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
<RandomFunc>:
  endbr64
  movzwl (%rdi),%ecx
  movzwl 0x2(%rdi),%edx
  test   %dx,%dx
  je     L2
  mov    $0x0,%eax
L16:
  add    $0x2,%eax
  cmp    %dx,%ax
  jb     L16
L2:
  lea    0x3ce2(%rcx),%eax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  {
    state[0UL] = input[0UL] + (unsigned short)2885;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      if (state[0UL] > local1) {
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] + 371143069UL;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<RandomFunc>:
  endbr64
  movzwl 0x2(%rdi),%edx
  movzwl (%rdi),%ecx
  test   %dx,%dx
  je     L0
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L13:
  add    $0x2,%eax
  cmp    %dx,%ax
  jb     L13
L0:
  lea    0x3ce2(%rcx),%eax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  {
    state[0UL] = input[0UL] + (unsigned short)2885;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      if (state[0UL] > local1) {
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] + 371143069UL;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<RandomFunc>:
  endbr64
  movzwl 0x2(%rdi),%edx
  movzwl (%rdi),%ecx
  test   %dx,%dx
  je     L13
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L9:
  add    $0x2,%eax
  cmp    %dx,%ax
  jb     L9
L13:
  lea    0x3ce2(%rcx),%eax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  {
    state[0UL] = input[0UL] + (unsigned short)2885;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      if (state[0UL] > local1) {
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] + 371143069UL;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<fc_malloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <assert@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L8
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <fc_fatal@plt>
L8:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

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

tool: D7: out of memory

assistant: ```c
void *fc_malloc(size_t size) {
  assert(size);
  void *mem = malloc(size);
  if (!mem)
    fc_fatal("out of memory");
  return mem;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<fc_malloc>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L19
L7:
  mov    %rbx,%rax
  pop    %rbx
  ret
L19:
  lea    D5(%rip),%rdi
  call   <fc_fatal@plt>
  jmp    L7
```

tool_calls: name: parse_data
arguments: data_label: D5
type_infered: string

tool: D5: out of memory

assistant: ```c
void *fc_malloc(size_t size) {
  assert(size);
  void *mem = malloc(size);
  if (!mem)
    fc_fatal("out of memory");
  return mem;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fc_malloc>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  call   <assert@plt>
  mov    %r12,%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L14
L0:
  mov    %r12,%rax
  pop    %r12
  ret
L14:
  lea    D5(%rip),%rdi
  call   <fc_fatal@plt>
  jmp    L0
```

tool_calls: name: extract_symbolic_data
arguments: tag: D5
type_guessed: string

tool: D5: out of memory

assistant: ```c
void *fc_malloc(size_t size) {
  assert(size);
  void *mem = malloc(size);
  if (!mem)
    fc_fatal("out of memory");
  return mem;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fc_malloc>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  call   <assert@plt>
  mov    %r12,%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L0
L18:
  mov    %r12,%rax
  pop    %r12
  ret
L0:
  lea    D8(%rip),%rdi
  call   <fc_fatal@plt>
  jmp    L18
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
type: string

tool: D8: out of memory

assistant: ```c
void *fc_malloc(size_t size) {
  assert(size);
  void *mem = malloc(size);
  if (!mem)
    fc_fatal("out of memory");
  return mem;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<get_miliseconds>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  lea    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <clock_gettime@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x18(%rbp),%rax
  imul   $0x3e8,%eax,%ecx
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x431bde83,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x12,%edx
  sar    $0x1f,%eax
  mov    %eax,%esi
  mov    %edx,%eax
  sub    %esi,%eax
  add    %ecx,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
uint64_t get_miliseconds() {
  struct timespec tp;
  assert(clock_gettime(CLOCK_MONOTONIC, &tp) == 0);
  return ((uint64_t)tp.tv_sec) * 1000 + tp.tv_nsec / 1000000;
}
```
user: Translate the below assembly code into C programming language:

```asm
<get_miliseconds>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    D5(%rip),%rax
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  test   %rax,%rax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  imul   $0x3e8,0x8(%rsp),%eax
  mov    (%rsp),%ecx
  movslq %ecx,%rdx
  imul   $0x431bde83,%rdx,%rdx
  sar    $0x32,%rdx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  add    %edx,%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t get_miliseconds() {
  struct timespec tp;
  assert(clock_gettime(CLOCK_MONOTONIC, &tp) == 0);
  return ((uint64_t)tp.tv_sec) * 1000 + tp.tv_nsec / 1000000;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<get_miliseconds>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D8(%rip),%rax
  mov    %rsp,%rsi
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  movslq (%rsp),%rdx
  imul   $0x3e8,0x8(%rsp),%eax
  mov    %rdx,%rcx
  imul   $0x431bde83,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x32,%rdx
  sub    %ecx,%edx
  add    %edx,%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t get_miliseconds() {
  struct timespec tp;
  assert(clock_gettime(CLOCK_MONOTONIC, &tp) == 0);
  return ((uint64_t)tp.tv_sec) * 1000 + tp.tv_nsec / 1000000;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<get_miliseconds>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D3(%rip),%rax
  mov    %rsp,%rsi
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  movslq (%rsp),%rdx
  imul   $0x3e8,0x8(%rsp),%eax
  mov    %rdx,%rcx
  imul   $0x431bde83,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x32,%rdx
  sub    %ecx,%edx
  add    %edx,%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t get_miliseconds() {
  struct timespec tp;
  assert(clock_gettime(CLOCK_MONOTONIC, &tp) == 0);
  return ((uint64_t)tp.tv_sec) * 1000 + tp.tv_nsec / 1000000;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<imin3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jge    L0
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x4(%rbp)
L0:
  mov    -0x1c(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jge    L16
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
L16:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int imin3(int a, int b, int c) {
  int ret = a;
  if (b < a)
    ret = b;
  if (c < a)
    ret = c;
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<imin3>:
  endbr64
  mov    %edx,%eax
  cmp    %esi,%edi
  cmovle %edi,%esi
  cmp    %edx,%edi
  cmovle %esi,%eax
  ret
```

assistant: ```c
int imin3(int a, int b, int c) {
  int ret = a;
  if (b < a)
    ret = b;
  if (c < a)
    ret = c;
  return ret;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<imin3>:
  endbr64
  cmp    %esi,%edi
  mov    %edx,%eax
  cmovle %edi,%esi
  cmp    %edx,%edi
  cmovle %esi,%eax
  ret
```

assistant: ```c
int imin3(int a, int b, int c) {
  int ret = a;
  if (b < a)
    ret = b;
  if (c < a)
    ret = c;
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<imin3>:
  endbr64
  cmp    %esi,%edi
  mov    %edx,%eax
  cmovle %edi,%esi
  cmp    %edx,%edi
  cmovle %esi,%eax
  ret
```

assistant: ```c
int imin3(int a, int b, int c) {
  int ret = a;
  if (b < a)
    ret = b;
  if (c < a)
    ret = c;
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<fact>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jg     L17
  mov    $0x1,%eax
  jmp    L2
L17:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <fact>
  imul   -0x4(%rbp),%eax
L2:
  leave
  ret
```

assistant: ```c
int fact(int n) {
  if (n <= 1)
    return 1;
  return fact(n - 1) * n;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fact>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  jle    L2
  push   %rbx
  mov    %edi,%ebx
  lea    -0x1(%rdi),%edi
  call   <fact>
  imul   %ebx,%eax
  pop    %rbx
  ret
L2:
  ret
```

assistant: ```c
int fact(int n) {
  if (n <= 1)
    return 1;
  return fact(n - 1) * n;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<fact>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  jle    L16
  xchg   %ax,%ax
L7:
  mov    %edi,%edx
  sub    $0x1,%edi
  imul   %edx,%eax
  cmp    $0x1,%edi
  jne    L7
L16:
  ret
```

assistant: ```c
int fact(int n) {
  if (n <= 1)
    return 1;
  return fact(n - 1) * n;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<fact>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  jle    L2
  xchg   %ax,%ax
L15:
  mov    %edi,%edx
  sub    $0x1,%edi
  imul   %edx,%eax
  cmp    $0x1,%edi
  jne    L15
L2:
  ret
```

assistant: ```c
int fact(int n) {
  if (n <= 1)
    return 1;
  return fact(n - 1) * n;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rax
  mov    (%rdx),%edx
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    0x4(%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%rdx
  add    $0x4,%rdx
  mov    0x4(%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  lea    0x4(%rax),%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    0x8(%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%rdx
  add    $0x8,%rdx
  mov    0x8(%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  lea    0x8(%rax),%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void swap(int **queue, int a, int b) {
  int unit;
  unit = queue[a][0];
  queue[a][0] = queue[b][0];
  queue[b][0] = unit;
  unit = queue[a][1];
  queue[a][1] = queue[b][1];
  queue[b][1] = unit;
  unit = queue[a][2];
  queue[a][2] = queue[b][2];
  queue[b][2] = unit;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<swap>:
  endbr64
  movslq %esi,%rsi
  lea    (%rdi,%rsi,8),%rcx
  mov    (%rcx),%rsi
  mov    (%rsi),%r8d
  movslq %edx,%rdx
  lea    (%rdi,%rdx,8),%rax
  mov    (%rax),%rdx
  mov    (%rdx),%edx
  mov    %edx,(%rsi)
  mov    (%rax),%rdx
  mov    %r8d,(%rdx)
  mov    (%rcx),%rdx
  mov    0x4(%rdx),%esi
  mov    (%rax),%rdi
  mov    0x4(%rdi),%edi
  mov    %edi,0x4(%rdx)
  mov    (%rax),%rdx
  mov    %esi,0x4(%rdx)
  mov    (%rcx),%rdx
  mov    0x8(%rdx),%ecx
  mov    (%rax),%rsi
  mov    0x8(%rsi),%esi
  mov    %esi,0x8(%rdx)
  mov    (%rax),%rax
  mov    %ecx,0x8(%rax)
  ret
```

assistant: ```c
void swap(int **queue, int a, int b) {
  int unit;
  unit = queue[a][0];
  queue[a][0] = queue[b][0];
  queue[b][0] = unit;
  unit = queue[a][1];
  queue[a][1] = queue[b][1];
  queue[b][1] = unit;
  unit = queue[a][2];
  queue[a][2] = queue[b][2];
  queue[b][2] = unit;
}
```
user: Translate the below assembly code into C programming language:

```asm
<swap>:
  endbr64
  movslq %esi,%rsi
  movslq %edx,%rdx
  mov    (%rdi,%rsi,8),%rcx
  mov    (%rdi,%rdx,8),%rax
  mov    (%rcx),%esi
  mov    (%rax),%edx
  mov    %edx,(%rcx)
  mov    %esi,(%rax)
  mov    0x4(%rax),%esi
  mov    0x4(%rcx),%edx
  mov    %esi,0x4(%rcx)
  mov    0x8(%rax),%esi
  mov    %edx,0x4(%rax)
  mov    0x8(%rcx),%edx
  mov    %esi,0x8(%rcx)
  mov    %edx,0x8(%rax)
  ret
```

assistant: ```c
void swap(int **queue, int a, int b) {
  int unit;
  unit = queue[a][0];
  queue[a][0] = queue[b][0];
  queue[b][0] = unit;
  unit = queue[a][1];
  queue[a][1] = queue[b][1];
  queue[b][1] = unit;
  unit = queue[a][2];
  queue[a][2] = queue[b][2];
  queue[b][2] = unit;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<swap>:
  endbr64
  movslq %esi,%rsi
  movslq %edx,%rdx
  mov    (%rdi,%rsi,8),%rcx
  mov    (%rdi,%rdx,8),%rax
  mov    (%rcx),%esi
  mov    (%rax),%edx
  mov    %edx,(%rcx)
  mov    %esi,(%rax)
  mov    0x4(%rax),%esi
  mov    0x4(%rcx),%edx
  mov    %esi,0x4(%rcx)
  mov    0x8(%rax),%esi
  mov    %edx,0x4(%rax)
  mov    0x8(%rcx),%edx
  mov    %esi,0x8(%rcx)
  mov    %edx,0x8(%rax)
  ret
```

assistant: ```c
void swap(int **queue, int a, int b) {
  int unit;
  unit = queue[a][0];
  queue[a][0] = queue[b][0];
  queue[b][0] = unit;
  unit = queue[a][1];
  queue[a][1] = queue[b][1];
  queue[b][1] = unit;
  unit = queue[a][2];
  queue[a][2] = queue[b][2];
  queue[b][2] = unit;
}
```
user: Translate the below assembly code into C programming language:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xb0,%rsp
  mov    %edi,-0xa4(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x9c(%rbp)
  movl   $0x7fffffff,-0x98(%rbp)
  cmpl   $0x0,-0xa4(%rbp)
  jle    L56
  mov    $0x1,%eax
  jmp    L52
L56:
  mov    $0xffffffff,%eax
L52:
  mov    %eax,-0x94(%rbp)
  mov    -0xa4(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  mov    %eax,-0xa4(%rbp)
  jmp    L8
L35:
  mov    -0xa4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    -0x98(%rbp),%eax
  sub    %ecx,%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
  cmp    %eax,-0x9c(%rbp)
  jle    L40
  mov    $0x0,%eax
  jmp    L25
L40:
  mov    -0x9c(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edi
  mov    -0xa4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  lea    (%rdi,%rcx,1),%eax
  mov    %eax,-0x9c(%rbp)
  mov    -0xa4(%rbp),%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,-0xa4(%rbp)
L8:
  cmpl   $0x0,-0xa4(%rbp)
  jne    L35
  mov    -0x9c(%rbp),%eax
  imul   -0x94(%rbp),%eax
L25:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

assistant: ```c
int reverse(int x) {
  int tag;
  int _x = 0;
  int array[32];
  int end;
  int max = (1 << 30) - 1 + (1 << 30);
  tag = x > 0 ? 1 : -1;
  x = x > 0 ? x : -x;
  while (x) {
    if (_x > (max - x % 10) / 10) {
      return 0;
    }
    _x = _x * 10 + x % 10;
    x = x / 10;
  }
  return _x * tag;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<reverse>:
  endbr64
  mov    %edi,%esi
  neg    %esi
  cmovs  %edi,%esi
  mov    $0x1,%eax
  test   %edi,%edi
  jle    L11
L2:
  movslq %esi,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %esi,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  mov    %esi,%edx
  sub    %ecx,%edx
  mov    $0x0,%edi
  mov    $0x7fffffff,%r8d
L45:
  lea    (%rdi,%rdi,4),%ecx
  lea    (%rdx,%rcx,2),%edi
  movslq %esi,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  sar    $0x1f,%esi
  sub    %esi,%ecx
  mov    %ecx,%esi
  je     L34
  movslq %ecx,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %ecx,%r9d
  sar    $0x1f,%r9d
  sub    %r9d,%edx
  lea    (%rdx,%rdx,4),%r9d
  add    %r9d,%r9d
  mov    %ecx,%edx
  sub    %r9d,%edx
  mov    %r8d,%r9d
  sub    %edx,%r9d
  movslq %r9d,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  sar    $0x1f,%r9d
  sub    %r9d,%ecx
  cmp    %edi,%ecx
  jge    L45
  mov    $0x0,%eax
  ret
L11:
  mov    %edi,%esi
  neg    %esi
  cmovs  %edi,%esi
  test   %edi,%edi
  je     L25
  mov    $0xffffffff,%eax
  jmp    L2
L25:
  mov    $0xffffffff,%eax
L34:
  imul   %edi,%eax
  ret
```

assistant: ```c
int reverse(int x) {
  int tag;
  int _x = 0;
  int array[32];
  int end;
  int max = (1 << 30) - 1 + (1 << 30);
  tag = x > 0 ? 1 : -1;
  x = x > 0 ? x : -x;
  while (x) {
    if (_x > (max - x % 10) / 10) {
      return 0;
    }
    _x = _x * 10 + x % 10;
    x = x / 10;
  }
  return _x * tag;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<reverse>:
  endbr64
  mov    %edi,%esi
  mov    $0x1,%eax
  neg    %esi
  cmovs  %edi,%esi
  test   %edi,%edi
  jle    L33
L1:
  mov    %esi,%ecx
  mov    $0xcccccccd,%r8d
  mov    $0x7fffffff,%r10d
  mov    %rcx,%rdx
  imul   %r8,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%edi
  add    %edi,%edi
  sub    %edi,%esi
  mov    %esi,%edx
  xor    %esi,%esi
  jmp    L53
  nopw   0x0(%rax,%rax,1)
L54:
  mov    %ecx,%r9d
  mov    %r9,%rdx
  imul   %r8,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%edi
  mov    %ecx,%edx
  mov    %r10d,%ecx
  add    %edi,%edi
  sub    %edi,%edx
  sub    %edx,%ecx
  imul   %r8,%rcx
  shr    $0x23,%rcx
  cmp    %esi,%ecx
  jl     L28
  mov    %r9,%rcx
L53:
  imul   %r8,%rcx
  lea    (%rsi,%rsi,4),%esi
  lea    (%rdx,%rsi,2),%esi
  shr    $0x23,%rcx
  test   %ecx,%ecx
  jne    L54
  imul   %esi,%eax
  ret
  nopl   0x0(%rax)
L28:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L33:
  je     L28
  mov    $0xffffffff,%eax
  jmp    L1
```

assistant: ```c
int reverse(int x) {
  int tag;
  int _x = 0;
  int array[32];
  int end;
  int max = (1 << 30) - 1 + (1 << 30);
  tag = x > 0 ? 1 : -1;
  x = x > 0 ? x : -x;
  while (x) {
    if (_x > (max - x % 10) / 10) {
      return 0;
    }
    _x = _x * 10 + x % 10;
    x = x / 10;
  }
  return _x * tag;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<reverse>:
  endbr64
  mov    %edi,%esi
  mov    $0x1,%eax
  neg    %esi
  cmovs  %edi,%esi
  test   %edi,%edi
  jle    L38
L33:
  mov    %esi,%ecx
  mov    $0xcccccccd,%r8d
  mov    $0x7fffffff,%r10d
  mov    %rcx,%rdx
  imul   %r8,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%edi
  add    %edi,%edi
  sub    %edi,%esi
  mov    %esi,%edx
  xor    %esi,%esi
  jmp    L26
  nopw   0x0(%rax,%rax,1)
L53:
  mov    %ecx,%r9d
  mov    %r9,%rdx
  imul   %r8,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%edi
  mov    %ecx,%edx
  mov    %r10d,%ecx
  add    %edi,%edi
  sub    %edi,%edx
  sub    %edx,%ecx
  imul   %r8,%rcx
  shr    $0x23,%rcx
  cmp    %esi,%ecx
  jl     L41
  mov    %r9,%rcx
L26:
  imul   %r8,%rcx
  lea    (%rsi,%rsi,4),%esi
  lea    (%rdx,%rsi,2),%esi
  shr    $0x23,%rcx
  test   %ecx,%ecx
  jne    L53
  imul   %esi,%eax
  ret
  nopl   0x0(%rax)
L41:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L38:
  je     L41
  mov    $0xffffffff,%eax
  jmp    L33
```

assistant: ```c
int reverse(int x) {
  int tag;
  int _x = 0;
  int array[32];
  int end;
  int max = (1 << 30) - 1 + (1 << 30);
  tag = x > 0 ? 1 : -1;
  x = x > 0 ? x : -x;
  while (x) {
    if (_x > (max - x % 10) / 10) {
      return 0;
    }
    _x = _x * 10 + x % 10;
    x = x / 10;
  }
  return _x * tag;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<printBWords>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L25
L18:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  cmp    $0x62,%al
  jne    L20
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L20:
  addl   $0x1,-0x4(%rbp)
L25:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L18
  nop
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D1
datatype: string

tool: D1: %s


assistant: ```c
void printBWords(char *toPrint[16], int arrLen) {
  int i;
  for (i = 0; i < arrLen; i++) {
    if (toPrint[i][0] == 'b')
      printf("%s\n", toPrint[i]);
  }
}
```