ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<prints>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r14
  mov    %edx,%r13d
  mov    %r8,%r12
  test   %edx,%edx
  jle    L61
  cmpb   $0x0,(%rsi)
  je     L102
  mov    %rsi,%rax
  mov    $0x1,%edx
  sub    %esi,%edx
L32:
  lea    (%rdx,%rax,1),%esi
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L32
  mov    $0x0,%r15d
  cmp    %esi,%r13d
  jg     L127
L131:
  mov    %ecx,%eax
  and    $0x2,%eax
  cmp    $0x1,%eax
  sbb    %eax,%eax
  and    $0xfffffff0,%eax
  add    $0x30,%eax
  mov    %eax,0xc(%rsp)
  and    $0x1,%ecx
  jne    L126
  test   %r15d,%r15d
  jle    L34
  mov    %r15d,%r13d
  movsbl %al,%ebx
L157:
  mov    %r12,%rdx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printchar@plt>
  sub    $0x1,%r13d
  jne    L157
L154:
  movzbl (%r14),%esi
  test   %sil,%sil
  jne    L132
  mov    %r15d,%r14d
  jmp    L40
L102:
  mov    $0x0,%esi
L127:
  mov    %r13d,%r15d
  sub    %esi,%r15d
  jmp    L131
L126:
  mov    %r15d,%r13d
  jmp    L182
L180:
  mov    $0x0,%r14d
  jmp    L71
L34:
  mov    %r15d,%r13d
  mov    %ecx,%r15d
  jmp    L154
L8:
  movl   $0x20,0xc(%rsp)
  jmp    L154
L61:
  and    $0x1,%ecx
  mov    %ecx,%r15d
  je     L8
  movl   $0x20,0xc(%rsp)
L182:
  movzbl (%r14),%esi
  mov    $0x0,%r15d
  test   %sil,%sil
  je     L180
L132:
  mov    %r14,%rbx
  mov    %r15d,%eax
  sub    %r14d,%eax
  lea    0x1(%rax),%r15d
L11:
  movsbl %sil,%esi
  mov    %r12,%rdx
  mov    %rbp,%rdi
  call   <printchar@plt>
  lea    (%r15,%rbx,1),%r14d
  add    $0x1,%rbx
  movzbl (%rbx),%esi
  test   %sil,%sil
  jne    L11
L71:
  test   %r13d,%r13d
  jle    L40
  mov    %r13d,%ebx
  movsbl 0xc(%rsp),%r15d
L170:
  mov    %r12,%rdx
  mov    %r15d,%esi
  mov    %rbp,%rdi
  call   <printchar@plt>
  sub    $0x1,%ebx
  jne    L170
  lea    (%r14,%r13,1),%r14d
L40:
  mov    %r14d,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int prints(char **out, const char *string, int width, int pad, char *buflimit) {
  register int pc = 0, padchar = ' ';
  if (width > 0) {
    register int len = 0;
    register const char *ptr;
    for (ptr = string; *ptr; ++ptr)
      ++len;
    if (len >= width)
      width = 0;
    else
      width -= len;
    if (pad & 2)
      padchar = '0';
  }
  if (!(pad & 1)) {
    for (; width > 0; --width) {
      printchar(out, padchar, buflimit);
      ++pc;
    }
  }
  for (; *string; ++string) {
    printchar(out, *string, buflimit);
    ++pc;
  }
  for (; width > 0; --width) {
    printchar(out, padchar, buflimit);
    ++pc;
  }
  return pc;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<prints>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %r8,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  movsbl (%rsi),%esi
  test   %edx,%edx
  jle    L23
  test   %sil,%sil
  je     L119
  mov    $0x1,%edx
  mov    %rbx,%rax
  sub    %ebx,%edx
  nopl   0x0(%rax)
L195:
  lea    (%rdx,%rax,1),%edi
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L195
  cmp    %r13d,%edi
  jl     L31
  mov    %ecx,%r13d
  and    $0x2,%r13d
  jne    L7
  and    $0x1,%ecx
  mov    %ecx,%r14d
  je     L153
  xor    %r14d,%r14d
L153:
  mov    $0x20,%r15d
  jmp    L94
  nopl   0x0(%rax)
L31:
  mov    %r13d,%r14d
  sub    %edi,%r14d
L83:
  mov    %ecx,%eax
  mov    %r14d,%r13d
  and    $0x2,%eax
  cmp    $0x1,%eax
  sbb    %r15d,%r15d
  and    $0xfffffff0,%r15d
  add    $0x30,%r15d
  and    $0x1,%ecx
  jne    L190
  test   %r14d,%r14d
  jle    L110
  cs nopw 0x0(%rax,%rax,1)
L17:
  mov    %r12,%rdx
  mov    %r15d,%esi
  mov    %rbp,%rdi
  call   <printchar@plt>
  sub    $0x1,%r13d
  jne    L17
  movsbl (%rbx),%esi
L148:
  test   %sil,%sil
  je     L34
L94:
  sub    %ebx,%r14d
  lea    0x1(%r14),%eax
  mov    %eax,0xc(%rsp)
  nopw   0x0(%rax,%rax,1)
L99:
  mov    %r12,%rdx
  mov    %rbp,%rdi
  call   <printchar@plt>
  mov    0xc(%rsp),%eax
  lea    (%rax,%rbx,1),%r14d
  movsbl 0x1(%rbx),%esi
  add    $0x1,%rbx
  test   %sil,%sil
  jne    L99
L112:
  test   %r13d,%r13d
  jle    L34
  mov    %r13d,%ebx
  nopl   0x0(%rax,%rax,1)
L138:
  mov    %r12,%rdx
  mov    %r15d,%esi
  mov    %rbp,%rdi
  call   <printchar@plt>
  sub    $0x1,%ebx
  jne    L138
  add    %r13d,%r14d
L34:
  add    $0x18,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L23:
  mov    $0x20,%r15d
  xor    %r14d,%r14d
  jmp    L148
  xchg   %ax,%ax
L7:
  mov    $0x30,%r15d
  xor    %r14d,%r14d
  xor    %r13d,%r13d
  jmp    L94
L190:
  xor    %r14d,%r14d
  test   %sil,%sil
  jne    L94
  jmp    L112
L119:
  mov    %edx,%r14d
  jmp    L83
L110:
  xor    %r14d,%r14d
  jmp    L148
```

assistant: ```c
int prints(char **out, const char *string, int width, int pad, char *buflimit) {
  register int pc = 0, padchar = ' ';
  if (width > 0) {
    register int len = 0;
    register const char *ptr;
    for (ptr = string; *ptr; ++ptr)
      ++len;
    if (len >= width)
      width = 0;
    else
      width -= len;
    if (pad & 2)
      padchar = '0';
  }
  if (!(pad & 1)) {
    for (; width > 0; --width) {
      printchar(out, padchar, buflimit);
      ++pc;
    }
  }
  for (; *string; ++string) {
    printchar(out, *string, buflimit);
    ++pc;
  }
  for (; width > 0; --width) {
    printchar(out, padchar, buflimit);
    ++pc;
  }
  return pc;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<prints>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %r8,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  movsbl (%rsi),%esi
  test   %edx,%edx
  jle    L4
  test   %sil,%sil
  je     L48
  mov    $0x1,%edx
  mov    %rbx,%rax
  sub    %ebx,%edx
  nopl   0x0(%rax)
L161:
  lea    (%rdx,%rax,1),%edi
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L161
  cmp    %edi,%r13d
  jg     L54
  mov    %ecx,%r13d
  and    $0x2,%r13d
  jne    L160
  and    $0x1,%ecx
  mov    %ecx,%r14d
  je     L90
  xor    %r14d,%r14d
L90:
  mov    $0x20,%r15d
  jmp    L169
  nopl   0x0(%rax)
L54:
  mov    %r13d,%r14d
  sub    %edi,%r14d
L101:
  mov    %ecx,%eax
  mov    %r14d,%r13d
  and    $0x2,%eax
  cmp    $0x1,%eax
  sbb    %r15d,%r15d
  and    $0xfffffff0,%r15d
  add    $0x30,%r15d
  and    $0x1,%ecx
  jne    L100
  test   %r14d,%r14d
  jle    L180
  cs nopw 0x0(%rax,%rax,1)
L177:
  mov    %r12,%rdx
  mov    %r15d,%esi
  mov    %rbp,%rdi
  call   <printchar@plt>
  sub    $0x1,%r13d
  jne    L177
  movsbl (%rbx),%esi
L56:
  test   %sil,%sil
  je     L125
L169:
  sub    %ebx,%r14d
  lea    0x1(%r14),%eax
  mov    %eax,0xc(%rsp)
  nopw   0x0(%rax,%rax,1)
L139:
  mov    %r12,%rdx
  mov    %rbp,%rdi
  call   <printchar@plt>
  mov    0xc(%rsp),%eax
  lea    (%rax,%rbx,1),%r14d
  movsbl 0x1(%rbx),%esi
  add    $0x1,%rbx
  test   %sil,%sil
  jne    L139
L43:
  test   %r13d,%r13d
  jle    L125
  mov    %r13d,%ebx
  nopl   0x0(%rax,%rax,1)
L24:
  mov    %r12,%rdx
  mov    %r15d,%esi
  mov    %rbp,%rdi
  call   <printchar@plt>
  sub    $0x1,%ebx
  jne    L24
  add    %r13d,%r14d
L125:
  add    $0x18,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  mov    $0x20,%r15d
  xor    %r14d,%r14d
  jmp    L56
  xchg   %ax,%ax
L160:
  mov    $0x30,%r15d
  xor    %r14d,%r14d
  xor    %r13d,%r13d
  jmp    L169
L100:
  xor    %r14d,%r14d
  test   %sil,%sil
  jne    L169
  jmp    L43
L48:
  mov    %edx,%r14d
  jmp    L101
L180:
  xor    %r14d,%r14d
  jmp    L56
```

assistant: ```c
int prints(char **out, const char *string, int width, int pad, char *buflimit) {
  register int pc = 0, padchar = ' ';
  if (width > 0) {
    register int len = 0;
    register const char *ptr;
    for (ptr = string; *ptr; ++ptr)
      ++len;
    if (len >= width)
      width = 0;
    else
      width -= len;
    if (pad & 2)
      padchar = '0';
  }
  if (!(pad & 1)) {
    for (; width > 0; --width) {
      printchar(out, padchar, buflimit);
      ++pc;
    }
  }
  for (; *string; ++string) {
    printchar(out, *string, buflimit);
    ++pc;
  }
  for (; width > 0; --width) {
    printchar(out, padchar, buflimit);
    ++pc;
  }
  return pc;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  add    $0x45,%eax
  mov    %al,-0x9(%rbp)
  movb   $0x0,-0xa(%rbp)
  jmp    L50
L45:
  movb   $0x0,-0xb(%rbp)
  jmp    L12
L67:
  movzbl -0x9(%rbp),%eax
  movzbl %al,%eax
  movzbl -0xb(%rbp),%ecx
  movzbl -0xa(%rbp),%edx
  add    %ecx,%edx
  cmp    %edx,%eax
  jle    L19
  movzbl -0x9(%rbp),%edx
  movzbl -0xa(%rbp),%eax
  add    %edx,%edx
  cltq
  mov    %dl,-0x9(%rbp,%rax,1)
  movzbl -0x9(%rbp),%edx
  movzbl -0xb(%rbp),%eax
  add    %edx,%edx
  cltq
  mov    %dl,-0x9(%rbp,%rax,1)
  jmp    L18
L19:
  movzbl -0xb(%rbp),%eax
  cltq
  movzbl -0x9(%rbp,%rax,1),%ecx
  movzbl -0xa(%rbp),%eax
  cltq
  movzbl -0x9(%rbp,%rax,1),%edx
  movzbl -0xb(%rbp),%eax
  add    %ecx,%edx
  cltq
  mov    %dl,-0x9(%rbp,%rax,1)
  movzbl -0xa(%rbp),%eax
  cltq
  movzbl -0x9(%rbp,%rax,1),%eax
  movzbl -0x9(%rbp),%edx
  sub    %edx,%eax
  mov    %al,-0x9(%rbp)
L18:
  addb   $0x2,-0xb(%rbp)
L12:
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    %al,-0xb(%rbp)
  jb     L67
  addb   $0x2,-0xa(%rbp)
L50:
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    %al,-0xa(%rbp)
  jb     L45
  movzbl -0x9(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  shl    $0x5,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L33
  call   <__stack_chk_fail@plt>
L33:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local2;
  unsigned char local1;
  {
    state[0UL] = input[0UL] + (unsigned char)69;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      local2 = 0UL;
      while (local2 < input[1UL]) {
        if (state[0UL] > local2 + local1) {
          state[local1] = state[0UL] + state[0UL];
          state[local2] = state[0UL] + state[0UL];
        } else {
          state[local2] += state[local1];
          state[0UL] = state[local1] - state[0UL];
        }
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] * 767205732UL) * (unsigned char)40;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  add    $0x45,%eax
  movzbl 0x1(%rdi),%r8d
  mov    $0x0,%r11d
  mov    $0x0,%r10d
  test   %r8b,%r8b
  je     L27
L22:
  mov    $0x0,%edx
  movzbl %r11b,%r9d
L0:
  movzbl %al,%edi
  movzbl %dl,%ecx
  add    %r9d,%ecx
  shl    $0x2,%eax
  cmp    %ecx,%edi
  cmovle %r10d,%eax
  add    $0x2,%edx
  cmp    %r8b,%dl
  jb     L0
  add    $0x2,%r11d
  cmp    %r8b,%r11b
  jb     L22
L27:
  lea    (%rax,%rax,4),%eax
  shl    $0x5,%eax
  mov    %al,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local2;
  unsigned char local1;
  {
    state[0UL] = input[0UL] + (unsigned char)69;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      local2 = 0UL;
      while (local2 < input[1UL]) {
        if (state[0UL] > local2 + local1) {
          state[local1] = state[0UL] + state[0UL];
          state[local2] = state[0UL] + state[0UL];
        } else {
          state[local2] += state[local1];
          state[0UL] = state[local1] - state[0UL];
        }
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] * 767205732UL) * (unsigned char)40;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  movzbl 0x1(%rdi),%r8d
  xor    %r11d,%r11d
  xor    %r10d,%r10d
  add    $0x45,%eax
  test   %r8b,%r8b
  je     L23
  nopw   0x0(%rax,%rax,1)
L18:
  movzbl %r11b,%r9d
  xor    %edx,%edx
  cs nopw 0x0(%rax,%rax,1)
L25:
  movzbl %dl,%ecx
  movzbl %al,%edi
  shl    $0x2,%eax
  add    %r9d,%ecx
  cmp    %ecx,%edi
  cmovle %r10d,%eax
  add    $0x2,%edx
  cmp    %r8b,%dl
  jb     L25
  add    $0x2,%r11d
  cmp    %r8b,%r11b
  jb     L18
L23:
  lea    (%rax,%rax,4),%eax
  shl    $0x5,%eax
  mov    %al,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local2;
  unsigned char local1;
  {
    state[0UL] = input[0UL] + (unsigned char)69;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      local2 = 0UL;
      while (local2 < input[1UL]) {
        if (state[0UL] > local2 + local1) {
          state[local1] = state[0UL] + state[0UL];
          state[local2] = state[0UL] + state[0UL];
        } else {
          state[local2] += state[local1];
          state[0UL] = state[local1] - state[0UL];
        }
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] * 767205732UL) * (unsigned char)40;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  movzbl 0x1(%rdi),%r8d
  add    $0x45,%eax
  test   %r8b,%r8b
  je     L17
  xor    %r11d,%r11d
  xor    %r10d,%r10d
  nopw   0x0(%rax,%rax,1)
L5:
  movzbl %r11b,%r9d
  xor    %edx,%edx
  cs nopw 0x0(%rax,%rax,1)
L4:
  movzbl %dl,%ecx
  movzbl %al,%edi
  shl    $0x2,%eax
  add    %r9d,%ecx
  cmp    %ecx,%edi
  cmovle %r10d,%eax
  add    $0x2,%edx
  cmp    %dl,%r8b
  ja     L4
  add    $0x2,%r11d
  cmp    %r8b,%r11b
  jb     L5
L17:
  lea    (%rax,%rax,4),%eax
  shl    $0x5,%eax
  mov    %al,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local2;
  unsigned char local1;
  {
    state[0UL] = input[0UL] + (unsigned char)69;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      local2 = 0UL;
      while (local2 < input[1UL]) {
        if (state[0UL] > local2 + local1) {
          state[local1] = state[0UL] + state[0UL];
          state[local2] = state[0UL] + state[0UL];
        } else {
          state[local2] += state[local1];
          state[0UL] = state[local1] - state[0UL];
        }
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] * 767205732UL) * (unsigned char)40;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D2(%rip),%rax
  mov    %rax,-0x8(%rbp)
  nop
  pop    %rbp
  ret
```

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

tool: D2: <xml><data Parent="%d" Name="%s"><Child Id="%d"/></data></xml>

assistant: ```c
void foo(void) {
  const char *a = "<xml>"
                  "<data Parent=\"%d\" Name=\"%s\">"
                  "<Child Id=\"%d\"/>"
                  "</data>"
                  "</xml>";
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<foo>:
  endbr64
  ret
```

assistant: ```c
void foo(void) {
  const char *a = "<xml>"
                  "<data Parent=\"%d\" Name=\"%s\">"
                  "<Child Id=\"%d\"/>"
                  "</data>"
                  "</xml>";
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<foo>:
  endbr64
  ret
```

assistant: ```c
void foo(void) {
  const char *a = "<xml>"
                  "<data Parent=\"%d\" Name=\"%s\">"
                  "<Child Id=\"%d\"/>"
                  "</data>"
                  "</xml>";
}
```
user: What is the c source code of the assembly code below?

```asm
<foo>:
  endbr64
  ret
```

assistant: ```c
void foo(void) {
  const char *a = "<xml>"
                  "<data Parent=\"%d\" Name=\"%s\">"
                  "<Child Id=\"%d\"/>"
                  "</data>"
                  "</xml>";
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xffffffff,-0x40(%rbp)
  movl   $0xffffffff,-0x3c(%rbp)
  lea    -0x3c(%rbp),%rcx
  lea    -0x40(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    $0x17,%esi
  mov    %rax,%rdi
  call   <enqueue@plt>
  lea    -0x3c(%rbp),%rcx
  lea    -0x40(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    $0x9,%esi
  mov    %rax,%rdi
  call   <enqueue@plt>
  lea    -0x3c(%rbp),%rcx
  lea    -0x40(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    $0xb,%esi
  mov    %rax,%rdi
  call   <enqueue@plt>
  lea    -0x3c(%rbp),%rcx
  lea    -0x40(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    $0xfffffff6,%esi
  mov    %rax,%rdi
  call   <enqueue@plt>
  lea    -0x3c(%rbp),%rcx
  lea    -0x40(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    $0x19,%esi
  mov    %rax,%rdi
  call   <enqueue@plt>
  lea    -0x3c(%rbp),%rcx
  lea    -0x40(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    $0x10,%esi
  mov    %rax,%rdi
  call   <enqueue@plt>
  lea    -0x3c(%rbp),%rcx
  lea    -0x40(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    $0x11,%esi
  mov    %rax,%rdi
  call   <enqueue@plt>
  lea    -0x3c(%rbp),%rcx
  lea    -0x40(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    $0x16,%esi
  mov    %rax,%rdi
  call   <enqueue@plt>
  lea    -0x3c(%rbp),%rcx
  lea    -0x40(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    $0x13,%esi
  mov    %rax,%rdi
  call   <enqueue@plt>
  lea    -0x3c(%rbp),%rcx
  lea    -0x40(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    $0x1e,%esi
  mov    %rax,%rdi
  call   <enqueue@plt>
  lea    -0x3c(%rbp),%rcx
  lea    -0x40(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    $0x20,%esi
  mov    %rax,%rdi
  call   <enqueue@plt>
  movl   $0x0,-0x38(%rbp)
  jmp    L17
L31:
  lea    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <empty@plt>
  test   %eax,%eax
  jne    L19
  lea    -0x3c(%rbp),%rdx
  lea    -0x40(%rbp),%rcx
  lea    -0x30(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <dequeue@plt>
  mov    %eax,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L19:
  addl   $0x1,-0x38(%rbp)
L17:
  cmpl   $0xa,-0x38(%rbp)
  jle    L31
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D1
type: string

tool: D1: Received Dequeued item: %d


assistant: ```c
int main() {
  int arr[10];
  int front = -1, rear = -1;
  enqueue(arr, 23, &front, &rear);
  enqueue(arr, 9, &front, &rear);
  enqueue(arr, 11, &front, &rear);
  enqueue(arr, -10, &front, &rear);
  enqueue(arr, 25, &front, &rear);
  enqueue(arr, 16, &front, &rear);
  enqueue(arr, 17, &front, &rear);
  enqueue(arr, 22, &front, &rear);
  enqueue(arr, 19, &front, &rear);
  enqueue(arr, 30, &front, &rear);
  enqueue(arr, 32, &front, &rear);
  int i;
  for (int idx = 0; idx < 10 + 1; idx++) {
    if (!empty(&front)) {
      i = dequeue(arr, &front, &rear);
      printf("Received Dequeued item: %d\n", i);
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0xffffffff,0x8(%rsp)
  movl   $0xffffffff,0xc(%rsp)
  lea    0xc(%rsp),%r12
  lea    0x8(%rsp),%rbp
  lea    0x10(%rsp),%rbx
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    $0x17,%esi
  mov    %rbx,%rdi
  call   <enqueue@plt>
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    $0x9,%esi
  mov    %rbx,%rdi
  call   <enqueue@plt>
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    $0xb,%esi
  mov    %rbx,%rdi
  call   <enqueue@plt>
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    $0xfffffff6,%esi
  mov    %rbx,%rdi
  call   <enqueue@plt>
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    $0x19,%esi
  mov    %rbx,%rdi
  call   <enqueue@plt>
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    $0x10,%esi
  mov    %rbx,%rdi
  call   <enqueue@plt>
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    $0x11,%esi
  mov    %rbx,%rdi
  call   <enqueue@plt>
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    $0x16,%esi
  mov    %rbx,%rdi
  call   <enqueue@plt>
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    $0x13,%esi
  mov    %rbx,%rdi
  call   <enqueue@plt>
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    $0x1e,%esi
  mov    %rbx,%rdi
  call   <enqueue@plt>
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    $0x20,%esi
  mov    %rbx,%rdi
  call   <enqueue@plt>
  mov    $0xb,%ebx
  lea    D9(%rip),%rbp
  jmp    L12
L39:
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  call   <dequeue@plt>
  mov    %eax,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
L5:
  sub    $0x1,%ebx
  je     L11
L12:
  lea    0x8(%rsp),%rdi
  call   <empty@plt>
  test   %eax,%eax
  jne    L5
  jmp    L39
L11:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D9
type_infered: string

tool: D9: Received Dequeued item: %d


assistant: ```c
int main() {
  int arr[10];
  int front = -1, rear = -1;
  enqueue(arr, 23, &front, &rear);
  enqueue(arr, 9, &front, &rear);
  enqueue(arr, 11, &front, &rear);
  enqueue(arr, -10, &front, &rear);
  enqueue(arr, 25, &front, &rear);
  enqueue(arr, 16, &front, &rear);
  enqueue(arr, 17, &front, &rear);
  enqueue(arr, 22, &front, &rear);
  enqueue(arr, 19, &front, &rear);
  enqueue(arr, 30, &front, &rear);
  enqueue(arr, 32, &front, &rear);
  int i;
  for (int idx = 0; idx < 10 + 1; idx++) {
    if (!empty(&front)) {
      i = dequeue(arr, &front, &rear);
      printf("Received Dequeued item: %d\n", i);
    }
  }
  return 0;
}
```