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
<e8back>:
  endbr64
  mov    %esi,%r9d
  xor    %eax,%eax
  lea    -0x5(%rsi),%esi
  mov    %rdi,%r8
  cmp    %esi,%eax
  jge    L17
L33:
  movslq %eax,%rdx
  lea    (%r8,%rdx,4),%rcx
  jmp    L6
  nopl   0x0(%rax)
L64:
  and    $0xfe,%edx
  cmp    $0xe8,%edx
  je     L12
L100:
  add    $0x4,%rcx
  cmp    %eax,%esi
  je     L78
L6:
  mov    (%rcx),%edx
  mov    %eax,%edi
  add    $0x1,%eax
  cmp    $0xf,%edx
  jne    L64
  mov    0x4(%rcx),%edx
  and    $0xf0,%edx
  cmp    $0x80,%edx
  jne    L100
  lea    0x2(%rdi),%ecx
L57:
  movslq %ecx,%rdx
  mov    %eax,%r10d
  lea    (%r8,%rdx,4),%rdi
  not    %r10d
  mov    (%rdi),%edx
  cmp    %r10d,%edx
  jl     L53
  test   %edx,%edx
  js     L23
  cmp    %r9d,%edx
  jge    L53
  sub    %ecx,%edx
  mov    %edx,(%rdi)
L53:
  add    $0x5,%eax
L11:
  cmp    %esi,%eax
  jl     L33
L17:
  ret
  nop
L78:
  ret
  nopl   0x0(%rax)
L23:
  add    %r9d,%edx
  add    $0x5,%eax
  mov    %edx,(%rdi)
  jmp    L11
L12:
  mov    %eax,%ecx
  mov    %edi,%eax
  jmp    L57
```

assistant: ```c
void e8back(uint8_t *buf, int32_t n) {
  int32_t i;
  int32_t *op;
  for (i = 0; i < n - 5;) {
    uint8_t b = buf[i];
    if ((b == 0xF) && (buf[i + 1] & 0xF0) == 0x80) {
      i++;
      b = 0xe8;
    }
    b &= 0xFE;
    i++;
    if (b == 0xe8) {
      op = (int32_t *)(buf + i);
      if (*op >= -i && *op < 0) {
        *op += n;
      } else if (*op >= 0 && *op < n) {
        *op -= i;
      }
      i += 4;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<e8back>:
  endbr64
  mov    %esi,%r9d
  xor    %eax,%eax
  lea    -0x5(%rsi),%esi
  mov    %rdi,%r8
  cmp    %esi,%eax
  jge    L4
L96:
  movslq %eax,%rdx
  lea    (%r8,%rdx,4),%rcx
  jmp    L27
  nopl   0x0(%rax)
L107:
  and    $0xfe,%edx
  cmp    $0xe8,%edx
  je     L36
L73:
  add    $0x4,%rcx
  cmp    %eax,%esi
  je     L57
L27:
  mov    (%rcx),%edx
  mov    %eax,%edi
  add    $0x1,%eax
  cmp    $0xf,%edx
  jne    L107
  mov    0x4(%rcx),%edx
  and    $0xf0,%edx
  cmp    $0x80,%edx
  jne    L73
  lea    0x2(%rdi),%ecx
L99:
  movslq %ecx,%rdx
  mov    %eax,%r10d
  lea    (%r8,%rdx,4),%rdi
  not    %r10d
  mov    (%rdi),%edx
  cmp    %r10d,%edx
  jl     L23
  test   %edx,%edx
  js     L9
  cmp    %r9d,%edx
  jge    L23
  sub    %ecx,%edx
  mov    %edx,(%rdi)
L23:
  add    $0x5,%eax
L34:
  cmp    %esi,%eax
  jl     L96
L4:
  ret
  nop
L57:
  ret
  nopl   0x0(%rax)
L9:
  add    %r9d,%edx
  add    $0x5,%eax
  mov    %edx,(%rdi)
  jmp    L34
L36:
  mov    %eax,%ecx
  mov    %edi,%eax
  jmp    L99
```

assistant: ```c
void e8back(uint8_t *buf, int32_t n) {
  int32_t i;
  int32_t *op;
  for (i = 0; i < n - 5;) {
    uint8_t b = buf[i];
    if ((b == 0xF) && (buf[i + 1] & 0xF0) == 0x80) {
      i++;
      b = 0xe8;
    }
    b &= 0xFE;
    i++;
    if (b == 0xe8) {
      op = (int32_t *)(buf + i);
      if (*op >= -i && *op < 0) {
        *op += n;
      } else if (*op >= 0 && *op < n) {
        *op -= i;
      }
      i += 4;
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<__put_unaligned_4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,%rbx
  lea    0x8(%rbx),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <__put_unaligned_2@plt>
  mov    -0x14(%rbp),%eax
  sar    $0x10,%eax
  mov    %rbx,%rsi
  mov    %eax,%edi
  call   <__put_unaligned_2@plt>
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
extern void __put_unaligned_4(__u32 __v, register __u8 *__p) {
  __put_unaligned_2(__v, __p + 2);
  __put_unaligned_2(__v >> 16, __p);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<__put_unaligned_4>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  lea    0x8(%rsi),%rsi
  call   <__put_unaligned_2@plt>
  sar    $0x10,%ebx
  mov    %ebx,%edi
  mov    %rbp,%rsi
  call   <__put_unaligned_2@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
extern void __put_unaligned_4(__u32 __v, register __u8 *__p) {
  __put_unaligned_2(__v, __p + 2);
  __put_unaligned_2(__v >> 16, __p);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<__put_unaligned_4>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  lea    0x8(%rsi),%rsi
  push   %rbx
  mov    %edi,%ebx
  sar    $0x10,%ebx
  sub    $0x8,%rsp
  call   <__put_unaligned_2@plt>
  add    $0x8,%rsp
  mov    %ebx,%edi
  mov    %rbp,%rsi
  pop    %rbx
  pop    %rbp
  jmp    1050 <__put_unaligned_2@plt>
```

assistant: ```c
extern void __put_unaligned_4(__u32 __v, register __u8 *__p) {
  __put_unaligned_2(__v, __p + 2);
  __put_unaligned_2(__v >> 16, __p);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<__put_unaligned_4>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  lea    0x8(%rsi),%rsi
  push   %rbx
  mov    %edi,%ebx
  sar    $0x10,%ebx
  sub    $0x8,%rsp
  call   <__put_unaligned_2@plt>
  add    $0x8,%rsp
  mov    %ebx,%edi
  mov    %rbp,%rsi
  pop    %rbx
  pop    %rbp
  jmp    1050 <__put_unaligned_2@plt>
```

assistant: ```c
extern void __put_unaligned_4(__u32 __v, register __u8 *__p) {
  __put_unaligned_2(__v, __p + 2);
  __put_unaligned_2(__v >> 16, __p);
}
```
user: Translate the below assembly code into C programming language:

```asm
<find_pattern_from_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <length@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <length@plt>
  sub    -0x8(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jle    L28
  cmpl   $0x0,-0x4(%rbp)
  jle    L28
  movl   $0x0,-0x14(%rbp)
  jmp    L14
L63:
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L48
L51:
  addl   $0x1,-0xc(%rbp)
L48:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jge    L75
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0xc(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L51
L75:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jne    L74
  mov    $0x0,%eax
  jmp    L61
L74:
  addl   $0x1,-0x14(%rbp)
L14:
  mov    -0x14(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L63
L28:
  mov    $0xffffffff,%eax
L61:
  leave
  ret
```

assistant: ```c
int find_pattern_from_str(char str[], char pattern[]) {
  int lnth, lnthdif, i, j, k;
  lnth = length(pattern);
  lnthdif = length(str) - lnth + 1;
  if (lnth > 0 && lnthdif > 0) {
    i = 0;
    while (i < lnthdif) {
      j = i;
      k = 0;
      while (k < lnth && str[j++] == pattern[k])
        k++;
      if (k == lnth) {
        return 0;
      } else {
        i++;
      }
    }
  }
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<find_pattern_from_str>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %rsi,%rdi
  call   <length@plt>
  mov    %eax,%ebp
  mov    %rbx,%rdi
  call   <length@plt>
  sub    %ebp,%eax
  test   %ebp,%ebp
  jle    L26
  test   %eax,%eax
  js     L26
  mov    %rbx,%rsi
  mov    $0x0,%edi
  jmp    L60
L77:
  cmp    %ecx,%ebp
  je     L1
  lea    0x1(%rdi),%edx
  add    $0x1,%rsi
  cmp    %edi,%eax
  jle    L67
  mov    %edx,%edi
L60:
  mov    $0x0,%edx
L24:
  mov    %edx,%ecx
  movzbl (%r12,%rdx,1),%ebx
  cmp    %bl,(%rsi,%rdx,1)
  jne    L77
  lea    0x1(%rdx),%ecx
  add    $0x1,%rdx
  cmp    %edx,%ebp
  jg     L24
  jmp    L77
L26:
  mov    $0xffffffff,%eax
  jmp    L91
L1:
  mov    $0x0,%eax
L91:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L67:
  mov    $0xffffffff,%eax
  jmp    L91
```

assistant: ```c
int find_pattern_from_str(char str[], char pattern[]) {
  int lnth, lnthdif, i, j, k;
  lnth = length(pattern);
  lnthdif = length(str) - lnth + 1;
  if (lnth > 0 && lnthdif > 0) {
    i = 0;
    while (i < lnthdif) {
      j = i;
      k = 0;
      while (k < lnth && str[j++] == pattern[k])
        k++;
      if (k == lnth) {
        return 0;
      } else {
        i++;
      }
    }
  }
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<find_pattern_from_str>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  call   <length@plt>
  mov    %rbx,%rdi
  mov    %eax,%ebp
  call   <length@plt>
  sub    %ebp,%eax
  test   %ebp,%ebp
  jle    L60
  test   %eax,%eax
  js     L60
  mov    %rbx,%rdi
  xor    %esi,%esi
  nop
L13:
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L56:
  movzbl (%r12,%rdx,1),%ebx
  mov    %edx,%ecx
  cmp    %bl,(%rdi,%rdx,1)
  jne    L16
  lea    0x1(%rdx),%ecx
  add    $0x1,%rdx
  cmp    %edx,%ebp
  jg     L56
L16:
  cmp    %ecx,%ebp
  je     L42
  lea    0x1(%rsi),%edx
  add    $0x1,%rdi
  cmp    %esi,%eax
  jle    L60
  mov    %edx,%esi
  jmp    L13
L60:
  pop    %rbx
  mov    $0xffffffff,%eax
  pop    %rbp
  pop    %r12
  ret
L42:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int find_pattern_from_str(char str[], char pattern[]) {
  int lnth, lnthdif, i, j, k;
  lnth = length(pattern);
  lnthdif = length(str) - lnth + 1;
  if (lnth > 0 && lnthdif > 0) {
    i = 0;
    while (i < lnthdif) {
      j = i;
      k = 0;
      while (k < lnth && str[j++] == pattern[k])
        k++;
      if (k == lnth) {
        return 0;
      } else {
        i++;
      }
    }
  }
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<find_pattern_from_str>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  call   <length@plt>
  mov    %rbx,%rdi
  mov    %eax,%ebp
  call   <length@plt>
  sub    %ebp,%eax
  test   %ebp,%ebp
  jle    L67
  test   %eax,%eax
  js     L67
  mov    %rbx,%rdi
  xor    %esi,%esi
  nop
L46:
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L17:
  movzbl (%r12,%rdx,1),%ebx
  mov    %edx,%ecx
  cmp    %bl,(%rdi,%rdx,1)
  jne    L61
  lea    0x1(%rdx),%ecx
  add    $0x1,%rdx
  cmp    %edx,%ebp
  jg     L17
L61:
  cmp    %ecx,%ebp
  je     L22
  lea    0x1(%rsi),%edx
  add    $0x1,%rdi
  cmp    %esi,%eax
  jle    L67
  mov    %edx,%esi
  jmp    L46
L67:
  pop    %rbx
  mov    $0xffffffff,%eax
  pop    %rbp
  pop    %r12
  ret
L22:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int find_pattern_from_str(char str[], char pattern[]) {
  int lnth, lnthdif, i, j, k;
  lnth = length(pattern);
  lnthdif = length(str) - lnth + 1;
  if (lnth > 0 && lnthdif > 0) {
    i = 0;
    while (i < lnthdif) {
      j = i;
      k = 0;
      while (k < lnth && str[j++] == pattern[k])
        k++;
      if (k == lnth) {
        return 0;
      } else {
        i++;
      }
    }
  }
  return -1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbx
  lea    D36(%rip),%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    D7(%rip),%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%eax
  lea    -0x1(%rax),%ecx
  cmp    $0xb,%ecx
  ja     L68
  mov    %ecx,%eax
  lea    D29(%rip),%rdx
  mov    (%rdx,%rax,4),%ebx
L7:
  mov    0xc(%rsp),%eax
  add    0x14(%rsp),%ebx
  mov    %ebx,%esi
  imul   $0xc28f5c29,%eax,%edx
  add    $0x51eb850,%edx
  ror    $0x4,%edx
  cmp    $0xa3d70a,%edx
  jbe    L12
  test   $0x3,%al
  jne    L23
L12:
  cmp    $0x1,%ecx
  jbe    L20
  add    $0x1,%esi
  lea    D39(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L41:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L64
  add    $0x20,%rsp
  pop    %rbx
  ret
L23:
  mov    $0x64,%edi
  cltd
  idiv   %edi
  test   %edx,%edx
  jne    L12
L20:
  lea    D39(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L41
L68:
  lea    D20(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x10(%rsp),%eax
  lea    -0x1(%rax),%ecx
  jmp    L7
L64:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D36
type: string
name: unpack_data
arguments: tag: D7
type: string
name: unpack_data
arguments: tag: D29
type: string
name: unpack_data
arguments: tag: D39
type: string
name: unpack_data
arguments: tag: D20
type: string

tool: D36: 
Please input the year,month,day:

tool: D7: %d,%d,%d

tool: 

tool: D39: the number of the day in this year is: %d


tool: D20: month error!

assistant: ```c
void main() {
  int year, month, day, sum, leap;
  printf("\nPlease input the year,month,day:\n");
  scanf("%d,%d,%d", &year, &month, &day);
  switch (month) {
  case 1:
    sum = 0;
    break;
  case 2:
    sum = 31;
    break;
  case 3:
    sum = 59;
    break;
  case 4:
    sum = 90;
    break;
  case 5:
    sum = 120;
    break;
  case 6:
    sum = 151;
    break;
  case 7:
    sum = 181;
    break;
  case 8:
    sum = 212;
    break;
  case 9:
    sum = 243;
    break;
  case 10:
    sum = 273;
    break;
  case 11:
    sum = 303;
    break;
  case 12:
    sum = 333;
    break;
  default:
    printf("month error!");
  }
  if ((0 == year % 400) || (0 == year % 4 || 0 != year % 100))
    leap = 1;
  else
    leap = 0;
  if (1 == month || 2 == month) {
    sum = sum + day;
    printf("the number of the day in this year is: %d\n", sum);
  } else {
    if (1 == leap) {
      sum = sum + day + 1;
      printf("the number of the day in this year is: %d\n", sum);
    } else {
      sum = sum + day;
      printf("the number of the day in this year is: %d\n", sum);
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D27(%rip),%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    D34(%rip),%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%eax
  lea    -0x1(%rax),%ecx
  cmp    $0xb,%ecx
  ja     L31
  mov    %ecx,%eax
  lea    D7(%rip),%rdx
  mov    (%rdx,%rax,4),%ebx
L65:
  mov    0xc(%rsp),%eax
  add    0x14(%rsp),%ebx
  mov    %ebx,%esi
  imul   $0xc28f5c29,%eax,%edx
  add    $0x51eb850,%edx
  ror    $0x4,%edx
  cmp    $0xa3d70a,%edx
  jbe    L0
  test   $0x3,%al
  jne    L76
L0:
  cmp    $0x1,%ecx
  jbe    L71
  add    $0x1,%esi
  lea    D24(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L29:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x20,%rsp
  pop    %rbx
  ret
L76:
  mov    $0x64,%edi
  cltd
  idiv   %edi
  test   %edx,%edx
  jne    L0
L71:
  lea    D24(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L29
L31:
  lea    D52(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x10(%rsp),%eax
  lea    -0x1(%rax),%ecx
  jmp    L65
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D27
type: string
name: extract_symbolic_data
arguments: label: D34
type: string
name: extract_symbolic_data
arguments: label: D7
type: string
name: extract_symbolic_data
arguments: label: D24
type: string
name: extract_symbolic_data
arguments: label: D52
type: string

tool: D27: 
Please input the year,month,day:

tool: D34: %d,%d,%d

tool: 

tool: D24: the number of the day in this year is: %d


tool: D52: month error!

assistant: ```c
void main() {
  int year, month, day, sum, leap;
  printf("\nPlease input the year,month,day:\n");
  scanf("%d,%d,%d", &year, &month, &day);
  switch (month) {
  case 1:
    sum = 0;
    break;
  case 2:
    sum = 31;
    break;
  case 3:
    sum = 59;
    break;
  case 4:
    sum = 90;
    break;
  case 5:
    sum = 120;
    break;
  case 6:
    sum = 151;
    break;
  case 7:
    sum = 181;
    break;
  case 8:
    sum = 212;
    break;
  case 9:
    sum = 243;
    break;
  case 10:
    sum = 273;
    break;
  case 11:
    sum = 303;
    break;
  case 12:
    sum = 333;
    break;
  default:
    printf("month error!");
  }
  if ((0 == year % 400) || (0 == year % 4 || 0 != year % 100))
    leap = 1;
  else
    leap = 0;
  if (1 == month || 2 == month) {
    sum = sum + day;
    printf("the number of the day in this year is: %d\n", sum);
  } else {
    if (1 == leap) {
      sum = sum + day + 1;
      printf("the number of the day in this year is: %d\n", sum);
    } else {
      sum = sum + day;
      printf("the number of the day in this year is: %d\n", sum);
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<array_entry_insert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  addl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  imul   %edx,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L1
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  neg    %eax
  jmp    L10
L1:
  mov    -0xc(%rbp),%eax
  sub    -0x14(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  imul   %edx,%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  cltq
  imul   -0x28(%rbp),%rax
  mov    %rax,%rcx
  mov    -0x8(%rbp),%rax
  add    %rax,%rcx
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  cltq
  imul   -0x28(%rbp),%rax
  mov    %rax,%rsi
  mov    -0x8(%rbp),%rax
  add    %rsi,%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memmove@plt>
  mov    -0x14(%rbp),%eax
  cltq
  imul   -0x28(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    -0x30(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x38(%rbp),%rax
  mov    -0xc(%rbp),%edx
  mov    %edx,(%rax)
L10:
  leave
  ret
```

assistant: ```c
int array_entry_insert(int index, void *elem, size_t elem_size, void **array_, int *array_count_) {
  void *array = *array_;
  int array_count = *array_count_;
  array_count++;
  array = realloc(array, array_count * elem_size);
  if (!array) {
    return -ENOMEM;
  }
  memmove(array + elem_size * (index + 1), array + elem_size * index, (array_count - index - 1) * elem_size);
  memcpy(array + elem_size * index, elem, elem_size);
  *array_ = array;
  *array_count_ = array_count;
}
```
user: What is the c source code of the assembly code below:

```asm
<array_entry_insert>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%r15d
  mov    %rsi,0x8(%rsp)
  mov    %rdx,%rbx
  mov    %rcx,%r14
  mov    %r8,%r13
  mov    (%rcx),%rdi
  mov    (%r8),%eax
  lea    0x1(%rax),%r12d
  mov    %edx,%esi
  imul   %r12d,%esi
  call   <realloc@plt>
  test   %rax,%rax
  je     L8
  mov    %rax,%rbp
  mov    %r15d,%eax
  movslq %r15d,%rdi
  imul   %rbx,%rdi
  lea    0x0(%rbp,%rdi,1),%r15
  mov    %r12d,%edx
  sub    %eax,%edx
  sub    $0x1,%edx
  imul   %ebx,%edx
  add    %rbx,%rdi
  add    %rbp,%rdi
  mov    %r15,%rsi
  call   <memmove@plt>
  mov    %rbx,%rdx
  mov    0x8(%rsp),%rsi
  mov    %r15,%rdi
  call   <memcpy@plt>
  mov    %rbp,(%r14)
  mov    %r12d,0x0(%r13)
L6:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L8:
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  neg    %eax
  jmp    L6
```

assistant: ```c
int array_entry_insert(int index, void *elem, size_t elem_size, void **array_, int *array_count_) {
  void *array = *array_;
  int array_count = *array_count_;
  array_count++;
  array = realloc(array, array_count * elem_size);
  if (!array) {
    return -ENOMEM;
  }
  memmove(array + elem_size * (index + 1), array + elem_size * index, (array_count - index - 1) * elem_size);
  memcpy(array + elem_size * index, elem, elem_size);
  *array_ = array;
  *array_count_ = array_count;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<array_entry_insert>:
  endbr64
  push   %r15
  mov    %rcx,%r15
  push   %r14
  mov    %r8,%r14
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    (%r8),%eax
  mov    %edi,0x8(%rsp)
  mov    (%rcx),%rdi
  lea    0x1(%rax),%ebx
  mov    %ebx,%esi
  imul   %edx,%esi
  call   <realloc@plt>
  mov    0x8(%rsp),%r9d
  test   %rax,%rax
  je     L0
  movslq %r9d,%rdi
  mov    %ebx,%edx
  mov    %rax,%r13
  imul   %r12,%rdi
  sub    %r9d,%edx
  sub    $0x1,%edx
  imul   %r12d,%edx
  lea    (%rax,%rdi,1),%r8
  add    %r12,%rdi
  mov    %r8,%rsi
  add    %rax,%rdi
  mov    %r8,0x8(%rsp)
  call   <memmove@plt>
  mov    0x8(%rsp),%rdi
  mov    %r12,%rdx
  mov    %rbp,%rsi
  call   <memcpy@plt>
  mov    %r13,(%r15)
  mov    %ebx,(%r14)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L0:
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  neg    %eax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int array_entry_insert(int index, void *elem, size_t elem_size, void **array_, int *array_count_) {
  void *array = *array_;
  int array_count = *array_count_;
  array_count++;
  array = realloc(array, array_count * elem_size);
  if (!array) {
    return -ENOMEM;
  }
  memmove(array + elem_size * (index + 1), array + elem_size * index, (array_count - index - 1) * elem_size);
  memcpy(array + elem_size * index, elem, elem_size);
  *array_ = array;
  *array_count_ = array_count;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<array_entry_insert>:
  endbr64
  push   %r15
  mov    %rcx,%r15
  push   %r14
  mov    %r8,%r14
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    (%r8),%eax
  mov    %edi,0x8(%rsp)
  mov    (%rcx),%rdi
  lea    0x1(%rax),%ebx
  mov    %ebx,%esi
  imul   %edx,%esi
  call   <realloc@plt>
  mov    0x8(%rsp),%r9d
  test   %rax,%rax
  je     L4
  movslq %r9d,%rdi
  mov    %ebx,%edx
  mov    %rax,%r13
  imul   %r12,%rdi
  sub    %r9d,%edx
  sub    $0x1,%edx
  imul   %r12d,%edx
  lea    (%rax,%rdi,1),%r8
  add    %r12,%rdi
  mov    %r8,%rsi
  add    %rax,%rdi
  mov    %r8,0x8(%rsp)
  call   <memmove@plt>
  mov    0x8(%rsp),%rdi
  mov    %r12,%rdx
  mov    %rbp,%rsi
  call   <memcpy@plt>
  mov    %r13,(%r15)
  mov    %ebx,(%r14)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L4:
  mov    D7(%rip),%rax
  mov    (%rax),%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  neg    %eax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int array_entry_insert(int index, void *elem, size_t elem_size, void **array_, int *array_count_) {
  void *array = *array_;
  int array_count = *array_count_;
  array_count++;
  array = realloc(array, array_count * elem_size);
  if (!array) {
    return -ENOMEM;
  }
  memmove(array + elem_size * (index + 1), array + elem_size * index, (array_count - index - 1) * elem_size);
  memcpy(array + elem_size * index, elem, elem_size);
  *array_ = array;
  *array_count_ = array_count;
}
```