ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<enc11>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movsd  %xmm0,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x30(%rbp),%rcx
  lea    -0x10(%rbp),%rax
  mov    $0x8,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x10(%rbp),%eax
  shr    $0x1e,%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  shr    $0x1c,%eax
  and    $0xc,%eax
  add    %edx,%eax
  add    $0x2a,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %dl,(%rax)
  movl   $0x0,-0x18(%rbp)
  jmp    L16
L41:
  movl   $0x0,-0x14(%rbp)
  jmp    L5
L21:
  mov    -0x18(%rbp),%eax
  cltq
  mov    -0x10(%rbp,%rax,4),%esi
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%ecx
  shr    %cl,%esi
  mov    %esi,%eax
  and    $0x3f,%eax
  lea    0x3b(%rax),%ecx
  mov    -0x18(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %eax,%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x14(%rbp)
L5:
  cmpl   $0x4,-0x14(%rbp)
  jle    L21
  addl   $0x1,-0x18(%rbp)
L16:
  cmpl   $0x1,-0x18(%rbp)
  jle    L41
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
void enc11(char *to, double x) {
  int i, j;
  unsigned int qw[2];
  memcpy(qw, &x, 8);
  to[0] = 42 + (qw[0] >> 30) + ((qw[1] >> 28) & 12);
  for (i = 0; i < 2; i++)
    for (j = 0; j < 5; j++)
      to[5 * i + j + 1] = 59 + ((qw[i] >> (6 * j)) & 63);
}
```
user: Translate the below assembly code into C programming language:

```asm
<enc11>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbx
  movsd  %xmm0,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  mov    $0x8,%edx
  call   <memcpy@plt>
  mov    0x10(%rsp),%edi
  mov    0x14(%rsp),%esi
  mov    %esi,%eax
  shr    $0x1c,%eax
  and    $0xc,%eax
  mov    %edi,%edx
  shr    $0x1e,%edx
  lea    0x2a(%rax,%rdx,1),%eax
  mov    %al,(%rbx)
  lea    0x1(%rbx),%rdx
  mov    $0x0,%ecx
L18:
  mov    %edi,%eax
  shr    %cl,%eax
  and    $0x3f,%eax
  add    $0x3b,%eax
  mov    %al,(%rdx)
  add    $0x6,%ecx
  add    $0x1,%rdx
  cmp    $0x1e,%ecx
  jne    L18
  lea    0x6(%rbx),%rdx
  mov    $0x0,%ecx
L17:
  mov    %esi,%eax
  shr    %cl,%eax
  and    $0x3f,%eax
  add    $0x3b,%eax
  mov    %al,(%rdx)
  add    $0x6,%ecx
  add    $0x1,%rdx
  cmp    $0x1e,%ecx
  jne    L17
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x20,%rsp
  pop    %rbx
  ret
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void enc11(char *to, double x) {
  int i, j;
  unsigned int qw[2];
  memcpy(qw, &x, 8);
  to[0] = 42 + (qw[0] >> 30) + ((qw[1] >> 28) & 12);
  for (i = 0; i < 2; i++)
    for (j = 0; j < 5; j++)
      to[5 * i + j + 1] = 59 + ((qw[i] >> (6 * j)) & 63);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<enc11>:
  endbr64
  push   %rbx
  mov    $0x8,%edx
  mov    %rdi,%rbx
  sub    $0x20,%rsp
  movsd  %xmm0,0x8(%rsp)
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <memcpy@plt>
  mov    0x14(%rsp),%esi
  mov    0x10(%rsp),%edi
  xor    %ecx,%ecx
  mov    %esi,%eax
  mov    %edi,%edx
  shr    $0x1c,%eax
  shr    $0x1e,%edx
  and    $0xc,%eax
  lea    0x2a(%rax,%rdx,1),%eax
  lea    0x1(%rbx),%rdx
  mov    %al,(%rbx)
L8:
  mov    %edi,%eax
  add    $0x1,%rdx
  shr    %cl,%eax
  add    $0x6,%ecx
  and    $0x3f,%eax
  add    $0x3b,%eax
  mov    %al,-0x1(%rdx)
  cmp    $0x1e,%ecx
  jne    L8
  lea    0x6(%rbx),%rdx
  xor    %ecx,%ecx
L15:
  mov    %esi,%eax
  add    $0x1,%rdx
  shr    %cl,%eax
  add    $0x6,%ecx
  and    $0x3f,%eax
  add    $0x3b,%eax
  mov    %al,-0x1(%rdx)
  cmp    $0x1e,%ecx
  jne    L15
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x20,%rsp
  pop    %rbx
  ret
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void enc11(char *to, double x) {
  int i, j;
  unsigned int qw[2];
  memcpy(qw, &x, 8);
  to[0] = 42 + (qw[0] >> 30) + ((qw[1] >> 28) & 12);
  for (i = 0; i < 2; i++)
    for (j = 0; j < 5; j++)
      to[5 * i + j + 1] = 59 + ((qw[i] >> (6 * j)) & 63);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<enc11>:
  endbr64
  push   %rbx
  mov    $0x8,%edx
  mov    %rdi,%rbx
  sub    $0x20,%rsp
  movsd  %xmm0,0x8(%rsp)
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <memcpy@plt>
  mov    0x14(%rsp),%edx
  mov    0x10(%rsp),%ecx
  movq   D0(%rip),%xmm0
  movq   D4(%rip),%xmm1
  mov    %edx,%eax
  mov    %ecx,%r10d
  movzbl %dl,%r11d
  mov    %ecx,%r9d
  shr    $0x6,%eax
  shr    $0x18,%r10d
  mov    %ecx,%r8d
  mov    %ecx,%edi
  movzbl %al,%eax
  movzbl %r10b,%r10d
  shr    $0x12,%r9d
  mov    %edx,%esi
  shl    $0x8,%rax
  movzbl %r9b,%r9d
  shr    $0xc,%r8d
  shr    $0x6,%edi
  or     %r11,%rax
  movzbl %r8b,%r8d
  shr    $0x1c,%esi
  shl    $0x8,%rax
  movzbl %dil,%edi
  movzbl %sil,%esi
  or     %r10,%rax
  shl    $0x8,%rax
  or     %r9,%rax
  shl    $0x8,%rax
  or     %r8,%rax
  shl    $0x8,%rax
  or     %rdi,%rax
  movzbl %cl,%edi
  shr    $0x1e,%ecx
  shl    $0x8,%rax
  add    $0x2a,%ecx
  or     %rdi,%rax
  or     $0x3b,%ch
  shl    $0x8,%rax
  pinsrw $0x0,%ecx,%xmm1
  or     %rsi,%rax
  movq   %rax,%xmm2
  mov    %edx,%eax
  shr    $0xc,%eax
  pand   %xmm2,%xmm0
  and    $0x3f,%eax
  paddb  %xmm1,%xmm0
  add    $0x3b,%eax
  mov    %al,0x8(%rbx)
  mov    %edx,%eax
  shr    $0x12,%eax
  movq   %xmm0,(%rbx)
  and    $0x3f,%eax
  add    $0x3b,%eax
  shr    $0x18,%edx
  and    $0x3f,%edx
  mov    %al,0x9(%rbx)
  add    $0x3b,%edx
  mov    %dl,0xa(%rbx)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x20,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D0
type: byte[8]
name: disassemble_data
arguments: label: D4
type: byte[8]

tool: {"D0": [0x0c,0x3f,0x3f,0x3f,0x3f,0x3f,0x3f,0x3f]}

tool: {"D4": [0x00,0x3b,0x3b,0x3b,0x3b,0x3b,0x3b,0x3b]}

assistant: ```c
void enc11(char *to, double x) {
  int i, j;
  unsigned int qw[2];
  memcpy(qw, &x, 8);
  to[0] = 42 + (qw[0] >> 30) + ((qw[1] >> 28) & 12);
  for (i = 0; i < 2; i++)
    for (j = 0; j < 5; j++)
      to[5 * i + j + 1] = 59 + ((qw[i] >> (6 * j)) & 63);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<intCompTrap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  movsd  %xmm0,-0x20(%rbp)
  mov    %esi,-0x24(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  movq   %rax,%xmm0
  call   <f@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x1,-0xc(%rbp)
  jmp    L9
L3:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movq   %rax,%xmm0
  call   <f@plt>
  addsd  %xmm0,%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L9:
  mov    -0xc(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L3
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movq   %rax,%xmm0
  call   <f@plt>
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x20(%rbp),%xmm0
  movsd  D0(%rip),%xmm1
  divsd  %xmm1,%xmm0
  mulsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
type_infered: string
name: extract_symbolic_data
arguments: tag: D0
type_infered: f32[2]

tool: D6: %d


tool: {"D0": [0.0,2.0]}

assistant: ```c
double intCompTrap(double *X, double h, int n) {
  double integral = f(X[0]);
  int i;
  for (i = 1; i < n; i++)
    integral += 2 * f(X[i]);
  printf("%d\n", i);
  integral += f(X[n]);
  return (integral * (h / 2));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<intCompTrap>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%r14
  movq   %xmm0,%r13
  mov    %esi,%r12d
  movsd  (%rdi),%xmm0
  call   <f@plt>
  movsd  %xmm0,0x8(%rsp)
  cmp    $0x1,%r12d
  jle    L23
  lea    0x8(%r14),%rbx
  lea    -0x2(%r12),%eax
  lea    0x10(%r14,%rax,8),%rbp
L20:
  movsd  (%rbx),%xmm0
  call   <f@plt>
  addsd  %xmm0,%xmm0
  addsd  0x8(%rsp),%xmm0
  movsd  %xmm0,0x8(%rsp)
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L20
  mov    %r12d,%esi
L1:
  lea    D14(%rip),%rdi
  call   <printf@plt>
  movslq %r12d,%r12
  movsd  (%r14,%r12,8),%xmm0
  call   <f@plt>
  movapd %xmm0,%xmm1
  movq   %r13,%xmm0
  mulsd  D1(%rip),%xmm0
  addsd  0x8(%rsp),%xmm1
  mulsd  %xmm1,%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L23:
  mov    $0x1,%esi
  jmp    L1
```

tool_calls: name: disassemble_data
arguments: data_tag: D14
type: string
name: disassemble_data
arguments: data_tag: D1
type: f32[2]

tool: D14: %d


tool: {"D1": [0.0,1.75]}

assistant: ```c
double intCompTrap(double *X, double h, int n) {
  double integral = f(X[0]);
  int i;
  for (i = 1; i < n; i++)
    integral += 2 * f(X[i]);
  printf("%d\n", i);
  integral += f(X[n]);
  return (integral * (h / 2));
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<intCompTrap>:
  endbr64
  push   %r14
  movq   %xmm0,%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  movslq %esi,%r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  movsd  (%rdi),%xmm0
  call   <f@plt>
  movsd  %xmm0,0x8(%rsp)
  cmp    $0x1,%r12d
  jle    L12
  lea    -0x2(%r12),%eax
  lea    0x8(%r13),%rbx
  lea    0x10(%r13,%rax,8),%rbp
  xchg   %ax,%ax
L18:
  movsd  (%rbx),%xmm0
  add    $0x8,%rbx
  call   <f@plt>
  addsd  %xmm0,%xmm0
  addsd  0x8(%rsp),%xmm0
  movsd  %xmm0,0x8(%rsp)
  cmp    %rbp,%rbx
  jne    L18
  mov    %r12d,%esi
L15:
  lea    D2(%rip),%rdi
  call   <printf@plt>
  movsd  0x0(%r13,%r12,8),%xmm0
  call   <f@plt>
  movapd %xmm0,%xmm1
  movq   %r14,%xmm0
  addsd  0x8(%rsp),%xmm1
  mulsd  D8(%rip),%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  mulsd  %xmm1,%xmm0
  ret
  nopl   (%rax)
L12:
  mov    $0x1,%esi
  jmp    L15
```

tool_calls: name: disassemble_data
arguments: tag: D2
type_guessed: string
name: disassemble_data
arguments: tag: D8
type_guessed: f32[2]

tool: D2: %d


tool: {"D8": [0.0,1.75]}

assistant: ```c
double intCompTrap(double *X, double h, int n) {
  double integral = f(X[0]);
  int i;
  for (i = 1; i < n; i++)
    integral += 2 * f(X[i]);
  printf("%d\n", i);
  integral += f(X[n]);
  return (integral * (h / 2));
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<intCompTrap>:
  endbr64
  push   %r14
  movq   %xmm0,%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  movslq %esi,%r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  movsd  (%rdi),%xmm0
  call   <f@plt>
  movsd  %xmm0,0x8(%rsp)
  cmp    $0x1,%r12d
  jle    L21
  lea    -0x2(%r12),%eax
  lea    0x8(%r13),%rbx
  lea    0x10(%r13,%rax,8),%rbp
  xchg   %ax,%ax
L19:
  movsd  (%rbx),%xmm0
  add    $0x8,%rbx
  call   <f@plt>
  addsd  %xmm0,%xmm0
  addsd  0x8(%rsp),%xmm0
  movsd  %xmm0,0x8(%rsp)
  cmp    %rbp,%rbx
  jne    L19
  mov    %r12d,%esi
L8:
  lea    D2(%rip),%rdi
  call   <printf@plt>
  movsd  0x0(%r13,%r12,8),%xmm0
  call   <f@plt>
  movapd %xmm0,%xmm1
  movq   %r14,%xmm0
  addsd  0x8(%rsp),%xmm1
  mulsd  D18(%rip),%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  mulsd  %xmm1,%xmm0
  ret
  nopl   (%rax)
L21:
  mov    $0x1,%esi
  jmp    L8
```

tool_calls: name: disassemble_data
arguments: tag: D2
data_type: string
name: disassemble_data
arguments: tag: D18
data_type: f32[2]

tool: D2: %d


tool: {"D18": [0.0,1.75]}

assistant: ```c
double intCompTrap(double *X, double h, int n) {
  double integral = f(X[0]);
  int i;
  for (i = 1; i < n; i++)
    integral += 2 * f(X[i]);
  printf("%d\n", i);
  integral += f(X[n]);
  return (integral * (h / 2));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x30,-0x30(%rbp)
  movl   $0x24,-0x2c(%rbp)
  movl   $0x44,-0x28(%rbp)
  movl   $0x48,-0x24(%rbp)
  movl   $0xc,-0x20(%rbp)
  movl   $0x30,-0x1c(%rbp)
  movl   $0x2,-0x18(%rbp)
  lea    -0x30(%rbp),%rax
  mov    $0x7,%esi
  mov    %rax,%rdi
  call   <selection_sort@plt>
  movl   $0x0,-0x34(%rbp)
  jmp    L20
L1:
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x34(%rbp)
L20:
  cmpl   $0x6,-0x34(%rbp)
  jle    L1
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D9
type_guessed: string

tool: D9: %d 

assistant: ```c
int main() {
  int i;
  int arr[7] = {48, 36, 68, 72, 12, 48, 02};
  selection_sort(arr, 7);
  for (i = 0; i < 7; i++) {
    printf("%d ", arr[i]);
  }
  printf("\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x30,(%rsp)
  movl   $0x24,0x4(%rsp)
  movl   $0x44,0x8(%rsp)
  movl   $0x48,0xc(%rsp)
  movl   $0xc,0x10(%rsp)
  movl   $0x30,0x14(%rsp)
  movl   $0x2,0x18(%rsp)
  mov    %rsp,%rbx
  mov    $0x7,%esi
  mov    %rbx,%rdi
  call   <selection_sort@plt>
  lea    0x1c(%rsp),%r12
  lea    D8(%rip),%rbp
L11:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L11
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
type: string

tool: D8: %d 

assistant: ```c
int main() {
  int i;
  int arr[7] = {48, 36, 68, 72, 12, 48, 02};
  selection_sort(arr, 7);
  for (i = 0; i < 7; i++) {
    printf("%d ", arr[i]);
  }
  printf("\n");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x7,%esi
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x2400000030,%rax
  mov    %rsp,%rbx
  lea    0x1c(%rsp),%r12
  movl   $0x2,0x18(%rsp)
  mov    %rbx,%rdi
  mov    %rax,(%rsp)
  movabs $0x4800000044,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x300000000c,%rax
  mov    %rax,0x10(%rsp)
  call   <selection_sort@plt>
  nopw   0x0(%rax,%rax,1)
L3:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L3
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

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

tool: D2: %d 

assistant: ```c
int main() {
  int i;
  int arr[7] = {48, 36, 68, 72, 12, 48, 02};
  selection_sort(arr, 7);
  for (i = 0; i < 7; i++) {
    printf("%d ", arr[i]);
  }
  printf("\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x7,%esi
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  sub    $0x30,%rsp
  movdqa D11(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x1c(%rsp),%r12
  mov    D13(%rip),%rax
  mov    %rbx,%rdi
  movl   $0x2,0x18(%rsp)
  mov    %rax,0x10(%rsp)
  movaps %xmm0,(%rsp)
  call   <selection_sort@plt>
  nopl   0x0(%rax,%rax,1)
L2:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L2
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D8
type_infered: string
name: unpack_data
arguments: label: D11
type_infered: byte[16]
name: unpack_data
arguments: label: D13
type_infered: byte[8]

tool: D8: %d 

tool: {"D11": [0x30,0x00,0x00,0x00,0x24,0x00,0x00,0x00,0x44,0x00,0x00,0x00,0x48,0x00,0x00,0x00]}

tool: {"D13": [0x0c,0x00,0x00,0x00,0x30,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int i;
  int arr[7] = {48, 36, 68, 72, 12, 48, 02};
  selection_sort(arr, 7);
  for (i = 0; i < 7; i++) {
    printf("%d ", arr[i]);
  }
  printf("\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<fixed_9_9_fract>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L7
L26:
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x4(%rbp)
L7:
  mov    -0x18(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x18(%rbp)
  test   %eax,%eax
  jne    L26
  mov    -0x14(%rbp),%eax
  and    $0x20000,%eax
  test   %eax,%eax
  je     L14
  negl   -0x14(%rbp)
L14:
  andl   $0x1ff,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  imul   -0x4(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0x1ff,%eax
  test   %eax,%eax
  je     L23
  mov    -0x14(%rbp),%eax
  sar    $0x9,%eax
  add    $0x1,%eax
  jmp    L25
L23:
  mov    -0x14(%rbp),%eax
  sar    $0x9,%eax
L25:
  pop    %rbp
  ret
```

assistant: ```c
int fixed_9_9_fract(int32_t x, int numDigits) {
  int multiplier = 1;
  while (numDigits--)
    multiplier *= 10;
  if (x & ((uint32_t)1 << 17))
    x = 0 - x;
  x &= (1 << 9) - 1;
  x *= multiplier;
  if (x & 0x1ff)
    return (x >> 9) + 1;
  else
    return x >> 9;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<fixed_9_9_fract>:
  endbr64
  lea    -0x1(%rsi),%edx
  test   %esi,%esi
  je     L28
  mov    $0x1,%eax
L13:
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    $0x1,%edx
  cmp    $0xffffffff,%edx
  jne    L13
L15:
  mov    %edi,%edx
  neg    %edx
  test   $0x20000,%edi
  cmovne %edx,%edi
  and    $0x1ff,%edi
  imul   %eax,%edi
  mov    %edi,%edx
  sar    $0x9,%edx
  add    $0x1,%edx
  mov    %edi,%eax
  sar    $0x9,%eax
  test   $0x1ff,%edi
  cmovne %edx,%eax
  ret
L28:
  mov    $0x1,%eax
  jmp    L15
```

assistant: ```c
int fixed_9_9_fract(int32_t x, int numDigits) {
  int multiplier = 1;
  while (numDigits--)
    multiplier *= 10;
  if (x & ((uint32_t)1 << 17))
    x = 0 - x;
  x &= (1 << 9) - 1;
  x *= multiplier;
  if (x & 0x1ff)
    return (x >> 9) + 1;
  else
    return x >> 9;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<fixed_9_9_fract>:
  endbr64
  lea    -0x1(%rsi),%edx
  mov    $0x1,%eax
  test   %esi,%esi
  je     L8
L2:
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    $0x1,%edx
  jae    L2
L8:
  mov    %edi,%edx
  neg    %edx
  test   $0x20000,%edi
  cmovne %edx,%edi
  and    $0x1ff,%edi
  imul   %eax,%edi
  mov    %edi,%eax
  and    $0x1ff,%edi
  sar    $0x9,%eax
  cmp    $0x1,%edi
  sbb    $0xffffffff,%eax
  ret
```

assistant: ```c
int fixed_9_9_fract(int32_t x, int numDigits) {
  int multiplier = 1;
  while (numDigits--)
    multiplier *= 10;
  if (x & ((uint32_t)1 << 17))
    x = 0 - x;
  x &= (1 << 9) - 1;
  x *= multiplier;
  if (x & 0x1ff)
    return (x >> 9) + 1;
  else
    return x >> 9;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fixed_9_9_fract>:
  endbr64
  test   %esi,%esi
  je     L39
  mov    %esi,%ecx
  cmp    $0x1,%esi
  je     L27
  movq   D5(%rip),%xmm0
  mov    %esi,%edx
  xor    %eax,%eax
  shr    %edx
  nopl   (%rax)
L11:
  movdqa %xmm0,%xmm1
  add    $0x1,%eax
  pslld  $0x2,%xmm1
  paddd  %xmm1,%xmm0
  pslld  $0x1,%xmm0
  cmp    %eax,%edx
  jne    L11
  movd   %xmm0,%esi
  pshufd $0xe5,%xmm0,%xmm2
  movd   %xmm2,%eax
  imul   %eax,%esi
  and    $0x1,%ecx
  je     L2
L27:
  lea    (%rsi,%rsi,4),%esi
  add    %esi,%esi
L2:
  mov    %edi,%eax
  neg    %eax
  test   $0x20000,%edi
  cmovne %eax,%edi
  and    $0x1ff,%edi
  imul   %esi,%edi
  mov    %edi,%eax
  and    $0x1ff,%edi
  sar    $0x9,%eax
  cmp    $0x1,%edi
  sbb    $0xffffffff,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L39:
  mov    $0x1,%esi
  jmp    L2
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
data_type: byte[8]

tool: {"D5": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int fixed_9_9_fract(int32_t x, int numDigits) {
  int multiplier = 1;
  while (numDigits--)
    multiplier *= 10;
  if (x & ((uint32_t)1 << 17))
    x = 0 - x;
  x &= (1 << 9) - 1;
  x *= multiplier;
  if (x & 0x1ff)
    return (x >> 9) + 1;
  else
    return x >> 9;
}
```
user: What is the c source code of the assembly code below:

```asm
<scroll1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x34(%rbp)
  mov    $0xfffe,%edi
  call   <delay@plt>
  mov    $0xfffe,%edi
  call   <delay@plt>
  mov    $0x7fff,%edi
  call   <delay@plt>
  jmp    L28
L23:
  mov    -0x34(%rbp),%eax
  add    $0x28,%eax
  mov    %eax,%edi
  call   <vdp_raddr@plt>
  lea    -0x30(%rbp),%rax
  mov    $0x28,%esi
  mov    %rax,%rdi
  call   <_vram_read@plt>
  mov    -0x34(%rbp),%eax
  mov    %eax,%edi
  call   <vdp_waddr@plt>
  lea    -0x30(%rbp),%rax
  mov    $0x28,%esi
  mov    %rax,%rdi
  call   <_vram_write@plt>
  addl   $0x28,-0x34(%rbp)
L28:
  cmpl   $0x397,-0x34(%rbp)
  jbe    L23
  mov    $0x28,%esi
  mov    $0x0,%edi
  call   <_vram_write_run@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  leave
  ret
```

assistant: ```c
void scroll1() {
  char buf[40];
  unsigned int addr = 0x0000;
  delay(65534);
  delay(65534);
  delay(32767);
  for (; addr < (0x0000 + 40 * (24 - 1)); addr += 40) {
    vdp_raddr(addr + 40);
    _vram_read(buf, sizeof(buf));
    vdp_waddr(addr);
    _vram_write(buf, sizeof(buf));
  }
  _vram_write_run(0, 40);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<scroll1>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    $0xfffe,%edi
  call   <delay@plt>
  mov    $0xfffe,%edi
  call   <delay@plt>
  mov    $0x7fff,%edi
  call   <delay@plt>
  mov    $0x0,%ebx
  mov    %rsp,%rbp
L13:
  mov    %ebx,%r12d
  add    $0x28,%ebx
  mov    %ebx,%edi
  call   <vdp_raddr@plt>
  mov    $0x28,%esi
  mov    %rbp,%rdi
  call   <_vram_read@plt>
  mov    %r12d,%edi
  call   <vdp_waddr@plt>
  mov    $0x28,%esi
  mov    %rbp,%rdi
  call   <_vram_write@plt>
  cmp    $0x398,%ebx
  jne    L13
  mov    $0x28,%esi
  mov    $0x0,%edi
  call   <_vram_write_run@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void scroll1() {
  char buf[40];
  unsigned int addr = 0x0000;
  delay(65534);
  delay(65534);
  delay(32767);
  for (; addr < (0x0000 + 40 * (24 - 1)); addr += 40) {
    vdp_raddr(addr + 40);
    _vram_read(buf, sizeof(buf));
    vdp_waddr(addr);
    _vram_write(buf, sizeof(buf));
  }
  _vram_write_run(0, 40);
}
```