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

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x7,%esi
  push   %rbp
  lea    D7(%rip),%rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x900000005,%rax
  mov    %rsp,%rbx
  lea    0x1c(%rsp),%r12
  movl   $0xa,0x18(%rsp)
  mov    %rbx,%rdi
  mov    %rax,(%rsp)
  movabs $0x600000001,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x900000002,%rax
  mov    %rax,0x10(%rsp)
  call   <heap_sort@plt>
  nopw   0x0(%rax,%rax,1)
L13:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L13
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D7
type: string

tool: D7: %d	

assistant: ```c
int main() {
  int nums[] = {5, 9, 1, 6, 2, 9, 10};
  int i;
  heap_sort(nums, sizeof(nums) / sizeof(nums[0]));
  for (i = 0; i < sizeof(nums) / sizeof(nums[0]); i++)
    printf("%d\t", nums[i]);
  printf("\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x7,%esi
  push   %rbp
  lea    D12(%rip),%rbp
  push   %rbx
  sub    $0x30,%rsp
  movdqa D23(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x1c(%rsp),%r12
  mov    D24(%rip),%rax
  mov    %rbx,%rdi
  movl   $0xa,0x18(%rsp)
  mov    %rax,0x10(%rsp)
  movaps %xmm0,(%rsp)
  call   <heap_sort@plt>
  nopl   0x0(%rax,%rax,1)
L12:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L12
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D12
datatype: string
name: extract_symbolic_data
arguments: data_label: D23
datatype: byte[16]
name: extract_symbolic_data
arguments: data_label: D24
datatype: byte[8]

tool: D12: %d	

tool: {"D23": [0x05,0x00,0x00,0x00,0x09,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x06,0x00,0x00,0x00]}

tool: {"D24": [0x02,0x00,0x00,0x00,0x09,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int nums[] = {5, 9, 1, 6, 2, 9, 10};
  int i;
  heap_sort(nums, sizeof(nums) / sizeof(nums[0]));
  for (i = 0; i < sizeof(nums) / sizeof(nums[0]); i++)
    printf("%d\t", nums[i]);
  printf("\n");
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<hextochar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  movb   $0x0,(%rax)
  movl   $0x0,-0x8(%rbp)
  jmp    L27
L26:
  mov    $0x7,%eax
  sub    -0x8(%rbp),%eax
  shl    $0x2,%eax
  mov    -0x14(%rbp),%edx
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  and    $0xf,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x9,-0x4(%rbp)
  jg     L38
  mov    -0x4(%rbp),%eax
  lea    0x30(%rax),%ecx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L11
L38:
  mov    -0x4(%rbp),%eax
  lea    0x37(%rax),%ecx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L11:
  addl   $0x1,-0x8(%rbp)
L27:
  cmpl   $0x7,-0x8(%rbp)
  jle    L26
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void hextochar(int n, char buffer[9]) {
  int i;
  buffer[8] = 0;
  for (i = 0; i < 8; ++i) {
    int nibble = 0xf & (n >> (4 * (7 - i)));
    if (nibble <= 9) {
      buffer[i] = nibble + '0';
    } else {
      buffer[i] = nibble - 10 + 'A';
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<hextochar>:
  endbr64
  movb   $0x0,0x8(%rsi)
  mov    $0x1c,%ecx
  jmp    L5
L3:
  add    $0x37,%eax
  mov    %al,(%rsi)
L26:
  sub    $0x4,%ecx
  add    $0x1,%rsi
  cmp    $0xfffffffc,%ecx
  je     L14
L5:
  mov    %edi,%eax
  sar    %cl,%eax
  and    $0xf,%eax
  cmp    $0x9,%eax
  jg     L3
  add    $0x30,%eax
  mov    %al,(%rsi)
  jmp    L26
L14:
  ret
```

assistant: ```c
void hextochar(int n, char buffer[9]) {
  int i;
  buffer[8] = 0;
  for (i = 0; i < 8; ++i) {
    int nibble = 0xf & (n >> (4 * (7 - i)));
    if (nibble <= 9) {
      buffer[i] = nibble + '0';
    } else {
      buffer[i] = nibble - 10 + 'A';
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<hextochar>:
  endbr64
  movb   $0x0,0x8(%rsi)
  mov    $0x1c,%ecx
  nopl   (%rax)
L2:
  mov    %edi,%eax
  sar    %cl,%eax
  and    $0xf,%eax
  lea    0x30(%rax),%r8d
  cmp    $0x9,%eax
  lea    0x37(%rax),%edx
  mov    %r8d,%eax
  cmovg  %edx,%eax
  sub    $0x4,%ecx
  add    $0x1,%rsi
  mov    %al,-0x1(%rsi)
  cmp    $0xfffffffc,%ecx
  jne    L2
  ret
```

assistant: ```c
void hextochar(int n, char buffer[9]) {
  int i;
  buffer[8] = 0;
  for (i = 0; i < 8; ++i) {
    int nibble = 0xf & (n >> (4 * (7 - i)));
    if (nibble <= 9) {
      buffer[i] = nibble + '0';
    } else {
      buffer[i] = nibble - 10 + 'A';
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<hextochar>:
  endbr64
  movb   $0x0,0x8(%rsi)
  mov    %rsi,%rdx
  mov    %edi,%esi
  mov    %edi,%eax
  shr    $0x1c,%esi
  cmp    $0xa,%esi
  lea    0x37(%rsi),%edi
  lea    0x30(%rsi),%ecx
  cmovge %edi,%ecx
  mov    %cl,(%rdx)
  mov    %eax,%ecx
  sar    $0x18,%ecx
  and    $0xf,%ecx
  cmp    $0xa,%ecx
  lea    0x30(%rcx),%esi
  lea    0x37(%rcx),%edi
  mov    %eax,%ecx
  cmovl  %esi,%edi
  sar    $0x14,%ecx
  and    $0xf,%ecx
  cmp    $0xa,%ecx
  lea    0x30(%rcx),%esi
  mov    %dil,0x1(%rdx)
  lea    0x37(%rcx),%edi
  mov    %eax,%ecx
  cmovl  %esi,%edi
  sar    $0x10,%ecx
  and    $0xf,%ecx
  mov    %dil,0x2(%rdx)
  cmp    $0xa,%ecx
  lea    0x30(%rcx),%esi
  lea    0x37(%rcx),%edi
  mov    %eax,%ecx
  cmovl  %esi,%edi
  sar    $0xc,%ecx
  and    $0xf,%ecx
  cmp    $0xa,%ecx
  lea    0x30(%rcx),%esi
  mov    %dil,0x3(%rdx)
  lea    0x37(%rcx),%edi
  mov    %eax,%ecx
  cmovl  %esi,%edi
  sar    $0x8,%ecx
  and    $0xf,%ecx
  mov    %dil,0x4(%rdx)
  cmp    $0xa,%ecx
  lea    0x30(%rcx),%esi
  lea    0x37(%rcx),%edi
  mov    %eax,%ecx
  cmovl  %esi,%edi
  sar    $0x4,%ecx
  and    $0xf,%ecx
  cmp    $0xa,%ecx
  lea    0x30(%rcx),%esi
  mov    %dil,0x5(%rdx)
  lea    0x37(%rcx),%edi
  cmovl  %esi,%edi
  and    $0xf,%eax
  lea    0x30(%rax),%esi
  lea    0x37(%rax),%ecx
  cmp    $0x9,%eax
  mov    %esi,%eax
  mov    %dil,0x6(%rdx)
  cmovg  %ecx,%eax
  mov    %al,0x7(%rdx)
  ret
```

assistant: ```c
void hextochar(int n, char buffer[9]) {
  int i;
  buffer[8] = 0;
  for (i = 0; i < 8; ++i) {
    int nibble = 0xf & (n >> (4 * (7 - i)));
    if (nibble <= 9) {
      buffer[i] = nibble + '0';
    } else {
      buffer[i] = nibble - 10 + 'A';
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<addRange>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  cmp    %rax,%rdx
  jge    L37
  lea    -0x20(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swapl@plt>
L37:
  mov    -0x20(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  sub    %rdx,%rax
  cmp    $0x3ff,%rax
  jle    L41
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  imul   %rdx,%rax
  mov    %rax,%rdx
  shr    $0x3f,%rdx
  add    %rdx,%rax
  sar    %rax
  mov    %rax,%rcx
  mov    -0x18(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  imul   %rdx,%rax
  mov    %rax,%rdx
  shr    $0x3f,%rdx
  add    %rdx,%rax
  sar    %rax
  neg    %rax
  add    %rcx,%rax
  jmp    L20
L41:
  movq   $0x0,-0x8(%rbp)
  jmp    L27
L46:
  mov    -0x20(%rbp),%rax
  add    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x20(%rbp)
L27:
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  cmp    %rax,%rdx
  jge    L46
  mov    -0x8(%rbp),%rax
L20:
  leave
  ret
```

assistant: ```c
long addRange(long l, long h) {
  if (h < l) {
    swapl(&l, &h);
  }
  if ((h - l) >= 1024)
    return (h * (h + 1) / 2) - ((l - 1) * l / 2);
  long acc = 0;
  while (h >= l) {
    acc += h;
    h--;
  }
  return acc;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<addRange>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %rsi,(%rsp)
  cmp    %rdi,%rsi
  jl     L66
  mov    %rdi,%rcx
  mov    %rsi,%rdx
  mov    %rsi,%rax
  sub    %rdi,%rax
  cmp    $0x3ff,%rax
  jg     L55
L23:
  mov    $0x0,%eax
L28:
  add    %rdx,%rax
  sub    $0x1,%rdx
  cmp    %rcx,%rdx
  jge    L28
L19:
  add    $0x18,%rsp
  ret
L66:
  mov    %rsp,%rsi
  lea    0x8(%rsp),%rdi
  call   <swapl@plt>
  mov    (%rsp),%rdx
  mov    0x8(%rsp),%rcx
  mov    %rdx,%rax
  sub    %rcx,%rax
  cmp    $0x3ff,%rax
  jg     L55
  mov    $0x0,%eax
  cmp    %rcx,%rdx
  jge    L23
  jmp    L19
L55:
  lea    0x1(%rdx),%rax
  imul   %rax,%rdx
  mov    %rdx,%rax
  shr    $0x3f,%rax
  add    %rdx,%rax
  sar    %rax
  lea    -0x1(%rcx),%rdx
  imul   %rdx,%rcx
  mov    %rcx,%rdx
  shr    $0x3f,%rdx
  add    %rcx,%rdx
  sar    %rdx
  sub    %rdx,%rax
  jmp    L19
```

assistant: ```c
long addRange(long l, long h) {
  if (h < l) {
    swapl(&l, &h);
  }
  if ((h - l) >= 1024)
    return (h * (h + 1) / 2) - ((l - 1) * l / 2);
  long acc = 0;
  while (h >= l) {
    acc += h;
    h--;
  }
  return acc;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<addRange>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %rsi,(%rsp)
  cmp    %rdi,%rsi
  jl     L27
  mov    %rsi,%rax
  mov    %rdi,%rcx
  mov    %rsi,%rdx
  sub    %rdi,%rax
  cmp    $0x3ff,%rax
  jg     L46
L58:
  xor    %eax,%eax
  nopl   0x0(%rax)
L0:
  add    %rdx,%rax
  sub    $0x1,%rdx
  cmp    %rcx,%rdx
  jge    L0
L55:
  add    $0x18,%rsp
  ret
  nopl   0x0(%rax)
L46:
  lea    0x1(%rdx),%rax
  add    $0x18,%rsp
  imul   %rax,%rdx
  mov    %rdx,%rax
  shr    $0x3f,%rax
  add    %rdx,%rax
  lea    -0x1(%rcx),%rdx
  imul   %rdx,%rcx
  sar    %rax
  mov    %rcx,%rdx
  shr    $0x3f,%rdx
  add    %rcx,%rdx
  sar    %rdx
  sub    %rdx,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L27:
  mov    %rsp,%rsi
  lea    0x8(%rsp),%rdi
  call   <swapl@plt>
  mov    (%rsp),%rdx
  mov    0x8(%rsp),%rcx
  mov    %rdx,%rax
  sub    %rcx,%rax
  cmp    $0x3ff,%rax
  jg     L46
  xor    %eax,%eax
  cmp    %rcx,%rdx
  jge    L58
  jmp    L55
```

assistant: ```c
long addRange(long l, long h) {
  if (h < l) {
    swapl(&l, &h);
  }
  if ((h - l) >= 1024)
    return (h * (h + 1) / 2) - ((l - 1) * l / 2);
  long acc = 0;
  while (h >= l) {
    acc += h;
    h--;
  }
  return acc;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<addRange>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %rsi,(%rsp)
  cmp    %rdi,%rsi
  jl     L33
  mov    %rsi,%rax
  mov    %rdi,%rcx
  mov    %rsi,%rdx
  sub    %rdi,%rax
  cmp    $0x3ff,%rax
  jg     L17
L24:
  xor    %eax,%eax
  nopl   0x0(%rax)
L40:
  add    %rdx,%rax
  sub    $0x1,%rdx
  cmp    %rdx,%rcx
  jle    L40
L45:
  add    $0x18,%rsp
  ret
  nopl   0x0(%rax)
L17:
  lea    0x1(%rdx),%rax
  add    $0x18,%rsp
  imul   %rax,%rdx
  mov    %rdx,%rax
  shr    $0x3f,%rax
  add    %rdx,%rax
  lea    -0x1(%rcx),%rdx
  imul   %rdx,%rcx
  sar    %rax
  mov    %rcx,%rdx
  shr    $0x3f,%rdx
  add    %rcx,%rdx
  sar    %rdx
  sub    %rdx,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L33:
  mov    %rsp,%rsi
  lea    0x8(%rsp),%rdi
  call   <swapl@plt>
  mov    (%rsp),%rdx
  mov    0x8(%rsp),%rcx
  mov    %rdx,%rax
  sub    %rcx,%rax
  cmp    $0x3ff,%rax
  jg     L17
  xor    %eax,%eax
  cmp    %rcx,%rdx
  jge    L24
  jmp    L45
```

assistant: ```c
long addRange(long l, long h) {
  if (h < l) {
    swapl(&l, &h);
  }
  if ((h - l) >= 1024)
    return (h * (h + 1) / 2) - ((l - 1) * l / 2);
  long acc = 0;
  while (h >= l) {
    acc += h;
    h--;
  }
  return acc;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ga_memcpy_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L6
L17:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x4(%rbp)
L6:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L17
  mov    -0x20(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
int *ga_memcpy_int(int *array1, int *array2, int n) {
  int a;
  for (a = 0; a < n; a++) {
    array2[a] = array1[a];
  }
  return array2;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ga_memcpy_int>:
  endbr64
  mov    %rsi,%rax
  test   %edx,%edx
  jle    L4
  mov    %edx,%edx
  mov    $0x0,%ecx
L13:
  mov    (%rdi,%rcx,4),%r8d
  mov    %r8d,(%rax,%rcx,4)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L13
L4:
  ret
```

assistant: ```c
int *ga_memcpy_int(int *array1, int *array2, int n) {
  int a;
  for (a = 0; a < n; a++) {
    array2[a] = array1[a];
  }
  return array2;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ga_memcpy_int>:
  endbr64
  mov    %rsi,%rax
  test   %edx,%edx
  jle    L19
  movslq %edx,%rdx
  xor    %ecx,%ecx
L15:
  mov    (%rdi,%rcx,4),%r8d
  mov    %r8d,(%rax,%rcx,4)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L15
L19:
  ret
```

assistant: ```c
int *ga_memcpy_int(int *array1, int *array2, int n) {
  int a;
  for (a = 0; a < n; a++) {
    array2[a] = array1[a];
  }
  return array2;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ga_memcpy_int>:
  endbr64
  mov    %rdi,%rcx
  mov    %rsi,%r8
  test   %edx,%edx
  jle    L29
  lea    0x4(%rdi),%rsi
  mov    %r8,%rax
  sub    %rsi,%rax
  cmp    $0x8,%rax
  jbe    L26
  lea    -0x1(%rdx),%eax
  cmp    $0x2,%eax
  jbe    L26
  mov    %edx,%esi
  xor    %eax,%eax
  shr    $0x2,%esi
  shl    $0x4,%rsi
  nopl   0x0(%rax)
L65:
  movdqu (%rcx,%rax,1),%xmm0
  movups %xmm0,(%r8,%rax,1)
  add    $0x10,%rax
  cmp    %rsi,%rax
  jne    L65
  mov    %edx,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%dl
  je     L29
  mov    %eax,%edi
  mov    (%rcx,%rdi,4),%r9d
  lea    0x0(,%rdi,4),%rsi
  mov    %r9d,(%r8,%rdi,4)
  lea    0x1(%rax),%edi
  cmp    %edi,%edx
  jle    L29
  mov    0x4(%rcx,%rsi,1),%edi
  add    $0x2,%eax
  mov    %edi,0x4(%r8,%rsi,1)
  cmp    %eax,%edx
  jle    L29
  mov    0x8(%rcx,%rsi,1),%eax
  mov    %eax,0x8(%r8,%rsi,1)
L29:
  mov    %r8,%rax
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L44:
  mov    (%rcx,%rax,4),%esi
  mov    %esi,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L44
  mov    %r8,%rax
  ret
```

assistant: ```c
int *ga_memcpy_int(int *array1, int *array2, int n) {
  int a;
  for (a = 0; a < n; a++) {
    array2[a] = array1[a];
  }
  return array2;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<next_mult>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x2,-0x4(%rbp)
  jmp    L4
L16:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L6
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %eax,-0x14(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  jmp    L29
L6:
  addl   $0x1,-0x4(%rbp)
L4:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L16
  mov    $0x1,%eax
L29:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
data_type: string

tool: D0: %d 

assistant: ```c
int next_mult(int number) {
  for (int i = 2; i <= number; i++) {
    if (number % i == 0) {
      number = number / i;
      printf("%d ", i);
      return number;
    }
  }
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<next_mult>:
  endbr64
  push   %rbx
  cmp    $0x1,%edi
  jle    L52
  test   $0x1,%dil
  je     L39
  mov    $0x2,%esi
L21:
  add    $0x1,%esi
  cmp    %esi,%edi
  jl     L27
  mov    %edi,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  jne    L21
L62:
  mov    %edi,%eax
  cltd
  idiv   %esi
  mov    %eax,%ebx
  lea    D2(%rip),%rdi
  call   <printf@plt>
L8:
  mov    %ebx,%eax
  pop    %rbx
  ret
L39:
  mov    $0x2,%esi
  jmp    L62
L27:
  mov    $0x1,%ebx
  jmp    L8
L52:
  mov    $0x1,%ebx
  jmp    L8
```

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

tool: D2: %d 

assistant: ```c
int next_mult(int number) {
  for (int i = 2; i <= number; i++) {
    if (number % i == 0) {
      number = number / i;
      printf("%d ", i);
      return number;
    }
  }
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<next_mult>:
  endbr64
  push   %r12
  cmp    $0x1,%edi
  jle    L26
  mov    $0x2,%esi
  test   $0x1,%dil
  jne    L2
  jmp    L40
  nopl   0x0(%rax,%rax,1)
L20:
  mov    %edi,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L40
L2:
  add    $0x1,%esi
  cmp    %esi,%edi
  jge    L20
L26:
  mov    $0x1,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
  nopl   0x0(%rax)
L40:
  mov    %edi,%eax
  lea    D4(%rip),%rdi
  cltd
  idiv   %esi
  mov    %eax,%r12d
  call   <printf@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
datatype: string

tool: D4: %d 

assistant: ```c
int next_mult(int number) {
  for (int i = 2; i <= number; i++) {
    if (number % i == 0) {
      number = number / i;
      printf("%d ", i);
      return number;
    }
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<next_mult>:
  endbr64
  push   %r12
  cmp    $0x1,%edi
  jle    L36
  mov    $0x2,%esi
  test   $0x1,%dil
  jne    L27
  jmp    L15
  nopl   0x0(%rax,%rax,1)
L42:
  mov    %edi,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L15
L27:
  add    $0x1,%esi
  cmp    %esi,%edi
  jge    L42
L36:
  mov    $0x1,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
  nopl   0x0(%rax)
L15:
  mov    %edi,%eax
  lea    D9(%rip),%rdi
  cltd
  idiv   %esi
  mov    %eax,%r12d
  call   <printf@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: label: D9
datatype: string

tool: D9: %d 

assistant: ```c
int next_mult(int number) {
  for (int i = 2; i <= number; i++) {
    if (number % i == 0) {
      number = number / i;
      printf("%d ", i);
      return number;
    }
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<xrop3_201_1_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x24(%rbp)
  mov    %al,-0x28(%rbp)
  movl   $0xff,-0x14(%rbp)
  movzbl -0x24(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x24(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x10(%rbp)
  movzbl -0x28(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x28(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  or     -0x14(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x10(%rbp),%eax
  xor    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  notl   -0x4(%rbp)
  mov    -0x10(%rbp),%eax
  not    %eax
  and    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  and    -0x10(%rbp),%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned xrop3_201_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;
  stk2 = T | D;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  return (stk1 & (~S)) | (D & S);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xrop3_201_1_0>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  movzbl %sil,%eax
  shl    $0x8,%eax
  or     %edi,%eax
  or     $0xff,%al
  not    %eax
  movzbl %dil,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_201_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;
  stk2 = T | D;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  return (stk1 & (~S)) | (D & S);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<xrop3_201_1_0>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  movzbl %sil,%eax
  shl    $0x8,%eax
  or     %edi,%eax
  movzbl %dil,%edi
  or     $0xff,%al
  not    %eax
  or     %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_201_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;
  stk2 = T | D;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  return (stk1 & (~S)) | (D & S);
}
```
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
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x38(%rbp),%rax
  mov    %rax,%rsi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L1
L49:
  mov    -0x38(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  mov    %rcx,%rsi
  sar    $0x3f,%rsi
  sub    %rsi,%rax
  mov    %rax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  mov    %edx,-0x3c(%rbp)
  mov    -0x3c(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  lea    0x1(%rax),%edx
  mov    -0x3c(%rbp),%eax
  cltq
  mov    %edx,-0x30(%rbp,%rax,4)
  mov    -0x38(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x38(%rbp)
L1:
  mov    -0x38(%rbp),%rax
  test   %rax,%rax
  jg     L49
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x44(%rbp)
  jmp    L19
L65:
  mov    -0x44(%rbp),%eax
  mov    %eax,%esi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x44(%rbp)
L19:
  cmpl   $0x9,-0x44(%rbp)
  jle    L65
  lea    D56(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x40(%rbp)
  jmp    L13
L39:
  mov    -0x40(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x40(%rbp)
L13:
  cmpl   $0x9,-0x40(%rbp)
  jle    L39
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L22
  call   <__stack_chk_fail@plt>
L22:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D22
type_guessed: string
name: unpack_data
arguments: data_tag: D33
type_guessed: string
name: unpack_data
arguments: data_tag: D5
type_guessed: string
name: unpack_data
arguments: data_tag: D30
type_guessed: string
name: unpack_data
arguments: data_tag: D56
type_guessed: string

tool: D22: enter a digit:

tool: D33: %ld

tool: D5: Digit:		

tool: D30: %2d

tool: D56: 
Occurrences:	

assistant: ```c
int main(void) {
  int digit, digit_occurrence[10] = {0};
  long n;
  printf("enter a digit:\n");
  scanf("%ld", &n);
  while (n > 0) {
    digit = n % 10;
    digit_occurrence[digit]++;
    n = n / 10;
  }
  printf("Digit:\t\t");
  for (int i = 0; i < 10; i++) {
    printf("%2d", i);
  }
  printf("\nOccurrences:\t");
  for (int i = 0; i < 10; i++) {
    printf("%2d", digit_occurrence[i]);
  }
  printf("\n");
  return 0;
}
```