ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <megaInit@plt>
  cmp    $0x2,%ebp
  je     L2
  lea    D27(%rip),%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%edi
  call   <exit@plt>
L2:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtoul@plt>
  lea    0x4(%rsp),%rsi
  mov    %rsp,%rdi
  mov    %eax,(%rsp)
  call   <RandomFunc@plt>
  mov    0x4(%rsp),%esi
  cmp    $0x5f363a57,%esi
  je     L37
L18:
  xor    %eax,%eax
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L37:
  lea    D18(%rip),%rdi
  call   <puts@plt>
  mov    0x4(%rsp),%esi
  jmp    L18
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D27
data_type: string
name: extract_symbolic_data
arguments: tag: D8
data_type: string
name: extract_symbolic_data
arguments: tag: D18
data_type: string

tool: D27: Call this program with %i arguments


tool: D8: %u


tool: D18: You win!

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned int input[1];
  unsigned int output[1];
  int randomFuns_i5;
  unsigned int randomFuns_value6;
  int randomFuns_main_i7;
  {
    megaInit();
    if (argc != 2) {
      printf("Call this program with %i arguments\n", 1);
      exit(-1);
    } else {
    }
    randomFuns_i5 = 0;
    while (randomFuns_i5 < 1) {
      randomFuns_value6 = (unsigned int)strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 1597389399U) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%u\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    D4(%rip),%rax
  mov    (%rax),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <setbuf@plt>
  jmp    L8
L0:
  addq   $0x8,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <do_cmd@plt>
L8:
  subl   $0x1,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jg     L0
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  setbuf(stdout, NULL);
  while (--argc > 0) {
    do_cmd(*++argv);
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %rsi,%rbx
  mov    $0x0,%esi
  mov    D9(%rip),%rax
  mov    (%rax),%edi
  call   <setbuf@plt>
  cmp    $0x1,%ebp
  jle    L8
  lea    -0x2(%rbp),%eax
  lea    0x8(%rbx,%rax,8),%rbp
L6:
  add    $0x8,%rbx
  mov    (%rbx),%rdi
  call   <do_cmd@plt>
  cmp    %rbp,%rbx
  jne    L6
L8:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  setbuf(stdout, NULL);
  while (--argc > 0) {
    do_cmd(*++argv);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  sub    $0x8,%rsp
  mov    D8(%rip),%rax
  mov    (%rax),%edi
  call   <setbuf@plt>
  cmp    $0x1,%ebp
  jle    L19
  lea    -0x2(%rbp),%eax
  lea    0x8(%rbx,%rax,8),%rbp
  nopl   0x0(%rax)
L2:
  mov    0x8(%rbx),%rdi
  add    $0x8,%rbx
  call   <do_cmd@plt>
  cmp    %rbp,%rbx
  jne    L2
L19:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  setbuf(stdout, NULL);
  while (--argc > 0) {
    do_cmd(*++argv);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  sub    $0x8,%rsp
  mov    D4(%rip),%rax
  mov    (%rax),%edi
  call   <setbuf@plt>
  cmp    $0x1,%ebp
  jle    L16
  lea    -0x2(%rbp),%eax
  lea    0x8(%rbx,%rax,8),%rbp
  nopl   0x0(%rax)
L6:
  mov    0x8(%rbx),%rdi
  add    $0x8,%rbx
  call   <do_cmd@plt>
  cmp    %rbp,%rbx
  jne    L6
L16:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  setbuf(stdout, NULL);
  while (--argc > 0) {
    do_cmd(*++argv);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<addparam>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x3c(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    0x2(%rax),%rbx
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %rbx,%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  mov    $0x3f,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L14
  cmpl   $0x0,-0x3c(%rbp)
  je     L3
  lea    D26(%rip),%rax
  jmp    L25
L3:
  lea    D6(%rip),%rax
  jmp    L25
L14:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  test   %rax,%rax
  jle    L6
  lea    D36(%rip),%rax
  jmp    L25
L6:
  lea    D6(%rip),%rax
L25:
  mov    -0x38(%rbp),%rcx
  mov    -0x30(%rbp),%rdx
  mov    -0x18(%rbp),%rdi
  mov    %rcx,%r8
  mov    %rdx,%rcx
  mov    %rax,%rdx
  lea    D4(%rip),%rax
  mov    %rax,%rsi
  call   <sprintf@plt>
  mov    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strrecat@plt>
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D26
type_guessed: string
name: unpack_data
arguments: data_tag: D6
type_guessed: string
name: unpack_data
arguments: data_tag: D36
type_guessed: string
name: unpack_data
arguments: data_tag: D4
type_guessed: string

tool: D26: ?

tool: 

tool: D36: &

tool: D4: %s%s=%s

assistant: ```c
char *addparam(char *orig, const char *parname, const char *parvalue, int addq) {
  char *append = malloc(1 + strlen(parname) + 1 + strlen(parvalue) + 1);
  sprintf(append, "%s%s=%s", (strchr(orig, '?') ? (addq ? "?" : "") : ((strlen(orig) > 0) ? "&" : "")), parname, parvalue);
  return strrecat(orig, append);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<addparam>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %rdx,%r13
  mov    %ecx,%r14d
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %rax,%rbp
  mov    %r13,%rdi
  call   <strlen@plt>
  lea    0x3(%rbp,%rax,1),%rdi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    $0x3f,%esi
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L19
  test   %r14d,%r14d
  lea    D36(%rip),%rdx
  lea    D32(%rip),%rax
  cmove  %rax,%rdx
L7:
  mov    %r13,%r8
  mov    %r12,%rcx
  lea    D37(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strrecat@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L19:
  mov    %rbx,%rdi
  call   <strlen@plt>
  test   %rax,%rax
  lea    D12(%rip),%rdx
  lea    D32(%rip),%rax
  cmovle %rax,%rdx
  jmp    L7
```

tool_calls: name: retrieve_data
arguments: label: D36
type_infered: string
name: retrieve_data
arguments: label: D32
type_infered: string
name: retrieve_data
arguments: label: D37
type_infered: string
name: retrieve_data
arguments: label: D12
type_infered: string

tool: D36: ?

tool: 

tool: D37: %s%s=%s

tool: D12: &

assistant: ```c
char *addparam(char *orig, const char *parname, const char *parvalue, int addq) {
  char *append = malloc(1 + strlen(parname) + 1 + strlen(parvalue) + 1);
  sprintf(append, "%s%s=%s", (strchr(orig, '?') ? (addq ? "?" : "") : ((strlen(orig) > 0) ? "&" : "")), parname, parvalue);
  return strrecat(orig, append);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<addparam>:
  endbr64
  push   %r14
  mov    %ecx,%r14d
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  push   %rbx
  mov    %rdx,%rbx
  call   <strlen@plt>
  mov    %rbx,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  lea    0x3(%r12,%rax,1),%rdi
  call   <malloc@plt>
  mov    $0x3f,%esi
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strchr@plt>
  test   %rax,%rax
  je     L11
  test   %r14d,%r14d
  lea    D39(%rip),%rdx
  lea    D11(%rip),%rax
  cmovne %rax,%rdx
L8:
  mov    %r12,%rdi
  mov    %rbx,%r8
  mov    %r13,%rcx
  lea    D30(%rip),%rsi
  call   <sprintf@plt>
  pop    %rbx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1090 <strrecat@plt>
  nopw   0x0(%rax,%rax,1)
L11:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    D39(%rip),%rdx
  test   %rax,%rax
  lea    D17(%rip),%rax
  cmovne %rax,%rdx
  jmp    L8
```

tool_calls: name: unpack_data
arguments: data_label: D39
type: string
name: unpack_data
arguments: data_label: D11
type: string
name: unpack_data
arguments: data_label: D30
type: string
name: unpack_data
arguments: data_label: D17
type: string

tool: 

tool: D11: ?

tool: D30: %s%s=%s

tool: D17: &

assistant: ```c
char *addparam(char *orig, const char *parname, const char *parvalue, int addq) {
  char *append = malloc(1 + strlen(parname) + 1 + strlen(parvalue) + 1);
  sprintf(append, "%s%s=%s", (strchr(orig, '?') ? (addq ? "?" : "") : ((strlen(orig) > 0) ? "&" : "")), parname, parvalue);
  return strrecat(orig, append);
}
```
user: Translate the below assembly code into C programming language:

```asm
<addparam>:
  endbr64
  push   %r14
  mov    %ecx,%r14d
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  push   %rbx
  mov    %rdx,%rbx
  call   <strlen@plt>
  mov    %rbx,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  lea    0x3(%r12,%rax,1),%rdi
  call   <malloc@plt>
  mov    $0x3f,%esi
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strchr@plt>
  test   %rax,%rax
  je     L4
  test   %r14d,%r14d
  lea    D19(%rip),%rdx
  lea    D29(%rip),%rax
  cmovne %rax,%rdx
L17:
  mov    %r12,%rdi
  mov    %rbx,%r8
  mov    %r13,%rcx
  lea    D6(%rip),%rsi
  call   <sprintf@plt>
  pop    %rbx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1090 <strrecat@plt>
  nopw   0x0(%rax,%rax,1)
L4:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    D19(%rip),%rdx
  test   %rax,%rax
  lea    D46(%rip),%rax
  cmovne %rax,%rdx
  jmp    L17
```

tool_calls: name: retrieve_data
arguments: tag: D19
type: string
name: retrieve_data
arguments: tag: D29
type: string
name: retrieve_data
arguments: tag: D6
type: string
name: retrieve_data
arguments: tag: D46
type: string

tool: 

tool: D29: ?

tool: D6: %s%s=%s

tool: D46: &

assistant: ```c
char *addparam(char *orig, const char *parname, const char *parvalue, int addq) {
  char *append = malloc(1 + strlen(parname) + 1 + strlen(parvalue) + 1);
  sprintf(append, "%s%s=%s", (strchr(orig, '?') ? (addq ? "?" : "") : ((strlen(orig) > 0) ? "&" : "")), parname, parvalue);
  return strrecat(orig, append);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
type: string
name: extract_symbolic_data
arguments: data_label: D4
type: string

tool: D5: Name:    Liangming Chen


tool: D4: Address: Kunming, Yunnan, China


assistant: ```c
int main(void) {
  printf("Name:    Liangming Chen\n");
  printf("Address: Kunming, Yunnan, China\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D15(%rip),%rdi
  call   <printf@plt>
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D15
type: string
name: parse_data
arguments: label: D9
type: string

tool: D15: Name:    Liangming Chen


tool: D9: Address: Kunming, Yunnan, China


assistant: ```c
int main(void) {
  printf("Name:    Liangming Chen\n");
  printf("Address: Kunming, Yunnan, China\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <printf@plt>
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D2
type_infered: string
name: read_data
arguments: tag: D8
type_infered: string

tool: D2: Name:    Liangming Chen


tool: D8: Address: Kunming, Yunnan, China


assistant: ```c
int main(void) {
  printf("Name:    Liangming Chen\n");
  printf("Address: Kunming, Yunnan, China\n");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D8
data_type: string
name: parse_data
arguments: tag: D1
data_type: string

tool: D8: Name:    Liangming Chen


tool: D1: Address: Kunming, Yunnan, China


assistant: ```c
int main(void) {
  printf("Name:    Liangming Chen\n");
  printf("Address: Kunming, Yunnan, China\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Invsqrt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movss  %xmm0,-0x14(%rbp)
  movss  -0x14(%rbp),%xmm1
  movss  D9(%rip),%xmm0
  mulss  %xmm1,%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  -0x14(%rbp),%xmm0
  movss  %xmm0,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  sar    %eax
  mov    %eax,%edx
  mov    $0x5f3759df,%eax
  sub    %edx,%eax
  mov    %eax,-0x8(%rbp)
  movss  -0x8(%rbp),%xmm1
  movss  -0x8(%rbp),%xmm0
  movaps %xmm0,%xmm2
  mulss  -0x4(%rbp),%xmm2
  movss  -0x8(%rbp),%xmm0
  mulss  %xmm0,%xmm2
  movss  D11(%rip),%xmm0
  subss  %xmm2,%xmm0
  mulss  %xmm1,%xmm0
  movss  %xmm0,-0x8(%rbp)
  movss  -0x8(%rbp),%xmm0
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D9
datatype: byte[4]
name: read_data
arguments: label: D11
datatype: byte[4]

tool: {"D9": [0x00,0x00,0x00,0x3f]}

tool: {"D11": [0x00,0x00,0xc0,0x3f]}

assistant: ```c
float Invsqrt(float x) {
  float xhalf = 0.5f * x;
  union {
    float x;
    int i;
  } u;
  u.x = x;
  u.i = 0x5f3759df - (u.i >> 1);
  u.x = u.x * (1.5f - xhalf * u.x * u.x);
  return u.x;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<Invsqrt>:
  endbr64
  movd   %xmm0,%edx
  sar    %edx
  mov    $0x5f3759df,%eax
  sub    %edx,%eax
  movd   %eax,%xmm2
  mulss  D14(%rip),%xmm0
  mulss  %xmm2,%xmm0
  mulss  %xmm2,%xmm0
  movss  D6(%rip),%xmm1
  subss  %xmm0,%xmm1
  mulss  %xmm2,%xmm1
  movaps %xmm1,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: label: D14
type: byte[4]
name: parse_data
arguments: label: D6
type: byte[4]

tool: {"D14": [0x00,0x00,0x00,0x3f]}

tool: {"D6": [0x00,0x00,0xc0,0x3f]}

assistant: ```c
float Invsqrt(float x) {
  float xhalf = 0.5f * x;
  union {
    float x;
    int i;
  } u;
  u.x = x;
  u.i = 0x5f3759df - (u.i >> 1);
  u.x = u.x * (1.5f - xhalf * u.x * u.x);
  return u.x;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<Invsqrt>:
  endbr64
  movd   %xmm0,%edx
  mov    $0x5f3759df,%eax
  movss  D12(%rip),%xmm1
  mulss  D3(%rip),%xmm0
  sar    %edx
  sub    %edx,%eax
  movd   %eax,%xmm2
  mulss  %xmm2,%xmm0
  mulss  %xmm2,%xmm0
  subss  %xmm0,%xmm1
  mulss  %xmm2,%xmm1
  movaps %xmm1,%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D12
type_guessed: byte[4]
name: extract_symbolic_data
arguments: tag: D3
type_guessed: byte[4]

tool: {"D12": [0x00,0x00,0xc0,0x3f]}

tool: {"D3": [0x00,0x00,0x00,0x3f]}

assistant: ```c
float Invsqrt(float x) {
  float xhalf = 0.5f * x;
  union {
    float x;
    int i;
  } u;
  u.x = x;
  u.i = 0x5f3759df - (u.i >> 1);
  u.x = u.x * (1.5f - xhalf * u.x * u.x);
  return u.x;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<Invsqrt>:
  endbr64
  movd   %xmm0,%edx
  mov    $0x5f3759df,%eax
  movss  D3(%rip),%xmm1
  mulss  D16(%rip),%xmm0
  sar    %edx
  sub    %edx,%eax
  movd   %eax,%xmm2
  mulss  %xmm2,%xmm0
  mulss  %xmm2,%xmm0
  subss  %xmm0,%xmm1
  mulss  %xmm2,%xmm1
  movaps %xmm1,%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D3
data_type: byte[4]
name: retrieve_data
arguments: tag: D16
data_type: byte[4]

tool: {"D3": [0x00,0x00,0xc0,0x3f]}

tool: {"D16": [0x00,0x00,0x00,0x3f]}

assistant: ```c
float Invsqrt(float x) {
  float xhalf = 0.5f * x;
  union {
    float x;
    int i;
  } u;
  u.x = x;
  u.i = 0x5f3759df - (u.i >> 1);
  u.x = u.x * (1.5f - xhalf * u.x * u.x);
  return u.x;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<solve_2750>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L45
L67:
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L63
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L53
L63:
  mov    -0x10(%rbp),%eax
  and    $0x3,%eax
  test   %eax,%eax
  je     L12
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%ecx
  mov    -0x10(%rbp),%eax
  lea    0x3(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x2,%eax
  add    $0x1,%eax
  mov    %ecx,%edx
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L53
L12:
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%ecx
  mov    -0x10(%rbp),%eax
  lea    0x3(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x2,%eax
  mov    %ecx,%edx
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L53:
  addl   $0x1,-0xc(%rbp)
L45:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jl     L67
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L50
  call   <__stack_chk_fail@plt>
L50:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D32
data_type: string
name: unpack_data
arguments: label: D14
data_type: string
name: unpack_data
arguments: label: D7
data_type: string

tool: D32: %d

tool: D14: 0 0

tool: D7: %d %d


assistant: ```c
int solve_2750() {
  int nCases, num, nFeet;
  scanf("%d", &nCases);
  for (num = 0; num < nCases; num++) {
    scanf("%d", &nFeet);
    if (nFeet % 2 != 0)
      printf("0 0\n");
    else if (nFeet % 4 != 0)
      printf("%d %d\n", nFeet / 4 + 1, nFeet / 2);
    else
      printf("%d %d\n", nFeet / 4, nFeet / 2);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<solve_2750>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,(%rsp)
  jle    L70
  mov    $0x0,%ebx
  lea    0x4(%rsp),%r12
  lea    D14(%rip),%rbp
  lea    D21(%rip),%r13
  lea    D25(%rip),%r14
  jmp    L24
L48:
  mov    %r14,%rdi
  call   <puts@plt>
  jmp    L25
L0:
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %eax,%edx
  sar    %edx
  lea    0x3(%rax),%esi
  test   %eax,%eax
  cmovns %eax,%esi
  sar    $0x2,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L25:
  add    $0x1,%ebx
  cmp    %ebx,(%rsp)
  jle    L70
L24:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   $0x1,%al
  jne    L48
  test   $0x3,%al
  je     L0
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %eax,%edx
  sar    %edx
  lea    0x3(%rax),%esi
  test   %eax,%eax
  cmovns %eax,%esi
  sar    $0x2,%esi
  add    $0x1,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L25
L70:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L67
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L67:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D14
datatype: string
name: read_data
arguments: label: D21
datatype: string
name: read_data
arguments: label: D25
datatype: string

tool: D14: %d

tool: D21: %d %d


tool: D25: 0 0

assistant: ```c
int solve_2750() {
  int nCases, num, nFeet;
  scanf("%d", &nCases);
  for (num = 0; num < nCases; num++) {
    scanf("%d", &nFeet);
    if (nFeet % 2 != 0)
      printf("0 0\n");
    else if (nFeet % 4 != 0)
      printf("%d %d\n", nFeet / 4 + 1, nFeet / 2);
    else
      printf("%d %d\n", nFeet / 4, nFeet / 2);
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<solve_2750>:
  endbr64
  push   %r14
  lea    D23(%rip),%r14
  push   %r13
  lea    D24(%rip),%r13
  push   %r12
  push   %rbp
  lea    D20(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  xor    %ebx,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    0x4(%rsp),%r12
  call   <scanf@plt>
  mov    (%rsp),%eax
  test   %eax,%eax
  jg     L60
  jmp    L34
L36:
  add    $0x1,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L3:
  add    $0x1,%ebx
  cmp    %ebx,(%rsp)
  jle    L34
L60:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   $0x1,%al
  jne    L54
  mov    %eax,%edx
  lea    0x3(%rax),%esi
  shr    $0x1f,%edx
  add    %eax,%edx
  sar    %edx
  test   %eax,%eax
  cmovns %eax,%esi
  sar    $0x2,%esi
  test   $0x3,%al
  jne    L36
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,(%rsp)
  jg     L60
L34:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L54:
  mov    %r14,%rdi
  call   <puts@plt>
  jmp    L3
L55:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D23
type: string
name: disassemble_data
arguments: data_tag: D24
type: string
name: disassemble_data
arguments: data_tag: D20
type: string

tool: D23: 0 0

tool: D24: %d %d


tool: D20: %d

assistant: ```c
int solve_2750() {
  int nCases, num, nFeet;
  scanf("%d", &nCases);
  for (num = 0; num < nCases; num++) {
    scanf("%d", &nFeet);
    if (nFeet % 2 != 0)
      printf("0 0\n");
    else if (nFeet % 4 != 0)
      printf("%d %d\n", nFeet / 4 + 1, nFeet / 2);
    else
      printf("%d %d\n", nFeet / 4, nFeet / 2);
  }
  return 0;
}
```