ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ieee_p_1242562249_sub_800139796931496790_1035706684>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r13
  mov    %rcx,%r12
  mov    %r8d,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    0xc(%rcx),%ecx
  lea    0x10(%rsp),%r14
  mov    %r14,%rsi
  call   <ieee_p_1242562249_sub_17126692536656888728_1035706684@plt>
  mov    %rax,%r8
  mov    %rsp,%rsi
  mov    %r14,%r9
  mov    %r12,%rcx
  mov    %r13,%rdx
  mov    %rbp,%rdi
  call   <ieee_p_1242562249_sub_3525738511873186323_1035706684@plt>
  mov    %rax,%r12
  mov    0xc(%rsp),%r13d
  mov    %r13d,%edi
  call   <xsi_get_transient_memory@plt>
  mov    %rax,%rbp
  mov    %r13d,%edx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    (%rsp),%ecx
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%edx
  mov    %ecx,(%rbx)
  mov    %eax,0x4(%rbx)
  mov    %edx,0x8(%rbx)
  sub    %ecx,%eax
  imul   %edx,%eax
  add    $0x1,%eax
  mov    %eax,0xc(%rbx)
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    %rbp,%rax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ieee_p_1242562249_sub_800139796931496790_1035706684(char *t1, char *t2, char *t3, char *t4, int t5) {
  char t7[24];
  char t12[16];
  char t13[16];
  char *t0;
  char *t8;
  unsigned char t9;
  char *t10;
  char *t11;
  char *t14;
  unsigned int t15;
  char *t16;
  char *t17;
  char *t18;
  unsigned int t19;
  unsigned int t20;
  char *t21;
  int t22;
  char *t23;
  int t24;
  char *t25;
  int t26;
  char *t27;
  char *t28;
  int t29;
  unsigned int t30;
LAB0:
  t8 = (t7 + 4U);
  t9 = (t3 != 0);
  if (t9 == 1)
    goto LAB3;
LAB2:
  t10 = (t7 + 12U);
  *((char **)t10) = t4;
  t11 = (t7 + 20U);
  *((int *)t11) = t5;
  t14 = (t4 + 12U);
  t15 = *((unsigned int *)t14);
  t16 = ieee_p_1242562249_sub_17126692536656888728_1035706684(t1, t13, t5, ((t15)));
  t17 = ieee_p_1242562249_sub_3525738511873186323_1035706684(t1, t12, t3, t4, t16, t13);
  t18 = (t12 + 12U);
  t19 = *((unsigned int *)t18);
  t20 = (1U * t19);
  t0 = xsi_get_transient_memory(t20);
  memcpy(t0, t17, t20);
  t21 = (t12 + 0U);
  t22 = *((int *)t21);
  t23 = (t12 + 4U);
  t24 = *((int *)t23);
  t25 = (t12 + 8U);
  t26 = *((int *)t25);
  t27 = (t2 + 0U);
  t28 = (t27 + 0U);
  *((int *)t28) = t22;
  t28 = (t27 + 4U);
  *((int *)t28) = t24;
  t28 = (t27 + 8U);
  *((int *)t28) = t26;
  t29 = (t24 - t22);
  t30 = (t29 * t26);
  t30 = (t30 + 1);
  t28 = (t27 + 12U);
  *((unsigned int *)t28) = t30;
LAB1:
  return t0;
LAB3:
  *((char **)t8) = t3;
  goto LAB2;
LAB4:;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ieee_p_1242562249_sub_800139796931496790_1035706684>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdx,%r13
  mov    %r8d,%edx
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x30,%rsp
  mov    0xc(%rcx),%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r14
  mov    %r14,%rsi
  call   <ieee_p_1242562249_sub_17126692536656888728_1035706684@plt>
  mov    %r12,%rcx
  mov    %r13,%rdx
  mov    %rsp,%rsi
  mov    %rax,%r8
  mov    %r14,%r9
  mov    %rbp,%rdi
  call   <ieee_p_1242562249_sub_3525738511873186323_1035706684@plt>
  mov    0xc(%rsp),%r13d
  mov    %rax,%rbp
  mov    %r13d,%edi
  call   <xsi_get_transient_memory@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memcpy@plt>
  mov    (%rsp),%ecx
  mov    0x4(%rsp),%edx
  mov    0x8(%rsp),%eax
  mov    %edx,0x4(%rbx)
  sub    %ecx,%edx
  imul   %eax,%edx
  mov    %ecx,(%rbx)
  mov    %eax,0x8(%rbx)
  add    $0x1,%edx
  mov    %edx,0xc(%rbx)
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x30,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ieee_p_1242562249_sub_800139796931496790_1035706684(char *t1, char *t2, char *t3, char *t4, int t5) {
  char t7[24];
  char t12[16];
  char t13[16];
  char *t0;
  char *t8;
  unsigned char t9;
  char *t10;
  char *t11;
  char *t14;
  unsigned int t15;
  char *t16;
  char *t17;
  char *t18;
  unsigned int t19;
  unsigned int t20;
  char *t21;
  int t22;
  char *t23;
  int t24;
  char *t25;
  int t26;
  char *t27;
  char *t28;
  int t29;
  unsigned int t30;
LAB0:
  t8 = (t7 + 4U);
  t9 = (t3 != 0);
  if (t9 == 1)
    goto LAB3;
LAB2:
  t10 = (t7 + 12U);
  *((char **)t10) = t4;
  t11 = (t7 + 20U);
  *((int *)t11) = t5;
  t14 = (t4 + 12U);
  t15 = *((unsigned int *)t14);
  t16 = ieee_p_1242562249_sub_17126692536656888728_1035706684(t1, t13, t5, ((t15)));
  t17 = ieee_p_1242562249_sub_3525738511873186323_1035706684(t1, t12, t3, t4, t16, t13);
  t18 = (t12 + 12U);
  t19 = *((unsigned int *)t18);
  t20 = (1U * t19);
  t0 = xsi_get_transient_memory(t20);
  memcpy(t0, t17, t20);
  t21 = (t12 + 0U);
  t22 = *((int *)t21);
  t23 = (t12 + 4U);
  t24 = *((int *)t23);
  t25 = (t12 + 8U);
  t26 = *((int *)t25);
  t27 = (t2 + 0U);
  t28 = (t27 + 0U);
  *((int *)t28) = t22;
  t28 = (t27 + 4U);
  *((int *)t28) = t24;
  t28 = (t27 + 8U);
  *((int *)t28) = t26;
  t29 = (t24 - t22);
  t30 = (t29 * t26);
  t30 = (t30 + 1);
  t28 = (t27 + 12U);
  *((unsigned int *)t28) = t30;
LAB1:
  return t0;
LAB3:
  *((char **)t8) = t3;
  goto LAB2;
LAB4:;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ieee_p_1242562249_sub_800139796931496790_1035706684>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdx,%r13
  mov    %r8d,%edx
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x30,%rsp
  mov    0xc(%rcx),%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r14
  mov    %r14,%rsi
  call   <ieee_p_1242562249_sub_17126692536656888728_1035706684@plt>
  mov    %r12,%rcx
  mov    %r13,%rdx
  mov    %rsp,%rsi
  mov    %rax,%r8
  mov    %r14,%r9
  mov    %rbp,%rdi
  call   <ieee_p_1242562249_sub_3525738511873186323_1035706684@plt>
  mov    0xc(%rsp),%r13d
  mov    %rax,%rbp
  mov    %r13d,%edi
  call   <xsi_get_transient_memory@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memcpy@plt>
  movq   (%rsp),%xmm0
  mov    0x8(%rsp),%eax
  movd   %xmm0,%ecx
  pshufd $0xe5,%xmm0,%xmm1
  movd   %xmm1,%edx
  mov    %eax,0x8(%rbx)
  sub    %ecx,%edx
  movq   %xmm0,(%rbx)
  imul   %eax,%edx
  add    $0x1,%edx
  mov    %edx,0xc(%rbx)
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x30,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ieee_p_1242562249_sub_800139796931496790_1035706684(char *t1, char *t2, char *t3, char *t4, int t5) {
  char t7[24];
  char t12[16];
  char t13[16];
  char *t0;
  char *t8;
  unsigned char t9;
  char *t10;
  char *t11;
  char *t14;
  unsigned int t15;
  char *t16;
  char *t17;
  char *t18;
  unsigned int t19;
  unsigned int t20;
  char *t21;
  int t22;
  char *t23;
  int t24;
  char *t25;
  int t26;
  char *t27;
  char *t28;
  int t29;
  unsigned int t30;
LAB0:
  t8 = (t7 + 4U);
  t9 = (t3 != 0);
  if (t9 == 1)
    goto LAB3;
LAB2:
  t10 = (t7 + 12U);
  *((char **)t10) = t4;
  t11 = (t7 + 20U);
  *((int *)t11) = t5;
  t14 = (t4 + 12U);
  t15 = *((unsigned int *)t14);
  t16 = ieee_p_1242562249_sub_17126692536656888728_1035706684(t1, t13, t5, ((t15)));
  t17 = ieee_p_1242562249_sub_3525738511873186323_1035706684(t1, t12, t3, t4, t16, t13);
  t18 = (t12 + 12U);
  t19 = *((unsigned int *)t18);
  t20 = (1U * t19);
  t0 = xsi_get_transient_memory(t20);
  memcpy(t0, t17, t20);
  t21 = (t12 + 0U);
  t22 = *((int *)t21);
  t23 = (t12 + 4U);
  t24 = *((int *)t23);
  t25 = (t12 + 8U);
  t26 = *((int *)t25);
  t27 = (t2 + 0U);
  t28 = (t27 + 0U);
  *((int *)t28) = t22;
  t28 = (t27 + 4U);
  *((int *)t28) = t24;
  t28 = (t27 + 8U);
  *((int *)t28) = t26;
  t29 = (t24 - t22);
  t30 = (t29 * t26);
  t30 = (t30 + 1);
  t28 = (t27 + 12U);
  *((unsigned int *)t28) = t30;
LAB1:
  return t0;
LAB3:
  *((char **)t8) = t3;
  goto LAB2;
LAB4:;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<init_hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <clear_user_hash_table@plt>
  mov    $0x0,%eax
  call   <clear_localuser_hash_table@plt>
  mov    $0x0,%eax
  call   <clear_channel_hash_table@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void init_hash(void) {
  clear_user_hash_table();
  clear_localuser_hash_table();
  clear_channel_hash_table();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<init_hash>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <clear_user_hash_table@plt>
  mov    $0x0,%eax
  call   <clear_localuser_hash_table@plt>
  mov    $0x0,%eax
  call   <clear_channel_hash_table@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void init_hash(void) {
  clear_user_hash_table();
  clear_localuser_hash_table();
  clear_channel_hash_table();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<init_hash>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <clear_user_hash_table@plt>
  xor    %eax,%eax
  call   <clear_localuser_hash_table@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1080 <clear_channel_hash_table@plt>
```

assistant: ```c
void init_hash(void) {
  clear_user_hash_table();
  clear_localuser_hash_table();
  clear_channel_hash_table();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<nni_strfree>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L6
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <nni_free@plt>
L6:
  nop
  leave
  ret
```

assistant: ```c
void nni_strfree(char *s) {
  if (s != NULL) {
    nni_free(s, strlen(s) + 1);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<nni_strfree>:
  endbr64
  test   %rdi,%rdi
  je     L9
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rsi
  mov    %rbx,%rdi
  call   <nni_free@plt>
  pop    %rbx
  ret
L9:
  ret
```

assistant: ```c
void nni_strfree(char *s) {
  if (s != NULL) {
    nni_free(s, strlen(s) + 1);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<nni_strfree>:
  endbr64
  test   %rdi,%rdi
  je     L1
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %rbp,%rdi
  pop    %rbp
  lea    0x1(%rax),%rsi
  jmp    1070 <nni_free@plt>
  nop
L1:
  ret
```

assistant: ```c
void nni_strfree(char *s) {
  if (s != NULL) {
    nni_free(s, strlen(s) + 1);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<nni_strfree>:
  endbr64
  test   %rdi,%rdi
  je     L4
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %rbp,%rdi
  pop    %rbp
  lea    0x1(%rax),%rsi
  jmp    1070 <nni_free@plt>
  nop
L4:
  ret
```

assistant: ```c
void nni_strfree(char *s) {
  if (s != NULL) {
    nni_free(s, strlen(s) + 1);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Partition>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %esi,%r15d
  mov    %rdx,(%rsp)
  test   %esi,%esi
  jle    L80
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rdx
  mov    $0x0,%r12d
L102:
  add    (%rax),%r12d
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L102
L94:
  mov    %r12d,%ebp
  shr    $0x1f,%ebp
  add    %r12d,%ebp
  sar    %ebp
  lea    0x1(%r15),%r14d
  lea    0x1(%rbp),%r13d
  mov    %r14d,%edi
  imul   %r13d,%edi
  shl    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  imul   %r13d,%r14d
  test   %r14d,%r14d
  jle    L134
  lea    -0x1(%r14),%edx
  lea    0x4(%rax,%rdx,4),%rdx
L148:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L148
L134:
  movl   $0x1,(%rbx)
  test   %r15d,%r15d
  jle    L67
  mov    0x8(%rsp),%rsi
  mov    %rsi,%rdi
  lea    -0x1(%r15),%eax
  lea    0x4(%rsi,%rax,4),%r10
  mov    %r13d,%esi
  mov    $0x0,%r8d
  jmp    L88
L80:
  mov    $0x0,%r12d
  jmp    L94
L26:
  mov    (%rdi),%r11d
  cmp    %eax,%r11d
  jg     L69
  mov    %eax,%ecx
  sub    %r11d,%ecx
  add    %r9d,%ecx
  movslq %ecx,%rcx
  cmpl   $0x0,(%rbx,%rcx,4)
  je     L69
  lea    (%rax,%rsi,1),%ecx
  movslq %ecx,%rcx
  movl   $0x1,(%rbx,%rcx,4)
L69:
  add    $0x1,%eax
  add    $0x4,%rdx
  cmp    %eax,%ebp
  jl     L136
L14:
  cmpl   $0x0,(%rdx)
  je     L26
  lea    (%rax,%rsi,1),%ecx
  movslq %ecx,%rcx
  movl   $0x1,(%rbx,%rcx,4)
  jmp    L69
L136:
  add    %r13d,%esi
  add    %r13d,%r8d
  add    $0x4,%rdi
  cmp    %r10,%rdi
  je     L67
L88:
  cmp    $0xffffffff,%r12d
  jl     L136
  mov    %r8d,%r9d
  movslq %r8d,%rax
  lea    (%rbx,%rax,4),%rdx
  mov    $0x0,%eax
  jmp    L14
L67:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    (%rsp),%rax
  mov    %rbx,(%rax)
  movslq %r14d,%r14
  mov    -0x4(%rbx,%r14,4),%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

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

tool: D4: 


assistant: ```c
int Partition(int *a, int dim, int **mat) {
  int s, i, j, *m;
  for (i = 0, s = 0; i < dim; i++)
    s += a[i];
  s /= 2;
  m = (int *)malloc(sizeof(int) * (dim + 1) * (s + 1));
  for (i = 0; i < (dim + 1) * (s + 1); i++)
    m[i] = 0;
  m[0] = 1;
  for (i = 1; i < dim + 1; i++) {
    for (j = 0; j < s + 1; j++) {
      if (m[(i - 1) * (s + 1) + j])
        m[i * (s + 1) + j] = 1;
      else if (j >= a[i - 1] && m[(i - 1) * (s + 1) + (j - a[i - 1])])
        m[i * (s + 1) + j] = 1;
    }
  }
  printf("\n");
  *mat = m;
  return m[(dim + 1) * (s + 1) - 1];
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<Partition>:
  endbr64
  push   %r15
  mov    %esi,%ecx
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdx,(%rsp)
  test   %esi,%esi
  jle    L114
  lea    -0x1(%rsi),%edx
  mov    %rdi,%rax
  xor    %r14d,%r14d
  lea    0x4(%rdi,%rdx,4),%rdx
  nopl   0x0(%rax)
L20:
  add    (%rax),%r14d
  add    $0x4,%rax
  cmp    %rax,%rdx
  jne    L20
  mov    %r14d,%r12d
  shr    $0x1f,%r12d
  add    %r14d,%r12d
  sar    %r12d
  lea    0x1(%r12),%r15d
  mov    %r15d,%r8d
L4:
  lea    0x1(%rcx),%ebp
  mov    %ecx,0xc(%rsp)
  mov    %ebp,%edi
  imul   %r15d,%ebp
  mov    %r8d,0x8(%rsp)
  imul   %r8d,%edi
  shl    $0x2,%edi
  call   <malloc@plt>
  test   %ebp,%ebp
  mov    0x8(%rsp),%r8d
  mov    0xc(%rsp),%ecx
  mov    %rax,%rbx
  jle    L113
  movslq %ebp,%rbp
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %ecx,0xc(%rsp)
  lea    0x0(,%rbp,4),%rdx
  mov    %r8d,0x8(%rsp)
  call   <memset@plt>
  mov    0x8(%rsp),%r8d
  mov    0xc(%rsp),%ecx
L45:
  movl   $0x1,(%rbx)
  test   %ecx,%ecx
  jle    L44
  lea    -0x1(%rcx),%eax
  mov    %r8d,%r10d
  xor    %r9d,%r9d
  lea    0x4(%r13,%rax,4),%r11
  nopw   0x0(%rax,%rax,1)
L56:
  cmp    $0xffffffff,%r14d
  jl     L9
  movslq %r9d,%rax
  mov    %r10d,%ecx
  lea    (%rbx,%rax,4),%rdx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L15:
  mov    (%rdx),%edi
  test   %edi,%edi
  jne    L108
  mov    0x0(%r13),%edi
  cmp    %eax,%edi
  jg     L59
  mov    %eax,%esi
  sub    %edi,%esi
  add    %r9d,%esi
  movslq %esi,%rsi
  mov    (%rbx,%rsi,4),%esi
  test   %esi,%esi
  je     L59
  nopw   0x0(%rax,%rax,1)
L108:
  movslq %ecx,%rsi
  movl   $0x1,(%rbx,%rsi,4)
L59:
  add    $0x1,%eax
  add    $0x4,%rdx
  add    $0x1,%ecx
  cmp    %r12d,%eax
  jle    L15
L9:
  add    $0x4,%r13
  add    %r15d,%r9d
  add    %r8d,%r10d
  cmp    %r13,%r11
  jne    L56
L44:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    (%rsp),%rax
  mov    %rbx,(%rax)
  mov    -0x4(%rbx,%rbp,4),%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L114:
  mov    $0x1,%r8d
  mov    $0x1,%r15d
  xor    %r12d,%r12d
  xor    %r14d,%r14d
  jmp    L4
L113:
  movslq %ebp,%rbp
  jmp    L45
```

tool_calls: name: disassemble_data
arguments: tag: D7
datatype: string

tool: D7: 


assistant: ```c
int Partition(int *a, int dim, int **mat) {
  int s, i, j, *m;
  for (i = 0, s = 0; i < dim; i++)
    s += a[i];
  s /= 2;
  m = (int *)malloc(sizeof(int) * (dim + 1) * (s + 1));
  for (i = 0; i < (dim + 1) * (s + 1); i++)
    m[i] = 0;
  m[0] = 1;
  for (i = 1; i < dim + 1; i++) {
    for (j = 0; j < s + 1; j++) {
      if (m[(i - 1) * (s + 1) + j])
        m[i * (s + 1) + j] = 1;
      else if (j >= a[i - 1] && m[(i - 1) * (s + 1) + (j - a[i - 1])])
        m[i * (s + 1) + j] = 1;
    }
  }
  printf("\n");
  *mat = m;
  return m[(dim + 1) * (s + 1) - 1];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<log_level_to_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x4,-0x4(%rbp)
  ja     L21
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D28(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D28(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D29(%rip),%rax
  jmp    L45
  lea    D7(%rip),%rax
  jmp    L45
  lea    D32(%rip),%rax
  jmp    L45
  lea    D59(%rip),%rax
  jmp    L45
  lea    D40(%rip),%rax
  jmp    L45
L21:
  mov    $0x0,%eax
L45:
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D29
data_type: string
name: extract_symbolic_data
arguments: data_label: D7
data_type: string
name: extract_symbolic_data
arguments: data_label: D32
data_type: string
name: extract_symbolic_data
arguments: data_label: D59
data_type: string
name: extract_symbolic_data
arguments: data_label: D40
data_type: string

tool: D29: error

tool: D7: warning

tool: D32: test

tool: D59: info

tool: D40: debug

assistant: ```c
const char *log_level_to_string(unsigned int level) {
  switch (level) {
  case 0:
    return "error";
    break;
  case 1:
    return "warning";
    break;
  case 2:
    return "test";
    break;
  case 3:
    return "info";
    break;
  case 4:
    return "debug";
    break;
  default:
    return NULL;
    break;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<log_level_to_string>:
  endbr64
  cmp    $0x4,%edi
  ja     L7
  mov    %edi,%edi
  lea    D15(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D33(%rip),%rax
  ret
  lea    D38(%rip),%rax
  ret
  lea    D41(%rip),%rax
  ret
  lea    D25(%rip),%rax
  ret
L7:
  mov    $0x0,%eax
  ret
  lea    D8(%rip),%rax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D33
type_infered: string
name: extract_symbolic_data
arguments: label: D38
type_infered: string
name: extract_symbolic_data
arguments: label: D41
type_infered: string
name: extract_symbolic_data
arguments: label: D25
type_infered: string
name: extract_symbolic_data
arguments: label: D8
type_infered: string

tool: D33: warning

tool: D38: test

tool: D41: info

tool: D25: debug

tool: D8: error

assistant: ```c
const char *log_level_to_string(unsigned int level) {
  switch (level) {
  case 0:
    return "error";
    break;
  case 1:
    return "warning";
    break;
  case 2:
    return "test";
    break;
  case 3:
    return "info";
    break;
  case 4:
    return "debug";
    break;
  default:
    return NULL;
    break;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<_skip_hostname>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x28(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L49
  mov    D25(%rip),%rax
  mov    (%rax),%eax
  neg    %eax
  jmp    L7
L49:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0xbf,%eax
  jle    L71
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  cmp    %rax,-0x8(%rbp)
  ja     L34
  mov    D25(%rip),%rax
  mov    (%rax),%eax
  neg    %eax
  jmp    L7
L34:
  mov    $0x2,%eax
  jmp    L7
L69:
  mov    -0xc(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x8(%rbp)
  ja     L71
  mov    D25(%rip),%rax
  mov    (%rax),%eax
  neg    %eax
  jmp    L7
L71:
  mov    -0xc(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L69
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
L7:
  pop    %rbp
  ret
```

assistant: ```c
ssize_t _skip_hostname(const uint8_t *buf, size_t len, const uint8_t *bufpos) {
  const uint8_t *buflim = buf + len;
  unsigned res = 0;
  if (bufpos >= buflim) {
    return -EBADMSG;
  }
  if (*bufpos >= 192) {
    if ((bufpos + 2) >= buflim) {
      return -EBADMSG;
    }
    return 2;
  }
  while (bufpos[res]) {
    res += bufpos[res] + 1;
    if ((&bufpos[res]) >= buflim) {
      return -EBADMSG;
    }
  }
  return res + 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<_skip_hostname>:
  endbr64
  lea    (%rdi,%rsi,4),%rsi
  cmp    %rdx,%rsi
  jbe    L9
  cmpl   $0xbf,(%rdx)
  jle    L12
  add    $0x8,%rdx
  mov    $0x2,%eax
  cmp    %rdx,%rsi
  ja     L36
  mov    D26(%rip),%rax
  mov    (%rax),%eax
  neg    %eax
  ret
L9:
  mov    D26(%rip),%rax
  mov    (%rax),%eax
  neg    %eax
  ret
L12:
  mov    $0x0,%eax
L32:
  mov    %eax,%ecx
  mov    (%rdx,%rcx,4),%ecx
  test   %ecx,%ecx
  je     L1
  lea    0x1(%rcx,%rax,1),%eax
  mov    %eax,%ecx
  lea    (%rdx,%rcx,4),%rcx
  cmp    %rcx,%rsi
  ja     L32
  mov    D26(%rip),%rax
  mov    (%rax),%eax
  neg    %eax
  ret
L1:
  add    $0x1,%eax
L36:
  ret
```

assistant: ```c
ssize_t _skip_hostname(const uint8_t *buf, size_t len, const uint8_t *bufpos) {
  const uint8_t *buflim = buf + len;
  unsigned res = 0;
  if (bufpos >= buflim) {
    return -EBADMSG;
  }
  if (*bufpos >= 192) {
    if ((bufpos + 2) >= buflim) {
      return -EBADMSG;
    }
    return 2;
  }
  while (bufpos[res]) {
    res += bufpos[res] + 1;
    if ((&bufpos[res]) >= buflim) {
      return -EBADMSG;
    }
  }
  return res + 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<_skip_hostname>:
  endbr64
  lea    (%rdi,%rsi,4),%rsi
  cmp    %rdx,%rsi
  jbe    L20
  mov    (%rdx),%ecx
  cmp    $0xbf,%ecx
  jg     L24
  xor    %eax,%eax
  jmp    L6
  nopl   0x0(%rax,%rax,1)
L32:
  lea    0x1(%rcx,%rax,1),%ecx
  mov    %rcx,%rax
  lea    (%rdx,%rcx,4),%rcx
  cmp    %rsi,%rcx
  jae    L20
  mov    (%rcx),%ecx
L6:
  test   %ecx,%ecx
  jne    L32
  add    $0x1,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L24:
  add    $0x8,%rdx
  cmp    %rdx,%rsi
  ja     L8
L20:
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  neg    %eax
  ret
  nopl   (%rax)
L8:
  mov    $0x2,%eax
  ret
```

assistant: ```c
ssize_t _skip_hostname(const uint8_t *buf, size_t len, const uint8_t *bufpos) {
  const uint8_t *buflim = buf + len;
  unsigned res = 0;
  if (bufpos >= buflim) {
    return -EBADMSG;
  }
  if (*bufpos >= 192) {
    if ((bufpos + 2) >= buflim) {
      return -EBADMSG;
    }
    return 2;
  }
  while (bufpos[res]) {
    res += bufpos[res] + 1;
    if ((&bufpos[res]) >= buflim) {
      return -EBADMSG;
    }
  }
  return res + 1;
}
```