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

```asm
<MakeDecoderHuffman>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <SizeTable@plt>
  mov    $0x0,%eax
  call   <CodeTable@plt>
  mov    $0x0,%eax
  call   <DecoderTables@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void MakeDecoderHuffman() {

  SizeTable();
  CodeTable();
  DecoderTables();
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<MakeDecoderHuffman>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <SizeTable@plt>
  mov    $0x0,%eax
  call   <CodeTable@plt>
  mov    $0x0,%eax
  call   <DecoderTables@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void MakeDecoderHuffman() {

  SizeTable();
  CodeTable();
  DecoderTables();
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<MakeDecoderHuffman>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <SizeTable@plt>
  xor    %eax,%eax
  call   <CodeTable@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1090 <DecoderTables@plt>
```

assistant: ```c
void MakeDecoderHuffman() {

  SizeTable();
  CodeTable();
  DecoderTables();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<MakeDecoderHuffman>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <SizeTable@plt>
  xor    %eax,%eax
  call   <CodeTable@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1090 <DecoderTables@plt>
```

assistant: ```c
void MakeDecoderHuffman() {

  SizeTable();
  CodeTable();
  DecoderTables();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<GetScale>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movsd  %xmm0,-0x28(%rbp)
  mov    %rdx,-0x30(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <FindAnimation@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L3
  mov    -0x30(%rbp),%rcx
  mov    -0x28(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  movq   %rdx,%xmm0
  mov    %rax,%rdi
  call   <_GetScale@plt>
  mov    $0x1,%eax
  jmp    L11
L3:
  mov    -0x30(%rbp),%rax
  lea    0x10(%rax),%rdx
  movsd  D0(%rip),%xmm0
  movsd  %xmm0,(%rdx)
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  movsd  (%rax),%xmm0
  mov    -0x30(%rbp),%rax
  movsd  %xmm0,(%rax)
  mov    $0x0,%eax
L11:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D0
type_infered: f64

tool: D0: 1

assistant: ```c
int GetScale(struct AnimationList *al, char *name, double time, double Sout[3]) {
  void *a = FindAnimation(name, al);
  if (a) {
    _GetScale(a, time, Sout);
    return 1;
  }
  Sout[0] = Sout[1] = Sout[2] = 1.0;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<GetScale>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rax
  mov    %rsi,%rdi
  movq   %xmm0,%rbp
  mov    %rdx,%rbx
  mov    %rax,%rsi
  call   <FindAnimation@plt>
  test   %rax,%rax
  jne    L7
  movsd  D5(%rip),%xmm0
  movsd  %xmm0,0x10(%rbx)
  movsd  %xmm0,0x8(%rbx)
  movsd  %xmm0,(%rbx)
  mov    $0x0,%eax
L5:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L7:
  mov    %rbx,%rsi
  movq   %rbp,%xmm0
  mov    %rax,%rdi
  call   <_GetScale@plt>
  mov    $0x1,%eax
  jmp    L5
```

tool_calls: name: unpack_data
arguments: data_label: D5
type_guessed: f64

tool: D5: 1

assistant: ```c
int GetScale(struct AnimationList *al, char *name, double time, double Sout[3]) {
  void *a = FindAnimation(name, al);
  if (a) {
    _GetScale(a, time, Sout);
    return 1;
  }
  Sout[0] = Sout[1] = Sout[2] = 1.0;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<GetScale>:
  endbr64
  push   %rbp
  mov    %rdi,%r8
  mov    %rsi,%rdi
  mov    %rdx,%rbp
  push   %rbx
  mov    %r8,%rsi
  movq   %xmm0,%rbx
  sub    $0x8,%rsp
  call   <FindAnimation@plt>
  test   %rax,%rax
  jne    L6
  movsd  D2(%rip),%xmm0
  xor    %eax,%eax
  movsd  %xmm0,0x10(%rbp)
  movsd  %xmm0,0x8(%rbp)
  movsd  %xmm0,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L6:
  mov    %rbp,%rsi
  movq   %rbx,%xmm0
  mov    %rax,%rdi
  call   <_GetScale@plt>
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: label: D2
type_guessed: f64

tool: D2: 1

assistant: ```c
int GetScale(struct AnimationList *al, char *name, double time, double Sout[3]) {
  void *a = FindAnimation(name, al);
  if (a) {
    _GetScale(a, time, Sout);
    return 1;
  }
  Sout[0] = Sout[1] = Sout[2] = 1.0;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<GetScale>:
  endbr64
  push   %rbp
  mov    %rdi,%r8
  mov    %rsi,%rdi
  mov    %rdx,%rbp
  push   %rbx
  mov    %r8,%rsi
  movq   %xmm0,%rbx
  sub    $0x8,%rsp
  call   <FindAnimation@plt>
  test   %rax,%rax
  jne    L9
  mov    D9(%rip),%rax
  movapd D9(%rip),%xmm0
  mov    %rax,0x10(%rbp)
  xor    %eax,%eax
  movups %xmm0,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L9:
  mov    %rbp,%rsi
  movq   %rbx,%xmm0
  mov    %rax,%rdi
  call   <_GetScale@plt>
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D9
type_infered: f64[2]

tool: {"D9": [1.0,1.0]}

assistant: ```c
int GetScale(struct AnimationList *al, char *name, double time, double Sout[3]) {
  void *a = FindAnimation(name, al);
  if (a) {
    _GetScale(a, time, Sout);
    return 1;
  }
  Sout[0] = Sout[1] = Sout[2] = 1.0;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cos_inverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movsd  %xmm0,-0x18(%rbp)
  movsd  %xmm1,-0x20(%rbp)
  mov    %rdi,-0x28(%rbp)
  mov    $0x1,%edi
  call   <atan@plt>
  shl    $0x2,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  addsd  %xmm0,%xmm0
  cvttsd2si %xmm0,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%edi
  movq   %rax,%xmm0
  call   <invmodDymola@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  addsd  %xmm0,%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x10(%rbp),%xmm1
  movsd  -0x20(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  comisd -0x10(%rbp),%xmm0
  jbe    L19
  movsd  -0x10(%rbp),%xmm0
  addsd  %xmm0,%xmm0
  movsd  %xmm0,-0x30(%rbp)
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <acos@plt>
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  movsd  -0x30(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  jmp    L8
L19:
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <acos@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
L8:
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm1
  mulsd  -0x10(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double cos_inverse(double x, double y, const char *ys) {
  double pi;
  double n;
  ;
  pi = 4 * atan(1);
  n = invmodDymola(y, 2 * pi);
  return (((y - n * 2 * pi) > pi) ? 2 * pi - acos(x) : acos(x)) + n * 2 * pi;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cos_inverse>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  movq   %xmm0,%rbp
  movq   %xmm1,%rbx
  mov    $0x1,%edi
  call   <atan@plt>
  shl    $0x2,%eax
  pxor   %xmm3,%xmm3
  cvtsi2sd %eax,%xmm3
  movsd  %xmm3,0x8(%rsp)
  addsd  %xmm3,%xmm3
  movsd  %xmm3,(%rsp)
  cvttsd2si %xmm3,%edi
  movq   %rbx,%xmm0
  call   <invmodDymola@plt>
  addsd  %xmm0,%xmm0
  movapd %xmm0,%xmm4
  movsd  0x8(%rsp),%xmm2
  mulsd  %xmm2,%xmm4
  movsd  %xmm4,0x8(%rsp)
  movq   %rbx,%xmm1
  subsd  %xmm4,%xmm1
  comisd %xmm2,%xmm1
  jbe    L17
  movq   %rbp,%xmm0
  call   <acos@plt>
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  movsd  (%rsp),%xmm0
  subsd  %xmm1,%xmm0
L16:
  addsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L17:
  movq   %rbp,%xmm0
  call   <acos@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  jmp    L16
```

assistant: ```c
double cos_inverse(double x, double y, const char *ys) {
  double pi;
  double n;
  ;
  pi = 4 * atan(1);
  n = invmodDymola(y, 2 * pi);
  return (((y - n * 2 * pi) > pi) ? 2 * pi - acos(x) : acos(x)) + n * 2 * pi;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<cos_inverse>:
  endbr64
  push   %r14
  mov    $0x1,%edi
  push   %rbx
  movq   %xmm1,%rbx
  sub    $0x18,%rsp
  movsd  %xmm0,0x8(%rsp)
  call   <atan@plt>
  pxor   %xmm1,%xmm1
  movq   %rbx,%xmm0
  shl    $0x2,%eax
  cvtsi2sd %eax,%xmm1
  movapd %xmm1,%xmm5
  movsd  %xmm1,(%rsp)
  addsd  %xmm1,%xmm5
  cvttsd2si %xmm5,%edi
  movq   %xmm5,%r14
  call   <invmodDymola@plt>
  movsd  (%rsp),%xmm1
  movq   %rbx,%xmm3
  movsd  0x8(%rsp),%xmm4
  addsd  %xmm0,%xmm0
  movapd %xmm0,%xmm2
  movapd %xmm4,%xmm0
  mulsd  %xmm1,%xmm2
  subsd  %xmm2,%xmm3
  movsd  %xmm2,(%rsp)
  comisd %xmm1,%xmm3
  jbe    L3
  call   <acos@plt>
  pxor   %xmm1,%xmm1
  movq   %r14,%xmm0
  movsd  (%rsp),%xmm2
  cvtsi2sd %eax,%xmm1
  add    $0x18,%rsp
  pop    %rbx
  pop    %r14
  subsd  %xmm1,%xmm0
  addsd  %xmm2,%xmm0
  ret
  nopl   0x0(%rax)
L3:
  call   <acos@plt>
  pxor   %xmm0,%xmm0
  movsd  (%rsp),%xmm2
  add    $0x18,%rsp
  cvtsi2sd %eax,%xmm0
  pop    %rbx
  pop    %r14
  addsd  %xmm2,%xmm0
  ret
```

assistant: ```c
double cos_inverse(double x, double y, const char *ys) {
  double pi;
  double n;
  ;
  pi = 4 * atan(1);
  n = invmodDymola(y, 2 * pi);
  return (((y - n * 2 * pi) > pi) ? 2 * pi - acos(x) : acos(x)) + n * 2 * pi;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cos_inverse>:
  endbr64
  push   %r14
  mov    $0x1,%edi
  push   %rbx
  movq   %xmm1,%rbx
  sub    $0x18,%rsp
  movsd  %xmm0,0x8(%rsp)
  call   <atan@plt>
  pxor   %xmm1,%xmm1
  movq   %rbx,%xmm0
  shl    $0x2,%eax
  cvtsi2sd %eax,%xmm1
  movapd %xmm1,%xmm5
  movsd  %xmm1,(%rsp)
  addsd  %xmm1,%xmm5
  cvttsd2si %xmm5,%edi
  movq   %xmm5,%r14
  call   <invmodDymola@plt>
  movsd  (%rsp),%xmm1
  movq   %rbx,%xmm3
  movsd  0x8(%rsp),%xmm4
  addsd  %xmm0,%xmm0
  movapd %xmm0,%xmm2
  movapd %xmm4,%xmm0
  mulsd  %xmm1,%xmm2
  subsd  %xmm2,%xmm3
  movsd  %xmm2,(%rsp)
  comisd %xmm1,%xmm3
  jbe    L4
  call   <acos@plt>
  pxor   %xmm1,%xmm1
  movq   %r14,%xmm0
  movsd  (%rsp),%xmm2
  cvtsi2sd %eax,%xmm1
  add    $0x18,%rsp
  pop    %rbx
  pop    %r14
  subsd  %xmm1,%xmm0
  addsd  %xmm2,%xmm0
  ret
  nopl   0x0(%rax)
L4:
  call   <acos@plt>
  pxor   %xmm0,%xmm0
  movsd  (%rsp),%xmm2
  add    $0x18,%rsp
  cvtsi2sd %eax,%xmm0
  pop    %rbx
  pop    %r14
  addsd  %xmm2,%xmm0
  ret
```

assistant: ```c
double cos_inverse(double x, double y, const char *ys) {
  double pi;
  double n;
  ;
  pi = 4 * atan(1);
  n = invmodDymola(y, 2 * pi);
  return (((y - n * 2 * pi) > pi) ? 2 * pi - acos(x) : acos(x)) + n * 2 * pi;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<arr_float_mark_nan_as_zero>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L2
L27:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  movd   %eax,%xmm0
  call   <isnanf@plt>
  test   %rax,%rax
  je     L14
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  movss  %xmm0,(%rax)
L14:
  addq   $0x1,-0x8(%rbp)
L2:
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jb     L27
  nop
  nop
  leave
  ret
```

assistant: ```c
void arr_float_mark_nan_as_zero(float *pSrc, unsigned long int num) {
  unsigned long int i;
  for (i = 0; i < num; i++) {
    if (isnanf(pSrc[i]))
      pSrc[i] = 0;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<arr_float_mark_nan_as_zero>:
  endbr64
  test   %rsi,%rsi
  je     L23
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    $0x0,%ebx
  jmp    L3
L5:
  add    $0x1,%rbx
  cmp    %rbx,%r12
  je     L33
L3:
  movss  0x0(%rbp,%rbx,4),%xmm0
  call   <isnanf@plt>
  test   %rax,%rax
  je     L5
  movl   $0x0,0x0(%rbp,%rbx,4)
  jmp    L5
L33:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L23:
  ret
```

assistant: ```c
void arr_float_mark_nan_as_zero(float *pSrc, unsigned long int num) {
  unsigned long int i;
  for (i = 0; i < num; i++) {
    if (isnanf(pSrc[i]))
      pSrc[i] = 0;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<arr_float_mark_nan_as_zero>:
  endbr64
  test   %rsi,%rsi
  je     L21
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopl   (%rax)
L29:
  movss  0x0(%rbp,%rbx,4),%xmm0
  call   <isnanf@plt>
  test   %rax,%rax
  je     L4
  movl   $0x0,0x0(%rbp,%rbx,4)
L4:
  add    $0x1,%rbx
  cmp    %rbx,%r12
  jne    L29
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L21:
  ret
```

assistant: ```c
void arr_float_mark_nan_as_zero(float *pSrc, unsigned long int num) {
  unsigned long int i;
  for (i = 0; i < num; i++) {
    if (isnanf(pSrc[i]))
      pSrc[i] = 0;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<arr_float_mark_nan_as_zero>:
  endbr64
  test   %rsi,%rsi
  je     L27
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopl   (%rax)
L25:
  movss  0x0(%rbp,%rbx,4),%xmm0
  call   <isnanf@plt>
  test   %rax,%rax
  je     L3
  movl   $0x0,0x0(%rbp,%rbx,4)
L3:
  add    $0x1,%rbx
  cmp    %rbx,%r12
  jne    L25
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L27:
  ret
```

assistant: ```c
void arr_float_mark_nan_as_zero(float *pSrc, unsigned long int num) {
  unsigned long int i;
  for (i = 0; i < num; i++) {
    if (isnanf(pSrc[i]))
      pSrc[i] = 0;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getche>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <getch@plt>
  mov    %al,-0x9(%rbp)
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  lea    -0x9(%rbp),%rcx
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  movzbl -0x9(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
char getche() {
  char c;
  c = getch();
  write(STDOUT_FILENO, &c, 1);
  return (c);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<getche>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <getch@plt>
  mov    %al,0x7(%rsp)
  lea    0x7(%rsp),%rsi
  mov    $0x1,%edx
  mov    D4(%rip),%rax
  mov    (%rax),%edi
  call   <write@plt>
  movzbl 0x7(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char getche() {
  char c;
  c = getch();
  write(STDOUT_FILENO, &c, 1);
  return (c);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getche>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <getch@plt>
  lea    0x7(%rsp),%rsi
  mov    $0x1,%edx
  mov    %al,0x7(%rsp)
  mov    D5(%rip),%rax
  mov    (%rax),%edi
  call   <write@plt>
  movzbl 0x7(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char getche() {
  char c;
  c = getch();
  write(STDOUT_FILENO, &c, 1);
  return (c);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<getche>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <getch@plt>
  lea    0x7(%rsp),%rsi
  mov    $0x1,%edx
  mov    %al,0x7(%rsp)
  mov    D5(%rip),%rax
  mov    (%rax),%edi
  call   <write@plt>
  movzbl 0x7(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char getche() {
  char c;
  c = getch();
  write(STDOUT_FILENO, &c, 1);
  return (c);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<piso>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movsd  %xmm0,-0x8(%rbp)
  lea    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <getNumber@plt>
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <floor@plt>
  leave
  ret
```

assistant: ```c
int piso(double a) {
  getNumber(&a);
  return floor(a);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<piso>:
  endbr64
  sub    $0x18,%rsp
  movsd  %xmm0,0x8(%rsp)
  lea    0x8(%rsp),%rdi
  call   <getNumber@plt>
  movsd  0x8(%rsp),%xmm0
  call   <floor@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int piso(double a) {
  getNumber(&a);
  return floor(a);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<piso>:
  endbr64
  sub    $0x18,%rsp
  lea    0x8(%rsp),%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <getNumber@plt>
  movsd  0x8(%rsp),%xmm0
  call   <floor@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int piso(double a) {
  getNumber(&a);
  return floor(a);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<piso>:
  endbr64
  sub    $0x18,%rsp
  lea    0x8(%rsp),%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <getNumber@plt>
  movsd  0x8(%rsp),%xmm0
  call   <floor@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int piso(double a) {
  getNumber(&a);
  return floor(a);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dob_to_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,-0x4(%rbp)
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x4(%rbp),%xmm1
  movsd  -0x18(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  comisd D0(%rip),%xmm0
  jb     L1
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  jmp    L11
L1:
  mov    -0x4(%rbp),%eax
L11:
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D0
type_infered: f64

tool: D0: 0.5

assistant: ```c
int dob_to_int(double x) {
  int res;
  res = (int)x;
  if (x - res >= 0.5) {
    return res + 1;
  } else {
    return res;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<dob_to_int>:
  endbr64
  cvttsd2si %xmm0,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  subsd  %xmm1,%xmm0
  comisd D2(%rip),%xmm0
  sbb    $0xffffffff,%eax
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D2
datatype: f64

tool: D2: 0.5

assistant: ```c
int dob_to_int(double x) {
  int res;
  res = (int)x;
  if (x - res >= 0.5) {
    return res + 1;
  } else {
    return res;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<dob_to_int>:
  endbr64
  cvttsd2si %xmm0,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  subsd  %xmm1,%xmm0
  comisd D9(%rip),%xmm0
  sbb    $0xffffffff,%eax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D9
datatype: f64

tool: D9: 0.5

assistant: ```c
int dob_to_int(double x) {
  int res;
  res = (int)x;
  if (x - res >= 0.5) {
    return res + 1;
  } else {
    return res;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dob_to_int>:
  endbr64
  cvttsd2si %xmm0,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  subsd  %xmm1,%xmm0
  comisd D0(%rip),%xmm0
  sbb    $0xffffffff,%eax
  ret
```

tool_calls: name: read_data
arguments: data_label: D0
type_guessed: f64

tool: D0: 0.5

assistant: ```c
int dob_to_int(double x) {
  int res;
  res = (int)x;
  if (x - res >= 0.5) {
    return res + 1;
  } else {
    return res;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpl   $0x3,-0x14(%rbp)
  je     L63
  cmpl   $0x3,-0x14(%rbp)
  jg     L79
  cmpl   $0x1,-0x14(%rbp)
  je     L6
  cmpl   $0x2,-0x14(%rbp)
  je     L52
  jmp    L79
L6:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <usage@plt>
  mov    $0x0,%eax
  call   <device_list@plt>
  jmp    L88
L52:
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <build@plt>
  jmp    L88
L63:
  mov    -0x20(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rbx
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %rbx,%rsi
  mov    %eax,%edi
  call   <build@plt>
  jmp    L88
L79:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <usage@plt>
  mov    $0x1,%eax
  jmp    L84
L88:
  mov    $0x0,%eax
L84:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  switch (argc) {
  case 1:
    usage(argv[0]);
    device_list();
    break;
  case 2:
    build(0, argv[1]);
    break;
  case 3:
    build(atoi(argv[1]), argv[2]);
    break;
  default:
    usage(argv[0]);
    return 1;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  cmp    $0x2,%edi
  je     L56
  cmp    $0x3,%edi
  je     L59
  cmp    $0x1,%edi
  je     L1
  mov    (%rsi),%rdi
  call   <usage@plt>
  mov    $0x1,%eax
  jmp    L27
L1:
  mov    (%rsi),%rdi
  call   <usage@plt>
  mov    $0x0,%eax
  call   <device_list@plt>
  mov    $0x0,%eax
L27:
  pop    %rbx
  ret
L56:
  mov    0x8(%rsi),%rsi
  mov    $0x0,%edi
  call   <build@plt>
  mov    $0x0,%eax
  jmp    L27
L59:
  mov    0x10(%rsi),%rbx
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
  mov    %rbx,%rsi
  call   <build@plt>
  mov    $0x0,%eax
  jmp    L27
```

assistant: ```c
int main(int argc, char *argv[]) {
  switch (argc) {
  case 1:
    usage(argv[0]);
    device_list();
    break;
  case 2:
    build(0, argv[1]);
    break;
  case 3:
    build(atoi(argv[1]), argv[2]);
    break;
  default:
    usage(argv[0]);
    return 1;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  cmp    $0x2,%edi
  je     L18
  cmp    $0x3,%edi
  je     L4
  cmp    $0x1,%edi
  mov    (%rsi),%rdi
  je     L5
  call   <usage@plt>
  mov    $0x1,%eax
  pop    %rbp
  ret
L18:
  mov    0x8(%rsi),%rsi
  xor    %edi,%edi
  call   <build@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <usage@plt>
  xor    %eax,%eax
  call   <device_list@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
L4:
  mov    0x8(%rsi),%rdi
  mov    0x10(%rsi),%rbp
  call   <atoi@plt>
  mov    %rbp,%rsi
  mov    %eax,%edi
  call   <build@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  switch (argc) {
  case 1:
    usage(argv[0]);
    device_list();
    break;
  case 2:
    build(0, argv[1]);
    break;
  case 3:
    build(atoi(argv[1]), argv[2]);
    break;
  default:
    usage(argv[0]);
    return 1;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  cmp    $0x2,%edi
  je     L17
  cmp    $0x3,%edi
  je     L6
  cmp    $0x1,%edi
  mov    (%rsi),%rdi
  je     L1
  call   <usage@plt>
  mov    $0x1,%eax
  pop    %rbp
  ret
L17:
  mov    0x8(%rsi),%rsi
  xor    %edi,%edi
  call   <build@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
L1:
  call   <usage@plt>
  xor    %eax,%eax
  call   <device_list@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
L6:
  mov    0x8(%rsi),%rdi
  mov    0x10(%rsi),%rbp
  call   <atoi@plt>
  mov    %rbp,%rsi
  mov    %eax,%edi
  call   <build@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  switch (argc) {
  case 1:
    usage(argv[0]);
    device_list();
    break;
  case 2:
    build(0, argv[1]);
    break;
  case 3:
    build(atoi(argv[1]), argv[2]);
    break;
  default:
    usage(argv[0]);
    return 1;
  }
  return 0;
}
```