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

```asm
<dup_vector>:
  endbr64
  movslq %esi,%rsi
  cmp    %rsi,%rdx
  jl     L18
  shl    $0x2,%rsi
  lea    0x4(,%rdx,4),%rax
  nopl   0x0(%rax,%rax,1)
L6:
  movss  (%rdi,%rsi,1),%xmm0
  movss  %xmm0,(%rcx,%rsi,1)
  add    $0x4,%rsi
  cmp    %rsi,%rax
  jne    L6
L18:
  ret
```

assistant: ```c
void dup_vector(float *src, long nl, long nh, float *dst) {
  int i;
  for (i = nl; i <= nh; i++) {
    dst[i] = src[i];
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<dup_vector>:
  endbr64
  movslq %esi,%rax
  cmp    %rax,%rdx
  jl     L26
  mov    %rdx,%r11
  mov    %rdx,%r9
  push   %rbx
  sub    %rax,%r11
  shl    $0x2,%rax
  lea    (%rcx,%rax,1),%r10
  lea    0x4(%rax),%rdx
  lea    (%rdi,%rdx,1),%rbx
  mov    %r10,%r8
  sub    %rbx,%r8
  cmp    $0x8,%r8
  jbe    L73
  cmp    $0x2,%r11
  jbe    L73
  add    $0x1,%r11
  lea    (%rdi,%rax,1),%rdx
  xor    %eax,%eax
  mov    %r11,%r8
  shr    $0x2,%r8
  shl    $0x4,%r8
  nopl   0x0(%rax)
L41:
  movups (%rdx,%rax,1),%xmm1
  movups %xmm1,(%r10,%rax,1)
  add    $0x10,%rax
  cmp    %r8,%rax
  jne    L41
  mov    %r11,%rax
  and    $0xfffffffffffffffc,%rax
  add    %eax,%esi
  cmp    %rax,%r11
  je     L56
  movslq %esi,%rax
  movss  (%rdi,%rax,4),%xmm0
  movss  %xmm0,(%rcx,%rax,4)
  lea    0x1(%rsi),%eax
  cltq
  cmp    %rax,%r9
  jl     L56
  movss  (%rdi,%rax,4),%xmm0
  add    $0x2,%esi
  lea    0x0(,%rax,4),%rdx
  movslq %esi,%rsi
  movss  %xmm0,(%rcx,%rax,4)
  cmp    %rsi,%r9
  jl     L56
  movss  0x4(%rdi,%rdx,1),%xmm0
  movss  %xmm0,0x4(%rcx,%rdx,1)
L56:
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L73:
  lea    0x4(,%r9,4),%rsi
  jmp    L25
  nopw   0x0(%rax,%rax,1)
L36:
  add    $0x4,%rdx
L25:
  movss  (%rdi,%rax,1),%xmm0
  movss  %xmm0,(%rcx,%rax,1)
  mov    %rdx,%rax
  cmp    %rsi,%rdx
  jne    L36
  pop    %rbx
  ret
L26:
  ret
```

assistant: ```c
void dup_vector(float *src, long nl, long nh, float *dst) {
  int i;
  for (i = nl; i <= nh; i++) {
    dst[i] = src[i];
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<hi_stacktrace_fd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void hi_stacktrace_fd(int fd) {
  if (fd > 0) {
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<hi_stacktrace_fd>:
  endbr64
  ret
```

assistant: ```c
void hi_stacktrace_fd(int fd) {
  if (fd > 0) {
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<hi_stacktrace_fd>:
  endbr64
  ret
```

assistant: ```c
void hi_stacktrace_fd(int fd) {
  if (fd > 0) {
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hi_stacktrace_fd>:
  endbr64
  ret
```

assistant: ```c
void hi_stacktrace_fd(int fd) {
  if (fd > 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
<hextochar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x14(%rbp)
  mov    %al,-0x18(%rbp)
  cmpb   $0x39,-0x14(%rbp)
  jle    L25
  movzbl -0x14(%rbp),%eax
  sub    $0x57,%eax
  jmp    L31
L25:
  movzbl -0x14(%rbp),%eax
  sub    $0x30,%eax
L31:
  mov    %al,-0x2(%rbp)
  cmpb   $0x39,-0x18(%rbp)
  jle    L36
  movzbl -0x18(%rbp),%eax
  sub    $0x57,%eax
  jmp    L1
L36:
  movzbl -0x18(%rbp),%eax
  sub    $0x30,%eax
L1:
  mov    %al,-0x1(%rbp)
  movzbl -0x2(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  movzbl -0x1(%rbp),%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned char hextochar(char a, char b) {
  unsigned char hn, ln;
  hn = a > '9' ? a - 'a' + 10 : a - '0';
  ln = b > '9' ? b - 'a' + 10 : b - '0';
  return ((hn << 4) | ln);
}
```
user: What is the c source code of the assembly code below?

```asm
<hextochar>:
  endbr64
  lea    -0x57(%rdi),%edx
  cmp    $0x3a,%dil
  lea    -0x30(%rdi),%eax
  cmovge %edx,%eax
  lea    -0x57(%rsi),%ecx
  lea    -0x30(%rsi),%edx
  cmp    $0x3a,%sil
  cmovge %ecx,%edx
  shl    $0x4,%eax
  or     %edx,%eax
  ret
```

assistant: ```c
unsigned char hextochar(char a, char b) {
  unsigned char hn, ln;
  hn = a > '9' ? a - 'a' + 10 : a - '0';
  ln = b > '9' ? b - 'a' + 10 : b - '0';
  return ((hn << 4) | ln);
}
```
user: Translate the below assembly code into C programming language:

```asm
<hextochar>:
  endbr64
  lea    -0x57(%rdi),%edx
  cmp    $0x3a,%dil
  lea    -0x30(%rdi),%eax
  cmovge %edx,%eax
  lea    -0x57(%rsi),%ecx
  lea    -0x30(%rsi),%edx
  cmp    $0x3a,%sil
  cmovge %ecx,%edx
  shl    $0x4,%eax
  or     %edx,%eax
  ret
```

assistant: ```c
unsigned char hextochar(char a, char b) {
  unsigned char hn, ln;
  hn = a > '9' ? a - 'a' + 10 : a - '0';
  ln = b > '9' ? b - 'a' + 10 : b - '0';
  return ((hn << 4) | ln);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<are_bytes_reversed>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movq   $0x1,(%rax)
  movzbl -0x10(%rbp),%eax
  test   %al,%al
  jne    L8
  mov    $0x1,%eax
  jmp    L21
L8:
  mov    $0x0,%eax
L21:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  leave
  ret
```

assistant: ```c
int are_bytes_reversed(void) {
  typedef unsigned long multi_t;
  unsigned char multibyte_integral_bytes[sizeof(multi_t)];
  multi_t *multibyte_integral;
  multibyte_integral = (multi_t *)(&multibyte_integral_bytes[0]);
  *multibyte_integral = 0x01;
  if (multibyte_integral_bytes[0] == 0)
    return 1;
  else
    return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<are_bytes_reversed>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int are_bytes_reversed(void) {
  typedef unsigned long multi_t;
  unsigned char multibyte_integral_bytes[sizeof(multi_t)];
  multi_t *multibyte_integral;
  multibyte_integral = (multi_t *)(&multibyte_integral_bytes[0]);
  *multibyte_integral = 0x01;
  if (multibyte_integral_bytes[0] == 0)
    return 1;
  else
    return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<are_bytes_reversed>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int are_bytes_reversed(void) {
  typedef unsigned long multi_t;
  unsigned char multibyte_integral_bytes[sizeof(multi_t)];
  multi_t *multibyte_integral;
  multibyte_integral = (multi_t *)(&multibyte_integral_bytes[0]);
  *multibyte_integral = 0x01;
  if (multibyte_integral_bytes[0] == 0)
    return 1;
  else
    return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<are_bytes_reversed>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int are_bytes_reversed(void) {
  typedef unsigned long multi_t;
  unsigned char multibyte_integral_bytes[sizeof(multi_t)];
  multi_t *multibyte_integral;
  multibyte_integral = (multi_t *)(&multibyte_integral_bytes[0]);
  *multibyte_integral = 0x01;
  if (multibyte_integral_bytes[0] == 0)
    return 1;
  else
    return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<str_trim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L23
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L7
L23:
  mov    $0xffffffff,%eax
  jmp    L12
L7:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <str_ltrim@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <str_rtrim@plt>
  mov    $0x0,%eax
L12:
  leave
  ret
```

assistant: ```c
int str_trim(char *s) {
  if (!s || !*s) {
    return -1;
  }
  str_ltrim(s);
  str_rtrim(s);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<str_trim>:
  endbr64
  test   %rdi,%rdi
  je     L10
  push   %rbx
  mov    %rdi,%rbx
  cmpb   $0x0,(%rdi)
  je     L15
  call   <str_ltrim@plt>
  mov    %rbx,%rdi
  call   <str_rtrim@plt>
  mov    $0x0,%eax
L22:
  pop    %rbx
  ret
L10:
  mov    $0xffffffff,%eax
  ret
L15:
  mov    $0xffffffff,%eax
  jmp    L22
```

assistant: ```c
int str_trim(char *s) {
  if (!s || !*s) {
    return -1;
  }
  str_ltrim(s);
  str_rtrim(s);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<str_trim>:
  endbr64
  test   %rdi,%rdi
  je     L14
  push   %rbp
  cmpb   $0x0,(%rdi)
  mov    %rdi,%rbp
  je     L13
  call   <str_ltrim@plt>
  mov    %rbp,%rdi
  call   <str_rtrim@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
L14:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int str_trim(char *s) {
  if (!s || !*s) {
    return -1;
  }
  str_ltrim(s);
  str_rtrim(s);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<str_trim>:
  endbr64
  test   %rdi,%rdi
  je     L13
  push   %rbp
  cmpb   $0x0,(%rdi)
  mov    %rdi,%rbp
  je     L7
  call   <str_ltrim@plt>
  mov    %rbp,%rdi
  call   <str_rtrim@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
L13:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int str_trim(char *s) {
  if (!s || !*s) {
    return -1;
  }
  str_ltrim(s);
  str_rtrim(s);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<example_enter>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
datatype: string

tool: D7: example_enter %p %d


assistant: ```c
void example_enter(void *node, void *object, struct node_operations *ops) {
  int *ctx = (int *)object;
  *ctx = *ctx + 1;
  printf("example_enter %p %d\n", node, *ctx);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<example_enter>:
  endbr64
  sub    $0x8,%rsp
  mov    (%rsi),%eax
  lea    0x1(%rax),%edx
  mov    %edx,(%rsi)
  mov    %rdi,%rsi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

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

tool: D0: example_enter %p %d


assistant: ```c
void example_enter(void *node, void *object, struct node_operations *ops) {
  int *ctx = (int *)object;
  *ctx = *ctx + 1;
  printf("example_enter %p %d\n", node, *ctx);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<example_enter>:
  endbr64
  mov    (%rsi),%eax
  lea    0x1(%rax),%edx
  mov    %edx,(%rsi)
  mov    %rdi,%rsi
  lea    D7(%rip),%rdi
  jmp    1050 <printf@plt>
```

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

tool: D7: example_enter %p %d


assistant: ```c
void example_enter(void *node, void *object, struct node_operations *ops) {
  int *ctx = (int *)object;
  *ctx = *ctx + 1;
  printf("example_enter %p %d\n", node, *ctx);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<example_enter>:
  endbr64
  mov    (%rsi),%eax
  lea    0x1(%rax),%edx
  mov    %edx,(%rsi)
  mov    %rdi,%rsi
  lea    D6(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: unpack_data
arguments: tag: D6
datatype: string

tool: D6: example_enter %p %d


assistant: ```c
void example_enter(void *node, void *object, struct node_operations *ops) {
  int *ctx = (int *)object;
  *ctx = *ctx + 1;
  printf("example_enter %p %d\n", node, *ctx);
}
```
user: What is the c source code of the assembly code below:

```asm
<unpackValueLongLong>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rcx
  lea    -0x10(%rbp),%rax
  mov    $0x8,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  lea    0x8(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
long long unpackValueLongLong(unsigned char const **data) {
  long long size;
  memcpy(&size, *data, sizeof(size));
  *data += sizeof(size);
  return size;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<unpackValueLongLong>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x8,%edx
  mov    (%rbx),%rsi
  call   <memcpy@plt>
  addq   $0x8,(%rbx)
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x10,%rsp
  pop    %rbx
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long unpackValueLongLong(unsigned char const **data) {
  long long size;
  memcpy(&size, *data, sizeof(size));
  *data += sizeof(size);
  return size;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<unpackValueLongLong>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x8,%edx
  sub    $0x10,%rsp
  mov    (%rbx),%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <memcpy@plt>
  addq   $0x8,(%rbx)
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x10,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long unpackValueLongLong(unsigned char const **data) {
  long long size;
  memcpy(&size, *data, sizeof(size));
  *data += sizeof(size);
  return size;
}
```
user: What is the c source code of the assembly code below?

```asm
<unpackValueLongLong>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x8,%edx
  sub    $0x10,%rsp
  mov    (%rbx),%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <memcpy@plt>
  addq   $0x8,(%rbx)
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x10,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long unpackValueLongLong(unsigned char const **data) {
  long long size;
  memcpy(&size, *data, sizeof(size));
  *data += sizeof(size);
  return size;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  shr    $0x3,%eax
  or     $0x61bab5,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  nop
  movl   $0x0,-0x10(%rbp)
  nop
  mov    -0xc(%rbp),%eax
  xor    $0x1377ff77,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  char copy12;
  {
    state[0UL] = (input[0UL] | 51238316UL) >> 3U;
    local1 = 0UL;
    while (local1 < 0U) {
      state[0UL] <<= ((state[0UL] >> 3U) & 7U) | 1UL;
      local1 += 2UL;
    }
    local1 = 0UL;
    while (local1 < 0U) {
      copy12 = *((char *)(&state[local1]) + 1);
      *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 3);
      *((char *)(&state[local1]) + 3) = copy12;
      state[local1] = state[0UL] & state[local1];
      local1++;
    }
    output[0UL] = state[0UL] ^ 326631287U;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  shr    $0x3,%eax
  or     $0x61bab5,%eax
  xor    $0x1377ff77,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  char copy12;
  {
    state[0UL] = (input[0UL] | 51238316UL) >> 3U;
    local1 = 0UL;
    while (local1 < 0U) {
      state[0UL] <<= ((state[0UL] >> 3U) & 7U) | 1UL;
      local1 += 2UL;
    }
    local1 = 0UL;
    while (local1 < 0U) {
      copy12 = *((char *)(&state[local1]) + 1);
      *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 3);
      *((char *)(&state[local1]) + 3) = copy12;
      state[local1] = state[0UL] & state[local1];
      local1++;
    }
    output[0UL] = state[0UL] ^ 326631287U;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  shr    $0x3,%eax
  or     $0x61bab5,%eax
  xor    $0x1377ff77,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  char copy12;
  {
    state[0UL] = (input[0UL] | 51238316UL) >> 3U;
    local1 = 0UL;
    while (local1 < 0U) {
      state[0UL] <<= ((state[0UL] >> 3U) & 7U) | 1UL;
      local1 += 2UL;
    }
    local1 = 0UL;
    while (local1 < 0U) {
      copy12 = *((char *)(&state[local1]) + 1);
      *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 3);
      *((char *)(&state[local1]) + 3) = copy12;
      state[local1] = state[0UL] & state[local1];
      local1++;
    }
    output[0UL] = state[0UL] ^ 326631287U;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  shr    $0x3,%eax
  or     $0x61bab5,%eax
  xor    $0x1377ff77,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  char copy12;
  {
    state[0UL] = (input[0UL] | 51238316UL) >> 3U;
    local1 = 0UL;
    while (local1 < 0U) {
      state[0UL] <<= ((state[0UL] >> 3U) & 7U) | 1UL;
      local1 += 2UL;
    }
    local1 = 0UL;
    while (local1 < 0U) {
      copy12 = *((char *)(&state[local1]) + 1);
      *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 3);
      *((char *)(&state[local1]) + 3) = copy12;
      state[local1] = state[0UL] & state[local1];
      local1++;
    }
    output[0UL] = state[0UL] ^ 326631287U;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xtensa_op2p1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  cmp    $0xf,%eax
  ja     L12
  mov    $0x1,%eax
  jmp    L15
L12:
  mov    $0x0,%eax
L15:
  pop    %rbp
  ret
```

assistant: ```c
int xtensa_op2p1(v)
int v;
{
  switch (v) {
  case 1:
  case 2:
  case 3:
  case 4:
  case 5:
  case 6:
  case 7:
  case 8:
  case 9:
  case 10:
  case 11:
  case 12:
  case 13:
  case 14:
  case 15:
  case 16: {
    return 1;
  } break;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<xtensa_op2p1>:
  endbr64
  sub    $0x1,%edi
  cmp    $0xf,%edi
  setbe  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int xtensa_op2p1(v)
int v;
{
  switch (v) {
  case 1:
  case 2:
  case 3:
  case 4:
  case 5:
  case 6:
  case 7:
  case 8:
  case 9:
  case 10:
  case 11:
  case 12:
  case 13:
  case 14:
  case 15:
  case 16: {
    return 1;
  } break;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<xtensa_op2p1>:
  endbr64
  sub    $0x1,%edi
  xor    %eax,%eax
  cmp    $0xf,%edi
  setbe  %al
  ret
```

assistant: ```c
int xtensa_op2p1(v)
int v;
{
  switch (v) {
  case 1:
  case 2:
  case 3:
  case 4:
  case 5:
  case 6:
  case 7:
  case 8:
  case 9:
  case 10:
  case 11:
  case 12:
  case 13:
  case 14:
  case 15:
  case 16: {
    return 1;
  } break;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<xtensa_op2p1>:
  endbr64
  sub    $0x1,%edi
  xor    %eax,%eax
  cmp    $0xf,%edi
  setbe  %al
  ret
```

assistant: ```c
int xtensa_op2p1(v)
int v;
{
  switch (v) {
  case 1:
  case 2:
  case 3:
  case 4:
  case 5:
  case 6:
  case 7:
  case 8:
  case 9:
  case 10:
  case 11:
  case 12:
  case 13:
  case 14:
  case 15:
  case 16: {
    return 1;
  } break;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D16(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    D4(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%edx
  mov    0x10(%rsp),%eax
  mov    %eax,%esi
  cmp    %eax,%edx
  je     L81
  mov    0x14(%rsp),%esi
  cmp    %esi,%edx
  je     L81
  cmp    %esi,%eax
  je     L81
  cmp    %esi,%edx
  setl   %cl
  cmp    %eax,%edx
  setg   %r8b
  test   %cl,%cl
  je     L65
  test   %r8b,%r8b
  jne    L84
L65:
  cmp    %eax,%edx
  setl   %dil
  cmp    %esi,%edx
  setg   %r9b
  test   %dil,%dil
  je     L64
  test   %r9b,%r9b
  jne    L124
L64:
  cmp    %esi,%eax
  setl   %dl
  test   %dil,%dil
  je     L115
  test   %dl,%dl
  jne    L41
L115:
  cmp    %esi,%eax
  setg   %dl
  test   %r8b,%r8b
  je     L81
  test   %dl,%dl
  je     L81
  mov    %eax,%esi
  jmp    L81
L84:
  mov    %edx,%esi
L81:
  lea    D27(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L116
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L124:
  mov    %edx,%esi
  jmp    L81
L41:
  mov    %eax,%esi
  jmp    L81
L116:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D16
data_type: string
name: retrieve_data
arguments: tag: D4
data_type: string
name: retrieve_data
arguments: tag: D27
data_type: string

tool: D16: Please enter 3 numbers separated by spaces > 

tool: D4: %d %d %d

tool: D27: %d is the median


assistant: ```c
int main() {
  int a, b, c, m;
  printf("Please enter 3 numbers separated by spaces > ");
  scanf("%d %d %d", &a, &b, &c);
  if (a == b || a == c)
    m = a;
  else if (b == c || b == a)
    m = b;
  else if (c == a || c == b)
    m = c;
  else {
    if ((a > b && a < c) || (a > c && a < b))
      m = a;
    else if ((b > a && b < c) || (b > c && b < a))
      m = b;
    else if ((c > a && c < b) || (c > b && c < a))
      m = c;
  }
  printf("%d is the median\n", m);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D16(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    0x14(%rsp),%rcx
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  mov    0x10(%rsp),%edx
  mov    %eax,%esi
  cmp    %edx,%eax
  je     L91
  mov    0x14(%rsp),%esi
  cmp    %esi,%eax
  je     L91
  cmp    %esi,%edx
  je     L91
  cmp    %esi,%eax
  setl   %cl
  cmp    %edx,%eax
  setg   %dil
  jle    L26
  test   %cl,%cl
  jne    L103
L26:
  cmp    %edx,%eax
  setl   %cl
  cmp    %esi,%eax
  setg   %r8b
  test   %cl,%cl
  je     L129
  test   %r8b,%r8b
  jne    L103
L129:
  cmp    %esi,%edx
  setl   %al
  test   %cl,%cl
  je     L92
  test   %al,%al
  jne    L64
L92:
  cmp    %esi,%edx
  setg   %al
  test   %dil,%dil
  je     L91
  test   %al,%al
  cmovne %edx,%esi
  nopw   0x0(%rax,%rax,1)
L91:
  xor    %eax,%eax
  lea    D26(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L103:
  mov    %eax,%esi
  jmp    L91
L64:
  mov    %edx,%esi
  jmp    L91
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D16
type_infered: string
name: parse_data
arguments: label: D25
type_infered: string
name: parse_data
arguments: label: D26
type_infered: string

tool: D16: Please enter 3 numbers separated by spaces > 

tool: D25: %d %d %d

tool: D26: %d is the median


assistant: ```c
int main() {
  int a, b, c, m;
  printf("Please enter 3 numbers separated by spaces > ");
  scanf("%d %d %d", &a, &b, &c);
  if (a == b || a == c)
    m = a;
  else if (b == c || b == a)
    m = b;
  else if (c == a || c == b)
    m = c;
  else {
    if ((a > b && a < c) || (a > c && a < b))
      m = a;
    else if ((b > a && b < c) || (b > c && b < a))
      m = b;
    else if ((c > a && c < b) || (c > b && c < a))
      m = c;
  }
  printf("%d is the median\n", m);
  return 0;
}
```