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

```asm
<approx1>:
  endbr64
  test   %edi,%edi
  jle    L29
  fld1
  add    $0x1,%edi
  mov    $0x1,%eax
  fldz
  flds   0xee6(%rip)        # 2000 <_fini+0xe9c>
  flds   0xee4(%rip)        # 2004 <_fini+0xea0>
  jmp    L51
  nopw   0x0(%rax,%rax,1)
L50:
  faddp  %st,%st(3)
  add    $0x1,%eax
  fadd   %st,%st(3)
  cmp    %eax,%edi
  je     L52
L51:
  fld    %st(3)
  fdivr  %st(2),%st
  test   $0x1,%al
  jne    L50
  fsubrp %st,%st(3)
  add    $0x1,%eax
  fadd   %st,%st(3)
  cmp    %eax,%edi
  jne    L51
  fstp   %st(0)
  fstp   %st(0)
  fstp   %st(1)
  jmp    L3
  xchg   %ax,%ax
L52:
  fstp   %st(0)
  fstp   %st(0)
  fstp   %st(1)
L3:
  ret
  nopw   0x0(%rax,%rax,1)
L29:
  fldz
  ret
```

assistant: ```c
long double approx1(int wut) {
  long double pi = 0;
  long double numer = 4;
  long double denom = 1;
  for (int i = 1; i <= wut; i++) {
    if (i % 2 != 0) {
      pi = pi + (numer / denom);
    } else {
      pi = pi - (numer / denom);
    }
    denom = denom + 2;
  }
  return pi;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<approx1>:
  endbr64
  test   %edi,%edi
  jle    L28
  fld1
  add    $0x1,%edi
  mov    $0x1,%eax
  fldz
  flds   0xee6(%rip)        # 2000 <_fini+0xe9c>
  flds   0xee4(%rip)        # 2004 <_fini+0xea0>
  jmp    L14
  nopw   0x0(%rax,%rax,1)
L52:
  faddp  %st,%st(3)
  add    $0x1,%eax
  fadd   %st,%st(3)
  cmp    %eax,%edi
  je     L47
L14:
  fld    %st(3)
  fdivr  %st(2),%st
  test   $0x1,%al
  jne    L52
  fsubrp %st,%st(3)
  add    $0x1,%eax
  fadd   %st,%st(3)
  cmp    %eax,%edi
  jne    L14
  fstp   %st(0)
  fstp   %st(0)
  fstp   %st(1)
  jmp    L11
  xchg   %ax,%ax
L47:
  fstp   %st(0)
  fstp   %st(0)
  fstp   %st(1)
L11:
  ret
  nopw   0x0(%rax,%rax,1)
L28:
  fldz
  ret
```

assistant: ```c
long double approx1(int wut) {
  long double pi = 0;
  long double numer = 4;
  long double denom = 1;
  for (int i = 1; i <= wut; i++) {
    if (i % 2 != 0) {
      pi = pi + (numer / denom);
    } else {
      pi = pi - (numer / denom);
    }
    denom = denom + 2;
  }
  return pi;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<radio_freq_cb>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  shr    $0x3,%eax
  mov    %al,-0x1(%rbp)
  movzbl -0x1(%rbp),%eax
  mov    $0x29,%edx
  mul    %dl
  shr    $0x8,%ax
  shr    $0x4,%al
  add    $0x30,%eax
  movzbl %al,%eax
  mov    %eax,%esi
  mov    $0x4,%edi
  call   <led_set_code@plt>
  movzbl -0x1(%rbp),%edx
  mov    $0x29,%ecx
  mov    %ecx,%eax
  mul    %dl
  shr    $0x8,%ax
  shr    $0x4,%al
  mov    $0x64,%ecx
  imul   %ecx,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    $0xffffffcd,%edx
  mul    %dl
  shr    $0x8,%ax
  shr    $0x3,%al
  add    $0x30,%eax
  movzbl %al,%eax
  mov    %eax,%esi
  mov    $0x3,%edi
  call   <led_set_code@plt>
  movzbl -0x1(%rbp),%ecx
  mov    $0x29,%edx
  mov    %edx,%eax
  mul    %cl
  shr    $0x8,%ax
  mov    %eax,%edx
  shr    $0x4,%dl
  mov    $0x64,%esi
  mov    %edx,%eax
  imul   %esi,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    $0xffffffcd,%ecx
  mov    %ecx,%eax
  mul    %dl
  shr    $0x8,%ax
  mov    %eax,%ecx
  shr    $0x3,%cl
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  lea    0x30(%rcx),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  mov    $0x2,%edi
  call   <led_set_code@plt>
  mov    -0x14(%rbp),%ecx
  mov    %ecx,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  mov    %eax,%edx
  shr    $0x3,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %dl,-0x1(%rbp)
  movzbl -0x1(%rbp),%eax
  add    $0x30,%eax
  movzbl %al,%eax
  mov    %eax,%esi
  mov    $0x1,%edi
  call   <led_set_code@plt>
  mov    $0x0,%eax
  call   <radio_get_stereo@plt>
  test   %rax,%rax
  je     L16
  mov    $0x5,%edi
  call   <led_set_dp@plt>
  jmp    L12
L16:
  mov    $0x5,%edi
  call   <led_clr_dp@plt>
L12:
  nop
  leave
  ret
```

assistant: ```c
void radio_freq_cb(unsigned int freq) {
  unsigned char tmp;
  tmp = (unsigned char)(freq / 10);
  led_set_code(4, tmp / 100 + 0x30);
  led_set_code(3, (tmp % 100) / 10 + 0x30);
  led_set_code(2, (tmp % 100) % 10 + 0x30);
  tmp = (unsigned char)(freq % 10);
  led_set_code(1, tmp + 0x30);
  if (radio_get_stereo()) {
    led_set_dp(5);
  } else {
    led_clr_dp(5);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<radio_freq_cb>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %edi,%r12d
  mov    $0xcccccccd,%eax
  imul   %rax,%r12
  shr    $0x23,%r12
  mov    $0x29,%ebx
  mov    %ebx,%eax
  mul    %r12b
  mov    %eax,%ebx
  shr    $0xc,%bx
  lea    0x30(%rbx),%esi
  movzbl %sil,%esi
  mov    $0x4,%edi
  call   <led_set_code@plt>
  mov    $0x64,%edx
  mov    %ebx,%eax
  imul   %edx,%eax
  mov    %r12d,%r13d
  sub    %eax,%r13d
  mov    $0xffffffcd,%ebx
  mov    %ebx,%eax
  mul    %r13b
  mov    %eax,%ebx
  shr    $0xb,%bx
  lea    0x30(%rbx),%esi
  movzbl %sil,%esi
  mov    $0x3,%edi
  call   <led_set_code@plt>
  lea    (%rbx,%rbx,4),%ebx
  add    %ebx,%ebx
  sub    %ebx,%r13d
  lea    0x30(%r13),%esi
  movzbl %sil,%esi
  mov    $0x2,%edi
  call   <led_set_code@plt>
  lea    (%r12,%r12,4),%eax
  add    %eax,%eax
  sub    %eax,%ebp
  lea    0x30(%rbp),%esi
  movzbl %sil,%esi
  mov    $0x1,%edi
  call   <led_set_code@plt>
  mov    $0x0,%eax
  call   <radio_get_stereo@plt>
  test   %rax,%rax
  je     L15
  mov    $0x5,%edi
  call   <led_set_dp@plt>
L2:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  mov    $0x5,%edi
  call   <led_clr_dp@plt>
  jmp    L2
```

assistant: ```c
void radio_freq_cb(unsigned int freq) {
  unsigned char tmp;
  tmp = (unsigned char)(freq / 10);
  led_set_code(4, tmp / 100 + 0x30);
  led_set_code(3, (tmp % 100) / 10 + 0x30);
  led_set_code(2, (tmp % 100) % 10 + 0x30);
  tmp = (unsigned char)(freq % 10);
  led_set_code(1, tmp + 0x30);
  if (radio_get_stereo()) {
    led_set_dp(5);
  } else {
    led_clr_dp(5);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<radio_freq_cb>:
  endbr64
  push   %r13
  mov    $0xcccccccd,%eax
  push   %r12
  mov    %edi,%r12d
  mov    $0x4,%edi
  push   %rbp
  mov    %r12,%rbp
  imul   %rax,%r12
  mov    $0x29,%eax
  push   %rbx
  shr    $0x23,%r12
  mul    %r12b
  sub    $0x8,%rsp
  mov    %r12d,%r13d
  mov    %eax,%ebx
  shr    $0xc,%bx
  lea    0x30(%rbx),%esi
  movzbl %sil,%esi
  call   <led_set_code@plt>
  mov    $0x64,%edx
  mov    %ebx,%eax
  mov    $0x3,%edi
  imul   %edx,%eax
  sub    %eax,%r13d
  mov    $0xffffffcd,%eax
  mul    %r13b
  mov    %eax,%ebx
  shr    $0xb,%bx
  lea    0x30(%rbx),%esi
  lea    (%rbx,%rbx,4),%ebx
  add    %ebx,%ebx
  movzbl %sil,%esi
  call   <led_set_code@plt>
  sub    %ebx,%r13d
  mov    $0x2,%edi
  lea    0x30(%r13),%esi
  movzbl %sil,%esi
  call   <led_set_code@plt>
  lea    (%r12,%r12,4),%eax
  mov    $0x1,%edi
  add    %eax,%eax
  sub    %eax,%ebp
  lea    0x30(%rbp),%esi
  movzbl %sil,%esi
  call   <led_set_code@plt>
  xor    %eax,%eax
  call   <radio_get_stereo@plt>
  mov    $0x5,%edi
  test   %rax,%rax
  je     L6
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    10b0 <led_set_dp@plt>
  nopw   0x0(%rax,%rax,1)
L6:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1090 <led_clr_dp@plt>
```

assistant: ```c
void radio_freq_cb(unsigned int freq) {
  unsigned char tmp;
  tmp = (unsigned char)(freq / 10);
  led_set_code(4, tmp / 100 + 0x30);
  led_set_code(3, (tmp % 100) / 10 + 0x30);
  led_set_code(2, (tmp % 100) % 10 + 0x30);
  tmp = (unsigned char)(freq % 10);
  led_set_code(1, tmp + 0x30);
  if (radio_get_stereo()) {
    led_set_dp(5);
  } else {
    led_clr_dp(5);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<radio_freq_cb>:
  endbr64
  push   %r13
  mov    $0xcccccccd,%eax
  push   %r12
  mov    %edi,%r12d
  mov    $0x4,%edi
  push   %rbp
  mov    %r12,%rbp
  imul   %rax,%r12
  mov    $0x29,%eax
  push   %rbx
  shr    $0x23,%r12
  mul    %r12b
  sub    $0x8,%rsp
  mov    %r12d,%r13d
  mov    %eax,%ebx
  shr    $0xc,%bx
  lea    0x30(%rbx),%esi
  movzbl %sil,%esi
  call   <led_set_code@plt>
  mov    $0x64,%edx
  mov    %ebx,%eax
  mov    $0x3,%edi
  imul   %edx,%eax
  sub    %eax,%r13d
  mov    $0xffffffcd,%eax
  mul    %r13b
  mov    %eax,%ebx
  shr    $0xb,%bx
  lea    0x30(%rbx),%esi
  lea    (%rbx,%rbx,4),%ebx
  add    %ebx,%ebx
  movzbl %sil,%esi
  call   <led_set_code@plt>
  sub    %ebx,%r13d
  mov    $0x2,%edi
  lea    0x30(%r13),%esi
  movzbl %sil,%esi
  call   <led_set_code@plt>
  lea    (%r12,%r12,4),%eax
  mov    $0x1,%edi
  add    %eax,%eax
  sub    %eax,%ebp
  lea    0x30(%rbp),%esi
  movzbl %sil,%esi
  call   <led_set_code@plt>
  xor    %eax,%eax
  call   <radio_get_stereo@plt>
  mov    $0x5,%edi
  test   %rax,%rax
  je     L2
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    10b0 <led_set_dp@plt>
  nopw   0x0(%rax,%rax,1)
L2:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1090 <led_clr_dp@plt>
```

assistant: ```c
void radio_freq_cb(unsigned int freq) {
  unsigned char tmp;
  tmp = (unsigned char)(freq / 10);
  led_set_code(4, tmp / 100 + 0x30);
  led_set_code(3, (tmp % 100) / 10 + 0x30);
  led_set_code(2, (tmp % 100) % 10 + 0x30);
  tmp = (unsigned char)(freq % 10);
  led_set_code(1, tmp + 0x30);
  if (radio_get_stereo()) {
    led_set_dp(5);
  } else {
    led_clr_dp(5);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<fibo_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x1,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L13
L33:
  cmpl   $0x1,-0x10(%rbp)
  jg     L38
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L29
L38:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L29:
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
L13:
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L33
  nop
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D0
type: string

tool: D0: %d


assistant: ```c
void fibo_1(int n) {
  int i, first = 0, second = 1, next;
  for (i = 0; i < n; i++) {
    if (i <= 1)
      next = i;
    else {
      next = first + second;
      first = second;
      second = next;
    }
    printf("%d\n", next);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<fibo_1>:
  endbr64
  test   %edi,%edi
  jle    L4
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r13d
  mov    $0x0,%ebx
  mov    $0x0,%esi
  mov    $0x1,%ebp
  mov    $0x0,%r12d
  lea    D2(%rip),%r14
  jmp    L46
L4:
  ret
L2:
  mov    %ebx,%esi
L46:
  mov    %r14,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,%r13d
  je     L18
  cmp    $0x1,%ebx
  jle    L2
  lea    (%r12,%rbp,1),%esi
  mov    %ebp,%r12d
  mov    %esi,%ebp
  jmp    L46
L18:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

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

tool: D2: %d


assistant: ```c
void fibo_1(int n) {
  int i, first = 0, second = 1, next;
  for (i = 0; i < n; i++) {
    if (i <= 1)
      next = i;
    else {
      next = first + second;
      first = second;
      second = next;
    }
    printf("%d\n", next);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<fibo_1>:
  endbr64
  test   %edi,%edi
  jle    L36
  push   %r14
  xor    %esi,%esi
  lea    D6(%rip),%r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  xor    %ebx,%ebx
L26:
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%r13d
  je     L48
L6:
  cmp    $0x1,%ebx
  jle    L9
  lea    (%r12,%rbp,1),%esi
  mov    %r14,%rdi
  add    $0x1,%ebx
  mov    %ebp,%r12d
  mov    %esi,%ebp
  call   <printf@plt>
  cmp    %ebx,%r13d
  jne    L6
L48:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L9:
  mov    $0x1,%esi
  jmp    L26
  nopw   0x0(%rax,%rax,1)
L36:
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
type: string

tool: D6: %d


assistant: ```c
void fibo_1(int n) {
  int i, first = 0, second = 1, next;
  for (i = 0; i < n; i++) {
    if (i <= 1)
      next = i;
    else {
      next = first + second;
      first = second;
      second = next;
    }
    printf("%d\n", next);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<fibo_1>:
  endbr64
  test   %edi,%edi
  jle    L47
  push   %r14
  xor    %esi,%esi
  lea    D9(%rip),%r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  xor    %ebx,%ebx
L25:
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%r13d
  je     L38
L19:
  cmp    $0x1,%ebx
  jle    L10
  lea    (%r12,%rbp,1),%esi
  mov    %r14,%rdi
  add    $0x1,%ebx
  mov    %ebp,%r12d
  mov    %esi,%ebp
  call   <printf@plt>
  cmp    %ebx,%r13d
  jne    L19
L38:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L10:
  mov    $0x1,%esi
  jmp    L25
  nopw   0x0(%rax,%rax,1)
L47:
  ret
```

tool_calls: name: read_data
arguments: label: D9
type_infered: string

tool: D9: %d


assistant: ```c
void fibo_1(int n) {
  int i, first = 0, second = 1, next;
  for (i = 0; i < n; i++) {
    if (i <= 1)
      next = i;
    else {
      next = first + second;
      first = second;
      second = next;
    }
    printf("%d\n", next);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<eval_set_time>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x8,%edi
  call   <hex2data@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <set_hmdl_time@plt>
  nop
  leave
  ret
```

assistant: ```c
void eval_set_time(const char in[]) {
  long time = hex2data(8, in);
  set_hmdl_time(time);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<eval_set_time>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  mov    $0x8,%edi
  call   <hex2data@plt>
  mov    %rax,%rdi
  call   <set_hmdl_time@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void eval_set_time(const char in[]) {
  long time = hex2data(8, in);
  set_hmdl_time(time);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<eval_set_time>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  mov    $0x8,%edi
  call   <hex2data@plt>
  add    $0x8,%rsp
  mov    %rax,%rdi
  jmp    1060 <set_hmdl_time@plt>
```

assistant: ```c
void eval_set_time(const char in[]) {
  long time = hex2data(8, in);
  set_hmdl_time(time);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<eval_set_time>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  mov    $0x8,%edi
  call   <hex2data@plt>
  add    $0x8,%rsp
  mov    %rax,%rdi
  jmp    1060 <set_hmdl_time@plt>
```

assistant: ```c
void eval_set_time(const char in[]) {
  long time = hex2data(8, in);
  set_hmdl_time(time);
}
```
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
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  jmp    L19
L26:
  movsd  -0x18(%rbp),%xmm0
  movsd  -0x10(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
L19:
  lea    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <llegeixx@plt>
  test   %rax,%rax
  je     L26
  movsd  -0x10(%rbp),%xmm0
  movapd %xmm0,%xmm1
  addsd  %xmm0,%xmm1
  movsd  D14(%rip),%xmm0
  addsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D14
type_infered: f32[2]
name: parse_data
arguments: tag: D19
type_infered: string

tool: {"D14": [0.0,1.875]}

tool: D19: %lf


assistant: ```c
int main(int argc, char *argv[]) {
  double num;
  double suma = 0.0;
  while (llegeixx(&num) == 0) {
    suma += num;
  }
  printf("%lf\n", 1 + 2 * suma);
  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
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D1(%rip),%rbx
  mov    %rsp,%rbp
  jmp    L28
L5:
  movq   %rbx,%xmm1
  addsd  (%rsp),%xmm1
  movq   %xmm1,%rbx
L28:
  mov    %rbp,%rdi
  call   <llegeixx@plt>
  test   %rax,%rax
  je     L5
  movq   %rbx,%xmm0
  addsd  %xmm0,%xmm0
  addsd  D2(%rip),%xmm0
  cvttsd2si %xmm0,%esi
  lea    D25(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type: f64
name: extract_symbolic_data
arguments: data_tag: D2
type: byte[8]
name: extract_symbolic_data
arguments: data_tag: D25
type: string

tool: D1: 0

tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: D25: %lf


assistant: ```c
int main(int argc, char *argv[]) {
  double num;
  double suma = 0.0;
  while (llegeixx(&num) == 0) {
    suma += num;
  }
  printf("%lf\n", 1 + 2 * suma);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbx
  pxor   %xmm0,%xmm0
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  jmp    L12
  nopl   0x0(%rax)
L25:
  addsd  0x10(%rsp),%xmm0
L12:
  mov    %rbx,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <llegeixx@plt>
  movsd  0x8(%rsp),%xmm0
  test   %rax,%rax
  je     L25
  addsd  %xmm0,%xmm0
  addsd  D6(%rip),%xmm0
  lea    D14(%rip),%rdi
  cvttsd2si %xmm0,%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D6
data_type: f32[2]
name: read_data
arguments: label: D14
data_type: string

tool: {"D6": [0.0,1.875]}

tool: D14: %lf


assistant: ```c
int main(int argc, char *argv[]) {
  double num;
  double suma = 0.0;
  while (llegeixx(&num) == 0) {
    suma += num;
  }
  printf("%lf\n", 1 + 2 * suma);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbx
  pxor   %xmm0,%xmm0
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  jmp    L18
  nopl   0x0(%rax)
L3:
  addsd  0x10(%rsp),%xmm0
L18:
  mov    %rbx,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <llegeixx@plt>
  movsd  0x8(%rsp),%xmm0
  test   %rax,%rax
  je     L3
  addsd  %xmm0,%xmm0
  addsd  D3(%rip),%xmm0
  lea    D4(%rip),%rdi
  cvttsd2si %xmm0,%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D3
datatype: f32[2]
name: parse_data
arguments: data_label: D4
datatype: string

tool: {"D3": [0.0,1.875]}

tool: D4: %lf


assistant: ```c
int main(int argc, char *argv[]) {
  double num;
  double suma = 0.0;
  while (llegeixx(&num) == 0) {
    suma += num;
  }
  printf("%lf\n", 1 + 2 * suma);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<fn_dynamic_long>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x28,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    $0x28,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <init_long@plt>
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <print_local_dynamic@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <print_long@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <print_long@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D0
type_guessed: string

tool: D0: [HEAP] 2 matrices gemelas tipo long dinamicas


assistant: ```c
void fn_dynamic_long(void) {
  long *lo1 = (long *)malloc(5 * sizeof(long));
  long *lo2 = (long *)malloc(5 * sizeof(long));
  init_long(lo1, lo2);
  printf("[HEAP] 2 matrices gemelas tipo long dinamicas\n");
  print_local_dynamic((unsigned long)lo1, (unsigned long)lo2);
  print_long(lo1);
  print_long(lo2);
  free(lo1);
  free(lo2);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<fn_dynamic_long>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x28,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    $0x28,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rax,%rsi
  mov    %rbp,%rdi
  call   <init_long@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <print_local_dynamic@plt>
  mov    %rbp,%rdi
  call   <print_long@plt>
  mov    %rbx,%rdi
  call   <print_long@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
data_type: string

tool: D1: [HEAP] 2 matrices gemelas tipo long dinamicas


assistant: ```c
void fn_dynamic_long(void) {
  long *lo1 = (long *)malloc(5 * sizeof(long));
  long *lo2 = (long *)malloc(5 * sizeof(long));
  init_long(lo1, lo2);
  printf("[HEAP] 2 matrices gemelas tipo long dinamicas\n");
  print_local_dynamic((unsigned long)lo1, (unsigned long)lo2);
  print_long(lo1);
  print_long(lo2);
  free(lo1);
  free(lo2);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<fn_dynamic_long>:
  endbr64
  push   %r12
  mov    $0x28,%edi
  push   %rbp
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    $0x28,%edi
  mov    %rax,%r12
  call   <malloc@plt>
  mov    %r12,%rdi
  mov    %rax,%rsi
  mov    %rax,%rbp
  call   <init_long@plt>
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <print_local_dynamic@plt>
  mov    %r12,%rdi
  call   <print_long@plt>
  mov    %rbp,%rdi
  call   <print_long@plt>
  mov    %r12,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    10b0 <free@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D5
type_guessed: string

tool: D5: [HEAP] 2 matrices gemelas tipo long dinamicas


assistant: ```c
void fn_dynamic_long(void) {
  long *lo1 = (long *)malloc(5 * sizeof(long));
  long *lo2 = (long *)malloc(5 * sizeof(long));
  init_long(lo1, lo2);
  printf("[HEAP] 2 matrices gemelas tipo long dinamicas\n");
  print_local_dynamic((unsigned long)lo1, (unsigned long)lo2);
  print_long(lo1);
  print_long(lo2);
  free(lo1);
  free(lo2);
}
```