ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<showbits_uc>:
  endbr64
  push   %r12
  lea    D0(%rip),%r12
  push   %rbp
  movzbl %dil,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nopl   (%rax)
L9:
  mov    %ebx,%ecx
  mov    %ebp,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  sar    %cl,%esi
  add    $0x1,%ebx
  and    $0x1,%esi
  call   <printf@plt>
  cmp    $0x8,%ebx
  jne    L9
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
```

tool_calls: name: parse_data
arguments: tag: D0
data_type: string

tool: D0: %d

assistant: ```c
void showbits_uc(const unsigned char bits) {
  int x;
  for (x = 0; x < sizeof bits << 3; x++)
    printf("%d", !!(bits & (1 << x)));
  printf("\n");
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<showbits_uc>:
  endbr64
  push   %r12
  lea    D2(%rip),%r12
  push   %rbp
  movzbl %dil,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nopl   (%rax)
L4:
  mov    %ebx,%ecx
  mov    %ebp,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  sar    %cl,%esi
  add    $0x1,%ebx
  and    $0x1,%esi
  call   <printf@plt>
  cmp    $0x8,%ebx
  jne    L4
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
data_type: string

tool: D2: %d

assistant: ```c
void showbits_uc(const unsigned char bits) {
  int x;
  for (x = 0; x < sizeof bits << 3; x++)
    printf("%d", !!(bits & (1 << x)));
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<xrop3_145_1_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x24(%rbp)
  mov    %al,-0x28(%rbp)
  movl   $0xff,-0x1c(%rbp)
  movzbl -0x24(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x24(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x18(%rbp)
  movzbl -0x28(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x28(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  not    %eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  or     -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  and    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  xor    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  notl   -0x4(%rbp)
  mov    -0x18(%rbp),%eax
  not    %eax
  and    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  and    -0x18(%rbp),%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned xrop3_145_1_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  unsigned stk4;
  stk4 = ~S;
  stk3 = T | stk4;
  stk2 = D & stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  return (stk1 & (~S)) | (D & S);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<xrop3_145_1_0>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  mov    %edi,%eax
  not    %eax
  or     %eax,%esi
  movzbl %sil,%eax
  or     %edi,%eax
  not    %eax
  movzbl %dil,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_145_1_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  unsigned stk4;
  stk4 = ~S;
  stk3 = T | stk4;
  stk2 = D & stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  return (stk1 & (~S)) | (D & S);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xrop3_145_1_0>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  mov    %edi,%eax
  not    %eax
  or     %eax,%esi
  movzbl %sil,%eax
  or     %edi,%eax
  movzbl %dil,%edi
  not    %eax
  or     %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_145_1_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  unsigned stk4;
  stk4 = ~S;
  stk3 = T | stk4;
  stk2 = D & stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  return (stk1 & (~S)) | (D & S);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<xrop3_145_1_0>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  mov    %edi,%eax
  not    %eax
  or     %eax,%esi
  movzbl %sil,%eax
  or     %edi,%eax
  movzbl %dil,%edi
  not    %eax
  or     %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_145_1_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  unsigned stk4;
  stk4 = ~S;
  stk3 = T | stk4;
  stk2 = D & stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  return (stk1 & (~S)) | (D & S);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x54(%rbp),%rdi
  lea    -0x58(%rbp),%rsi
  lea    -0x40(%rbp),%rcx
  lea    -0x48(%rbp),%rdx
  lea    -0x50(%rbp),%rax
  mov    %rdi,%r9
  mov    %rsi,%r8
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x58(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  movsd  -0x50(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  D72(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x38(%rbp)
  mov    -0x58(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  movsd  -0x48(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  D72(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x30(%rbp)
  mov    -0x58(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  movsd  -0x40(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  D72(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x28(%rbp)
  mov    -0x54(%rbp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  -0x50(%rbp),%xmm2
  movapd %xmm0,%xmm1
  divsd  %xmm2,%xmm1
  movsd  D72(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  mov    -0x54(%rbp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  -0x48(%rbp),%xmm2
  movapd %xmm0,%xmm1
  divsd  %xmm2,%xmm1
  movsd  D72(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  mov    -0x54(%rbp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  -0x40(%rbp),%xmm2
  movapd %xmm0,%xmm1
  divsd  %xmm2,%xmm1
  movsd  D72(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x28(%rbp),%xmm1
  movsd  -0x30(%rbp),%xmm0
  mov    -0x38(%rbp),%rax
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movsd  -0x10(%rbp),%xmm1
  movsd  -0x18(%rbp),%xmm0
  mov    -0x20(%rbp),%rax
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D3
datatype: string
name: retrieve_data
arguments: label: D72
datatype: byte[8]
name: retrieve_data
arguments: label: D42
datatype: string

tool: D3: %lf%lf%lf%d%d

tool: {"D72": [0x00,0x00,0x00,0x00,0x00,0x00,0x59,0x40]}

tool: D42: %.2lf %.2lf %.2lf


assistant: ```c
int main() {
  double a, b, c, d, e, f, g, h, i;
  int x, y;
  scanf("%lf%lf%lf%d%d", &a, &b, &c, &x, &y);
  d = x * a / 100;
  e = x * b / 100;
  f = x * c / 100;
  g = y / a * 100;
  h = y / b * 100;
  i = y / c * 100;
  printf("%.2lf %.2lf %.2lf\n", d, e, f);
  printf("%.2lf %.2lf %.2lf\n", g, h, i);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rcx
  lea    0x28(%rsp),%rdx
  lea    0x20(%rsp),%rsi
  lea    0x1c(%rsp),%r9
  lea    0x18(%rsp),%r8
  lea    D23(%rip),%rdi
  call   <scanf@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sdl 0x18(%rsp),%xmm0
  mov    0x20(%rsp),%rbp
  mov    0x28(%rsp),%r12
  mov    0x30(%rsp),%r13
  pxor   %xmm3,%xmm3
  cvtsi2sdl 0x1c(%rsp),%xmm3
  movsd  %xmm3,0x8(%rsp)
  movq   %r13,%xmm2
  mulsd  %xmm0,%xmm2
  movq   %r12,%xmm1
  mulsd  %xmm0,%xmm1
  movq   %rbp,%xmm4
  mulsd  %xmm4,%xmm0
  divsd  D39(%rip),%xmm0
  divsd  D39(%rip),%xmm2
  divsd  D39(%rip),%xmm1
  lea    D68(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm3
  movapd %xmm3,%xmm2
  movq   %r13,%xmm5
  divsd  %xmm5,%xmm2
  movapd %xmm3,%xmm1
  movq   %r12,%xmm6
  divsd  %xmm6,%xmm1
  movq   %rbp,%xmm7
  divsd  %xmm7,%xmm3
  movapd %xmm3,%xmm0
  mulsd  D39(%rip),%xmm0
  mulsd  D39(%rip),%xmm2
  mulsd  D39(%rip),%xmm1
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D23
type_infered: string
name: retrieve_data
arguments: label: D39
type_infered: byte[8]
name: retrieve_data
arguments: label: D68
type_infered: string

tool: D23: %lf%lf%lf%d%d

tool: {"D39": [0x00,0x00,0x00,0x00,0x00,0x00,0x59,0x40]}

tool: D68: %.2lf %.2lf %.2lf


assistant: ```c
int main() {
  double a, b, c, d, e, f, g, h, i;
  int x, y;
  scanf("%lf%lf%lf%d%d", &a, &b, &c, &x, &y);
  d = x * a / 100;
  e = x * b / 100;
  f = x * c / 100;
  g = y / a * 100;
  h = y / b * 100;
  i = y / c * 100;
  printf("%.2lf %.2lf %.2lf\n", d, e, f);
  printf("%.2lf %.2lf %.2lf\n", g, h, i);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D11(%rip),%rdi
  lea    D4(%rip),%rbp
  sub    $0x50,%rsp
  lea    0x40(%rsp),%rcx
  lea    0x38(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rsi
  lea    0x2c(%rsp),%r9
  lea    0x28(%rsp),%r8
  call   <scanf@plt>
  pxor   %xmm0,%xmm0
  pxor   %xmm3,%xmm3
  mov    %rbp,%rdi
  cvtsi2sdl 0x28(%rsp),%xmm0
  movsd  0x30(%rsp),%xmm5
  movsd  0x38(%rsp),%xmm6
  movsd  0x40(%rsp),%xmm7
  movsd  D24(%rip),%xmm4
  cvtsi2sdl 0x2c(%rsp),%xmm3
  movsd  %xmm6,0x8(%rsp)
  movsd  %xmm7,0x10(%rsp)
  movapd %xmm0,%xmm2
  movapd %xmm0,%xmm1
  movsd  %xmm5,(%rsp)
  mulsd  %xmm7,%xmm2
  movsd  %xmm3,0x18(%rsp)
  mulsd  %xmm6,%xmm1
  mulsd  %xmm5,%xmm0
  divsd  %xmm4,%xmm2
  divsd  %xmm4,%xmm0
  divsd  %xmm4,%xmm1
  call   <printf@plt>
  movsd  0x18(%rsp),%xmm3
  movsd  (%rsp),%xmm5
  mov    %rbp,%rdi
  movsd  0x10(%rsp),%xmm7
  movsd  0x8(%rsp),%xmm6
  movapd %xmm3,%xmm2
  movapd %xmm3,%xmm1
  mov    D24(%rip),%rax
  divsd  %xmm5,%xmm3
  movq   %rax,%xmm4
  divsd  %xmm7,%xmm2
  mulsd  %xmm4,%xmm3
  movapd %xmm3,%xmm0
  divsd  %xmm6,%xmm1
  mulsd  %xmm4,%xmm2
  mulsd  %xmm4,%xmm1
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x50,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D11
type_infered: string
name: unpack_data
arguments: data_label: D4
type_infered: string
name: unpack_data
arguments: data_label: D24
type_infered: byte[8]

tool: D11: %lf%lf%lf%d%d

tool: D4: %.2lf %.2lf %.2lf


tool: {"D24": [0x00,0x00,0x00,0x00,0x00,0x00,0x59,0x40]}

assistant: ```c
int main() {
  double a, b, c, d, e, f, g, h, i;
  int x, y;
  scanf("%lf%lf%lf%d%d", &a, &b, &c, &x, &y);
  d = x * a / 100;
  e = x * b / 100;
  f = x * c / 100;
  g = y / a * 100;
  h = y / b * 100;
  i = y / c * 100;
  printf("%.2lf %.2lf %.2lf\n", d, e, f);
  printf("%.2lf %.2lf %.2lf\n", g, h, i);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D21(%rip),%rdi
  lea    D0(%rip),%rbp
  sub    $0x50,%rsp
  lea    0x40(%rsp),%rcx
  lea    0x38(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rsi
  lea    0x2c(%rsp),%r9
  lea    0x28(%rsp),%r8
  call   <scanf@plt>
  pxor   %xmm0,%xmm0
  pxor   %xmm3,%xmm3
  mov    %rbp,%rdi
  cvtsi2sdl 0x28(%rsp),%xmm0
  movsd  0x30(%rsp),%xmm5
  movsd  0x38(%rsp),%xmm6
  movsd  0x40(%rsp),%xmm7
  movsd  D39(%rip),%xmm4
  cvtsi2sdl 0x2c(%rsp),%xmm3
  movsd  %xmm6,0x8(%rsp)
  movsd  %xmm7,0x10(%rsp)
  movapd %xmm0,%xmm2
  movapd %xmm0,%xmm1
  movsd  %xmm5,(%rsp)
  mulsd  %xmm7,%xmm2
  movsd  %xmm3,0x18(%rsp)
  mulsd  %xmm6,%xmm1
  mulsd  %xmm5,%xmm0
  divsd  %xmm4,%xmm2
  divsd  %xmm4,%xmm0
  divsd  %xmm4,%xmm1
  call   <printf@plt>
  movsd  0x18(%rsp),%xmm3
  movsd  (%rsp),%xmm5
  mov    %rbp,%rdi
  movsd  0x10(%rsp),%xmm7
  movsd  0x8(%rsp),%xmm6
  movapd %xmm3,%xmm2
  movapd %xmm3,%xmm1
  mov    D39(%rip),%rax
  divsd  %xmm5,%xmm3
  movq   %rax,%xmm4
  divsd  %xmm7,%xmm2
  mulsd  %xmm4,%xmm3
  movapd %xmm3,%xmm0
  divsd  %xmm6,%xmm1
  mulsd  %xmm4,%xmm2
  mulsd  %xmm4,%xmm1
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x50,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D21
data_type: string
name: parse_data
arguments: data_label: D0
data_type: string
name: parse_data
arguments: data_label: D39
data_type: byte[8]

tool: D21: %lf%lf%lf%d%d

tool: D0: %.2lf %.2lf %.2lf


tool: {"D39": [0x00,0x00,0x00,0x00,0x00,0x00,0x59,0x40]}

assistant: ```c
int main() {
  double a, b, c, d, e, f, g, h, i;
  int x, y;
  scanf("%lf%lf%lf%d%d", &a, &b, &c, &x, &y);
  d = x * a / 100;
  e = x * b / 100;
  f = x * c / 100;
  g = y / a * 100;
  h = y / b * 100;
  i = y / c * 100;
  printf("%.2lf %.2lf %.2lf\n", d, e, f);
  printf("%.2lf %.2lf %.2lf\n", g, h, i);
  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
  mov    %rsp,%rbp
  mov    $0xad,%edi
  call   <preload_seq@plt>
  mov    $0x35,%edi
  call   <preload_seq@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void main(void) {
  preload_seq(173);
  preload_seq(53);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xad,%edi
  call   <preload_seq@plt>
  mov    $0x35,%edi
  call   <preload_seq@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void main(void) {
  preload_seq(173);
  preload_seq(53);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xad,%edi
  call   <preload_seq@plt>
  mov    $0x35,%edi
  add    $0x8,%rsp
  jmp    1050 <preload_seq@plt>
```

assistant: ```c
void main(void) {
  preload_seq(173);
  preload_seq(53);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xad,%edi
  call   <preload_seq@plt>
  mov    $0x35,%edi
  add    $0x8,%rsp
  jmp    1050 <preload_seq@plt>
```

assistant: ```c
void main(void) {
  preload_seq(173);
  preload_seq(53);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<insert_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x1,-0xc(%rbp)
  jmp    L14
L3:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L44
L37:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  add    $0x1,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  subl   $0x1,-0x8(%rbp)
L44:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jge    L38
  cmpl   $0x0,-0x8(%rbp)
  jns    L37
L38:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L14:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L3
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int insert_sort(int num, int *list) {
  int key;
  for (int i = 1; i < num; i++) {
    key = *(list + i);
    int j = i - 1;
    while ((key < *(list + j)) && (j >= 0)) {
      *(list + j + 1) = *(list + j);
      j--;
    }
    *(list + j) = key;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<insert_sort>:
  endbr64
  cmp    $0x1,%edi
  jle    L8
  lea    -0x1(%rdi),%r10d
  mov    $0x0,%r9d
  jmp    L29
L7:
  mov    %edi,(%r8)
  add    $0x1,%r9
  add    $0x4,%rsi
  cmp    %r10,%r9
  je     L8
L29:
  mov    %rsi,%r8
  mov    0x4(%rsi),%edi
  mov    %r9d,%ecx
  mov    (%rsi),%edx
  test   %r9d,%r9d
  js     L7
  cmp    %edx,%edi
  jge    L7
  lea    -0x4(%rsi),%rax
L72:
  mov    %rax,%r8
  mov    %edx,0x8(%rax)
  mov    (%rax),%edx
  sub    $0x4,%rax
  sub    $0x1,%ecx
  js     L7
  cmp    %edi,%edx
  jg     L72
  jmp    L7
L8:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int insert_sort(int num, int *list) {
  int key;
  for (int i = 1; i < num; i++) {
    key = *(list + i);
    int j = i - 1;
    while ((key < *(list + j)) && (j >= 0)) {
      *(list + j + 1) = *(list + j);
      j--;
    }
    *(list + j) = key;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<insert_sort>:
  endbr64
  cmp    $0x1,%edi
  jle    L6
  lea    -0x1(%rdi),%r10d
  xor    %r9d,%r9d
L5:
  mov    0x4(%rsi),%edi
  mov    (%rsi),%edx
  mov    %rsi,%r8
  mov    %r9d,%ecx
  cmp    %edx,%edi
  jge    L49
  lea    -0x4(%rsi),%rax
  jmp    L32
  nopl   (%rax)
L46:
  cmp    %edi,%edx
  jle    L49
L32:
  sub    $0x1,%ecx
  mov    %edx,0x8(%rax)
  mov    %rax,%r8
  mov    (%rax),%edx
  sub    $0x4,%rax
  cmp    $0xffffffff,%ecx
  jne    L46
L49:
  add    $0x1,%r9
  mov    %edi,(%r8)
  add    $0x4,%rsi
  cmp    %r10,%r9
  jne    L5
L6:
  xor    %eax,%eax
  ret
```

assistant: ```c
int insert_sort(int num, int *list) {
  int key;
  for (int i = 1; i < num; i++) {
    key = *(list + i);
    int j = i - 1;
    while ((key < *(list + j)) && (j >= 0)) {
      *(list + j + 1) = *(list + j);
      j--;
    }
    *(list + j) = key;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<insert_sort>:
  endbr64
  cmp    $0x1,%edi
  jle    L47
  lea    -0x1(%rdi),%r10d
  xor    %r9d,%r9d
L22:
  mov    0x4(%rsi),%edi
  mov    (%rsi),%edx
  mov    %rsi,%r8
  mov    %r9d,%ecx
  cmp    %edx,%edi
  jge    L36
  lea    -0x4(%rsi),%rax
  jmp    L0
  nopl   (%rax)
L4:
  cmp    %edi,%edx
  jle    L36
L0:
  sub    $0x1,%ecx
  mov    %edx,0x8(%rax)
  mov    %rax,%r8
  mov    (%rax),%edx
  sub    $0x4,%rax
  cmp    $0xffffffff,%ecx
  jne    L4
L36:
  add    $0x1,%r9
  mov    %edi,(%r8)
  add    $0x4,%rsi
  cmp    %r9,%r10
  jne    L22
L47:
  xor    %eax,%eax
  ret
```

assistant: ```c
int insert_sort(int num, int *list) {
  int key;
  for (int i = 1; i < num; i++) {
    key = *(list + i);
    int j = i - 1;
    while ((key < *(list + j)) && (j >= 0)) {
      *(list + j + 1) = *(list + j);
      j--;
    }
    *(list + j) = key;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<AgencyTypeToString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x8,-0x4(%rbp)
  je     L84
  cmpl   $0x8,-0x4(%rbp)
  jg     L104
  cmpl   $0x4,-0x4(%rbp)
  je     L79
  cmpl   $0x4,-0x4(%rbp)
  jg     L104
  cmpl   $0x1,-0x4(%rbp)
  je     L14
  cmpl   $0x2,-0x4(%rbp)
  je     L9
  jmp    L104
L14:
  lea    D6(%rip),%rax
  jmp    L65
L9:
  lea    D1(%rip),%rax
  jmp    L65
L79:
  lea    D2(%rip),%rax
  jmp    L65
L84:
  lea    D14(%rip),%rax
  jmp    L65
L104:
  lea    D0(%rip),%rax
L65:
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D6
type_guessed: string
name: read_data
arguments: data_label: D1
type_guessed: string
name: read_data
arguments: data_label: D2
type_guessed: string
name: read_data
arguments: data_label: D14
type_guessed: string
name: read_data
arguments: data_label: D0
type_guessed: string

tool: D6: SYNTACTIC PARSE

tool: D1: SEMANTIC PARSE

tool: D2: ANAPHORIC PARSE

tool: D14: UNDERSTANDING

tool: D0: ??

assistant: ```c
char *AgencyTypeToString(int rep_type) {
  switch (rep_type) {
  case 0x0001:
    return ("SYNTACTIC PARSE");
  case 0x0002:
    return ("SEMANTIC PARSE");
  case 0x0004:
    return ("ANAPHORIC PARSE");
  case 0x0008:
    return ("UNDERSTANDING");
  default:
    return ("??");
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<AgencyTypeToString>:
  endbr64
  lea    D47(%rip),%rax
  cmp    $0x4,%edi
  je     L14
  jg     L4
  lea    D21(%rip),%rax
  cmp    $0x1,%edi
  je     L14
  cmp    $0x2,%edi
  lea    D38(%rip),%rax
  lea    D43(%rip),%rdx
  cmove  %rdx,%rax
  ret
L4:
  cmp    $0x8,%edi
  lea    D38(%rip),%rax
  lea    D40(%rip),%rdx
  cmove  %rdx,%rax
L14:
  ret
```

tool_calls: name: read_data
arguments: data_label: D47
type: string
name: read_data
arguments: data_label: D21
type: string
name: read_data
arguments: data_label: D38
type: string
name: read_data
arguments: data_label: D43
type: string
name: read_data
arguments: data_label: D40
type: string

tool: D47: ANAPHORIC PARSE

tool: D21: SYNTACTIC PARSE

tool: D38: ??

tool: D43: SEMANTIC PARSE

tool: D40: UNDERSTANDING

assistant: ```c
char *AgencyTypeToString(int rep_type) {
  switch (rep_type) {
  case 0x0001:
    return ("SYNTACTIC PARSE");
  case 0x0002:
    return ("SEMANTIC PARSE");
  case 0x0004:
    return ("ANAPHORIC PARSE");
  case 0x0008:
    return ("UNDERSTANDING");
  default:
    return ("??");
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<AgencyTypeToString>:
  endbr64
  lea    D41(%rip),%rax
  cmp    $0x4,%edi
  je     L4
  jg     L7
  lea    D2(%rip),%rax
  cmp    $0x1,%edi
  je     L4
  cmp    $0x2,%edi
  lea    D59(%rip),%rdx
  lea    D33(%rip),%rax
  cmove  %rdx,%rax
  ret
  nopl   0x0(%rax)
L7:
  cmp    $0x8,%edi
  lea    D33(%rip),%rax
  lea    D52(%rip),%rdx
  cmove  %rdx,%rax
L4:
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D41
type: string
name: retrieve_data
arguments: data_tag: D2
type: string
name: retrieve_data
arguments: data_tag: D59
type: string
name: retrieve_data
arguments: data_tag: D33
type: string
name: retrieve_data
arguments: data_tag: D52
type: string

tool: D41: ANAPHORIC PARSE

tool: D2: SYNTACTIC PARSE

tool: D59: SEMANTIC PARSE

tool: D33: ??

tool: D52: UNDERSTANDING

assistant: ```c
char *AgencyTypeToString(int rep_type) {
  switch (rep_type) {
  case 0x0001:
    return ("SYNTACTIC PARSE");
  case 0x0002:
    return ("SEMANTIC PARSE");
  case 0x0004:
    return ("ANAPHORIC PARSE");
  case 0x0008:
    return ("UNDERSTANDING");
  default:
    return ("??");
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<AgencyTypeToString>:
  endbr64
  lea    D6(%rip),%rax
  cmp    $0x4,%edi
  je     L23
  jg     L7
  lea    D37(%rip),%rax
  cmp    $0x1,%edi
  je     L23
  cmp    $0x2,%edi
  lea    D5(%rip),%rdx
  lea    D38(%rip),%rax
  cmove  %rdx,%rax
  ret
  nopl   0x0(%rax)
L7:
  cmp    $0x8,%edi
  lea    D38(%rip),%rax
  lea    D59(%rip),%rdx
  cmove  %rdx,%rax
L23:
  ret
```

tool_calls: name: disassemble_data
arguments: label: D6
type_infered: string
name: disassemble_data
arguments: label: D37
type_infered: string
name: disassemble_data
arguments: label: D5
type_infered: string
name: disassemble_data
arguments: label: D38
type_infered: string
name: disassemble_data
arguments: label: D59
type_infered: string

tool: D6: ANAPHORIC PARSE

tool: D37: SYNTACTIC PARSE

tool: D5: SEMANTIC PARSE

tool: D38: ??

tool: D59: UNDERSTANDING

assistant: ```c
char *AgencyTypeToString(int rep_type) {
  switch (rep_type) {
  case 0x0001:
    return ("SYNTACTIC PARSE");
  case 0x0002:
    return ("SEMANTIC PARSE");
  case 0x0004:
    return ("ANAPHORIC PARSE");
  case 0x0008:
    return ("UNDERSTANDING");
  default:
    return ("??");
  }
}
```