ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_value_from_response>:
  endbr64
  push   %r15
  push   %r14
  mov    %rcx,%r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  mov    %rsi,%rdi
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %rdx,0x8(%rsp)
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <strdup@plt>
  test   %rax,%rax
  je     L34
  lea    D8(%rip),%r12
  mov    %rax,%rdi
  mov    %rax,%r13
  mov    %r12,%rsi
  call   <strtok@plt>
  mov    %rax,%r15
  test   %rax,%rax
  jne    L12
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L66:
  mov    %r12,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %rax,%r15
  test   %rax,%rax
  je     L18
L12:
  mov    %rbp,%rdx
  mov    %r15,%rsi
  mov    %rbx,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L66
  add    %r15,%rbp
  sub    $0x1,%r14
  mov    $0xfffffffe,%r12d
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %r14,%rax
  ja     L74
  mov    0x8(%rsp),%rdi
  mov    %rbp,%rsi
  mov    $0x1,%r12d
  call   <strcpy@plt>
  jmp    L74
  nopl   0x0(%rax)
L18:
  mov    $0xffffffff,%r12d
L74:
  mov    %r13,%rdi
  call   <free@plt>
L10:
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L34:
  mov    $0xfffffffd,%r12d
  jmp    L10
```

tool_calls: name: retrieve_data
arguments: tag: D8
datatype: string

tool: D8: &,


assistant: ```c
int get_value_from_response(const char *buf, const char *key, char *retbuf, size_t retbuflen) {
  int ret;
  char *tokens, *kv_pair;
  size_t keylen = strlen(key);
  tokens = strdup(buf);
  if (tokens == NULL)
    return -3;
  ret = -1;
  kv_pair = strtok(tokens, "&,\r\n");
  for (; kv_pair != NULL; kv_pair = strtok(NULL, "&,\r\n")) {
    if (strncmp(key, kv_pair, keylen) == 0) {
      char *val = &kv_pair[keylen];
      if (strlen(val) > retbuflen - 1) {
        ret = -2;
      } else {
        strcpy(retbuf, val);
        ret = 1;
      }
      break;
    }
  }
  free(tokens);
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  movl   $0x1,-0x14(%rbp)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
void f() {
  int a;
  int test15 = sizeof(struct { char qq[(_Bool)&a]; });
  ;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<f>:
  endbr64
  ret
```

assistant: ```c
void f() {
  int a;
  int test15 = sizeof(struct { char qq[(_Bool)&a]; });
  ;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<f>:
  endbr64
  ret
```

assistant: ```c
void f() {
  int a;
  int test15 = sizeof(struct { char qq[(_Bool)&a]; });
  ;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<f>:
  endbr64
  ret
```

assistant: ```c
void f() {
  int a;
  int test15 = sizeof(struct { char qq[(_Bool)&a]; });
  ;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<putstk>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  mov    $0x0,%eax
  call   <mnem_pop@plt>
  cmpb   $0x1,-0x4(%rbp)
  jne    L12
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <ol@plt>
  jmp    L11
L12:
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <ol@plt>
L11:
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D15
data_type: string
name: retrieve_data
arguments: data_label: D4
data_type: string

tool: D15: MOV  [BX],AL

tool: D4: MOV  [BX],AX

assistant: ```c
putstk(typeobj) char typeobj;
{
  mnem_pop();
  if (typeobj == 1)
    ol("MOV  [BX],AL");
  else
    ol("MOV  [BX],AX");
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<putstk>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0x0,%eax
  call   <mnem_pop@plt>
  cmp    $0x1,%bl
  je     L2
  lea    D12(%rip),%rdi
  mov    $0x0,%eax
  call   <ol@plt>
L6:
  pop    %rbx
  ret
L2:
  lea    D17(%rip),%rdi
  mov    $0x0,%eax
  call   <ol@plt>
  jmp    L6
```

tool_calls: name: disassemble_data
arguments: label: D12
datatype: string
name: disassemble_data
arguments: label: D17
datatype: string

tool: D12: MOV  [BX],AX

tool: D17: MOV  [BX],AL

assistant: ```c
putstk(typeobj) char typeobj;
{
  mnem_pop();
  if (typeobj == 1)
    ol("MOV  [BX],AL");
  else
    ol("MOV  [BX],AX");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<putstk>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  mov    %edi,%ebx
  call   <mnem_pop@plt>
  cmp    $0x1,%bl
  je     L7
  lea    D4(%rip),%rdi
  xor    %eax,%eax
  pop    %rbx
  jmp    1060 <ol@plt>
  nopw   0x0(%rax,%rax,1)
L7:
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  pop    %rbx
  jmp    1060 <ol@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D4
type_infered: string
name: retrieve_data
arguments: data_label: D15
type_infered: string

tool: D4: MOV  [BX],AX

tool: D15: MOV  [BX],AL

assistant: ```c
putstk(typeobj) char typeobj;
{
  mnem_pop();
  if (typeobj == 1)
    ol("MOV  [BX],AL");
  else
    ol("MOV  [BX],AX");
}
```
user: What is the c source code of the assembly code below?

```asm
<putstk>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  mov    %edi,%ebx
  call   <mnem_pop@plt>
  cmp    $0x1,%bl
  je     L0
  lea    D2(%rip),%rdi
  xor    %eax,%eax
  pop    %rbx
  jmp    1060 <ol@plt>
  nopw   0x0(%rax,%rax,1)
L0:
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  pop    %rbx
  jmp    1060 <ol@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D2
type_guessed: string
name: unpack_data
arguments: data_tag: D7
type_guessed: string

tool: D2: MOV  [BX],AX

tool: D7: MOV  [BX],AL

assistant: ```c
putstk(typeobj) char typeobj;
{
  mnem_pop();
  if (typeobj == 1)
    ol("MOV  [BX],AL");
  else
    ol("MOV  [BX],AX");
}
```
user: Translate the below assembly code into C programming language:

```asm
<is_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x3,-0x14(%rbp)
  jg     L10
  mov    $0x1,%eax
  jmp    L37
L10:
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L73
  mov    $0x0,%eax
  jmp    L37
L73:
  movl   $0x3,-0x4(%rbp)
  jmp    L34
L21:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L3
  mov    $0x0,%eax
  jmp    L37
L3:
  addl   $0x1,-0x4(%rbp)
L34:
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x14(%rbp),%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  cvttsd2si %xmm0,%eax
  cmp    %eax,-0x4(%rbp)
  jle    L21
  mov    $0x1,%eax
L37:
  leave
  ret
```

assistant: ```c
int is_prime(int n) {
  if (n < 4)
    return 1;
  if (!(n % 2))
    return 0;
  for (int a = 3; a < (int)(sqrt(n)) + 1; a++)
    if (!(n % a))
      return 0;
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<is_prime>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x3,%edi
  jle    L31
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    $0x0,%eax
  mov    $0x3,%ebx
  test   $0x1,%dil
  jne    L37
L70:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L20:
  add    $0x1,%ebx
L37:
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebp,%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L27
  sqrtsd %xmm0,%xmm0
L75:
  cvttsd2si %xmm0,%eax
  cmp    %ebx,%eax
  jl     L23
  mov    %ebp,%eax
  cltd
  idiv   %ebx
  mov    %edx,%eax
  test   %edx,%edx
  jne    L20
  jmp    L70
L27:
  call   <sqrt@plt>
  jmp    L75
L23:
  mov    $0x1,%eax
  jmp    L70
L31:
  ret
```

assistant: ```c
int is_prime(int n) {
  if (n < 4)
    return 1;
  if (!(n % 2))
    return 0;
  for (int a = 3; a < (int)(sqrt(n)) + 1; a++)
    if (!(n % a))
      return 0;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_prime>:
  endbr64
  cmp    $0x3,%edi
  jle    L10
  test   $0x1,%dil
  je     L50
  push   %rbx
  pxor   %xmm1,%xmm1
  mov    $0x3,%ebx
  pxor   %xmm2,%xmm2
  cvtsi2sd %edi,%xmm1
  sub    $0x10,%rsp
  jmp    L33
  nopw   0x0(%rax,%rax,1)
L59:
  mov    %edi,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  je     L30
  add    $0x1,%ebx
L33:
  ucomisd %xmm1,%xmm2
  ja     L20
  movapd %xmm1,%xmm0
  sqrtsd %xmm0,%xmm0
L19:
  cvttsd2si %xmm0,%eax
  cmp    %ebx,%eax
  jge    L59
  add    $0x10,%rsp
  mov    $0x1,%eax
  pop    %rbx
  ret
  nopl   (%rax)
L30:
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L50:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L10:
  mov    $0x1,%eax
  ret
L20:
  movapd %xmm1,%xmm0
  mov    %edi,0xc(%rsp)
  movsd  %xmm1,(%rsp)
  call   <sqrt@plt>
  mov    0xc(%rsp),%edi
  movsd  (%rsp),%xmm1
  pxor   %xmm2,%xmm2
  jmp    L19
```

assistant: ```c
int is_prime(int n) {
  if (n < 4)
    return 1;
  if (!(n % 2))
    return 0;
  for (int a = 3; a < (int)(sqrt(n)) + 1; a++)
    if (!(n % a))
      return 0;
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<is_prime>:
  endbr64
  cmp    $0x3,%edi
  jle    L5
  test   $0x1,%dil
  je     L84
  pxor   %xmm1,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %edi,%xmm1
  ucomisd %xmm1,%xmm0
  ja     L59
  sqrtsd %xmm1,%xmm1
  mov    $0x3,%ecx
  cvttsd2si %xmm1,%esi
  jmp    L7
L0:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L84
  add    $0x1,%ecx
L7:
  cmp    %ecx,%esi
  jge    L0
L5:
  mov    $0x1,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L84:
  xor    %eax,%eax
  ret
L59:
  push   %rbx
  mov    $0x3,%ebx
  sub    $0x10,%rsp
  jmp    L37
L12:
  mov    0xc(%rsp),%edi
  mov    %edi,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  je     L80
  movsd  (%rsp),%xmm1
  add    $0x1,%ebx
L37:
  movapd %xmm1,%xmm0
  mov    %edi,0xc(%rsp)
  movsd  %xmm1,(%rsp)
  call   <sqrt@plt>
  cvttsd2si %xmm0,%eax
  cmp    %eax,%ebx
  jle    L12
  mov    $0x1,%eax
L47:
  add    $0x10,%rsp
  pop    %rbx
  ret
L80:
  xor    %eax,%eax
  jmp    L47
```

assistant: ```c
int is_prime(int n) {
  if (n < 4)
    return 1;
  if (!(n % 2))
    return 0;
  for (int a = 3; a < (int)(sqrt(n)) + 1; a++)
    if (!(n % a))
      return 0;
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<leading_zeros>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L40
  jmp    L38
L27:
  shrl   -0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L4
  movl   $0x1,-0x4(%rbp)
  jmp    L38
L4:
  addl   $0x1,-0x8(%rbp)
L38:
  cmpl   $0x0,-0x4(%rbp)
  je     L27
L40:
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int leading_zeros(unsigned int code) {
  unsigned int count = 0;
  int done = 0;
  if (code) {
    while (!done) {
      code >>= 1;
      if (!code)
        done = 1;
      else
        count++;
    }
  }
  return count;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<leading_zeros>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  je     L14
  mov    $0x0,%eax
  jmp    L1
L26:
  add    $0x1,%eax
L1:
  shr    %edi
  jne    L26
L14:
  ret
```

assistant: ```c
unsigned int leading_zeros(unsigned int code) {
  unsigned int count = 0;
  int done = 0;
  if (code) {
    while (!done) {
      code >>= 1;
      if (!code)
        done = 1;
      else
        count++;
    }
  }
  return count;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<leading_zeros>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  je     L4
  shr    %eax
  mov    %eax,%edx
  mov    $0x0,%eax
  je     L4
  nopl   (%rax)
L27:
  add    $0x1,%eax
  shr    %edx
  jne    L27
L4:
  ret
```

assistant: ```c
unsigned int leading_zeros(unsigned int code) {
  unsigned int count = 0;
  int done = 0;
  if (code) {
    while (!done) {
      code >>= 1;
      if (!code)
        done = 1;
      else
        count++;
    }
  }
  return count;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<leading_zeros>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  je     L14
  shr    %eax
  mov    %eax,%edx
  mov    $0x0,%eax
  je     L14
  nopl   (%rax)
L13:
  add    $0x1,%eax
  shr    %edx
  jne    L13
L14:
  ret
```

assistant: ```c
unsigned int leading_zeros(unsigned int code) {
  unsigned int count = 0;
  int done = 0;
  if (code) {
    while (!done) {
      code >>= 1;
      if (!code)
        done = 1;
      else
        count++;
    }
  }
  return count;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x64,-0x28(%rbp)
  movss  D17(%rip),%xmm0
  movss  %xmm0,-0x24(%rbp)
  movsd  D30(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movb   $0x41,-0x29(%rbp)
  movl   $0x64655a,-0x11(%rbp)
  movl   $0x77616853,-0xd(%rbp)
  movb   $0x0,-0x9(%rbp)
  mov    -0x28(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  pxor   %xmm1,%xmm1
  cvtss2sd -0x24(%rbp),%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  lea    D84(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    -0x20(%rbp),%rax
  movq   %rax,%xmm0
  lea    D65(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movsbl -0x29(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x11(%rbp),%rax
  mov    %rax,%rsi
  lea    D83(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xd(%rbp),%rax
  mov    %rax,%rsi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movsbl -0x29(%rbp),%edx
  lea    -0xd(%rbp),%rcx
  lea    -0x11(%rbp),%rax
  mov    %rax,%rsi
  lea    D76(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D17
type_infered: byte[4]
name: extract_symbolic_data
arguments: tag: D30
type_infered: f64
name: extract_symbolic_data
arguments: tag: D8
type_infered: string
name: extract_symbolic_data
arguments: tag: D84
type_infered: string
name: extract_symbolic_data
arguments: tag: D65
type_infered: string
name: extract_symbolic_data
arguments: tag: D7
type_infered: string
name: extract_symbolic_data
arguments: tag: D83
type_infered: string
name: extract_symbolic_data
arguments: tag: D20
type_infered: string
name: extract_symbolic_data
arguments: tag: D76
type_infered: string

tool: {"D17": [0x7b,0x14,0x16,0x40]}

tool: D30: 56677.4532

tool: D8: You are %d miles away.


tool: D84: You have %f levels of power.


tool: D65: You have %f awesome super power.


tool: D7: I have an initial %c


tool: D83: I have a firtst name %s


tool: D20: I have a last name %s


tool: D76: My whole name is %s %c. %s.


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56677.4532;
  char initial = 'A';
  char first_name[] = "Zed";
  char last_name[] = "Shaw";
  printf("You are %d miles away.\n", distance);
  printf("You have %f levels of power.\n", power);
  printf("You have %f awesome super power.\n", super_power);
  printf("I have an initial %c\n", initial);
  printf("I have a firtst name %s\n", first_name);
  printf("I have a last name %s\n", last_name);
  printf("My whole name is %s %c. %s.\n", first_name, initial, last_name);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x64655a,0xf(%rsp)
  movl   $0x77616853,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  mov    $0x64,%esi
  lea    D62(%rip),%rdi
  call   <printf@plt>
  movsd  D68(%rip),%xmm0
  lea    D44(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movsd  D35(%rip),%xmm0
  lea    D36(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x41,%esi
  lea    D45(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xf(%rsp),%rbx
  mov    %rbx,%rsi
  lea    D69(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x13(%rsp),%rbp
  mov    %rbp,%rsi
  lea    D54(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rcx
  mov    $0x41,%edx
  mov    %rbx,%rsi
  lea    D22(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D62
type_infered: string
name: unpack_data
arguments: label: D68
type_infered: byte[8]
name: unpack_data
arguments: label: D44
type_infered: string
name: unpack_data
arguments: label: D35
type_infered: f64
name: unpack_data
arguments: label: D36
type_infered: string
name: unpack_data
arguments: label: D45
type_infered: string
name: unpack_data
arguments: label: D69
type_infered: string
name: unpack_data
arguments: label: D54
type_infered: string
name: unpack_data
arguments: label: D22
type_infered: string

tool: D62: You are %d miles away.


tool: {"D68": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D44: You have %f levels of power.


tool: D35: 56677.4532

tool: D36: You have %f awesome super power.


tool: D45: I have an initial %c


tool: D69: I have a firtst name %s


tool: D54: I have a last name %s


tool: D22: My whole name is %s %c. %s.


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56677.4532;
  char initial = 'A';
  char first_name[] = "Zed";
  char last_name[] = "Shaw";
  printf("You are %d miles away.\n", distance);
  printf("You have %f levels of power.\n", power);
  printf("You have %f awesome super power.\n", super_power);
  printf("I have an initial %c\n", initial);
  printf("I have a firtst name %s\n", first_name);
  printf("I have a last name %s\n", last_name);
  printf("My whole name is %s %c. %s.\n", first_name, initial, last_name);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x64,%esi
  lea    D66(%rip),%rdi
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x64655a,0xf(%rsp)
  lea    0xf(%rsp),%rbp
  lea    0x13(%rsp),%r12
  movl   $0x77616853,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  call   <printf@plt>
  movsd  D87(%rip),%xmm0
  mov    $0x1,%eax
  lea    D29(%rip),%rdi
  call   <printf@plt>
  movsd  D40(%rip),%xmm0
  mov    $0x1,%eax
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x41,%esi
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D22(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rcx
  mov    $0x41,%edx
  mov    %rbp,%rsi
  lea    D24(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D66
datatype: string
name: retrieve_data
arguments: data_label: D87
datatype: byte[8]
name: retrieve_data
arguments: data_label: D29
datatype: string
name: retrieve_data
arguments: data_label: D40
datatype: f64
name: retrieve_data
arguments: data_label: D4
datatype: string
name: retrieve_data
arguments: data_label: D38
datatype: string
name: retrieve_data
arguments: data_label: D9
datatype: string
name: retrieve_data
arguments: data_label: D22
datatype: string
name: retrieve_data
arguments: data_label: D24
datatype: string

tool: D66: You are %d miles away.


tool: {"D87": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D29: You have %f levels of power.


tool: D40: 56677.4532

tool: D4: You have %f awesome super power.


tool: D38: I have an initial %c


tool: D9: I have a firtst name %s


tool: D22: I have a last name %s


tool: D24: My whole name is %s %c. %s.


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56677.4532;
  char initial = 'A';
  char first_name[] = "Zed";
  char last_name[] = "Shaw";
  printf("You are %d miles away.\n", distance);
  printf("You have %f levels of power.\n", power);
  printf("You have %f awesome super power.\n", super_power);
  printf("I have an initial %c\n", initial);
  printf("I have a firtst name %s\n", first_name);
  printf("I have a last name %s\n", last_name);
  printf("My whole name is %s %c. %s.\n", first_name, initial, last_name);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x64,%esi
  lea    D36(%rip),%rdi
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x64655a,0xf(%rsp)
  lea    0xf(%rsp),%rbp
  lea    0x13(%rsp),%r12
  movl   $0x77616853,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  call   <printf@plt>
  movsd  D52(%rip),%xmm0
  mov    $0x1,%eax
  lea    D30(%rip),%rdi
  call   <printf@plt>
  movsd  D8(%rip),%xmm0
  mov    $0x1,%eax
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x41,%esi
  lea    D35(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D70(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rcx
  mov    $0x41,%edx
  mov    %rbp,%rsi
  lea    D82(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D36
data_type: string
name: read_data
arguments: label: D52
data_type: byte[8]
name: read_data
arguments: label: D30
data_type: string
name: read_data
arguments: label: D8
data_type: f64
name: read_data
arguments: label: D2
data_type: string
name: read_data
arguments: label: D35
data_type: string
name: read_data
arguments: label: D70
data_type: string
name: read_data
arguments: label: D6
data_type: string
name: read_data
arguments: label: D82
data_type: string

tool: D36: You are %d miles away.


tool: {"D52": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D30: You have %f levels of power.


tool: D8: 56677.4532

tool: D2: You have %f awesome super power.


tool: D35: I have an initial %c


tool: D70: I have a firtst name %s


tool: D6: I have a last name %s


tool: D82: My whole name is %s %c. %s.


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56677.4532;
  char initial = 'A';
  char first_name[] = "Zed";
  char last_name[] = "Shaw";
  printf("You are %d miles away.\n", distance);
  printf("You have %f levels of power.\n", power);
  printf("You have %f awesome super power.\n", super_power);
  printf("I have an initial %c\n", initial);
  printf("I have a firtst name %s\n", first_name);
  printf("I have a last name %s\n", last_name);
  printf("My whole name is %s %c. %s.\n", first_name, initial, last_name);
  return 0;
}
```
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
  mov    $0x5,%edi
  call   <exampleHeader@plt>
  mov    $0x5,%edi
  call   <displaySumSmallest@plt>
  mov    $0x3,%edi
  call   <displaySumSmallest@plt>
  mov    $0x8,%edi
  call   <displaySumSmallest@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  exampleHeader(5);
  displaySumSmallest(5);
  displaySumSmallest(3);
  displaySumSmallest(8);
  return 0;
}
```