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

```asm
<BSP_Button2_Init>:
  endbr64
  mov    0x400fe608,%eax
  or     $0x8,%eax
  mov    %eax,0x400fe608
L2:
  mov    0x400fea08,%eax
  test   $0x8,%al
  je     L2
  movl   $0x4c4f434b,0x40007520
  movl   $0xff,0x40007524
  mov    0x40007528,%eax
  and    $0x7f,%al
  mov    %eax,0x40007528
  mov    0x4000752c,%eax
  and    $0xfffffff,%eax
  mov    %eax,0x4000752c
  mov    0x40007400,%eax
  and    $0x7f,%al
  mov    %eax,0x40007400
  mov    0x40007420,%eax
  and    $0x7f,%al
  mov    %eax,0x40007420
  mov    0x40007510,%eax
  and    $0x7f,%al
  mov    %eax,0x40007510
  mov    0x4000751c,%eax
  or     $0x80,%al
  mov    %eax,0x4000751c
  ret
```

assistant: ```c
void BSP_Button2_Init(void) {
  (*((volatile uint32_t *)0x400FE608)) |= 0x00000008;
  while (((*((volatile uint32_t *)0x400FEA08)) & 0x08) == 0) {
  };
  (*((volatile uint32_t *)0x40007520)) = 0x4C4F434B;
  (*((volatile uint32_t *)0x40007524)) = 0xFF;
  (*((volatile uint32_t *)0x40007528)) &= ~0x80;
  (*((volatile uint32_t *)0x4000752C)) = ((*((volatile uint32_t *)0x4000752C)) & 0x0FFFFFFF) + 0x00000000;
  (*((volatile uint32_t *)0x40007400)) &= ~0x80;
  (*((volatile uint32_t *)0x40007420)) &= ~0x80;
  (*((volatile uint32_t *)0x40007510)) &= ~0x80;
  (*((volatile uint32_t *)0x4000751C)) |= 0x80;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<BSP_Button2_Init>:
  endbr64
  mov    0x400fe608,%eax
  or     $0x8,%eax
  mov    %eax,0x400fe608
  nopl   (%rax)
L5:
  mov    0x400fea08,%eax
  test   $0x8,%al
  je     L5
  movl   $0x4c4f434b,0x40007520
  movl   $0xff,0x40007524
  mov    0x40007528,%eax
  and    $0x7f,%al
  mov    %eax,0x40007528
  mov    0x4000752c,%eax
  and    $0xfffffff,%eax
  mov    %eax,0x4000752c
  mov    0x40007400,%eax
  and    $0x7f,%al
  mov    %eax,0x40007400
  mov    0x40007420,%eax
  and    $0x7f,%al
  mov    %eax,0x40007420
  mov    0x40007510,%eax
  and    $0x7f,%al
  mov    %eax,0x40007510
  mov    0x4000751c,%eax
  or     $0x80,%al
  mov    %eax,0x4000751c
  ret
```

assistant: ```c
void BSP_Button2_Init(void) {
  (*((volatile uint32_t *)0x400FE608)) |= 0x00000008;
  while (((*((volatile uint32_t *)0x400FEA08)) & 0x08) == 0) {
  };
  (*((volatile uint32_t *)0x40007520)) = 0x4C4F434B;
  (*((volatile uint32_t *)0x40007524)) = 0xFF;
  (*((volatile uint32_t *)0x40007528)) &= ~0x80;
  (*((volatile uint32_t *)0x4000752C)) = ((*((volatile uint32_t *)0x4000752C)) & 0x0FFFFFFF) + 0x00000000;
  (*((volatile uint32_t *)0x40007400)) &= ~0x80;
  (*((volatile uint32_t *)0x40007420)) &= ~0x80;
  (*((volatile uint32_t *)0x40007510)) &= ~0x80;
  (*((volatile uint32_t *)0x4000751C)) |= 0x80;
}
```
user: Translate the below assembly code into C programming language:

```asm
<BSP_Button2_Init>:
  endbr64
  mov    0x400fe608,%eax
  or     $0x8,%eax
  mov    %eax,0x400fe608
  nopl   (%rax)
L2:
  mov    0x400fea08,%eax
  test   $0x8,%al
  je     L2
  movl   $0x4c4f434b,0x40007520
  movl   $0xff,0x40007524
  mov    0x40007528,%eax
  and    $0x7f,%al
  mov    %eax,0x40007528
  mov    0x4000752c,%eax
  and    $0xfffffff,%eax
  mov    %eax,0x4000752c
  mov    0x40007400,%eax
  and    $0x7f,%al
  mov    %eax,0x40007400
  mov    0x40007420,%eax
  and    $0x7f,%al
  mov    %eax,0x40007420
  mov    0x40007510,%eax
  and    $0x7f,%al
  mov    %eax,0x40007510
  mov    0x4000751c,%eax
  or     $0x80,%al
  mov    %eax,0x4000751c
  ret
```

assistant: ```c
void BSP_Button2_Init(void) {
  (*((volatile uint32_t *)0x400FE608)) |= 0x00000008;
  while (((*((volatile uint32_t *)0x400FEA08)) & 0x08) == 0) {
  };
  (*((volatile uint32_t *)0x40007520)) = 0x4C4F434B;
  (*((volatile uint32_t *)0x40007524)) = 0xFF;
  (*((volatile uint32_t *)0x40007528)) &= ~0x80;
  (*((volatile uint32_t *)0x4000752C)) = ((*((volatile uint32_t *)0x4000752C)) & 0x0FFFFFFF) + 0x00000000;
  (*((volatile uint32_t *)0x40007400)) &= ~0x80;
  (*((volatile uint32_t *)0x40007420)) &= ~0x80;
  (*((volatile uint32_t *)0x40007510)) &= ~0x80;
  (*((volatile uint32_t *)0x4000751C)) |= 0x80;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<bpp_walk_left>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x2c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L52
L59:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jge    L54
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%eax
  shl    $0x3,%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
L54:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jle    L28
  addl   $0x1,-0x14(%rbp)
  jmp    L35
L28:
  movl   $0x0,-0x14(%rbp)
L35:
  mov    -0x14(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jle    L53
  mov    -0xc(%rbp),%eax
  jmp    L29
L53:
  subl   $0x1,-0x10(%rbp)
L52:
  cmpl   $0x0,-0x10(%rbp)
  jns    L59
  mov    $0x0,%eax
L29:
  pop    %rbp
  ret
```

assistant: ```c
int bpp_walk_left(int *H, int x0, int tol) {
  int downcount = 0;
  int x;
  int peakx, cmax, mt = 0;
  cmax = H[x0];
  peakx = x0;
  for (x = x0 - 1; x >= 0; x--) {
    if (H[x] > cmax) {
      cmax = H[x];
      peakx = x;
      mt = (8 * cmax) / 10;
    }
    if (H[x] < mt)
      ++downcount;
    else
      downcount = 0;
    if (downcount > tol)
      return peakx;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bpp_walk_left>:
  endbr64
  mov    %rdi,%r10
  mov    %esi,%eax
  mov    %edx,%esi
  movslq %eax,%rdx
  mov    (%rdi,%rdx,4),%r9d
  mov    %eax,%ecx
  sub    $0x1,%ecx
  js     L11
  movslq %ecx,%rcx
  mov    $0x0,%edi
  mov    $0x0,%edx
  mov    $0x0,%r11d
  jmp    L44
L37:
  add    $0x1,%edx
  cmp    %edi,%r8d
  cmovge %r11d,%edx
  cmp    %esi,%edx
  jg     L1
  sub    $0x1,%rcx
  test   %ecx,%ecx
  js     L42
L44:
  mov    (%r10,%rcx,4),%r8d
  cmp    %r9d,%r8d
  jle    L37
  lea    0x0(,%r8,8),%eax
  movslq %eax,%rdi
  imul   $0x66666667,%rdi,%rdi
  sar    $0x22,%rdi
  sar    $0x1f,%eax
  sub    %eax,%edi
  mov    %r8d,%r9d
  mov    %ecx,%eax
  jmp    L37
L42:
  mov    $0x0,%eax
  ret
L11:
  mov    $0x0,%eax
L1:
  ret
```

assistant: ```c
int bpp_walk_left(int *H, int x0, int tol) {
  int downcount = 0;
  int x;
  int peakx, cmax, mt = 0;
  cmax = H[x0];
  peakx = x0;
  for (x = x0 - 1; x >= 0; x--) {
    if (H[x] > cmax) {
      cmax = H[x];
      peakx = x;
      mt = (8 * cmax) / 10;
    }
    if (H[x] < mt)
      ++downcount;
    else
      downcount = 0;
    if (downcount > tol)
      return peakx;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bpp_walk_left>:
  endbr64
  mov    %esi,%eax
  mov    %edx,%esi
  mov    %rdi,%r10
  movslq %eax,%rdx
  mov    %eax,%ecx
  mov    (%rdi,%rdx,4),%r9d
  sub    $0x1,%ecx
  js     L15
  movslq %ecx,%rcx
  xor    %edi,%edi
  xor    %edx,%edx
  xor    %r11d,%r11d
  jmp    L7
  nopl   (%rax)
L1:
  sub    $0x1,%rcx
  test   %ecx,%ecx
  js     L15
L7:
  mov    (%r10,%rcx,4),%r8d
  cmp    %r9d,%r8d
  jle    L4
  lea    0x0(,%r8,8),%eax
  mov    %r8d,%r9d
  movslq %eax,%rdi
  sar    $0x1f,%eax
  imul   $0x66666667,%rdi,%rdi
  sar    $0x22,%rdi
  sub    %eax,%edi
  mov    %ecx,%eax
L4:
  add    $0x1,%edx
  cmp    %edi,%r8d
  cmovge %r11d,%edx
  cmp    %esi,%edx
  jle    L1
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  xor    %eax,%eax
  ret
```

assistant: ```c
int bpp_walk_left(int *H, int x0, int tol) {
  int downcount = 0;
  int x;
  int peakx, cmax, mt = 0;
  cmax = H[x0];
  peakx = x0;
  for (x = x0 - 1; x >= 0; x--) {
    if (H[x] > cmax) {
      cmax = H[x];
      peakx = x;
      mt = (8 * cmax) / 10;
    }
    if (H[x] < mt)
      ++downcount;
    else
      downcount = 0;
    if (downcount > tol)
      return peakx;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<bpp_walk_left>:
  endbr64
  mov    %esi,%eax
  mov    %edx,%esi
  mov    %rdi,%r10
  movslq %eax,%rdx
  mov    %eax,%ecx
  mov    (%rdi,%rdx,4),%r9d
  sub    $0x1,%ecx
  js     L20
  movslq %ecx,%rcx
  xor    %edi,%edi
  xor    %edx,%edx
  xor    %r11d,%r11d
  jmp    L42
  nopl   (%rax)
L4:
  sub    $0x1,%rcx
  test   %ecx,%ecx
  js     L20
L42:
  mov    (%r10,%rcx,4),%r8d
  cmp    %r9d,%r8d
  jle    L30
  lea    0x0(,%r8,8),%eax
  mov    %r8d,%r9d
  movslq %eax,%rdi
  sar    $0x1f,%eax
  imul   $0x66666667,%rdi,%rdi
  sar    $0x22,%rdi
  sub    %eax,%edi
  mov    %ecx,%eax
L30:
  add    $0x1,%edx
  cmp    %edi,%r8d
  cmovge %r11d,%edx
  cmp    %esi,%edx
  jle    L4
  ret
  nopl   0x0(%rax,%rax,1)
L20:
  xor    %eax,%eax
  ret
```

assistant: ```c
int bpp_walk_left(int *H, int x0, int tol) {
  int downcount = 0;
  int x;
  int peakx, cmax, mt = 0;
  cmax = H[x0];
  peakx = x0;
  for (x = x0 - 1; x >= 0; x--) {
    if (H[x] > cmax) {
      cmax = H[x];
      peakx = x;
      mt = (8 * cmax) / 10;
    }
    if (H[x] < mt)
      ++downcount;
    else
      downcount = 0;
    if (downcount > tol)
      return peakx;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<pe_as_lit_id_map_streq>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  je     L43
  mov    $0x0,%eax
  jmp    L44
L46:
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x10(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <pe_as_lit_id_map_chareq@plt>
  test   %eax,%eax
  jne    L43
  mov    $0x0,%eax
  jmp    L44
L43:
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  test   %rax,%rax
  jne    L46
  mov    $0x1,%eax
L44:
  leave
  ret
```

assistant: ```c
int pe_as_lit_id_map_streq(size_t al, char const *a, size_t bl, char const *b) {
  if (al != bl)
    return 0;
  while (al-- != 0) {
    if (!pe_as_lit_id_map_chareq(*a++, *b++))
      return 0;
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<pe_as_lit_id_map_streq>:
  endbr64
  mov    $0x0,%eax
  cmp    %rdx,%rdi
  jne    L34
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    %rcx,%rbp
  lea    (%rsi,%rdi,1),%r12
L17:
  cmp    %r12,%rbx
  je     L23
  add    $0x1,%rbp
  add    $0x1,%rbx
  movsbl -0x1(%rbp),%esi
  movsbl -0x1(%rbx),%edi
  call   <pe_as_lit_id_map_chareq@plt>
  test   %eax,%eax
  jne    L17
  jmp    L1
L23:
  mov    $0x1,%eax
L1:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L34:
  ret
```

assistant: ```c
int pe_as_lit_id_map_streq(size_t al, char const *a, size_t bl, char const *b) {
  if (al != bl)
    return 0;
  while (al-- != 0) {
    if (!pe_as_lit_id_map_chareq(*a++, *b++))
      return 0;
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<pe_as_lit_id_map_streq>:
  endbr64
  cmp    %rdx,%rdi
  je     L28
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L28:
  push   %r12
  lea    (%rsi,%rdi,1),%r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L39
L33:
  movsbl 0x0(%rbp),%esi
  movsbl (%rbx),%edi
  add    $0x1,%rbp
  add    $0x1,%rbx
  call   <pe_as_lit_id_map_chareq@plt>
  test   %eax,%eax
  je     L1
L39:
  cmp    %r12,%rbx
  jne    L33
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int pe_as_lit_id_map_streq(size_t al, char const *a, size_t bl, char const *b) {
  if (al != bl)
    return 0;
  while (al-- != 0) {
    if (!pe_as_lit_id_map_chareq(*a++, *b++))
      return 0;
  }
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<pe_as_lit_id_map_streq>:
  endbr64
  cmp    %rdx,%rdi
  je     L30
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L30:
  push   %r12
  lea    (%rsi,%rdi,1),%r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L12
L3:
  movsbl 0x0(%rbp),%esi
  movsbl (%rbx),%edi
  add    $0x1,%rbp
  add    $0x1,%rbx
  call   <pe_as_lit_id_map_chareq@plt>
  test   %eax,%eax
  je     L17
L12:
  cmp    %r12,%rbx
  jne    L3
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int pe_as_lit_id_map_streq(size_t al, char const *a, size_t bl, char const *b) {
  if (al != bl)
    return 0;
  while (al-- != 0) {
    if (!pe_as_lit_id_map_chareq(*a++, *b++))
      return 0;
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <report_start@plt>
  mov    $0x0,%eax
  call   <base_test@plt>
  mov    $0x80000b00,%edi
  call   <greth_test@plt>
  mov    $0x80000c00,%edi
  call   <greth_test@plt>
  mov    $0x0,%eax
  call   <report_end@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  report_start();
  base_test();
  greth_test(0x80000b00);
  greth_test(0x80000c00);
  report_end();
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <report_start@plt>
  mov    $0x0,%eax
  call   <base_test@plt>
  mov    $0x80000b00,%edi
  call   <greth_test@plt>
  mov    $0x80000c00,%edi
  call   <greth_test@plt>
  mov    $0x0,%eax
  call   <report_end@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  report_start();
  base_test();
  greth_test(0x80000b00);
  greth_test(0x80000c00);
  report_end();
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <report_start@plt>
  xor    %eax,%eax
  call   <base_test@plt>
  mov    $0x80000b00,%edi
  call   <greth_test@plt>
  mov    $0x80000c00,%edi
  call   <greth_test@plt>
  xor    %eax,%eax
  call   <report_end@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  report_start();
  base_test();
  greth_test(0x80000b00);
  greth_test(0x80000c00);
  report_end();
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <report_start@plt>
  xor    %eax,%eax
  call   <base_test@plt>
  mov    $0x80000b00,%edi
  call   <greth_test@plt>
  mov    $0x80000c00,%edi
  call   <greth_test@plt>
  xor    %eax,%eax
  call   <report_end@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  report_start();
  base_test();
  greth_test(0x80000b00);
  greth_test(0x80000c00);
  report_end();
}
```
user: What is the c source code of the assembly code below:

```asm
<switch_prob>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  sub    $0x3c,%rax
  cmp    $0x5,%rax
  ja     L11
  lea    0x0(,%rax,4),%rdx
  lea    D8(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D8(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L13
  mov    -0x18(%rbp),%rax
  sar    $0x3,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L13
  mov    -0x18(%rbp),%rax
  mov    %eax,%edx
  mov    $0x4,%eax
  sub    %edx,%eax
  mov    %eax,%ecx
  shlq   %cl,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  imul   %rax,%rax
  mov    %rax,-0x18(%rbp)
L11:
  mov    -0x18(%rbp),%rax
  add    $0x4b,%rax
  mov    %rax,-0x8(%rbp)
L13:
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long switch_prob(long x, long n) {
  long result = x;
  switch (n) {
  case 60:
  case 62:
    result = x * 8;
    break;
  case 63:
    result = x >> 3;
    break;
  case 64:
    x = x << 4 - x;
  case 65:
    x = x * x;
  default:
    result = x + 0x4B;
  }
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<switch_prob>:
  endbr64
  sub    $0x3c,%rsi
  cmp    $0x5,%rsi
  ja     L0
  lea    D9(%rip),%rdx
  movslq (%rdx,%rsi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    0x0(,%rdi,8),%rax
  ret
  mov    %rdi,%rax
  sar    $0x3,%rax
  ret
  mov    $0x4,%ecx
  sub    %edi,%ecx
  shl    %cl,%rdi
  imul   %rdi,%rdi
L0:
  lea    0x4b(%rdi),%rax
  ret
```

assistant: ```c
long switch_prob(long x, long n) {
  long result = x;
  switch (n) {
  case 60:
  case 62:
    result = x * 8;
    break;
  case 63:
    result = x >> 3;
    break;
  case 64:
    x = x << 4 - x;
  case 65:
    x = x * x;
  default:
    result = x + 0x4B;
  }
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<switch_prob>:
  endbr64
  sub    $0x3c,%rsi
  cmp    $0x5,%rsi
  ja     L7
  lea    D2(%rip),%rdx
  movslq (%rdx,%rsi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nop
  mov    $0x4,%ecx
  sub    %edi,%ecx
  shl    %cl,%rdi
  imul   %rdi,%rdi
L7:
  lea    0x4b(%rdi),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0x0(,%rdi,8),%rax
  ret
  nopl   0x0(%rax)
  mov    %rdi,%rax
  sar    $0x3,%rax
  ret
```

assistant: ```c
long switch_prob(long x, long n) {
  long result = x;
  switch (n) {
  case 60:
  case 62:
    result = x * 8;
    break;
  case 63:
    result = x >> 3;
    break;
  case 64:
    x = x << 4 - x;
  case 65:
    x = x * x;
  default:
    result = x + 0x4B;
  }
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<switch_prob>:
  endbr64
  sub    $0x3c,%rsi
  cmp    $0x5,%rsi
  ja     L3
  lea    D4(%rip),%rdx
  movslq (%rdx,%rsi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nop
  mov    $0x4,%ecx
  sub    %edi,%ecx
  shl    %cl,%rdi
  imul   %rdi,%rdi
L3:
  lea    0x4b(%rdi),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    0x0(,%rdi,8),%rax
  ret
  nopl   0x0(%rax)
  mov    %rdi,%rax
  sar    $0x3,%rax
  ret
```

assistant: ```c
long switch_prob(long x, long n) {
  long result = x;
  switch (n) {
  case 60:
  case 62:
    result = x * 8;
    break;
  case 63:
    result = x >> 3;
    break;
  case 64:
    x = x << 4 - x;
  case 65:
    x = x * x;
  default:
    result = x + 0x4B;
  }
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <ghdl_main@plt>
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

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

tool: D0: ghdl_main: %d


assistant: ```c
int main(int argc, char **argv) {
  printf("ghdl_main: %d\n", ghdl_main(argc, argv));
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  call   <ghdl_main@plt>
  mov    %eax,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D4
type_infered: string

tool: D4: ghdl_main: %d


assistant: ```c
int main(int argc, char **argv) {
  printf("ghdl_main: %d\n", ghdl_main(argc, argv));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  call   <ghdl_main@plt>
  lea    D0(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D0
type_guessed: string

tool: D0: ghdl_main: %d


assistant: ```c
int main(int argc, char **argv) {
  printf("ghdl_main: %d\n", ghdl_main(argc, argv));
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  call   <ghdl_main@plt>
  lea    D7(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
type_guessed: string

tool: D7: ghdl_main: %d


assistant: ```c
int main(int argc, char **argv) {
  printf("ghdl_main: %d\n", ghdl_main(argc, argv));
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<write16>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    -0xc(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    -0xc(%rbp),%edx
  sar    $0x8,%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void write16(uint8_t *p, uint16_t value) {
  p[0] = value;
  p[1] = value >> 8;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<write16>:
  endbr64
  mov    %esi,(%rdi)
  sar    $0x8,%esi
  mov    %esi,0x4(%rdi)
  ret
```

assistant: ```c
void write16(uint8_t *p, uint16_t value) {
  p[0] = value;
  p[1] = value >> 8;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<write16>:
  endbr64
  mov    %esi,(%rdi)
  sar    $0x8,%esi
  mov    %esi,0x4(%rdi)
  ret
```

assistant: ```c
void write16(uint8_t *p, uint16_t value) {
  p[0] = value;
  p[1] = value >> 8;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<write16>:
  endbr64
  mov    %esi,%eax
  movd   %esi,%xmm0
  sar    $0x8,%eax
  movd   %eax,%xmm1
  punpckldq %xmm1,%xmm0
  movq   %xmm0,(%rdi)
  ret
```

assistant: ```c
void write16(uint8_t *p, uint16_t value) {
  p[0] = value;
  p[1] = value >> 8;
}
```