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
<str_trim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  cmpq   $0x0,-0x30(%rbp)
  jne    L18
  mov    -0x28(%rbp),%rax
  jmp    L48
L18:
  movq   $0x0,-0x10(%rbp)
  jmp    L16
L91:
  addq   $0x1,-0x10(%rbp)
L16:
  mov    -0x30(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L28
  mov    -0x30(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L91
L28:
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  jmp    L62
L8:
  subq   $0x1,-0x8(%rbp)
L62:
  cmpq   $0x0,-0x8(%rbp)
  je     L87
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L8
L87:
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L95
L77:
  mov    -0x30(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x28(%rbp)
  addq   $0x1,-0x18(%rbp)
L95:
  mov    -0x18(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L77
  mov    -0x28(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x28(%rbp),%rax
L48:
  leave
  ret
```

assistant: ```c
char *str_trim(char *dest, const char *src) {
  if (src == NULL)
    return dest;
  size_t i;
  size_t start;
  size_t end;
  start = 0;
  while (src[start] != '\0' && isspace((unsigned char)src[start])) {
    ++start;
  }
  end = strlen(src);
  while (end > 0 && isspace((unsigned char)src[end - 1])) {
    --end;
  }
  for (i = start; i < end; i++) {
    *dest = src[i];
    ++dest;
  }
  *dest = '\0';
  return dest;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<str_trim>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  test   %rsi,%rsi
  je     L44
  mov    %rsi,%r13
  movzbl (%rsi),%edi
  mov    $0x0,%ebp
  test   %dil,%dil
  je     L77
L75:
  movzbl %dil,%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L77
  add    $0x1,%rbp
  movzbl 0x0(%r13,%rbp,1),%edi
  test   %dil,%dil
  jne    L75
L77:
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %rax,%rbx
  jmp    L16
L47:
  mov    %r14,%rbx
L16:
  test   %rbx,%rbx
  je     L83
  lea    -0x1(%rbx),%r14
  movzbl -0x1(%r13,%rbx,1),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L47
  cmp    %rbp,%rbx
  jbe    L83
  mov    %rbp,%rax
  mov    %r12,%rcx
  sub    %rbp,%rcx
L87:
  movzbl 0x0(%r13,%rax,1),%edx
  mov    %dl,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rbx
  jne    L87
  sub    %rbp,%rbx
  add    %rbx,%r12
L83:
  movb   $0x0,(%r12)
  mov    %r12,%rax
L91:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L44:
  mov    %rdi,%rax
  jmp    L91
```

assistant: ```c
char *str_trim(char *dest, const char *src) {
  if (src == NULL)
    return dest;
  size_t i;
  size_t start;
  size_t end;
  start = 0;
  while (src[start] != '\0' && isspace((unsigned char)src[start])) {
    ++start;
  }
  end = strlen(src);
  while (end > 0 && isspace((unsigned char)src[end - 1])) {
    --end;
  }
  for (i = start; i < end; i++) {
    *dest = src[i];
    ++dest;
  }
  *dest = '\0';
  return dest;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<str_trim>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %rsi,%rsi
  je     L30
  movzbl (%rsi),%edi
  mov    %rsi,%r13
  xor    %ebp,%ebp
  test   %dil,%dil
  jne    L10
  jmp    L23
  nopw   0x0(%rax,%rax,1)
L4:
  add    $0x1,%rbp
  movzbl 0x0(%r13,%rbp,1),%edi
  test   %dil,%dil
  je     L23
L10:
  call   <isspace@plt>
  test   %rax,%rax
  jne    L4
L23:
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %rax,%rbx
  jmp    L89
  cs nopw 0x0(%rax,%rax,1)
L81:
  movzbl -0x1(%r13,%rbx,1),%edi
  lea    -0x1(%rbx),%r14
  call   <isspace@plt>
  test   %rax,%rax
  je     L78
  mov    %r14,%rbx
L89:
  test   %rbx,%rbx
  jne    L81
L64:
  mov    %r12,%rax
L101:
  movb   $0x0,(%rax)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L78:
  cmp    %rbp,%rbx
  jbe    L64
  mov    %r12,%rcx
  mov    %rbp,%rax
  sub    %rbp,%rcx
  xchg   %ax,%ax
L100:
  movzbl 0x0(%r13,%rax,1),%edx
  mov    %dl,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rbx
  jne    L100
  sub    %rbp,%rbx
  lea    (%r12,%rbx,1),%rax
  jmp    L101
  nopl   0x0(%rax,%rax,1)
L30:
  pop    %rbx
  mov    %rdi,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *str_trim(char *dest, const char *src) {
  if (src == NULL)
    return dest;
  size_t i;
  size_t start;
  size_t end;
  start = 0;
  while (src[start] != '\0' && isspace((unsigned char)src[start])) {
    ++start;
  }
  end = strlen(src);
  while (end > 0 && isspace((unsigned char)src[end - 1])) {
    --end;
  }
  for (i = start; i < end; i++) {
    *dest = src[i];
    ++dest;
  }
  *dest = '\0';
  return dest;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<net2short>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %ax,-0x2(%rbp)
  movswl -0x2(%rbp),%eax
  shl    $0x8,%eax
  mov    %ax,-0x2(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  or     %ax,-0x2(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  lea    0x2(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  movzwl -0x2(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned short int net2short(unsigned char **bufp) {
  short int i;
  i = **bufp;
  i <<= 8;
  i |= *(*bufp + 1);
  *bufp += 2;
  return i;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<net2short>:
  endbr64
  mov    (%rdi),%rdx
  movzbl (%rdx),%ecx
  shl    $0x8,%ecx
  movzbl 0x1(%rdx),%eax
  or     %ecx,%eax
  add    $0x2,%rdx
  mov    %rdx,(%rdi)
  ret
```

assistant: ```c
unsigned short int net2short(unsigned char **bufp) {
  short int i;
  i = **bufp;
  i <<= 8;
  i |= *(*bufp + 1);
  *bufp += 2;
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<net2short>:
  endbr64
  mov    (%rdi),%rax
  movzwl (%rax),%r8d
  add    $0x2,%rax
  mov    %rax,(%rdi)
  rol    $0x8,%r8w
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned short int net2short(unsigned char **bufp) {
  short int i;
  i = **bufp;
  i <<= 8;
  i |= *(*bufp + 1);
  *bufp += 2;
  return i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<net2short>:
  endbr64
  mov    (%rdi),%rax
  movzwl (%rax),%r8d
  add    $0x2,%rax
  mov    %rax,(%rdi)
  rol    $0x8,%r8w
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned short int net2short(unsigned char **bufp) {
  short int i;
  i = **bufp;
  i <<= 8;
  i |= *(*bufp + 1);
  *bufp += 2;
  return i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<zalloc_align>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dx,-0x4(%rbp)
  mov    %ax,-0x8(%rbp)
  mov    $0x0,%edi
  call   <alloc@plt>
  movzwl -0x4(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %eax,%eax
  je     L1
  mov    $0x0,%edi
  call   <alloc@plt>
  movzwl -0x4(%rbp),%ecx
  mov    $0x0,%edx
  div    %ecx
  mov    %edx,%eax
  mov    %eax,%edx
  movzwl -0x4(%rbp),%eax
  sub    %edx,%eax
  movzwl %ax,%eax
  mov    %eax,%edi
  call   <alloc@plt>
L1:
  movzwl -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <zalloc@plt>
  leave
  ret
```

assistant: ```c
void *zalloc_align(unsigned short boundary, unsigned short size) {
  if ((int)alloc(0) % boundary)
    alloc(boundary - ((unsigned int)alloc(0) % boundary));
  return zalloc(size);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<zalloc_align>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    $0x0,%edi
  call   <alloc@plt>
  movzwl %bp,%ecx
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L14
L4:
  movzwl %bx,%edi
  call   <zalloc@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L14:
  mov    $0x0,%edi
  call   <alloc@plt>
  movzwl %bp,%ecx
  mov    $0x0,%edx
  div    %ecx
  sub    %edx,%ebp
  movzwl %bp,%edi
  call   <alloc@plt>
  jmp    L4
```

assistant: ```c
void *zalloc_align(unsigned short boundary, unsigned short size) {
  if ((int)alloc(0) % boundary)
    alloc(boundary - ((unsigned int)alloc(0) % boundary));
  return zalloc(size);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<zalloc_align>:
  endbr64
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  xor    %edi,%edi
  push   %rbx
  mov    %esi,%ebx
  movzwl %bp,%r12d
  call   <alloc@plt>
  cltd
  idiv   %r12d
  test   %edx,%edx
  jne    L3
  movzwl %bx,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1070 <zalloc@plt>
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %edi,%edi
  call   <alloc@plt>
  xor    %edx,%edx
  div    %r12d
  sub    %edx,%ebp
  movzwl %bp,%edi
  call   <alloc@plt>
  movzwl %bx,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1070 <zalloc@plt>
```

assistant: ```c
void *zalloc_align(unsigned short boundary, unsigned short size) {
  if ((int)alloc(0) % boundary)
    alloc(boundary - ((unsigned int)alloc(0) % boundary));
  return zalloc(size);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<zalloc_align>:
  endbr64
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  xor    %edi,%edi
  push   %rbx
  mov    %esi,%ebx
  movzwl %bp,%r12d
  call   <alloc@plt>
  cltd
  idiv   %r12d
  test   %edx,%edx
  jne    L0
  movzwl %bx,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1070 <zalloc@plt>
  nopl   0x0(%rax,%rax,1)
L0:
  xor    %edi,%edi
  call   <alloc@plt>
  xor    %edx,%edx
  div    %r12d
  sub    %edx,%ebp
  movzwl %bp,%edi
  call   <alloc@plt>
  movzwl %bx,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1070 <zalloc@plt>
```

assistant: ```c
void *zalloc_align(unsigned short boundary, unsigned short size) {
  if ((int)alloc(0) % boundary)
    alloc(boundary - ((unsigned int)alloc(0) % boundary));
  return zalloc(size);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getTailCommand>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x1,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L0
L18:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x10(%rbp)
  addl   $0x1,-0xc(%rbp)
L0:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x10(%rbp)
  jl     L18
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *getTailCommand(char *input) {
  char *temp = (char *)malloc(sizeof(char) * (strlen(input) - 1));
  int i, j;
  for (i = 1, j = 0; i < strlen(input); i++, j++)
    temp[j] = input[i];
  return temp;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<getTailCommand>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    -0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    $0x0,%ebx
  jmp    L8
L7:
  movzbl 0x1(%rbp,%rbx,1),%edx
  mov    %dl,(%r12,%rbx,1)
  mov    %rax,%rbx
L8:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  lea    0x1(%rbx),%rax
  cmp    %eax,%edx
  jg     L7
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *getTailCommand(char *input) {
  char *temp = (char *)malloc(sizeof(char) * (strlen(input) - 1));
  int i, j;
  for (i = 1, j = 0; i < strlen(input); i++, j++)
    temp[j] = input[i];
  return temp;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<getTailCommand>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  call   <strlen@plt>
  lea    -0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  jmp    L5
  nop
L14:
  movzbl 0x1(%rbp,%rbx,1),%edx
  mov    %dl,(%r12,%rbx,1)
  mov    %rax,%rbx
L5:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%r8d
  lea    0x1(%rbx),%rax
  cmp    %eax,%r8d
  jg     L14
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

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

```asm
<getTailCommand>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  call   <strlen@plt>
  lea    -0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  jmp    L16
  nop
L18:
  movzbl 0x1(%rbp,%rbx,1),%edx
  mov    %dl,(%r12,%rbx,1)
  mov    %rax,%rbx
L16:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%r8d
  lea    0x1(%rbx),%rax
  cmp    %eax,%r8d
  jg     L18
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *getTailCommand(char *input) {
  char *temp = (char *)malloc(sizeof(char) * (strlen(input) - 1));
  int i, j;
  for (i = 1, j = 0; i < strlen(input); i++, j++)
    temp[j] = input[i];
  return temp;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xfd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D11(%rip),%rax
  mov    %rax,%rsi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0xfb8(%rbp)
  movl   $0x0,-0xfd0(%rbp)
  jmp    L34
L73:
  lea    -0xfb0(%rbp),%rdx
  mov    -0xfd0(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rax,%rdx
  mov    -0xfb8(%rbp),%rax
  lea    D36(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <fscanf@plt>
  mov    -0xfb8(%rbp),%rax
  mov    %rax,%rdi
  call   <feof@plt>
  test   %rax,%rax
  jne    L97
  addl   $0x1,-0xfd0(%rbp)
L34:
  cmpl   $0x3e7,-0xfd0(%rbp)
  jle    L73
  jmp    L117
L97:
  nop
L117:
  mov    -0xfd0(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xfcc(%rbp)
  jmp    L11
L88:
  mov    -0xfcc(%rbp),%eax
  mov    %eax,-0xfc8(%rbp)
  movl   $0x0,-0xfc4(%rbp)
  jmp    L75
L104:
  mov    -0xfc4(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%edx
  mov    -0xfc8(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  jle    L38
  mov    -0xfc4(%rbp),%eax
  mov    %eax,-0xfc8(%rbp)
L38:
  addl   $0x1,-0xfc4(%rbp)
L75:
  mov    -0xfc4(%rbp),%eax
  cmp    -0xfcc(%rbp),%eax
  jl     L104
  mov    -0xfcc(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  mov    %eax,-0xfbc(%rbp)
  mov    -0xfc8(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%edx
  mov    -0xfcc(%rbp),%eax
  cltq
  mov    %edx,-0xfb0(%rbp,%rax,4)
  mov    -0xfc8(%rbp),%eax
  cltq
  mov    -0xfbc(%rbp),%edx
  mov    %edx,-0xfb0(%rbp,%rax,4)
  subl   $0x1,-0xfcc(%rbp)
L11:
  cmpl   $0x0,-0xfcc(%rbp)
  jns    L88
  movl   $0x0,-0xfc0(%rbp)
  jmp    L59
L30:
  mov    -0xfc0(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0xfc0(%rbp)
L59:
  mov    -0xfc0(%rbp),%eax
  cmp    -0xfd0(%rbp),%eax
  jl     L30
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L105
  call   <__stack_chk_fail@plt>
L105:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D11
type_guessed: string
name: extract_symbolic_data
arguments: label: D34
type_guessed: string
name: extract_symbolic_data
arguments: label: D36
type_guessed: string
name: extract_symbolic_data
arguments: label: D30
type_guessed: string

tool: D11: r

tool: D34: res/input.txt

tool: D36: %d

tool: D30: %d


assistant: ```c
int main(void) {
  FILE *fp = fopen("res/input.txt", "r");
  int data[1000], size;
  for (size = 0; size < 1000; ++size) {
    fscanf(fp, "%d", &data[size]);
    if (feof(fp))
      break;
  }
  for (int i = size - 1; i >= 0; --i) {
    int max = i;
    for (int j = 0; j < i; ++j)
      if (data[j] > data[max])
        max = j;
    int tmp = data[i];
    data[i] = data[max];
    data[max] = tmp;
  }
  for (int i = 0; i < size; ++i)
    printf("%d\n", data[i]);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xfb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  lea    D6(%rip),%rsi
  lea    D2(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%r13
  mov    $0x0,%r12d
  lea    D13(%rip),%r14
L86:
  mov    %r12d,%ebp
  mov    %rsp,%rbx
  lea    (%rbx,%r12,4),%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <fscanf@plt>
  mov    %r13,%rdi
  call   <feof@plt>
  test   %rax,%rax
  jne    L82
  lea    0x1(%r12),%ebp
  add    $0x1,%r12
  cmp    $0x3e8,%r12
  jne    L86
L82:
  mov    %ebp,%edi
  sub    $0x1,%edi
  js     L56
  lea    -0x2(%rbp),%r9d
  movslq %ebp,%rax
  lea    (%rsp,%rax,4),%r8
  mov    $0x0,%r10d
  jmp    L97
L51:
  mov    %rcx,%rax
L75:
  movslq %edx,%rcx
  mov    (%rsp,%rcx,4),%ecx
  cmp    %ecx,(%rsp,%rax,4)
  cmovg  %eax,%edx
  lea    0x1(%rax),%rcx
  cmp    %rsi,%rax
  jne    L51
  mov    -0x4(%r8),%eax
  movslq %edx,%rdx
  mov    (%rsp,%rdx,4),%ecx
  mov    %ecx,-0x4(%r8)
  mov    %eax,(%rsp,%rdx,4)
L29:
  sub    $0x1,%edi
  sub    $0x1,%r9d
  sub    $0x4,%r8
L97:
  mov    %edi,%edx
  mov    %r9d,%esi
  mov    %r10,%rax
  test   %edi,%edi
  jg     L75
  test   %r9d,%r9d
  jns    L29
L56:
  test   %ebp,%ebp
  jle    L2
  lea    -0x1(%rbp),%eax
  lea    0x4(%rbx,%rax,4),%r12
  lea    D16(%rip),%rbp
L48:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L48
L2:
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L66
  mov    $0x0,%eax
  add    $0xfb0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L66:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D6
type: string
name: unpack_data
arguments: tag: D2
type: string
name: unpack_data
arguments: tag: D13
type: string
name: unpack_data
arguments: tag: D16
type: string

tool: D6: r

tool: D2: res/input.txt

tool: D13: %d

tool: D16: %d


assistant: ```c
int main(void) {
  FILE *fp = fopen("res/input.txt", "r");
  int data[1000], size;
  for (size = 0; size < 1000; ++size) {
    fscanf(fp, "%d", &data[size]);
    if (feof(fp))
      break;
  }
  for (int i = size - 1; i >= 0; --i) {
    int max = i;
    for (int j = 0; j < i; ++j)
      if (data[j] > data[max])
        max = j;
    int tmp = data[i];
    data[i] = data[max];
    data[max] = tmp;
  }
  for (int i = 0; i < size; ++i)
    printf("%d\n", data[i]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  lea    D35(%rip),%rsi
  lea    D33(%rip),%rdi
  push   %r13
  lea    D18(%rip),%r14
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0xfb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <fopen@plt>
  mov    %rax,%r13
  jmp    L38
  nopl   0x0(%rax,%rax,1)
L18:
  add    $0x1,%r12
  cmp    $0x3e8,%r12
  je     L78
L38:
  mov    %r13,%rdi
  lea    (%rbx,%r12,4),%rdx
  mov    %r14,%rsi
  mov    %r12d,%ebp
  call   <fscanf@plt>
  mov    %r13,%rdi
  call   <feof@plt>
  test   %rax,%rax
  je     L18
  lea    -0x1(%r12),%esi
  test   %r12d,%r12d
  je     L8
L122:
  movslq %esi,%rax
  lea    (%rbx,%rax,4),%r8
  nopl   0x0(%rax)
L112:
  mov    (%r8),%r10d
  lea    -0x1(%rsi),%edi
  xor    %eax,%eax
  mov    %rdi,%r9
  mov    %r10d,%ecx
  test   %esi,%esi
  jg     L60
  jmp    L6
  nopl   0x0(%rax)
L9:
  mov    %rdx,%rax
L60:
  mov    (%rbx,%rax,4),%edx
  cmp    %ecx,%edx
  jle    L14
  mov    %edx,%ecx
  mov    %eax,%esi
L14:
  lea    0x1(%rax),%rdx
  cmp    %rdi,%rax
  jne    L9
  movslq %esi,%rsi
  mov    %ecx,(%r8)
  mov    %r10d,(%rsp,%rsi,4)
L31:
  sub    $0x4,%r8
  mov    %r9d,%esi
  jmp    L112
  nopw   0x0(%rax,%rax,1)
L6:
  cmp    $0xffffffff,%r9d
  jne    L31
  xor    %r12d,%r12d
  lea    D24(%rip),%r13
L41:
  mov    (%rbx,%r12,4),%esi
  mov    %r13,%rdi
  add    $0x1,%r12
  call   <printf@plt>
  cmp    %r12d,%ebp
  jg     L41
L8:
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L116
  add    $0xfb0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L78:
  mov    $0x3e8,%ebp
  mov    $0x3e7,%esi
  jmp    L122
L116:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D35
type: string
name: retrieve_data
arguments: data_tag: D33
type: string
name: retrieve_data
arguments: data_tag: D18
type: string
name: retrieve_data
arguments: data_tag: D24
type: string

tool: D35: r

tool: D33: res/input.txt

tool: D18: %d

tool: D24: %d


assistant: ```c
int main(void) {
  FILE *fp = fopen("res/input.txt", "r");
  int data[1000], size;
  for (size = 0; size < 1000; ++size) {
    fscanf(fp, "%d", &data[size]);
    if (feof(fp))
      break;
  }
  for (int i = size - 1; i >= 0; --i) {
    int max = i;
    for (int j = 0; j < i; ++j)
      if (data[j] > data[max])
        max = j;
    int tmp = data[i];
    data[i] = data[max];
    data[max] = tmp;
  }
  for (int i = 0; i < size; ++i)
    printf("%d\n", data[i]);
}
```