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

```asm
<raid6_collect>:
  endbr64
  test   %edi,%edi
  jle    L105
  push   %r14
  xor    %eax,%eax
  push   %r13
  movslq %edi,%r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L3
  nopl   0x0(%rax)
L49:
  movl   $0xffffffff,(%r9,%rax,4)
  mov    %r10d,%edi
L84:
  test   %dil,%dil
  je     L27
  mov    D9(%rip),%rsi
  mov    (%rsi,%rcx,8),%rdx
  sub    (%rsi,%r11,8),%edx
  lea    0xff(%rdx),%esi
  cmovs  %esi,%edx
  mov    %edx,(%r9,%rax,4)
L27:
  or     %r11,%rcx
  jne    L67
  movl   $0xffffff01,(%r9,%rax,4)
L67:
  add    $0x1,%rax
  cmp    %r13,%rax
  je     L37
L3:
  movsbq (%r12,%rax,1),%rsi
  movsbq (%r8,%rax,1),%r14
  mov    0x0(%rbp,%rax,8),%r10
  mov    (%rbx,%rax,8),%rdx
  mov    %rsi,%r11
  mov    %r14,%rcx
  xor    %r10,%rcx
  xor    %rdx,%r11
  cmp    %rdx,%rsi
  setne  %dil
  cmp    %r10,%r14
  setne  %r10b
  jne    L82
  test   %dil,%dil
  jne    L49
L82:
  cmp    %rdx,%rsi
  jne    L20
  test   %r10b,%r10b
  je     L20
  movl   $0xfffffffe,(%r9,%rax,4)
  jmp    L84
  nopw   0x0(%rax,%rax,1)
L37:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L105:
  ret
L20:
  and    %r10d,%edi
  jmp    L84
```

assistant: ```c
void raid6_collect(int chunk_size, uint8_t *p, uint8_t *q, char *chunkP, char *chunkQ, int *results) {
  int i;
  int data_id;
  uint8_t Px, Qx;
  extern uint8_t raid6_gflog[];
  for (i = 0; i < chunk_size; i++) {
    Px = (uint8_t)chunkP[i] ^ (uint8_t)p[i];
    Qx = (uint8_t)chunkQ[i] ^ (uint8_t)q[i];
    if ((Px != 0) && (Qx == 0))
      results[i] = -1;
    if ((Px == 0) && (Qx != 0))
      results[i] = -2;
    if ((Px != 0) && (Qx != 0)) {
      data_id = (raid6_gflog[Qx] - raid6_gflog[Px]);
      if (data_id < 0)
        data_id += 255;
      results[i] = data_id;
    }
    if ((Px == 0) && (Qx == 0))
      results[i] = -255;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<raid6_collect>:
  endbr64
  test   %edi,%edi
  jle    L11
  push   %r14
  xor    %eax,%eax
  push   %r13
  movslq %edi,%r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L14
  nopl   0x0(%rax)
L13:
  movl   $0xffffffff,(%r9,%rax,4)
  mov    %r10d,%edi
L47:
  test   %dil,%dil
  je     L87
  mov    D1(%rip),%rsi
  mov    (%rsi,%rcx,8),%rdx
  sub    (%rsi,%r11,8),%edx
  lea    0xff(%rdx),%esi
  cmovs  %esi,%edx
  mov    %edx,(%r9,%rax,4)
L87:
  or     %r11,%rcx
  jne    L46
  movl   $0xffffff01,(%r9,%rax,4)
L46:
  add    $0x1,%rax
  cmp    %r13,%rax
  je     L34
L14:
  movsbq (%r12,%rax,1),%rsi
  movsbq (%r8,%rax,1),%r14
  mov    0x0(%rbp,%rax,8),%r10
  mov    (%rbx,%rax,8),%rdx
  mov    %rsi,%r11
  mov    %r14,%rcx
  xor    %r10,%rcx
  xor    %rdx,%r11
  cmp    %rdx,%rsi
  setne  %dil
  cmp    %r10,%r14
  setne  %r10b
  jne    L96
  test   %dil,%dil
  jne    L13
L96:
  cmp    %rdx,%rsi
  jne    L62
  test   %r10b,%r10b
  je     L62
  movl   $0xfffffffe,(%r9,%rax,4)
  jmp    L47
  nopw   0x0(%rax,%rax,1)
L34:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L11:
  ret
L62:
  and    %r10d,%edi
  jmp    L47
```

assistant: ```c
void raid6_collect(int chunk_size, uint8_t *p, uint8_t *q, char *chunkP, char *chunkQ, int *results) {
  int i;
  int data_id;
  uint8_t Px, Qx;
  extern uint8_t raid6_gflog[];
  for (i = 0; i < chunk_size; i++) {
    Px = (uint8_t)chunkP[i] ^ (uint8_t)p[i];
    Qx = (uint8_t)chunkQ[i] ^ (uint8_t)q[i];
    if ((Px != 0) && (Qx == 0))
      results[i] = -1;
    if ((Px == 0) && (Qx != 0))
      results[i] = -2;
    if ((Px != 0) && (Qx != 0)) {
      data_id = (raid6_gflog[Qx] - raid6_gflog[Px]);
      if (data_id < 0)
        data_id += 255;
      results[i] = data_id;
    }
    if ((Px == 0) && (Qx == 0))
      results[i] = -255;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<socket_send>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    -0x18(%rbp),%edx
  mov    -0x20(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  mov    $0x0,%ecx
  mov    %eax,%edi
  call   <send@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int socket_send(int sock, const char *buff, int buff_len) {
  int rv = send(sock, buff, buff_len, 0);
  return rv;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<socket_send>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%ecx
  call   <send@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int socket_send(int sock, const char *buff, int buff_len) {
  int rv = send(sock, buff, buff_len, 0);
  return rv;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<socket_send>:
  endbr64
  xor    %ecx,%ecx
  jmp    1050 <send@plt>
```

assistant: ```c
int socket_send(int sock, const char *buff, int buff_len) {
  int rv = send(sock, buff, buff_len, 0);
  return rv;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<socket_send>:
  endbr64
  xor    %ecx,%ecx
  jmp    1050 <send@plt>
```

assistant: ```c
int socket_send(int sock, const char *buff, int buff_len) {
  int rv = send(sock, buff, buff_len, 0);
  return rv;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<crnl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <cr@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <nl@plt>
  nop
  leave
  ret
```

assistant: ```c
void crnl(struct terminal *term) {
  cr(term);
  nl(term);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<crnl>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <cr@plt>
  mov    %rbx,%rdi
  call   <nl@plt>
  pop    %rbx
  ret
```

assistant: ```c
void crnl(struct terminal *term) {
  cr(term);
  nl(term);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<crnl>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <cr@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <nl@plt>
```

assistant: ```c
void crnl(struct terminal *term) {
  cr(term);
  nl(term);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<crnl>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <cr@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <nl@plt>
```

assistant: ```c
void crnl(struct terminal *term) {
  cr(term);
  nl(term);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%edi
  call   <nativeInvert@plt>
  test   %rax,%rax
  je     L13
  mov    $0x0,%eax
  jmp    L4
L13:
  mov    $0x1,%eax
L4:
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  if (nativeInvert(false)) {
    return 0;
  } else {
    return 1;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <nativeInvert@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  if (nativeInvert(false)) {
    return 0;
  } else {
    return 1;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <nativeInvert@plt>
  test   %rax,%rax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int main() {
  if (nativeInvert(false)) {
    return 0;
  } else {
    return 1;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <nativeInvert@plt>
  test   %rax,%rax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int main() {
  if (nativeInvert(false)) {
    return 0;
  } else {
    return 1;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<append_buf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    %r8,-0x28(%rbp)
  cmpq   $0x80000,-0x10(%rbp)
  setbe  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  cmp    $0x80000,%rax
  jbe    L12
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <flush_buf@plt>
  test   %rax,%rax
  jns    L12
  mov    $0xffffffff,%eax
  jmp    L8
L12:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,(%rax)
  mov    $0x0,%eax
L8:
  leave
  ret
```

assistant: ```c
int append_buf(const char *str, size_t len, int outfd, char *outbuf, size_t *outlen) {
  assert(len <= 524288);
  if (*outlen + len > 524288)
    if (flush_buf(outfd, outbuf, outlen) < 0)
      return -1;
  memcpy(outbuf + *outlen, str, len);
  *outlen += len;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<append_buf>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    %edx,%r14d
  mov    %rcx,%r12
  mov    %r8,%rbx
  cmp    $0x80000,%rsi
  setbe  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbp,%rax
  add    (%rbx),%rax
  cmp    $0x80000,%rax
  jbe    L16
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %r14d,%edi
  call   <flush_buf@plt>
  test   %rax,%rax
  js     L20
L16:
  mov    %r12,%rdi
  add    (%rbx),%rdi
  mov    %rbp,%rdx
  mov    %r13,%rsi
  call   <memcpy@plt>
  add    %rbp,(%rbx)
  mov    $0x0,%eax
L10:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L20:
  mov    $0xffffffff,%eax
  jmp    L10
```

assistant: ```c
int append_buf(const char *str, size_t len, int outfd, char *outbuf, size_t *outlen) {
  assert(len <= 524288);
  if (*outlen + len > 524288)
    if (flush_buf(outfd, outbuf, outlen) < 0)
      return -1;
  memcpy(outbuf + *outlen, str, len);
  *outlen += len;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<append_buf>:
  endbr64
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %rdi,%r13
  xor    %edi,%edi
  cmp    $0x80000,%rsi
  push   %r12
  setbe  %dil
  mov    %rcx,%r12
  push   %rbp
  mov    %r8,%rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <assert@plt>
  mov    0x0(%rbp),%rax
  lea    (%rax,%rbx,1),%rdx
  cmp    $0x80000,%rdx
  jbe    L24
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %r14d,%edi
  call   <flush_buf@plt>
  test   %rax,%rax
  js     L26
  mov    0x0(%rbp),%rax
L24:
  lea    (%r12,%rax,1),%rdi
  mov    %rbx,%rdx
  mov    %r13,%rsi
  call   <memcpy@plt>
  add    %rbx,0x0(%rbp)
  xor    %eax,%eax
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L26:
  mov    $0xffffffff,%eax
  jmp    L9
```

assistant: ```c
int append_buf(const char *str, size_t len, int outfd, char *outbuf, size_t *outlen) {
  assert(len <= 524288);
  if (*outlen + len > 524288)
    if (flush_buf(outfd, outbuf, outlen) < 0)
      return -1;
  memcpy(outbuf + *outlen, str, len);
  *outlen += len;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<append_buf>:
  endbr64
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %rdi,%r13
  xor    %edi,%edi
  cmp    $0x80000,%rsi
  push   %r12
  setbe  %dil
  mov    %rcx,%r12
  push   %rbp
  mov    %r8,%rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <assert@plt>
  mov    0x0(%rbp),%rax
  lea    (%rax,%rbx,1),%rdx
  cmp    $0x80000,%rdx
  jbe    L7
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %r14d,%edi
  call   <flush_buf@plt>
  test   %rax,%rax
  js     L17
  mov    0x0(%rbp),%rax
L7:
  lea    (%r12,%rax,1),%rdi
  mov    %rbx,%rdx
  mov    %r13,%rsi
  call   <memcpy@plt>
  add    %rbx,0x0(%rbp)
  xor    %eax,%eax
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L17:
  mov    $0xffffffff,%eax
  jmp    L9
```

assistant: ```c
int append_buf(const char *str, size_t len, int outfd, char *outbuf, size_t *outlen) {
  assert(len <= 524288);
  if (*outlen + len > 524288)
    if (flush_buf(outfd, outbuf, outlen) < 0)
      return -1;
  memcpy(outbuf + *outlen, str, len);
  *outlen += len;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<populate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  jmp    L17
L4:
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    -0xc(%rbp),%edx
  mov    %edx,(%rax)
L17:
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  test   %eax,%eax
  jg     L4
  nop
  pop    %rbp
  ret
```

assistant: ```c
populate(int *p, int n) {
  while (n-- > 0)
    *p++ = n;
}
```
user: What is the c source code of the assembly code below?

```asm
<populate>:
  endbr64
  lea    -0x1(%rsi),%eax
  test   %esi,%esi
  jle    L0
L17:
  add    $0x4,%rdi
  mov    %eax,-0x4(%rdi)
  sub    $0x1,%eax
  cmp    $0xffffffff,%eax
  jne    L17
L0:
  ret
```

assistant: ```c
populate(int *p, int n) {
  while (n-- > 0)
    *p++ = n;
}
```
user: What is the c source code of the assembly code below:

```asm
<populate>:
  endbr64
  lea    -0x1(%rsi),%eax
  test   %esi,%esi
  jle    L9
  nopl   0x0(%rax,%rax,1)
L5:
  add    $0x4,%rdi
  mov    %eax,-0x4(%rdi)
  sub    $0x1,%eax
  jae    L5
L9:
  ret
```

assistant: ```c
populate(int *p, int n) {
  while (n-- > 0)
    *p++ = n;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<populate>:
  endbr64
  lea    -0x1(%rsi),%edx
  test   %esi,%esi
  jle    L39
  cmp    $0x2,%edx
  jbe    L50
  mov    %esi,%ecx
  movd   %esi,%xmm3
  movdqa D12(%rip),%xmm2
  mov    %rdi,%rax
  shr    $0x2,%ecx
  pshufd $0x0,%xmm3,%xmm0
  paddd  D7(%rip),%xmm0
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rdi,%rcx,1),%rcx
  nopl   (%rax)
L41:
  movdqa %xmm0,%xmm1
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  movups %xmm1,-0x10(%rax)
  cmp    %rcx,%rax
  jne    L41
  mov    %esi,%eax
  and    $0xfffffffc,%eax
  mov    %eax,%ecx
  sub    %eax,%edx
  lea    (%rdi,%rcx,4),%rdi
  cmp    %eax,%esi
  je     L39
L50:
  mov    %edx,(%rdi)
  lea    -0x1(%rdx),%eax
  test   %edx,%edx
  jle    L39
  mov    %eax,0x4(%rdi)
  sub    $0x2,%edx
  test   %eax,%eax
  jle    L39
  mov    %edx,0x8(%rdi)
L39:
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D12
type: byte[16]
name: retrieve_data
arguments: tag: D7
type: byte[16]

tool: {"D12": [0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff]}

tool: {"D7": [0xff,0xff,0xff,0xff,0xfe,0xff,0xff,0xff,0xfd,0xff,0xff,0xff,0xfc,0xff,0xff,0xff]}

assistant: ```c
populate(int *p, int n) {
  while (n-- > 0)
    *p++ = n;
}
```
user: Translate the below assembly code into C programming language:

```asm
<_ht_hash_ptr>:
  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    -0x20(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <_ht_hash_int@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
unsigned int _ht_hash_ptr(void *ht, void *d) {
  union {
    int i;
    void *p;
  } u;
  u.p = d;
  return _ht_hash_int(ht, &u.i);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<_ht_hash_ptr>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsi,(%rsp)
  mov    %rsp,%rsi
  call   <_ht_hash_int@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned int _ht_hash_ptr(void *ht, void *d) {
  union {
    int i;
    void *p;
  } u;
  u.p = d;
  return _ht_hash_int(ht, &u.i);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<_ht_hash_ptr>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsi,(%rsp)
  mov    %rsp,%rsi
  call   <_ht_hash_int@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned int _ht_hash_ptr(void *ht, void *d) {
  union {
    int i;
    void *p;
  } u;
  u.p = d;
  return _ht_hash_int(ht, &u.i);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<_ht_hash_ptr>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsi,(%rsp)
  mov    %rsp,%rsi
  call   <_ht_hash_int@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned int _ht_hash_ptr(void *ht, void *d) {
  union {
    int i;
    void *p;
  } u;
  u.p = d;
  return _ht_hash_int(ht, &u.i);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L14
L78:
  movl   $0x0,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L74
  movl   $0x1,-0x14(%rbp)
  jmp    L65
L11:
  mov    -0x18(%rbp),%eax
  cltd
  idivl  -0x14(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L44
  addl   $0x1,-0x10(%rbp)
L44:
  addl   $0x1,-0x14(%rbp)
L65:
  mov    -0x14(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L11
  cmpl   $0x2,-0x10(%rbp)
  jne    L74
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L74:
  addl   $0x1,-0x18(%rbp)
L14:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jle    L78
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L33
  call   <__stack_chk_fail@plt>
L33:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D0
datatype: string
name: unpack_data
arguments: tag: D9
datatype: string
name: unpack_data
arguments: tag: D8
datatype: string
name: unpack_data
arguments: tag: D31
datatype: string

tool: D0: 
Enter a number:

tool: D9: %d

tool: D8: 
%d is a prime factor

tool: D31: 
%d

assistant: ```c
int main() {
  int num, i = 1, j, k, c = 0;
  printf("\nEnter a number:");
  scanf("%d", &num);
  while (i <= num) {
    k = 0;
    if (num % i == 0) {
      j = 1;
      while (j <= i) {
        if (i % j == 0)
          k++;
        j++;
      }
      if (k == 2) {
        printf("\n%d is a prime factor", i);
        c++;
      }
    }
    i++;
  }
  printf("\n%d", c);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D11(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D33(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L56
  mov    $0x0,%ebp
  mov    $0x1,%ebx
  lea    D2(%rip),%r12
  jmp    L46
L14:
  add    $0x1,%ebx
  mov    0x4(%rsp),%eax
  cmp    %ebx,%eax
  jl     L35
L46:
  cltd
  idiv   %ebx
  mov    %edx,%esi
  test   %edx,%edx
  jne    L14
  test   %ebx,%ebx
  jle    L14
  mov    $0x1,%ecx
  lea    0x1(%rbx),%edi
L17:
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  cmp    $0x1,%edx
  adc    $0x0,%esi
  add    $0x1,%ecx
  cmp    %edi,%ecx
  jne    L17
  cmp    $0x2,%esi
  jne    L14
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebp
  jmp    L14
L56:
  mov    $0x0,%ebp
L35:
  mov    %ebp,%esi
  lea    D30(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D11
data_type: string
name: disassemble_data
arguments: data_label: D33
data_type: string
name: disassemble_data
arguments: data_label: D2
data_type: string
name: disassemble_data
arguments: data_label: D30
data_type: string

tool: D11: 
Enter a number:

tool: D33: %d

tool: D2: 
%d is a prime factor

tool: D30: 
%d

assistant: ```c
int main() {
  int num, i = 1, j, k, c = 0;
  printf("\nEnter a number:");
  scanf("%d", &num);
  while (i <= num) {
    k = 0;
    if (num % i == 0) {
      j = 1;
      while (j <= i) {
        if (i % j == 0)
          k++;
        j++;
      }
      if (k == 2) {
        printf("\n%d is a prime factor", i);
        c++;
      }
    }
    i++;
  }
  printf("\n%d", c);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r12
  lea    D28(%rip),%rdi
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D0(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r8d
  test   %r8d,%r8d
  jle    L65
  mov    $0x1,%esi
  lea    D19(%rip),%r12
  jmp    L54
  nopl   0x0(%rax,%rax,1)
L35:
  mov    %ebx,%esi
  cmp    %ebx,%r8d
  jl     L65
L54:
  mov    %r8d,%eax
  lea    0x1(%rsi),%ebx
  cltd
  idiv   %esi
  mov    %edx,%edi
  test   %edx,%edx
  jne    L35
  mov    $0x1,%ecx
  nopl   0x0(%rax,%rax,1)
L60:
  mov    %esi,%eax
  cltd
  idiv   %ecx
  cmp    $0x1,%edx
  adc    $0x0,%edi
  add    $0x1,%ecx
  cmp    %ebx,%ecx
  jne    L60
  cmp    $0x2,%edi
  jne    L35
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <printf@plt>
  mov    0x4(%rsp),%r8d
  jmp    L35
  nopl   0x0(%rax,%rax,1)
L65:
  xor    %eax,%eax
  mov    %ebp,%esi
  lea    D32(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D28
datatype: string
name: retrieve_data
arguments: data_tag: D0
datatype: string
name: retrieve_data
arguments: data_tag: D19
datatype: string
name: retrieve_data
arguments: data_tag: D32
datatype: string

tool: D28: 
Enter a number:

tool: D0: %d

tool: D19: 
%d is a prime factor

tool: D32: 
%d

assistant: ```c
int main() {
  int num, i = 1, j, k, c = 0;
  printf("\nEnter a number:");
  scanf("%d", &num);
  while (i <= num) {
    k = 0;
    if (num % i == 0) {
      j = 1;
      while (j <= i) {
        if (i % j == 0)
          k++;
        j++;
      }
      if (k == 2) {
        printf("\n%d is a prime factor", i);
        c++;
      }
    }
    i++;
  }
  printf("\n%d", c);
  return 0;
}
```