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

```asm
<reverseWords>:
  endbr64
  mov    %rdi,%rcx
  test   %rdi,%rdi
  je     L156
  movzbl (%rdi),%eax
  test   %al,%al
  je     L156
  lea    0x1(%rdi),%rsi
  mov    $0x20,%edi
  mov    $0xffffffff,%edx
  jmp    L188
L59:
  add    $0x1,%edx
  movslq %edx,%rdi
  mov    %al,(%rcx,%rdi,1)
  mov    %eax,%edi
L40:
  add    $0x1,%rsi
  movzbl -0x1(%rsi),%eax
  test   %al,%al
  je     L85
L188:
  cmp    $0x20,%al
  jne    L59
  cmp    $0x20,%dil
  je     L40
  add    $0x1,%edx
  movslq %edx,%rdi
  movb   $0x20,(%rcx,%rdi,1)
  mov    %eax,%edi
  jmp    L40
L85:
  movslq %edx,%rax
  add    %rcx,%rax
  cmpb   $0x20,(%rax)
  je     L29
  add    $0x1,%edx
  movslq %edx,%rax
  movb   $0x0,(%rcx,%rax,1)
L65:
  mov    %edx,%r8d
  shr    $0x1f,%r8d
  add    %edx,%r8d
  sar    %r8d
  cmp    $0x1,%edx
  jle    L151
  movslq %edx,%rdx
  lea    -0x1(%rcx,%rdx,1),%rdx
  mov    $0x0,%eax
L119:
  movzbl (%rcx,%rax,1),%esi
  movzbl (%rdx),%edi
  mov    %dil,(%rcx,%rax,1)
  mov    %sil,(%rdx)
  add    $0x1,%rax
  sub    $0x1,%rdx
  cmp    %eax,%r8d
  jg     L119
L151:
  movzbl (%rcx),%eax
  test   %al,%al
  je     L156
  mov    $0x0,%edi
  lea    -0x2(%rcx),%r10
  jmp    L56
L29:
  movb   $0x0,(%rax)
  jmp    L65
L100:
  add    $0x1,%edi
L130:
  movslq %edi,%rax
  movzbl (%rcx,%rax,1),%eax
  test   %al,%al
  je     L156
L56:
  cmp    $0x20,%al
  je     L100
  movslq %edi,%rdx
  add    %rcx,%rdx
  lea    0x1(%rdi),%eax
  cltq
  testb  $0xdf,(%rdx)
  je     L130
L103:
  mov    %eax,%r9d
  movzbl (%rcx,%rax,1),%esi
  add    $0x1,%rax
  test   $0xdf,%sil
  jne    L103
  mov    %r9d,%eax
  sub    %edi,%eax
  mov    %eax,%esi
  shr    $0x1f,%esi
  add    %eax,%esi
  sar    %esi
  lea    (%rsi,%rdi,1),%eax
  cmp    %edi,%eax
  jle    L46
  movslq %r9d,%r8
  lea    -0x1(%rcx,%r8,1),%rax
  add    %r10,%r8
  lea    -0x1(%rsi),%esi
  sub    %rsi,%r8
L10:
  movzbl (%rdx),%esi
  movzbl (%rax),%edi
  mov    %dil,(%rdx)
  mov    %sil,(%rax)
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %rax,%r8
  jne    L10
  mov    %r9d,%edi
  jmp    L130
L46:
  mov    %r9d,%edi
  jmp    L130
L156:
  ret
```

assistant: ```c
void reverseWords(char *s) {
  if (!s || !s[0])
    return;
  int i, j, l;
  char prev;
  prev = ' ';
  for (l = -1, i = 0; s[i]; i++) {
    if (s[i] != ' ') {
      s[++l] = s[i];
      prev = s[i];
      continue;
    }
    if (prev == ' ') {
      continue;
    } else {
      s[++l] = ' ';
      prev = ' ';
    }
  }
  if (s[l] == ' ')
    s[l] = 0;
  else
    s[++l] = 0;
  for (i = 0; i < l / 2; i++) {
    prev = s[i];
    s[i] = s[l - i - 1];
    s[l - i - 1] = prev;
  }
  for (i = 0; s[i];) {
    if (s[i] == ' ') {
      i++;
      continue;
    }
    j = i;
    while (s[i] && s[i] != ' ')
      i++;
    for (l = j; l < j + (i - j) / 2; l++) {
      prev = s[l];
      s[l] = s[i - l + j - 1];
      s[i - l + j - 1] = prev;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<reverseWords>:
  endbr64
  mov    %rdi,%r8
  test   %rdi,%rdi
  je     L48
  movzbl (%rdi),%eax
  test   %al,%al
  je     L48
  lea    0x1(%rdi),%rdx
  mov    $0x20,%esi
  mov    $0xffffffff,%ecx
  jmp    L194
  nopl   0x0(%rax,%rax,1)
L106:
  add    $0x1,%ecx
  movslq %ecx,%rsi
  mov    %al,(%r8,%rsi,1)
  mov    %eax,%esi
L159:
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L185
L194:
  cmp    $0x20,%al
  jne    L106
  cmp    $0x20,%sil
  je     L159
  add    $0x1,%ecx
  add    $0x1,%rdx
  mov    $0x20,%esi
  movslq %ecx,%rax
  movb   $0x20,(%r8,%rax,1)
  movzbl -0x1(%rdx),%eax
  test   %al,%al
  jne    L194
L185:
  movslq %ecx,%rax
  lea    (%r8,%rax,1),%rdx
  cmpb   $0x20,(%rdx)
  je     L206
  add    $0x1,%ecx
  movslq %ecx,%rax
  movb   $0x0,(%r8,%rax,1)
L43:
  mov    %ecx,%edi
  lea    -0x1(%r8,%rax,1),%rdx
  xor    %eax,%eax
  shr    $0x1f,%edi
  add    %ecx,%edi
  sar    %edi
  cmp    $0x1,%ecx
  jle    L94
  nopl   (%rax)
L135:
  movzbl (%rdx),%esi
  movzbl (%r8,%rax,1),%ecx
  sub    $0x1,%rdx
  mov    %sil,(%r8,%rax,1)
  add    $0x1,%rax
  mov    %cl,0x1(%rdx)
  cmp    %eax,%edi
  jg     L135
L94:
  movzbl (%r8),%esi
  xor    %r11d,%r11d
  test   %sil,%sil
  je     L62
  nopl   0x0(%rax,%rax,1)
L47:
  cmp    $0x20,%sil
  je     L132
  movslq %r11d,%rdx
  lea    (%r8,%rdx,1),%rax
  movzbl (%rax),%esi
  test   $0xdf,%sil
  je     L21
  lea    0x1(%r8,%rdx,1),%rdx
  lea    0x1(%r11),%r10d
  sub    %edx,%r10d
  cs nopw 0x0(%rax,%rax,1)
L100:
  movzbl (%rdx),%edi
  lea    (%r10,%rdx,1),%r9d
  mov    %rdx,%rcx
  lea    0x1(%rdx),%rdx
  test   $0xdf,%dil
  jne    L100
  mov    %r9d,%r10d
  sub    %r11d,%r10d
  sar    %r10d
  lea    (%r10,%r11,1),%edx
  cmp    %r11d,%edx
  jle    L32
  movslq %r9d,%rdx
  sub    $0x1,%r10d
  lea    -0x1(%r8,%rdx,1),%rdx
  add    %rax,%r10
  jmp    L19
  nopw   0x0(%rax,%rax,1)
L134:
  movzbl 0x1(%rax),%esi
  add    $0x1,%rax
L19:
  movzbl (%rdx),%edi
  sub    $0x1,%rdx
  mov    %dil,(%rax)
  mov    %sil,0x1(%rdx)
  cmp    %rax,%r10
  jne    L134
  movzbl (%rcx),%esi
  mov    %r9d,%r11d
L21:
  test   %sil,%sil
  jne    L47
L48:
  ret
  nopl   0x0(%rax,%rax,1)
L132:
  add    $0x1,%r11d
  movslq %r11d,%rax
  movzbl (%r8,%rax,1),%esi
  jmp    L21
L206:
  movb   $0x0,(%rdx)
  jmp    L43
L62:
  ret
L32:
  mov    %edi,%esi
  mov    %r9d,%r11d
  jmp    L21
```

assistant: ```c
void reverseWords(char *s) {
  if (!s || !s[0])
    return;
  int i, j, l;
  char prev;
  prev = ' ';
  for (l = -1, i = 0; s[i]; i++) {
    if (s[i] != ' ') {
      s[++l] = s[i];
      prev = s[i];
      continue;
    }
    if (prev == ' ') {
      continue;
    } else {
      s[++l] = ' ';
      prev = ' ';
    }
  }
  if (s[l] == ' ')
    s[l] = 0;
  else
    s[++l] = 0;
  for (i = 0; i < l / 2; i++) {
    prev = s[i];
    s[i] = s[l - i - 1];
    s[l - i - 1] = prev;
  }
  for (i = 0; s[i];) {
    if (s[i] == ' ') {
      i++;
      continue;
    }
    j = i;
    while (s[i] && s[i] != ' ')
      i++;
    for (l = j; l < j + (i - j) / 2; l++) {
      prev = s[l];
      s[l] = s[i - l + j - 1];
      s[i - l + j - 1] = prev;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<reverseWords>:
  endbr64
  mov    %rdi,%r9
  test   %rdi,%rdi
  je     L175
  movzbl (%rdi),%eax
  test   %al,%al
  je     L175
  lea    0x1(%rdi),%rdx
  mov    $0x20,%esi
  mov    $0xffffffff,%ecx
  jmp    L64
  nopl   0x0(%rax,%rax,1)
L62:
  add    $0x1,%ecx
  movslq %ecx,%rsi
  mov    %al,(%r9,%rsi,1)
  mov    %eax,%esi
L150:
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L120
L64:
  cmp    $0x20,%al
  jne    L62
  cmp    $0x20,%sil
  je     L150
  add    $0x1,%ecx
  add    $0x1,%rdx
  mov    $0x20,%esi
  movslq %ecx,%rax
  movb   $0x20,(%r9,%rax,1)
  movzbl -0x1(%rdx),%eax
  test   %al,%al
  jne    L64
L120:
  movslq %ecx,%rax
  lea    (%r9,%rax,1),%rdx
  cmpb   $0x20,(%rdx)
  je     L198
  add    $0x1,%ecx
  movslq %ecx,%rax
  movb   $0x0,(%r9,%rax,1)
L94:
  mov    %ecx,%edi
  lea    -0x1(%r9,%rax,1),%rdx
  xor    %eax,%eax
  shr    $0x1f,%edi
  add    %ecx,%edi
  sar    %edi
  cmp    $0x1,%ecx
  jle    L41
  nopl   (%rax)
L88:
  movzbl (%rdx),%esi
  movzbl (%r9,%rax,1),%ecx
  sub    $0x1,%rdx
  mov    %sil,(%r9,%rax,1)
  add    $0x1,%rax
  mov    %cl,0x1(%rdx)
  cmp    %eax,%edi
  jg     L88
L41:
  movzbl (%r9),%esi
  xor    %r8d,%r8d
  lea    -0x2(%r9),%r11
  test   %sil,%sil
  je     L113
  nop
L178:
  cmp    $0x20,%sil
  je     L138
  test   $0xdf,%sil
  je     L178
  movslq %r8d,%rdx
  lea    0x1(%r8),%edi
  lea    0x1(%r9,%rdx,1),%rax
  sub    %eax,%edi
  xchg   %ax,%ax
L22:
  movzbl (%rax),%esi
  lea    (%rdi,%rax,1),%r10d
  mov    %rax,%rcx
  lea    0x1(%rax),%rax
  test   $0xdf,%sil
  jne    L22
  mov    %r10d,%edi
  sub    %r8d,%edi
  sar    %edi
  lea    (%rdi,%r8,1),%eax
  cmp    %r8d,%eax
  jle    L56
  movslq %r10d,%r8
  lea    -0x1(%rdi),%esi
  add    %r9,%rdx
  lea    -0x1(%r9,%r8,1),%rax
  add    %r11,%r8
  sub    %rsi,%r8
  nopl   0x0(%rax)
L115:
  movzbl (%rdx),%esi
  movzbl (%rax),%edi
  sub    $0x1,%rax
  add    $0x1,%rdx
  mov    %dil,-0x1(%rdx)
  mov    %sil,0x1(%rax)
  cmp    %rax,%r8
  jne    L115
  movzbl (%rcx),%esi
  mov    %r10d,%r8d
L173:
  test   %sil,%sil
  jne    L178
L175:
  ret
  nopl   0x0(%rax,%rax,1)
L138:
  add    $0x1,%r8d
  movslq %r8d,%rax
  movzbl (%r9,%rax,1),%esi
  jmp    L173
L198:
  movb   $0x0,(%rdx)
  jmp    L94
L113:
  ret
L56:
  mov    %r10d,%r8d
  jmp    L173
```

assistant: ```c
void reverseWords(char *s) {
  if (!s || !s[0])
    return;
  int i, j, l;
  char prev;
  prev = ' ';
  for (l = -1, i = 0; s[i]; i++) {
    if (s[i] != ' ') {
      s[++l] = s[i];
      prev = s[i];
      continue;
    }
    if (prev == ' ') {
      continue;
    } else {
      s[++l] = ' ';
      prev = ' ';
    }
  }
  if (s[l] == ' ')
    s[l] = 0;
  else
    s[++l] = 0;
  for (i = 0; i < l / 2; i++) {
    prev = s[i];
    s[i] = s[l - i - 1];
    s[l - i - 1] = prev;
  }
  for (i = 0; s[i];) {
    if (s[i] == ' ') {
      i++;
      continue;
    }
    j = i;
    while (s[i] && s[i] != ' ')
      i++;
    for (l = j; l < j + (i - j) / 2; l++) {
      prev = s[l];
      s[l] = s[i - l + j - 1];
      s[i - l + j - 1] = prev;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<mem_get_le32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x18,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x10,%eax
  or     %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  or     %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  or     %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int mem_get_le32(const void *vmem) {
  unsigned int val;
  const unsigned char *mem = (const unsigned char *)vmem;
  val = mem[3] << 24;
  val |= mem[2] << 16;
  val |= mem[1] << 8;
  val |= mem[0];
  return val;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<mem_get_le32>:
  endbr64
  movzbl 0x3(%rdi),%edx
  shl    $0x18,%edx
  movzbl 0x2(%rdi),%eax
  shl    $0x10,%eax
  or     %eax,%edx
  movzbl 0x1(%rdi),%eax
  shl    $0x8,%eax
  or     %eax,%edx
  movzbl (%rdi),%eax
  or     %edx,%eax
  ret
```

assistant: ```c
unsigned int mem_get_le32(const void *vmem) {
  unsigned int val;
  const unsigned char *mem = (const unsigned char *)vmem;
  val = mem[3] << 24;
  val |= mem[2] << 16;
  val |= mem[1] << 8;
  val |= mem[0];
  return val;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<mem_get_le32>:
  endbr64
  mov    (%rdi),%eax
  ret
```

assistant: ```c
unsigned int mem_get_le32(const void *vmem) {
  unsigned int val;
  const unsigned char *mem = (const unsigned char *)vmem;
  val = mem[3] << 24;
  val |= mem[2] << 16;
  val |= mem[1] << 8;
  val |= mem[0];
  return val;
}
```
user: Translate the below assembly code into C programming language:

```asm
<CmdOPEN>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <OpenConnect@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cmp    $0xffffffff,%eax
  jne    L36
  mov    $0xffffffff,%eax
  jmp    L64
L36:
  mov    -0xc(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <RecvMessage_@plt>
  cmp    $0xffffffff,%eax
  jne    L32
  mov    -0xc(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <CmdQUIT@plt>
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <CmdCLOSE@plt>
  mov    $0xffffffff,%eax
  jmp    L64
L32:
  mov    -0x18(%rbp),%rax
  lea    D0(%rip),%rdx
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <OkResultCheck@plt>
  cmp    $0xffffffff,%eax
  jne    L31
  mov    -0xc(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <CmdQUIT@plt>
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <CmdCLOSE@plt>
  mov    $0xffffffff,%eax
  jmp    L64
L31:
  mov    -0xc(%rbp),%eax
L64:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L30
  call   <__stack_chk_fail@plt>
L30:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D0
datatype: string

tool: D0: 220

assistant: ```c
int CmdOPEN(struct ftpsub_buf *fbf, const char *rhost, int iport) {
  int sockfd;
  if ((sockfd = OpenConnect(fbf, rhost, iport)) == -1) {
    return (-1);
  };
  if (RecvMessage_(fbf, sockfd) == -1) {
    CmdQUIT(fbf, sockfd);
    CmdCLOSE(fbf, &sockfd);
    return (-1);
  }
  if (OkResultCheck(fbf, 1, "220") == -1) {
    CmdQUIT(fbf, sockfd);
    CmdCLOSE(fbf, &sockfd);
    return (-1);
  }
  return (sockfd);
}
```
user: What is the c source code of the assembly code below?

```asm
<CmdOPEN>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <OpenConnect@plt>
  mov    %eax,%ebx
  mov    %eax,0x4(%rsp)
  cmp    $0xffffffff,%eax
  je     L43
  mov    %eax,%esi
  mov    %rbp,%rdi
  call   <RecvMessage_@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  je     L48
  lea    D2(%rip),%rdx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <OkResultCheck@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  je     L54
  mov    0x4(%rsp),%ebx
L43:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  mov    %ebx,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L48:
  mov    0x4(%rsp),%esi
  mov    %rbp,%rdi
  call   <CmdQUIT@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <CmdCLOSE@plt>
  jmp    L43
L54:
  mov    0x4(%rsp),%esi
  mov    %rbp,%rdi
  call   <CmdQUIT@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <CmdCLOSE@plt>
  jmp    L43
L39:
  call   <__stack_chk_fail@plt>
```

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

tool: D2: 220

assistant: ```c
int CmdOPEN(struct ftpsub_buf *fbf, const char *rhost, int iport) {
  int sockfd;
  if ((sockfd = OpenConnect(fbf, rhost, iport)) == -1) {
    return (-1);
  };
  if (RecvMessage_(fbf, sockfd) == -1) {
    CmdQUIT(fbf, sockfd);
    CmdCLOSE(fbf, &sockfd);
    return (-1);
  }
  if (OkResultCheck(fbf, 1, "220") == -1) {
    CmdQUIT(fbf, sockfd);
    CmdCLOSE(fbf, &sockfd);
    return (-1);
  }
  return (sockfd);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<CmdOPEN>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <OpenConnect@plt>
  mov    %eax,0x4(%rsp)
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L11
  mov    %eax,%esi
  mov    %rbp,%rdi
  call   <RecvMessage_@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L12
  lea    D6(%rip),%rdx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <OkResultCheck@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L12
  mov    0x4(%rsp),%r12d
L11:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
L12:
  mov    0x4(%rsp),%esi
  mov    %rbp,%rdi
  call   <CmdQUIT@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <CmdCLOSE@plt>
  jmp    L11
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D6
type_guessed: string

tool: D6: 220

assistant: ```c
int CmdOPEN(struct ftpsub_buf *fbf, const char *rhost, int iport) {
  int sockfd;
  if ((sockfd = OpenConnect(fbf, rhost, iport)) == -1) {
    return (-1);
  };
  if (RecvMessage_(fbf, sockfd) == -1) {
    CmdQUIT(fbf, sockfd);
    CmdCLOSE(fbf, &sockfd);
    return (-1);
  }
  if (OkResultCheck(fbf, 1, "220") == -1) {
    CmdQUIT(fbf, sockfd);
    CmdCLOSE(fbf, &sockfd);
    return (-1);
  }
  return (sockfd);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<CmdOPEN>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <OpenConnect@plt>
  mov    %eax,0x4(%rsp)
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L15
  mov    %eax,%esi
  mov    %rbp,%rdi
  call   <RecvMessage_@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L46
  lea    D5(%rip),%rdx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <OkResultCheck@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L46
  mov    0x4(%rsp),%r12d
L15:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
L46:
  mov    0x4(%rsp),%esi
  mov    %rbp,%rdi
  call   <CmdQUIT@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <CmdCLOSE@plt>
  jmp    L15
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D5
type: string

tool: D5: 220

assistant: ```c
int CmdOPEN(struct ftpsub_buf *fbf, const char *rhost, int iport) {
  int sockfd;
  if ((sockfd = OpenConnect(fbf, rhost, iport)) == -1) {
    return (-1);
  };
  if (RecvMessage_(fbf, sockfd) == -1) {
    CmdQUIT(fbf, sockfd);
    CmdCLOSE(fbf, &sockfd);
    return (-1);
  }
  if (OkResultCheck(fbf, 1, "220") == -1) {
    CmdQUIT(fbf, sockfd);
    CmdCLOSE(fbf, &sockfd);
    return (-1);
  }
  return (sockfd);
}
```
user: What is the c source code of the assembly code below:

```asm
<get_last_winner_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    $0x28,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L86
  cmpl   $0x2,-0x14(%rbp)
  jne    L78
  mov    -0x8(%rbp),%rax
  lea    D47(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  jmp    L66
L78:
  cmpl   $0x0,-0x14(%rbp)
  jne    L13
  mov    -0x8(%rbp),%rax
  lea    D22(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  jmp    L66
L13:
  mov    -0x8(%rbp),%rax
  lea    D11(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  jmp    L66
L86:
  cmpl   $0x2,-0x14(%rbp)
  jne    L84
  mov    -0x8(%rbp),%rax
  lea    D47(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  jmp    L66
L84:
  cmpl   $0x0,-0x14(%rbp)
  jne    L45
  mov    -0x8(%rbp),%rax
  lea    D9(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  jmp    L66
L45:
  mov    -0x8(%rbp),%rax
  lea    D53(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L66:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D47
type: string
name: parse_data
arguments: tag: D22
type: string
name: parse_data
arguments: tag: D11
type: string
name: parse_data
arguments: tag: D9
type: string
name: parse_data
arguments: tag: D53
type: string

tool: D47: Waiting

tool: D22: You Won

tool: D11: They won

tool: D9: They Won

tool: D53: You won

assistant: ```c
char *get_last_winner_str(int last, char *name, char *u1) {
  char *last_str = malloc(40);
  if (strcmp(name, u1) == 0) {
    if (last == 2)
      strcpy(last_str, "Waiting");
    else if (last == 0)
      strcpy(last_str, "You Won");
    else
      strcpy(last_str, "They won");
  } else {
    if (last == 2)
      strcpy(last_str, "Waiting");
    else if (last == 0)
      strcpy(last_str, "They Won");
    else
      strcpy(last_str, "You won");
  }
  return last_str;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_last_winner_str>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %rsi,%r12
  mov    %rdx,%r13
  mov    $0x28,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L89
  cmp    $0x2,%ebp
  je     L21
  test   %ebp,%ebp
  jne    L98
  lea    D31(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
L60:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L21:
  lea    D36(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  jmp    L60
L98:
  lea    D42(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  jmp    L60
L89:
  cmp    $0x2,%ebp
  je     L54
  test   %ebp,%ebp
  jne    L26
  lea    D1(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  jmp    L60
L54:
  lea    D36(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  jmp    L60
L26:
  lea    D10(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  jmp    L60
```

tool_calls: name: unpack_data
arguments: data_tag: D31
type_guessed: string
name: unpack_data
arguments: data_tag: D36
type_guessed: string
name: unpack_data
arguments: data_tag: D42
type_guessed: string
name: unpack_data
arguments: data_tag: D1
type_guessed: string
name: unpack_data
arguments: data_tag: D10
type_guessed: string

tool: D31: You Won

tool: D36: Waiting

tool: D42: They won

tool: D1: They Won

tool: D10: You won

assistant: ```c
char *get_last_winner_str(int last, char *name, char *u1) {
  char *last_str = malloc(40);
  if (strcmp(name, u1) == 0) {
    if (last == 2)
      strcpy(last_str, "Waiting");
    else if (last == 0)
      strcpy(last_str, "You Won");
    else
      strcpy(last_str, "They won");
  } else {
    if (last == 2)
      strcpy(last_str, "Waiting");
    else if (last == 0)
      strcpy(last_str, "They Won");
    else
      strcpy(last_str, "You won");
  }
  return last_str;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_last_winner_str>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edi,%ebx
  mov    $0x28,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L13
  cmp    $0x2,%ebx
  je     L0
  test   %ebx,%ebx
  jne    L6
  mov    %r12,%rdi
  lea    D6(%rip),%rsi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L13:
  cmp    $0x2,%ebx
  je     L0
  test   %ebx,%ebx
  jne    L49
  mov    %r12,%rdi
  lea    D48(%rip),%rsi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L0:
  mov    %r12,%rdi
  lea    D15(%rip),%rsi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L6:
  mov    %r12,%rdi
  lea    D21(%rip),%rsi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L49:
  mov    %r12,%rdi
  lea    D39(%rip),%rsi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
datatype: string
name: unpack_data
arguments: data_tag: D48
datatype: string
name: unpack_data
arguments: data_tag: D15
datatype: string
name: unpack_data
arguments: data_tag: D21
datatype: string
name: unpack_data
arguments: data_tag: D39
datatype: string

tool: D6: You Won

tool: D48: They Won

tool: D15: Waiting

tool: D21: They won

tool: D39: You won

assistant: ```c
char *get_last_winner_str(int last, char *name, char *u1) {
  char *last_str = malloc(40);
  if (strcmp(name, u1) == 0) {
    if (last == 2)
      strcpy(last_str, "Waiting");
    else if (last == 0)
      strcpy(last_str, "You Won");
    else
      strcpy(last_str, "They won");
  } else {
    if (last == 2)
      strcpy(last_str, "Waiting");
    else if (last == 0)
      strcpy(last_str, "They Won");
    else
      strcpy(last_str, "You won");
  }
  return last_str;
}
```