ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<num_digits>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x9,%edi
  jg     L6
  ret
L6:
  sub    $0x8,%rsp
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%edi
  sub    %edi,%eax
  mov    %eax,%edi
  call   <num_digits>
  add    $0x1,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int num_digits(int n) {
  if (n < 10)
    return 1;
  else
    return 1 + num_digits(n / 10);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<num_digits>:
  endbr64
  mov    $0x1,%edx
  mov    $0xcccccccd,%esi
  cmp    $0x9,%edi
  jle    L18
  nopl   0x0(%rax,%rax,1)
L9:
  mov    %edi,%edi
  add    $0x1,%edx
  mov    %rdi,%rcx
  imul   %rsi,%rdi
  shr    $0x23,%rdi
  cmp    $0x63,%ecx
  jg     L9
  mov    %edx,%eax
  ret
L18:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int num_digits(int n) {
  if (n < 10)
    return 1;
  else
    return 1 + num_digits(n / 10);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<num_digits>:
  endbr64
  mov    $0x1,%edx
  mov    $0xcccccccd,%esi
  cmp    $0x9,%edi
  jle    L4
  nopl   0x0(%rax,%rax,1)
L18:
  mov    %edi,%edi
  add    $0x1,%edx
  mov    %rdi,%rcx
  imul   %rsi,%rdi
  shr    $0x23,%rdi
  cmp    $0x63,%ecx
  jg     L18
  mov    %edx,%eax
  ret
L4:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int num_digits(int n) {
  if (n < 10)
    return 1;
  else
    return 1 + num_digits(n / 10);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<mode_grant>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %rdx,-0x18(%rbp)
  cmpl   $0x2,-0xc(%rbp)
  jg     L15
  lea    D5(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <logprintf@plt>
  mov    $0x14,%eax
  jmp    L14
L15:
  mov    -0x18(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sqlite_add_permission@plt>
L14:
  leave
  ret
```

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

tool: D5: Not enough arguments for grant



assistant: ```c
int mode_grant(sqlite3 *db, int argc, char **argv) {
  if (argc < 3) {
    logprintf(0, "Not enough arguments for grant\n\n");
    return 20;
  }
  return sqlite_add_permission(db, argv[1], argv[2]);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<mode_grant>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x2,%esi
  jle    L4
  mov    0x10(%rdx),%rax
  mov    0x8(%rdx),%rsi
  mov    %rax,%rdx
  call   <sqlite_add_permission@plt>
L1:
  add    $0x8,%rsp
  ret
L4:
  lea    D7(%rip),%rsi
  mov    $0x0,%edi
  call   <logprintf@plt>
  mov    $0x14,%eax
  jmp    L1
```

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

tool: D7: Not enough arguments for grant



assistant: ```c
int mode_grant(sqlite3 *db, int argc, char **argv) {
  if (argc < 3) {
    logprintf(0, "Not enough arguments for grant\n\n");
    return 20;
  }
  return sqlite_add_permission(db, argv[1], argv[2]);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<mode_grant>:
  endbr64
  cmp    $0x2,%esi
  jle    L7
  mov    0x10(%rdx),%r8
  mov    0x8(%rdx),%rsi
  mov    %r8,%rdx
  jmp    1070 <sqlite_add_permission@plt>
  nopl   0x0(%rax)
L7:
  sub    $0x8,%rsp
  lea    D4(%rip),%rsi
  xor    %edi,%edi
  call   <logprintf@plt>
  mov    $0x14,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D4
datatype: string

tool: D4: Not enough arguments for grant



assistant: ```c
int mode_grant(sqlite3 *db, int argc, char **argv) {
  if (argc < 3) {
    logprintf(0, "Not enough arguments for grant\n\n");
    return 20;
  }
  return sqlite_add_permission(db, argv[1], argv[2]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mode_grant>:
  endbr64
  cmp    $0x2,%esi
  jle    L2
  mov    0x10(%rdx),%r8
  mov    0x8(%rdx),%rsi
  mov    %r8,%rdx
  jmp    1070 <sqlite_add_permission@plt>
  nopl   0x0(%rax)
L2:
  sub    $0x8,%rsp
  lea    D1(%rip),%rsi
  xor    %edi,%edi
  call   <logprintf@plt>
  mov    $0x14,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D1
type: string

tool: D1: Not enough arguments for grant



assistant: ```c
int mode_grant(sqlite3 *db, int argc, char **argv) {
  if (argc < 3) {
    logprintf(0, "Not enough arguments for grant\n\n");
    return 20;
  }
  return sqlite_add_permission(db, argv[1], argv[2]);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D9
type_infered: string

tool: D9: Usage: %s [tty path]
Runs the FBUS protocol on a given terminal.


assistant: ```c
void usage(const char *path) {
  printf("Usage: %s [tty path]\n"
         "Runs the FBUS protocol on a given terminal.\n",
         path);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

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

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

tool: D0: Usage: %s [tty path]
Runs the FBUS protocol on a given terminal.


assistant: ```c
void usage(const char *path) {
  printf("Usage: %s [tty path]\n"
         "Runs the FBUS protocol on a given terminal.\n",
         path);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<usage>:
  endbr64
  mov    %rdi,%rsi
  lea    D2(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: label: D2
type: string

tool: D2: Usage: %s [tty path]
Runs the FBUS protocol on a given terminal.


assistant: ```c
void usage(const char *path) {
  printf("Usage: %s [tty path]\n"
         "Runs the FBUS protocol on a given terminal.\n",
         path);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<usage>:
  endbr64
  mov    %rdi,%rsi
  lea    D7(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: unpack_data
arguments: label: D7
data_type: string

tool: D7: Usage: %s [tty path]
Runs the FBUS protocol on a given terminal.


assistant: ```c
void usage(const char *path) {
  printf("Usage: %s [tty path]\n"
         "Runs the FBUS protocol on a given terminal.\n",
         path);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<thread_job>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x30,%rsp
  mov    %rdi,-0x2028(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x2018(%rbp)
  lea    -0x2010(%rbp),%rax
  mov    $0x2000,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0x0,-0x201c(%rbp)
  jmp    L72
L6:
  mov    $0x0,%eax
  call   <random@plt>
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  test   %eax,%eax
  je     L71
  cmp    $0x1,%eax
  je     L69
  jmp    L63
L71:
  mov    $0x0,%eax
  call   <random@plt>
  cltd
  shr    $0x16,%edx
  add    %edx,%eax
  and    $0x3ff,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  lea    -0x2010(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <alloc@plt>
  jmp    L36
L69:
  mov    $0x0,%eax
  call   <random@plt>
  cltd
  shr    $0x16,%edx
  add    %edx,%eax
  and    $0x3ff,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  lea    -0x2010(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <release@plt>
  jmp    L36
L63:
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  call   <abort@plt>
L36:
  addl   $0x1,-0x201c(%rbp)
L72:
  cmpl   $0x459,-0x201c(%rbp)
  jle    L6
  movl   $0x0,-0x201c(%rbp)
  jmp    L31
L91:
  mov    -0x201c(%rbp),%eax
  cltq
  mov    -0x2010(%rbp,%rax,8),%rax
  test   %rax,%rax
  je     L67
  addq   $0x1,-0x2018(%rbp)
L67:
  addl   $0x1,-0x201c(%rbp)
L31:
  cmpl   $0x3ff,-0x201c(%rbp)
  jle    L91
  mov    -0x2018(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x2018(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L92
  call   <__stack_chk_fail@plt>
L92:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D3
data_type: string
name: extract_symbolic_data
arguments: tag: D10
data_type: string

tool: D3: EER

tool: D10: Still allocated: %lu


assistant: ```c
void *thread_job(void *arg) {
  void *pool[1024];
  int i;
  unsigned long count = 0;
  memset(pool, 0, sizeof(void *) * 1024);
  for (i = 0; i < 1114; i++) {
    switch (random() % 2) {
    case 0:
      alloc(pool, random() % 1024);
      break;
    case 1:
      release(pool, random() % 1024);
      break;
    default:
      printf("EER\n");
      abort();
    }
  }
  for (i = 0; i < 1024; i++) {
    if (pool[i] != NULL) {
      count++;
    }
  }
  printf("Still allocated: %lu\n", count);
  return (void *)count;
}
```
user: Translate the below assembly code into C programming language:

```asm
<thread_job>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x2008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x2000,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  mov    $0x45a,%ebx
  lea    D6(%rip),%rbp
  jmp    L48
L4:
  mov    $0x0,%eax
  call   <random@plt>
  cltd
  shr    $0x16,%edx
  lea    (%rax,%rdx,1),%esi
  and    $0x3ff,%esi
  sub    %edx,%esi
  mov    %rsp,%rdi
  call   <alloc@plt>
L34:
  sub    $0x1,%ebx
  je     L19
L48:
  mov    $0x0,%eax
  call   <random@plt>
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  lea    (%rax,%rcx,1),%edx
  and    $0x1,%edx
  sub    %ecx,%edx
  test   $0x1,%al
  je     L4
  cmp    $0x1,%edx
  je     L54
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L34
L54:
  mov    $0x0,%eax
  call   <random@plt>
  cltd
  shr    $0x16,%edx
  lea    (%rax,%rdx,1),%esi
  and    $0x3ff,%esi
  sub    %edx,%esi
  mov    %rsp,%rdi
  call   <release@plt>
  jmp    L34
L19:
  mov    %rsp,%rax
  lea    0x2000(%rsp),%rdx
  mov    $0x0,%ebx
L2:
  cmpq   $0x1,(%rax)
  sbb    $0xffffffffffffffff,%rbx
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L2
  mov    %rbx,%rsi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x2008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L64
  mov    %rbx,%rax
  add    $0x2018,%rsp
  pop    %rbx
  pop    %rbp
  ret
L64:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
type_guessed: string
name: retrieve_data
arguments: data_tag: D4
type_guessed: string

tool: D6: EER

tool: D4: Still allocated: %lu


assistant: ```c
void *thread_job(void *arg) {
  void *pool[1024];
  int i;
  unsigned long count = 0;
  memset(pool, 0, sizeof(void *) * 1024);
  for (i = 0; i < 1114; i++) {
    switch (random() % 2) {
    case 0:
      alloc(pool, random() % 1024);
      break;
    case 1:
      release(pool, random() % 1024);
      break;
    default:
      printf("EER\n");
      abort();
    }
  }
  for (i = 0; i < 1024; i++) {
    if (pool[i] != NULL) {
      count++;
    }
  }
  printf("Still allocated: %lu\n", count);
  return (void *)count;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<thread_job>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x10,%rsp
  mov    $0x2000,%edx
  xor    %esi,%esi
  mov    $0x45a,%ebp
  lea    D15(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x2008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <memset@plt>
  jmp    L22
  nopw   0x0(%rax,%rax,1)
L55:
  mov    %r12,%rdi
  call   <puts@plt>
  xor    %eax,%eax
  call   <abort@plt>
L48:
  sub    $0x1,%ebp
  je     L45
L22:
  xor    %eax,%eax
  call   <random@plt>
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  lea    (%rax,%rcx,1),%edx
  and    $0x1,%edx
  sub    %ecx,%edx
  test   $0x1,%al
  je     L79
  cmp    $0x1,%edx
  jne    L55
  xor    %eax,%eax
  call   <random@plt>
  mov    %rbx,%rdi
  cltd
  shr    $0x16,%edx
  lea    (%rax,%rdx,1),%esi
  and    $0x3ff,%esi
  sub    %edx,%esi
  call   <release@plt>
  sub    $0x1,%ebp
  jne    L22
L45:
  lea    0x2000(%rsp),%rax
  xor    %r12d,%r12d
  nop
L11:
  cmpq   $0x1,(%rbx)
  sbb    $0xffffffffffffffff,%r12
  add    $0x8,%rbx
  cmp    %rax,%rbx
  jne    L11
  xor    %eax,%eax
  mov    %r12,%rsi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    0x2008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L62
  add    $0x2010,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L79:
  xor    %eax,%eax
  call   <random@plt>
  mov    %rbx,%rdi
  cltd
  shr    $0x16,%edx
  lea    (%rax,%rdx,1),%esi
  and    $0x3ff,%esi
  sub    %edx,%esi
  call   <alloc@plt>
  jmp    L48
L62:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D15
type_guessed: string
name: extract_symbolic_data
arguments: tag: D6
type_guessed: string

tool: D15: EER

tool: D6: Still allocated: %lu


assistant: ```c
void *thread_job(void *arg) {
  void *pool[1024];
  int i;
  unsigned long count = 0;
  memset(pool, 0, sizeof(void *) * 1024);
  for (i = 0; i < 1114; i++) {
    switch (random() % 2) {
    case 0:
      alloc(pool, random() % 1024);
      break;
    case 1:
      release(pool, random() % 1024);
      break;
    default:
      printf("EER\n");
      abort();
    }
  }
  for (i = 0; i < 1024; i++) {
    if (pool[i] != NULL) {
      count++;
    }
  }
  printf("Still allocated: %lu\n", count);
  return (void *)count;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<thread_job>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x10,%rsp
  mov    $0x2000,%edx
  xor    %esi,%esi
  mov    $0x45a,%ebp
  lea    D13(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x2008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <memset@plt>
  jmp    L5
  nopw   0x0(%rax,%rax,1)
L15:
  mov    %r12,%rdi
  call   <puts@plt>
  xor    %eax,%eax
  call   <abort@plt>
L10:
  sub    $0x1,%ebp
  je     L37
L5:
  xor    %eax,%eax
  call   <random@plt>
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  lea    (%rax,%rcx,1),%edx
  and    $0x1,%edx
  sub    %ecx,%edx
  test   $0x1,%al
  je     L33
  cmp    $0x1,%edx
  jne    L15
  xor    %eax,%eax
  call   <random@plt>
  mov    %rbx,%rdi
  cltd
  shr    $0x16,%edx
  lea    (%rax,%rdx,1),%esi
  and    $0x3ff,%esi
  sub    %edx,%esi
  call   <release@plt>
  sub    $0x1,%ebp
  jne    L5
L37:
  lea    0x2000(%rsp),%rax
  xor    %r12d,%r12d
  nop
L34:
  cmpq   $0x1,(%rbx)
  sbb    $0xffffffffffffffff,%r12
  add    $0x8,%rbx
  cmp    %rax,%rbx
  jne    L34
  xor    %eax,%eax
  mov    %r12,%rsi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    0x2008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L76
  add    $0x2010,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L33:
  xor    %eax,%eax
  call   <random@plt>
  mov    %rbx,%rdi
  cltd
  shr    $0x16,%edx
  lea    (%rax,%rdx,1),%esi
  and    $0x3ff,%esi
  sub    %edx,%esi
  call   <alloc@plt>
  jmp    L10
L76:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D13
type: string
name: disassemble_data
arguments: data_tag: D6
type: string

tool: D13: EER

tool: D6: Still allocated: %lu


assistant: ```c
void *thread_job(void *arg) {
  void *pool[1024];
  int i;
  unsigned long count = 0;
  memset(pool, 0, sizeof(void *) * 1024);
  for (i = 0; i < 1114; i++) {
    switch (random() % 2) {
    case 0:
      alloc(pool, random() % 1024);
      break;
    case 1:
      release(pool, random() % 1024);
      break;
    default:
      printf("EER\n");
      abort();
    }
  }
  for (i = 0; i < 1024; i++) {
    if (pool[i] != NULL) {
      count++;
    }
  }
  printf("Still allocated: %lu\n", count);
  return (void *)count;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<md_kernel>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,-0x20(%rsp)
  mov    %rsi,-0x18(%rsp)
  mov    %rdx,-0x10(%rsp)
  mov    %rcx,%r12
  mov    0x38(%rsp),%rdx
  mov    0x40(%rsp),%eax
  movslq %eax,%rbx
  mov    %rbx,-0x8(%rsp)
  mov    (%rcx,%rbx,4),%r15d
  mov    (%r8,%rbx,4),%r14d
  mov    (%r9,%rbx,4),%ebx
  mov    %ebx,-0x2c(%rsp)
  shl    $0x5,%eax
  cltq
  shl    $0x2,%rax
  lea    (%rdx,%rax,1),%r11
  lea    0x80(%rdx,%rax,1),%rax
  mov    %rax,-0x28(%rsp)
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  mov    $0x0,%r13d
L3:
  movslq (%r11),%rax
  mov    %r15d,%esi
  sub    (%r12,%rax,4),%esi
  mov    %r14d,%ecx
  sub    (%r8,%rax,4),%ecx
  mov    -0x2c(%rsp),%edx
  sub    (%r9,%rax,4),%edx
  mov    %esi,%edi
  imul   %esi,%edi
  mov    %ecx,%eax
  imul   %ecx,%eax
  add    %edi,%eax
  mov    %edx,%edi
  imul   %edx,%edi
  add    %edi,%eax
  mov    %eax,%r10d
  imul   %eax,%r10d
  imul   %eax,%r10d
  mov    %r10d,%edi
  shl    $0x4,%edi
  sub    %r10d,%edi
  sub    $0x14,%edi
  imul   %r10d,%eax
  imul   %edi,%eax
  imul   %eax,%esi
  add    %esi,%r13d
  imul   %eax,%ecx
  add    %ecx,%ebp
  imul   %eax,%edx
  add    %edx,%ebx
  add    $0x4,%r11
  cmp    -0x28(%rsp),%r11
  jne    L3
  mov    -0x20(%rsp),%rcx
  mov    -0x8(%rsp),%rax
  mov    %r13d,(%rcx,%rax,4)
  mov    -0x18(%rsp),%rsi
  mov    %ebp,(%rsi,%rax,4)
  mov    -0x10(%rsp),%rcx
  mov    %ebx,(%rcx,%rax,4)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void md_kernel(int d_force_x[32], int d_force_y[32], int d_force_z[32], int position_x[32], int position_y[32], int position_z[32], int NL[32 * 32], int i) {
  int j, jidx;
  int delx, dely, delz, r2inv, r2invTEMP, r2invTEMP2, r2invTEMP3, t1, t2, t3;
  int i_x = position_x[i];
  int i_y = position_y[i];
  int i_z = position_z[i];
  int fx = 0;
  int fy = 0;
  int fz = 0;
loop_j:
  for (j = 0; j < 32; j++) {
    jidx = NL[i * 32 + j];
    int j_x = position_x[jidx];
    int j_y = position_y[jidx];
    int j_z = position_z[jidx];
    delx = i_x - j_x;
    dely = i_y - j_y;
    delz = i_z - j_z;
    r2invTEMP = delx * delx;
    r2invTEMP2 = dely * dely;
    r2invTEMP3 = delz * delz;
    t1 = r2invTEMP + r2invTEMP2;
    t2 = t1 + r2invTEMP3;
    r2inv = t2;
    int r6inv = r2inv * r2inv * r2inv;
    int force = r2inv * r6inv * (15 * r6inv - 20);
    fx += delx * force;
    fy += dely * force;
    fz += delz * force;
  }
  d_force_x[i] = fx;
  d_force_y[i] = fy;
  d_force_z[i] = fz;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<md_kernel>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  movslq 0x40(%rsp),%rbx
  mov    %rbx,%rax
  mov    (%rcx,%rbx,4),%r15d
  mov    (%r8,%rbx,4),%r14d
  shl    $0x5,%eax
  mov    %rdx,-0x10(%rsp)
  mov    0x38(%rsp),%rdx
  cltq
  mov    %rbx,-0x8(%rsp)
  mov    (%r9,%rbx,4),%ebx
  shl    $0x2,%rax
  mov    %rdi,-0x20(%rsp)
  lea    (%rdx,%rax,1),%r11
  lea    0x80(%rdx,%rax,1),%rax
  mov    %rsi,-0x18(%rsp)
  mov    %rax,-0x28(%rsp)
  mov    %ebx,-0x2c(%rsp)
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L2:
  movslq (%r11),%rax
  mov    %r15d,%esi
  mov    %r14d,%ecx
  mov    -0x2c(%rsp),%edx
  add    $0x4,%r11
  sub    0x0(%r13,%rax,4),%esi
  sub    (%r8,%rax,4),%ecx
  sub    (%r9,%rax,4),%edx
  mov    %esi,%edi
  mov    %ecx,%eax
  imul   %esi,%edi
  imul   %ecx,%eax
  add    %edi,%eax
  mov    %edx,%edi
  imul   %edx,%edi
  add    %edi,%eax
  mov    %eax,%r10d
  imul   %eax,%r10d
  imul   %eax,%r10d
  mov    %r10d,%edi
  imul   %r10d,%eax
  shl    $0x4,%edi
  sub    %r10d,%edi
  sub    $0x14,%edi
  imul   %edi,%eax
  imul   %eax,%esi
  imul   %eax,%ecx
  imul   %eax,%edx
  add    %esi,%r12d
  add    %ecx,%ebp
  add    %edx,%ebx
  cmp    %r11,-0x28(%rsp)
  jne    L2
  mov    -0x8(%rsp),%rax
  mov    -0x20(%rsp),%rcx
  mov    -0x18(%rsp),%rsi
  mov    %r12d,(%rcx,%rax,4)
  mov    -0x10(%rsp),%rcx
  mov    %ebp,(%rsi,%rax,4)
  mov    %ebx,(%rcx,%rax,4)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void md_kernel(int d_force_x[32], int d_force_y[32], int d_force_z[32], int position_x[32], int position_y[32], int position_z[32], int NL[32 * 32], int i) {
  int j, jidx;
  int delx, dely, delz, r2inv, r2invTEMP, r2invTEMP2, r2invTEMP3, t1, t2, t3;
  int i_x = position_x[i];
  int i_y = position_y[i];
  int i_z = position_z[i];
  int fx = 0;
  int fy = 0;
  int fz = 0;
loop_j:
  for (j = 0; j < 32; j++) {
    jidx = NL[i * 32 + j];
    int j_x = position_x[jidx];
    int j_y = position_y[jidx];
    int j_z = position_z[jidx];
    delx = i_x - j_x;
    dely = i_y - j_y;
    delz = i_z - j_z;
    r2invTEMP = delx * delx;
    r2invTEMP2 = dely * dely;
    r2invTEMP3 = delz * delz;
    t1 = r2invTEMP + r2invTEMP2;
    t2 = t1 + r2invTEMP3;
    r2inv = t2;
    int r6inv = r2inv * r2inv * r2inv;
    int force = r2inv * r6inv * (15 * r6inv - 20);
    fx += delx * force;
    fy += dely * force;
    fz += delz * force;
  }
  d_force_x[i] = fx;
  d_force_y[i] = fy;
  d_force_z[i] = fz;
}
```
user: Translate the below assembly code into C programming language:

```asm
<md_kernel>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  movslq 0x40(%rsp),%rbx
  mov    %rbx,%rax
  mov    (%rcx,%rbx,4),%r15d
  mov    (%r8,%rbx,4),%r14d
  shl    $0x5,%eax
  mov    %rdx,-0x10(%rsp)
  mov    0x38(%rsp),%rdx
  cltq
  mov    %rbx,-0x8(%rsp)
  mov    (%r9,%rbx,4),%ebx
  shl    $0x2,%rax
  mov    %rdi,-0x20(%rsp)
  lea    (%rdx,%rax,1),%r11
  lea    0x80(%rdx,%rax,1),%rax
  mov    %rsi,-0x18(%rsp)
  mov    %rax,-0x28(%rsp)
  mov    %ebx,-0x2c(%rsp)
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L0:
  movslq (%r11),%rax
  mov    %r15d,%esi
  mov    %r14d,%ecx
  mov    -0x2c(%rsp),%edx
  add    $0x4,%r11
  sub    0x0(%r13,%rax,4),%esi
  sub    (%r8,%rax,4),%ecx
  sub    (%r9,%rax,4),%edx
  mov    %esi,%edi
  mov    %ecx,%eax
  imul   %esi,%edi
  imul   %ecx,%eax
  add    %edi,%eax
  mov    %edx,%edi
  imul   %edx,%edi
  add    %edi,%eax
  mov    %eax,%r10d
  imul   %eax,%r10d
  imul   %eax,%r10d
  mov    %r10d,%edi
  imul   %r10d,%eax
  shl    $0x4,%edi
  sub    %r10d,%edi
  sub    $0x14,%edi
  imul   %edi,%eax
  imul   %eax,%esi
  imul   %eax,%ecx
  imul   %eax,%edx
  add    %esi,%r12d
  add    %ecx,%ebp
  add    %edx,%ebx
  cmp    %r11,-0x28(%rsp)
  jne    L0
  mov    -0x8(%rsp),%rax
  mov    -0x20(%rsp),%rcx
  mov    -0x18(%rsp),%rsi
  mov    %r12d,(%rcx,%rax,4)
  mov    -0x10(%rsp),%rcx
  mov    %ebp,(%rsi,%rax,4)
  mov    %ebx,(%rcx,%rax,4)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void md_kernel(int d_force_x[32], int d_force_y[32], int d_force_z[32], int position_x[32], int position_y[32], int position_z[32], int NL[32 * 32], int i) {
  int j, jidx;
  int delx, dely, delz, r2inv, r2invTEMP, r2invTEMP2, r2invTEMP3, t1, t2, t3;
  int i_x = position_x[i];
  int i_y = position_y[i];
  int i_z = position_z[i];
  int fx = 0;
  int fy = 0;
  int fz = 0;
loop_j:
  for (j = 0; j < 32; j++) {
    jidx = NL[i * 32 + j];
    int j_x = position_x[jidx];
    int j_y = position_y[jidx];
    int j_z = position_z[jidx];
    delx = i_x - j_x;
    dely = i_y - j_y;
    delz = i_z - j_z;
    r2invTEMP = delx * delx;
    r2invTEMP2 = dely * dely;
    r2invTEMP3 = delz * delz;
    t1 = r2invTEMP + r2invTEMP2;
    t2 = t1 + r2invTEMP3;
    r2inv = t2;
    int r6inv = r2inv * r2inv * r2inv;
    int force = r2inv * r6inv * (15 * r6inv - 20);
    fx += delx * force;
    fy += dely * force;
    fz += delz * force;
  }
  d_force_x[i] = fx;
  d_force_y[i] = fy;
  d_force_z[i] = fz;
}
```