12)
  je     L80
L9:
  mov    %r13d,(%r12)
L16:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L28:
  ret
```

assistant: ```c
void wrout(const char *p) {
  const char *q;
  int saved_errno;
  if (!p)
    return;
  q = p;
  while (*q)
    q++;
  if (q == p)
    return;
  saved_errno = errno;
  while (p < q) {
    ssize_t n = write(STDOUT_FILENO, p, (size_t)(q - p));
    if (n > 0)
      p += n;
    else if (n != (ssize_t)-1 || errno != EINTR)
      break;
  }
  errno = saved_errno;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<wrout>:
  endbr64
  test   %rdi,%rdi
  je     L81
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  cmpb   $0x0,(%rdi)
  mov    %rdi,%rbx
  je     L19
  mov    %rdi,%r14
L53:
  add    $0x1,%r14
  cmpb   $0x0,(%r14)
  jne    L53
  cmp    %r14,%rbx
  je     L19
  mov    D8(%rip),%r12
  mov    (%r12),%r13d
  jae    L83
  mov    D5(%rip),%rbp
  jmp    L72
  nopl   (%rax)
L14:
  add    %rax,%rbx
L63:
  cmp    %rbx,%r14
  jbe    L83
L72:
  mov    %r14,%rdx
  mov    0x0(%rbp),%edi
  mov    %rbx,%rsi
  sub    %rbx,%rdx
  call   <write@plt>
  test   %rax,%rax
  jg     L14
  cmp    $0xffffffffffffffff,%rax
  jne    L83
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,(%r12)
  je     L63
L83:
  mov    %r13d,(%r12)
L19:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L81:
  ret
```

assistant: ```c
void wrout(const char *p) {
  const char *q;
  int saved_errno;
  if (!p)
    return;
  q = p;
  while (*q)
    q++;
  if (q == p)
    return;
  saved_errno = errno;
  while (p < q) {
    ssize_t n = write(STDOUT_FILENO, p, (size_t)(q - p));
    if (n > 0)
      p += n;
    else if (n != (ssize_t)-1 || errno != EINTR)
      break;
  }
  errno = saved_errno;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sub_time>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x10(%rbp),%rax
  mov    0x8(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    0x8(%rax),%rax
  sub    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,0x8(%rax)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  sub    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <normalize_time@plt>
  nop
  leave
  ret
```

assistant: ```c
void sub_time(struct timespec *result, struct timespec *x, struct timespec *y) {
  result->tv_sec = x->tv_sec - y->tv_sec;
  result->tv_nsec = x->tv_nsec - y->tv_nsec;
  normalize_time(result);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sub_time>:
  endbr64
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rax
  sub    0x8(%rdx),%rax
  mov    %rax,0x8(%rdi)
  mov    (%rsi),%rax
  sub    (%rdx),%rax
  mov    %rax,(%rdi)
  call   <normalize_time@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void sub_time(struct timespec *result, struct timespec *x, struct timespec *y) {
  result->tv_sec = x->tv_sec - y->tv_sec;
  result->tv_nsec = x->tv_nsec - y->tv_nsec;
  normalize_time(result);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sub_time>:
  endbr64
  mov    0x8(%rsi),%rax
  sub    0x8(%rdx),%rax
  mov    %rax,0x8(%rdi)
  mov    (%rsi),%rax
  sub    (%rdx),%rax
  mov    %rax,(%rdi)
  jmp    1050 <normalize_time@plt>
```

assistant: ```c
void sub_time(struct timespec *result, struct timespec *x, struct timespec *y) {
  result->tv_sec = x->tv_sec - y->tv_sec;
  result->tv_nsec = x->tv_nsec - y->tv_nsec;
  normalize_time(result);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<sub_time>:
  endbr64
  movdqu (%rsi),%xmm0
  movdqu (%rdx),%xmm1
  psubq  %xmm1,%xmm0
  movups %xmm0,(%rdi)
  jmp    1050 <normalize_time@plt>
```

assistant: ```c
void sub_time(struct timespec *result, struct timespec *x, struct timespec *y) {
  result->tv_sec = x->tv_sec - y->tv_sec;
  result->tv_nsec = x->tv_nsec - y->tv_nsec;
  normalize_time(result);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ft_build_cmd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <ft_strquote@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <ft_echap_cmd@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0xffffffff,-0xc(%rbp)
  jmp    L11
L9:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x1,%al
  jne    L11
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  neg    %eax
  mov    %eax,%ecx
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L11:
  addl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L9
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *ft_build_cmd(char *cmd) {
  int i;
  char *echap;
  cmd = ft_strquote(cmd);
  echap = ft_echap_cmd(cmd);
  i = -1;
  while (cmd[++i]) {
    if (echap[i] == 1)
      cmd[i] = -cmd[i];
  }
  free(echap);
  return (cmd);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ft_build_cmd>:
  endbr64
  push   %rbx
  call   <ft_strquote@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <ft_echap_cmd@plt>
  mov    %rax,%rdi
  movzbl (%rbx),%ecx
  test   %cl,%cl
  je     L27
  mov    $0x0,%edx
  jmp    L4
L26:
  add    $0x1,%rdx
  movzbl (%rbx,%rdx,1),%ecx
  test   %cl,%cl
  je     L27
L4:
  cmpb   $0x1,(%rdi,%rdx,1)
  jne    L26
  neg    %ecx
  mov    %cl,(%rbx,%rdx,1)
  jmp    L26
L27:
  call   <free@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *ft_build_cmd(char *cmd) {
  int i;
  char *echap;
  cmd = ft_strquote(cmd);
  echap = ft_echap_cmd(cmd);
  i = -1;
  while (cmd[++i]) {
    if (echap[i] == 1)
      cmd[i] = -cmd[i];
  }
  free(echap);
  return (cmd);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ft_build_cmd>:
  endbr64
  push   %r12
  call   <ft_strquote@plt>
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <ft_echap_cmd@plt>
  movzbl (%r12),%ecx
  mov    %rax,%rdi
  test   %cl,%cl
  je     L28
  xor    %edx,%edx
  nopl   0x0(%rax)
L9:
  cmpb   $0x1,(%rdi,%rdx,1)
  jne    L3
  neg    %ecx
  mov    %cl,(%r12,%rdx,1)
L3:
  add    $0x1,%rdx
  movzbl (%r12,%rdx,1),%ecx
  test   %cl,%cl
  jne    L9
L28:
  call   <free@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *ft_build_cmd(char *cmd) {
  int i;
  char *echap;
  cmd = ft_strquote(cmd);
  echap = ft_echap_cmd(cmd);
  i = -1;
  while (cmd[++i]) {
    if (echap[i] == 1)
      cmd[i] = -cmd[i];
  }
  free(echap);
  return (cmd);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ft_build_cmd>:
  endbr64
  push   %r12
  call   <ft_strquote@plt>
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <ft_echap_cmd@plt>
  movzbl (%r12),%ecx
  mov    %rax,%rdi
  test   %cl,%cl
  je     L32
  xor    %edx,%edx
  nopl   0x0(%rax)
L37:
  cmpb   $0x1,(%rdi,%rdx,1)
  je     L26
  add    $0x1,%rdx
  movzbl (%r12,%rdx,1),%ecx
  test   %cl,%cl
  jne    L37
L32:
  call   <free@plt>
  mov    %r12,%rax
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L26:
  neg    %ecx
  mov    %cl,(%r12,%rdx,1)
  add    $0x1,%rdx
  movzbl (%r12,%rdx,1),%ecx
  test   %cl,%cl
  jne    L37
  call   <free@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *ft_build_cmd(char *cmd) {
  int i;
  char *echap;
  cmd = ft_strquote(cmd);
  echap = ft_echap_cmd(cmd);
  i = -1;
  while (cmd[++i]) {
    if (echap[i] == 1)
      cmd[i] = -cmd[i];
  }
  free(echap);
  return (cmd);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<qsort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  cmpl   $0x8,-0x20(%rbp)
  je     L6
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0xa,%edi
  call   <exit@plt>
L6:
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L2
L41:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x10(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  comisd %xmm1,%xmm0
  jbe    L5
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x10(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  movl   $0x1,-0xc(%rbp)
L5:
  addl   $0x1,-0x10(%rbp)
L2:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x10(%rbp)
  jl     L41
  cmpl   $0x0,-0xc(%rbp)
  jne    L6
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D3
type: string

tool: D3: failed in qsort!

assistant: ```c
qsort(array, n, l, f) int n, l, (*f)();
double array[];
{
  int j, change;
  double temp;
  if (l != sizeof(double)) {
    printf("failed in qsort!\n");
    exit(10);
  } else {
    do {
      change = 0;
      for (j = 0; j < (n - 1); j++) {
        if (array[j] > array[j + 1]) {
          temp = array[j];
          array[j] = array[j + 1];
          array[j + 1] = temp;
          change = 1;
        }
      }
    } while (change);
  }
}
```