ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r13
  xor    %eax,%eax
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq %edi,%rbp
  lea    D5(%rip),%rdi
  push   %rbx
  mov    %ebp,%esi
  sub    $0x8,%rsp
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L13
  xor    %ebx,%ebx
  lea    D18(%rip),%r13
  nopl   0x0(%rax)
L4:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L4
L13:
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D21(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
data_type: string
name: extract_symbolic_data
arguments: data_tag: D18
data_type: string
name: extract_symbolic_data
arguments: data_tag: D21
data_type: string

tool: D5: The command line has %d arguments: 


tool: D18: %d: %s


tool: D21: Done

assistant: ```c
int main(int argc, char *argv[]) {
  int count;
  printf("The command line has %d arguments: \n", argc);
  for (count = 0; count < argc; count++)
    printf("%d: %s\n", count, argv[count]);
  printf("\n");
  puts("Done");
  getchar();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<A2HexByte>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x24(%rbp)
  mov    %al,-0x28(%rbp)
  movsbl -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <ToHex@plt>
  shl    $0x4,%eax
  mov    %eax,%ebx
  movsbl -0x28(%rbp),%eax
  mov    %eax,%edi
  call   <ToHex@plt>
  add    %ebx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int A2HexByte(char A1, char A2) {
  int HexByte;
  HexByte = 16 * ToHex(A1) + ToHex(A2);
  return HexByte;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<A2HexByte>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%ebp
  movsbl %dil,%edi
  call   <ToHex@plt>
  mov    %eax,%ebx
  movsbl %bpl,%edi
  call   <ToHex@plt>
  shl    $0x4,%ebx
  add    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int A2HexByte(char A1, char A2) {
  int HexByte;
  HexByte = 16 * ToHex(A1) + ToHex(A2);
  return HexByte;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<A2HexByte>:
  endbr64
  push   %rbp
  movsbl %dil,%edi
  mov    %esi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <ToHex@plt>
  movsbl %bpl,%edi
  mov    %eax,%ebx
  call   <ToHex@plt>
  shl    $0x4,%ebx
  add    $0x8,%rsp
  add    %ebx,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int A2HexByte(char A1, char A2) {
  int HexByte;
  HexByte = 16 * ToHex(A1) + ToHex(A2);
  return HexByte;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<A2HexByte>:
  endbr64
  push   %rbp
  movsbl %dil,%edi
  mov    %esi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <ToHex@plt>
  movsbl %bpl,%edi
  mov    %eax,%ebx
  call   <ToHex@plt>
  shl    $0x4,%ebx
  add    $0x8,%rsp
  add    %ebx,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int A2HexByte(char A1, char A2) {
  int HexByte;
  HexByte = 16 * ToHex(A1) + ToHex(A2);
  return HexByte;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<unknown_option>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <print_usage@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D6
type_infered: string

tool: D6: Unknown option: %s



assistant: ```c
void unknown_option(char *unknown, char *prog_name) {
  printf("Unknown option: %s\n\n", unknown);
  print_usage(prog_name);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<unknown_option>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  mov    %rdi,%rsi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <print_usage@plt>
  pop    %rbx
  ret
```

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

tool: D8: Unknown option: %s



assistant: ```c
void unknown_option(char *unknown, char *prog_name) {
  printf("Unknown option: %s\n\n", unknown);
  print_usage(prog_name);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<unknown_option>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  mov    %rdi,%rsi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <print_usage@plt>
```

tool_calls: name: parse_data
arguments: tag: D8
data_type: string

tool: D8: Unknown option: %s



assistant: ```c
void unknown_option(char *unknown, char *prog_name) {
  printf("Unknown option: %s\n\n", unknown);
  print_usage(prog_name);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<unknown_option>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  mov    %rdi,%rsi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <print_usage@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D7
type_infered: string

tool: D7: Unknown option: %s



assistant: ```c
void unknown_option(char *unknown, char *prog_name) {
  printf("Unknown option: %s\n\n", unknown);
  print_usage(prog_name);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<rec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x100000(%rsp),%r11
L39:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L39
  sub    $0x30,%rsp
  mov    %edi,-0x100024(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x0,-0x100024(%rbp)
  jne    L48
  mov    $0x0,%eax
  jmp    L57
L48:
  movl   $0x0,-0x100018(%rbp)
  jmp    L46
L62:
  mov    -0x100018(%rbp),%eax
  imul   -0x100024(%rbp),%eax
  mov    %eax,%edx
  mov    -0x100018(%rbp),%eax
  cltq
  mov    %edx,-0x100010(%rbp,%rax,4)
  addl   $0x1,-0x100018(%rbp)
L46:
  mov    -0x100018(%rbp),%eax
  cmp    $0x3ffff,%eax
  jbe    L62
  mov    -0x100024(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <rec>
  mov    %eax,-0x100014(%rbp)
  movl   $0x0,-0x100018(%rbp)
  jmp    L76
L40:
  mov    -0x100018(%rbp),%eax
  cltq
  mov    -0x100010(%rbp,%rax,4),%eax
  add    %eax,-0x100014(%rbp)
  addl   $0x1,-0x100018(%rbp)
L76:
  mov    -0x100018(%rbp),%eax
  cmp    $0x3ffff,%eax
  jbe    L40
  mov    -0x100014(%rbp),%eax
L57:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L30
  call   <__stack_chk_fail@plt>
L30:
  leave
  ret
```

assistant: ```c
int rec(int depth) {
  int i, zzz;
  int arr[(1048576 / sizeof(int))];
  if (depth == 0)
    return 0;
  for (i = 0; i < (1048576 / sizeof(int)); i++)
    arr[i] = i * depth;
  zzz = rec(depth - 1);
  for (i = 0; i < (1048576 / sizeof(int)); i++)
    zzz += arr[i];
  return zzz;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<rec>:
  endbr64
  push   %rbp
  push   %rbx
  lea    -0x100000(%rsp),%r11
L17:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L17
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x100008(%rsp)
  xor    %eax,%eax
  mov    %edi,%eax
  test   %edi,%edi
  je     L23
  mov    %rsp,%rbx
  lea    0x100000(%rsp),%rbp
  mov    %rbx,%rax
  mov    $0x0,%edx
L39:
  mov    %edx,(%rax)
  add    %edi,%edx
  add    $0x4,%rax
  cmp    %rbp,%rax
  jne    L39
  sub    $0x1,%edi
  call   <rec>
L43:
  add    (%rbx),%eax
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L43
L23:
  mov    0x100008(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L37
  add    $0x100018,%rsp
  pop    %rbx
  pop    %rbp
  ret
L37:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int rec(int depth) {
  int i, zzz;
  int arr[(1048576 / sizeof(int))];
  if (depth == 0)
    return 0;
  for (i = 0; i < (1048576 / sizeof(int)); i++)
    arr[i] = i * depth;
  zzz = rec(depth - 1);
  for (i = 0; i < (1048576 / sizeof(int)); i++)
    zzz += arr[i];
  return zzz;
}
```
user: Translate the below assembly code into C programming language:

```asm
<rec>:
  endbr64
  push   %rbp
  push   %rbx
  lea    -0x100000(%rsp),%r11
L36:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L36
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x100008(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  je     L0
  mov    %rsp,%rbx
  lea    0x100000(%rsp),%rbp
  xor    %edx,%edx
  mov    %rbx,%rax
  nopw   0x0(%rax,%rax,1)
L13:
  mov    %edx,(%rax)
  add    $0x4,%rax
  add    %edi,%edx
  cmp    %rbp,%rax
  jne    L13
  sub    $0x1,%edi
  call   <rec>
  nopl   (%rax)
L41:
  add    (%rbx),%eax
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L41
L0:
  mov    0x100008(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L49
  add    $0x100018,%rsp
  pop    %rbx
  pop    %rbp
  ret
L49:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int rec(int depth) {
  int i, zzz;
  int arr[(1048576 / sizeof(int))];
  if (depth == 0)
    return 0;
  for (i = 0; i < (1048576 / sizeof(int)); i++)
    arr[i] = i * depth;
  zzz = rec(depth - 1);
  for (i = 0; i < (1048576 / sizeof(int)); i++)
    zzz += arr[i];
  return zzz;
}
```
user: What is the c source code of the assembly code below:

```asm
<double_transposed_direct_form_2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  movsd  %xmm0,-0x30(%rbp)
  mov    %rsi,-0x38(%rbp)
  mov    %rdx,-0x40(%rbp)
  mov    %ecx,-0x44(%rbp)
  mov    %r8d,-0x48(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x38(%rbp),%rax
  add    $0x8,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x48(%rbp),%edx
  mov    -0x44(%rbp),%eax
  cmp    %eax,%edx
  cmovge %edx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movsd  (%rax),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x30(%rbp),%xmm1
  mov    -0x28(%rbp),%rax
  movsd  (%rax),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x38(%rbp),%rax
  movsd  (%rax),%xmm1
  movsd  -0x8(%rbp),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L16
L22:
  mov    -0x20(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x44(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x20(%rbp)
  jge    L14
  mov    -0x18(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movsd  (%rax),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x8(%rbp),%xmm1
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
L14:
  mov    -0x48(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x20(%rbp)
  jge    L15
  mov    -0x10(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movsd  (%rax),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x30(%rbp),%xmm1
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
L15:
  addl   $0x1,-0x20(%rbp)
L16:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x20(%rbp)
  jl     L22
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double double_transposed_direct_form_2(double w[], double x, double a[], double b[], int Na, int Nb) {
  double *a_ptr, *b_ptr;
  double yout = 0;
  a_ptr = &a[1];
  b_ptr = &b[0];
  int Nw = Na > Nb ? Na : Nb;
  yout = (*b_ptr++ * x) + w[0];
  yout = yout / a[0];
  int j;
  for (j = 0; j < Nw - 1; j++) {
    w[j] = w[j + 1];
    if (j < Na - 1) {
      w[j] -= *a_ptr++ * yout;
    }
    if (j < Nb - 1) {
      w[j] += *b_ptr++ * x;
    }
  }
  return yout;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<double_transposed_direct_form_2>:
  endbr64
  mov    %rdi,%rax
  movapd %xmm0,%xmm3
  mov    %rsi,%r11
  mov    %ecx,%esi
  lea    0x8(%r11),%rdi
  cmp    %ecx,%r8d
  mov    %ecx,%r10d
  cmovge %r8d,%r10d
  lea    0x8(%rdx),%r9
  mulsd  (%rdx),%xmm0
  addsd  (%rax),%xmm0
  divsd  (%r11),%xmm0
  cmp    $0x1,%r10d
  jle    L9
  sub    $0x1,%r10d
  mov    $0x0,%edx
  sub    $0x1,%esi
  sub    $0x1,%r8d
  jmp    L11
L44:
  add    $0x1,%edx
  add    $0x8,%rax
  cmp    %r10d,%edx
  je     L9
L11:
  mov    %rax,%rcx
  movsd  0x8(%rax),%xmm1
  movsd  %xmm1,(%rax)
  cmp    %edx,%esi
  jle    L55
  movapd %xmm0,%xmm2
  mulsd  (%rdi),%xmm2
  subsd  %xmm2,%xmm1
  movsd  %xmm1,(%rax)
  lea    0x8(%rdi),%rdi
L55:
  cmp    %edx,%r8d
  jle    L44
  movapd %xmm3,%xmm1
  mulsd  (%r9),%xmm1
  addsd  (%rcx),%xmm1
  movsd  %xmm1,(%rcx)
  lea    0x8(%r9),%r9
  jmp    L44
L9:
  ret
```

assistant: ```c
double double_transposed_direct_form_2(double w[], double x, double a[], double b[], int Na, int Nb) {
  double *a_ptr, *b_ptr;
  double yout = 0;
  a_ptr = &a[1];
  b_ptr = &b[0];
  int Nw = Na > Nb ? Na : Nb;
  yout = (*b_ptr++ * x) + w[0];
  yout = yout / a[0];
  int j;
  for (j = 0; j < Nw - 1; j++) {
    w[j] = w[j + 1];
    if (j < Na - 1) {
      w[j] -= *a_ptr++ * yout;
    }
    if (j < Nb - 1) {
      w[j] += *b_ptr++ * x;
    }
  }
  return yout;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<double_transposed_direct_form_2>:
  endbr64
  movapd %xmm0,%xmm3
  movsd  (%rdx),%xmm0
  mov    %rsi,%r9
  mov    %rdi,%rax
  cmp    %ecx,%r8d
  lea    0x8(%rsi),%rsi
  lea    0x8(%rdx),%rdi
  mulsd  %xmm3,%xmm0
  addsd  (%rax),%xmm0
  divsd  (%r9),%xmm0
  mov    %ecx,%r9d
  cmovge %r8d,%r9d
  sub    $0x1,%r9d
  test   %r9d,%r9d
  jle    L25
  sub    $0x1,%ecx
  sub    $0x1,%r8d
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L18:
  movsd  0x8(%rax),%xmm1
  movsd  %xmm1,(%rax)
  cmp    %ecx,%edx
  jge    L29
  movsd  (%rsi),%xmm2
  add    $0x8,%rsi
  mulsd  %xmm0,%xmm2
  subsd  %xmm2,%xmm1
  movsd  %xmm1,(%rax)
L29:
  cmp    %r8d,%edx
  jge    L39
  movsd  (%rdi),%xmm1
  add    $0x8,%rdi
  mulsd  %xmm3,%xmm1
  addsd  (%rax),%xmm1
  movsd  %xmm1,(%rax)
L39:
  add    $0x1,%edx
  add    $0x8,%rax
  cmp    %r9d,%edx
  jne    L18
L25:
  ret
```

assistant: ```c
double double_transposed_direct_form_2(double w[], double x, double a[], double b[], int Na, int Nb) {
  double *a_ptr, *b_ptr;
  double yout = 0;
  a_ptr = &a[1];
  b_ptr = &b[0];
  int Nw = Na > Nb ? Na : Nb;
  yout = (*b_ptr++ * x) + w[0];
  yout = yout / a[0];
  int j;
  for (j = 0; j < Nw - 1; j++) {
    w[j] = w[j + 1];
    if (j < Na - 1) {
      w[j] -= *a_ptr++ * yout;
    }
    if (j < Nb - 1) {
      w[j] += *b_ptr++ * x;
    }
  }
  return yout;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<double_transposed_direct_form_2>:
  endbr64
  cmp    %ecx,%r8d
  mov    %ecx,%r10d
  movapd %xmm0,%xmm3
  movsd  (%rdx),%xmm0
  cmovge %r8d,%r10d
  mov    %rsi,%r9
  lea    0x8(%rdx),%rsi
  mulsd  %xmm3,%xmm0
  sub    $0x1,%r10d
  addsd  (%rdi),%xmm0
  divsd  (%r9),%xmm0
  test   %r10d,%r10d
  jle    L68
  sub    $0x1,%ecx
  sub    $0x1,%r8d
  test   %ecx,%ecx
  jle    L2
  cmp    %ecx,%r10d
  cmovle %r10d,%ecx
  xor    %eax,%eax
  nop
L14:
  movsd  0x8(%rdi,%rax,8),%xmm1
  movsd  %xmm1,(%rdi,%rax,8)
  movsd  0x8(%r9,%rax,8),%xmm2
  mulsd  %xmm0,%xmm2
  subsd  %xmm2,%xmm1
  movsd  %xmm1,(%rdi,%rax,8)
  cmp    %eax,%r8d
  jle    L29
  movsd  (%rsi),%xmm2
  add    $0x8,%rsi
  mulsd  %xmm3,%xmm2
  addsd  %xmm1,%xmm2
  movsd  %xmm2,(%rdi,%rax,8)
L29:
  lea    0x1(%rax),%edx
  add    $0x1,%rax
  cmp    %eax,%ecx
  jg     L14
  cmp    %r10d,%edx
  jge    L58
L38:
  movslq %edx,%rax
  nopl   0x0(%rax)
L26:
  movsd  0x8(%rdi,%rax,8),%xmm1
  movsd  %xmm1,(%rdi,%rax,8)
  cmp    %eax,%r8d
  jle    L46
  movsd  (%rsi),%xmm2
  add    $0x8,%rsi
  mulsd  %xmm3,%xmm2
  addsd  %xmm1,%xmm2
  movsd  %xmm2,(%rdi,%rax,8)
L46:
  add    $0x1,%rax
  cmp    %eax,%r10d
  jg     L26
L68:
  ret
  nop
L2:
  xor    %edx,%edx
  jmp    L38
  nopl   0x0(%rax)
L58:
  ret
```

assistant: ```c
double double_transposed_direct_form_2(double w[], double x, double a[], double b[], int Na, int Nb) {
  double *a_ptr, *b_ptr;
  double yout = 0;
  a_ptr = &a[1];
  b_ptr = &b[0];
  int Nw = Na > Nb ? Na : Nb;
  yout = (*b_ptr++ * x) + w[0];
  yout = yout / a[0];
  int j;
  for (j = 0; j < Nw - 1; j++) {
    w[j] = w[j + 1];
    if (j < Na - 1) {
      w[j] -= *a_ptr++ * yout;
    }
    if (j < Nb - 1) {
      w[j] += *b_ptr++ * x;
    }
  }
  return yout;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<get_choice>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <get_first@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L13
L26:
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <get_first@plt>
  mov    %eax,-0x4(%rbp)
L13:
  cmpl   $0x60,-0x4(%rbp)
  jle    L5
  cmpl   $0x63,-0x4(%rbp)
  jle    L9
L5:
  cmpl   $0x71,-0x4(%rbp)
  jne    L26
L9:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D7
type_infered: string
name: read_data
arguments: label: D28
type_infered: string
name: read_data
arguments: label: D0
type_infered: string
name: read_data
arguments: label: D36
type_infered: string

tool: D7: Enter the letter of your choice: 


tool: D28: a. advice             b. bell


tool: D0: c. count              q. quit


tool: D36: Please respond with a, b, c or q.


assistant: ```c
char get_choice(void) {
  int ch;
  printf("Enter the letter of your choice: \n");
  printf("a. advice             b. bell\n");
  printf("c. count              q. quit\n");
  ch = get_first();
  while ((ch < 'a' || ch > 'c') && ch != 'q') {
    printf("Please respond with a, b, c or q.\n");
    ch = get_first();
  }
  return ch;
}
```
user: Translate the below assembly code into C programming language:

```asm
<get_choice>:
  endbr64
  push   %rbx
  lea    D5(%rip),%rdi
  call   <printf@plt>
  lea    D19(%rip),%rdi
  call   <printf@plt>
  lea    D34(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <get_first@plt>
  lea    -0x61(%rax),%edx
  cmp    $0x2,%edx
  jbe    L28
  cmp    $0x71,%eax
  je     L28
  lea    D36(%rip),%rbx
L36:
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <get_first@plt>
  lea    -0x61(%rax),%edx
  cmp    $0x2,%edx
  jbe    L28
  cmp    $0x71,%eax
  jne    L36
L28:
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
type: string
name: retrieve_data
arguments: data_tag: D19
type: string
name: retrieve_data
arguments: data_tag: D34
type: string
name: retrieve_data
arguments: data_tag: D36
type: string

tool: D5: Enter the letter of your choice: 


tool: D19: a. advice             b. bell


tool: D34: c. count              q. quit


tool: D36: Please respond with a, b, c or q.


assistant: ```c
char get_choice(void) {
  int ch;
  printf("Enter the letter of your choice: \n");
  printf("a. advice             b. bell\n");
  printf("c. count              q. quit\n");
  ch = get_first();
  while ((ch < 'a' || ch > 'c') && ch != 'q') {
    printf("Please respond with a, b, c or q.\n");
    ch = get_first();
  }
  return ch;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_choice>:
  endbr64
  push   %rbx
  lea    D35(%rip),%rdi
  call   <printf@plt>
  lea    D26(%rip),%rdi
  call   <printf@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <get_first@plt>
  lea    -0x61(%rax),%edx
  cmp    $0x2,%edx
  jbe    L9
  cmp    $0x71,%eax
  je     L9
  lea    D39(%rip),%rbx
  nopl   0x0(%rax)
L33:
  mov    %rbx,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <get_first@plt>
  lea    -0x61(%rax),%edx
  cmp    $0x2,%edx
  jbe    L9
  cmp    $0x71,%eax
  jne    L33
L9:
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D35
type_guessed: string
name: unpack_data
arguments: data_label: D26
type_guessed: string
name: unpack_data
arguments: data_label: D0
type_guessed: string
name: unpack_data
arguments: data_label: D39
type_guessed: string

tool: D35: Enter the letter of your choice: 


tool: D26: a. advice             b. bell


tool: D0: c. count              q. quit


tool: D39: Please respond with a, b, c or q.


assistant: ```c
char get_choice(void) {
  int ch;
  printf("Enter the letter of your choice: \n");
  printf("a. advice             b. bell\n");
  printf("c. count              q. quit\n");
  ch = get_first();
  while ((ch < 'a' || ch > 'c') && ch != 'q') {
    printf("Please respond with a, b, c or q.\n");
    ch = get_first();
  }
  return ch;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_choice>:
  endbr64
  push   %rbx
  lea    D1(%rip),%rdi
  call   <printf@plt>
  lea    D29(%rip),%rdi
  call   <printf@plt>
  lea    D10(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <get_first@plt>
  lea    -0x61(%rax),%edx
  cmp    $0x2,%edx
  jbe    L6
  cmp    $0x71,%eax
  je     L6
  lea    D12(%rip),%rbx
  nopl   0x0(%rax)
L34:
  mov    %rbx,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <get_first@plt>
  lea    -0x61(%rax),%edx
  cmp    $0x2,%edx
  jbe    L6
  cmp    $0x71,%eax
  jne    L34
L6:
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: tag: D1
type: string
name: read_data
arguments: tag: D29
type: string
name: read_data
arguments: tag: D10
type: string
name: read_data
arguments: tag: D12
type: string

tool: D1: Enter the letter of your choice: 


tool: D29: a. advice             b. bell


tool: D10: c. count              q. quit


tool: D12: Please respond with a, b, c or q.


assistant: ```c
char get_choice(void) {
  int ch;
  printf("Enter the letter of your choice: \n");
  printf("a. advice             b. bell\n");
  printf("c. count              q. quit\n");
  ch = get_first();
  while ((ch < 'a' || ch > 'c') && ch != 'q') {
    printf("Please respond with a, b, c or q.\n");
    ch = get_first();
  }
  return ch;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<search>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x9,-0x8(%rbp)
  jmp    L61
L6:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x1c(%rbp)
  jge    L11
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L61
L11:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x1c(%rbp)
  jle    L49
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L61
L49:
  mov    -0x4(%rbp),%eax
  jmp    L26
L61:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L6
  mov    $0xffffffff,%eax
L26:
  pop    %rbp
  ret
```

assistant: ```c
int search(int A[10], int x) {
  int baixo = 0, alto = 10 - 1, meio;
  while (baixo <= alto) {
    meio = (baixo + alto) / 2;
    if (x < A[meio])
      alto = meio - 1;
    else if (x > A[meio])
      baixo = meio + 1;
    else
      return meio;
  }
  return -1;
}
```