ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<pri_switch2str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0xa,-0x4(%rbp)
  ja     L86
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D114(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D114(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D15(%rip),%rax
  jmp    L69
  lea    D95(%rip),%rax
  jmp    L69
  lea    D79(%rip),%rax
  jmp    L69
  lea    D63(%rip),%rax
  jmp    L69
  lea    D112(%rip),%rax
  jmp    L69
  lea    D60(%rip),%rax
  jmp    L69
  lea    D94(%rip),%rax
  jmp    L69
  lea    D68(%rip),%rax
  jmp    L69
  lea    D42(%rip),%rax
  jmp    L69
L86:
  lea    D78(%rip),%rax
L69:
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D15
type_infered: string
name: parse_data
arguments: data_tag: D95
type_infered: string
name: parse_data
arguments: data_tag: D79
type_infered: string
name: parse_data
arguments: data_tag: D63
type_infered: string
name: parse_data
arguments: data_tag: D112
type_infered: string
name: parse_data
arguments: data_tag: D60
type_infered: string
name: parse_data
arguments: data_tag: D94
type_infered: string
name: parse_data
arguments: data_tag: D68
type_infered: string
name: parse_data
arguments: data_tag: D42
type_infered: string
name: parse_data
arguments: data_tag: D78
type_infered: string

tool: D15: National ISDN

tool: D95: Nortel DMS100

tool: D79: Lucent 5E

tool: D63: AT&T 4ESS

tool: D112: National ISDN 1

tool: D60: EuroISDN

tool: D94: GR303 EOC

tool: D68: GR303 TMC

tool: D42: Q.SIG switch

tool: D78: Unknown switchtype

assistant: ```c
char *pri_switch2str(int sw) {
  switch (sw) {
  case 1:
    return "National ISDN";
  case 2:
    return "Nortel DMS100";
  case 3:
    return "Lucent 5E";
  case 4:
    return "AT&T 4ESS";
  case 7:
    return "National ISDN 1";
  case 5:
    return "EuroISDN";
  case 8:
    return "GR303 EOC";
  case 9:
    return "GR303 TMC";
  case 10:
    return "Q.SIG switch";
  default:
    return "Unknown switchtype";
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<pri_switch2str>:
  endbr64
  cmp    $0xa,%edi
  ja     L3
  mov    %edi,%edi
  lea    D109(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D82(%rip),%rax
  ret
  lea    D89(%rip),%rax
  ret
  lea    D12(%rip),%rax
  ret
  lea    D25(%rip),%rax
  ret
  lea    D103(%rip),%rax
  ret
  lea    D60(%rip),%rax
  ret
  lea    D85(%rip),%rax
  ret
  lea    D21(%rip),%rax
  ret
L3:
  lea    D6(%rip),%rax
  ret
  lea    D107(%rip),%rax
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D82
type_guessed: string
name: unpack_data
arguments: data_label: D89
type_guessed: string
name: unpack_data
arguments: data_label: D12
type_guessed: string
name: unpack_data
arguments: data_label: D25
type_guessed: string
name: unpack_data
arguments: data_label: D103
type_guessed: string
name: unpack_data
arguments: data_label: D60
type_guessed: string
name: unpack_data
arguments: data_label: D85
type_guessed: string
name: unpack_data
arguments: data_label: D21
type_guessed: string
name: unpack_data
arguments: data_label: D6
type_guessed: string
name: unpack_data
arguments: data_label: D107
type_guessed: string

tool: D82: Nortel DMS100

tool: D89: Lucent 5E

tool: D12: AT&T 4ESS

tool: D25: National ISDN 1

tool: D103: EuroISDN

tool: D60: GR303 EOC

tool: D85: GR303 TMC

tool: D21: Q.SIG switch

tool: D6: Unknown switchtype

tool: D107: National ISDN

assistant: ```c
char *pri_switch2str(int sw) {
  switch (sw) {
  case 1:
    return "National ISDN";
  case 2:
    return "Nortel DMS100";
  case 3:
    return "Lucent 5E";
  case 4:
    return "AT&T 4ESS";
  case 7:
    return "National ISDN 1";
  case 5:
    return "EuroISDN";
  case 8:
    return "GR303 EOC";
  case 9:
    return "GR303 TMC";
  case 10:
    return "Q.SIG switch";
  default:
    return "Unknown switchtype";
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<tersini_bul>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L12
L17:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x4(%rbp)
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
L12:
  cmpl   $0x0,-0x14(%rbp)
  jg     L17
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int tersini_bul(int x) {
  int S = 0;
  while (x > 0) {
    int b = x % 10;
    S = (S * 10) + b;
    x /= 10;
  }
  return S;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<tersini_bul>:
  endbr64
  test   %edi,%edi
  jle    L28
  mov    $0x0,%edx
L16:
  lea    (%rdx,%rdx,4),%esi
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  mov    %edi,%ecx
  sub    %edx,%ecx
  lea    (%rcx,%rsi,2),%edx
  mov    %edi,%ecx
  mov    %eax,%edi
  cmp    $0x9,%ecx
  jg     L16
L0:
  mov    %edx,%eax
  ret
L28:
  mov    $0x0,%edx
  jmp    L0
```

assistant: ```c
int tersini_bul(int x) {
  int S = 0;
  while (x > 0) {
    int b = x % 10;
    S = (S * 10) + b;
    x /= 10;
  }
  return S;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<tersini_bul>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  jle    L10
  mov    $0xcccccccd,%r9d
  nopl   0x0(%rax)
L11:
  mov    %edi,%eax
  mov    %edi,%ecx
  lea    (%r8,%r8,4),%esi
  imul   %r9,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%ecx
  mov    %edi,%edx
  mov    %eax,%edi
  lea    (%rcx,%rsi,2),%r8d
  cmp    $0x9,%edx
  jg     L11
L10:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int tersini_bul(int x) {
  int S = 0;
  while (x > 0) {
    int b = x % 10;
    S = (S * 10) + b;
    x /= 10;
  }
  return S;
}
```
user: Translate the below assembly code into C programming language:

```asm
<tersini_bul>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  jle    L4
  mov    $0xcccccccd,%r9d
  nopl   0x0(%rax)
L17:
  mov    %edi,%eax
  mov    %edi,%ecx
  lea    (%r8,%r8,4),%esi
  imul   %r9,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%ecx
  mov    %edi,%edx
  mov    %eax,%edi
  lea    (%rcx,%rsi,2),%r8d
  cmp    $0x9,%edx
  jg     L17
L4:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int tersini_bul(int x) {
  int S = 0;
  while (x > 0) {
    int b = x % 10;
    S = (S * 10) + b;
    x /= 10;
  }
  return S;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x1,%ebx
  mov    $0xffffffff,%ecx
  shl    %cl,%ebx
  push   %rbx
  push   $0x0
  push   $0x0
  push   $0x2
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x2,%ecx
  mov    $0x0,%edx
  mov    $0x200,%esi
  lea    D49(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x3,%eax
  mov    $0xffffffff,%ecx
  sar    %cl,%eax
  mov    %eax,%ecx
  add    $0x20,%rsp
  mov    $0x0,%edx
  mov    $0x1,%esi
  lea    D39(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %ebx,%ebx
  jne    L13
L6:
  mov    $0xb,%esi
  lea    D40(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xb,%edi
  call   <exit@plt>
  lea    D10(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
L13:
  mov    $0x8,%esi
  lea    D40(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x8,%edi
  call   <exit@plt>
  jmp    L6
```

tool_calls: name: retrieve_data
arguments: tag: D49
data_type: string
name: retrieve_data
arguments: tag: D39
data_type: string
name: retrieve_data
arguments: tag: D40
data_type: string
name: retrieve_data
arguments: tag: D10
data_type: string

tool: D49: a=%d, b=%d, c=%d, d=%d:%u, e=%d:%u, f=%d, g=%d


tool: D39: aa=%d, bb=%d, cc=%d


tool: D40: Error %d


tool: D10: Success

assistant: ```c
int main() {
  int x = 257;
  int aa, bb, cc;
  int a = ((char)1) << 9;
  char b = ((char)1) << 9;
  int c = 1 << ((char)257);
  unsigned long long d = 1 << 33;
  unsigned long long e = ((unsigned long long)1) << 33;
  int f = 1 << 64;
  int g = 1 << -1;
  printf("a=%d, b=%d, c=%d, d=%d:%u, e=%d:%u, f=%d, g=%d\n", a, b, c, (int)(d >> 32), (unsigned int)d, (int)(e >> 32), (unsigned int)e, f, g);
  aa = ((unsigned long long)0x200000000LL) >> 33;
  bb = ((unsigned long long)0x200000000LL) >> 65;
  cc = 3 >> -1;
  printf("aa=%d, bb=%d, cc=%d\n", aa, bb, cc);
  if (a != 512) {
    printf("Error %d\n", (int)2);
    exit(2);
  };
  if (b != 0) {
    printf("Error %d\n", (int)3);
    exit(3);
  };
  if (c != 2) {
    printf("Error %d\n", (int)4);
    exit(4);
  };
  if (d != 0) {
    printf("Error %d\n", (int)5);
    exit(5);
  };
  if (e != ((unsigned long long)0x200000000)) {
    printf("Error %d\n", (int)6);
    exit(6);
  };
  if (f != 0) {
    printf("Error %d\n", (int)7);
    exit(7);
  };
  if (g != 0) {
    printf("Error %d\n", (int)8);
    exit(8);
  };
  if (aa != 1) {
    printf("Error %d\n", (int)9);
    exit(9);
  };
  if (bb != 0) {
    printf("Error %d\n", (int)10);
    exit(10);
  };
  if (cc != 6) {
    printf("Error %d\n", (int)11);
    exit(11);
  };
  {
    printf("Success\n");
    exit(0);
  };
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0xffffffff,%ecx
  mov    $0x1,%ebx
  xor    %edx,%edx
  shl    %cl,%ebx
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  mov    $0x2,%ecx
  push   %rbx
  mov    $0x200,%esi
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  push   $0x0
  push   $0x0
  push   $0x2
  call   <printf@plt>
  mov    $0xffffffff,%ecx
  add    $0x20,%rsp
  xor    %edx,%edx
  mov    $0x3,%eax
  mov    $0x1,%esi
  lea    D45(%rip),%rdi
  sar    %cl,%eax
  mov    %eax,%ecx
  xor    %eax,%eax
  call   <printf@plt>
  test   %ebx,%ebx
  jne    L3
L6:
  mov    $0xb,%esi
  lea    D19(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xb,%edi
  call   <exit@plt>
  lea    D41(%rip),%rdi
  call   <puts@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
L3:
  mov    $0x8,%esi
  lea    D19(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x8,%edi
  call   <exit@plt>
  jmp    L6
```

tool_calls: name: disassemble_data
arguments: data_label: D0
type_infered: string
name: disassemble_data
arguments: data_label: D45
type_infered: string
name: disassemble_data
arguments: data_label: D19
type_infered: string
name: disassemble_data
arguments: data_label: D41
type_infered: string

tool: D0: a=%d, b=%d, c=%d, d=%d:%u, e=%d:%u, f=%d, g=%d


tool: D45: aa=%d, bb=%d, cc=%d


tool: D19: Error %d


tool: D41: Success

assistant: ```c
int main() {
  int x = 257;
  int aa, bb, cc;
  int a = ((char)1) << 9;
  char b = ((char)1) << 9;
  int c = 1 << ((char)257);
  unsigned long long d = 1 << 33;
  unsigned long long e = ((unsigned long long)1) << 33;
  int f = 1 << 64;
  int g = 1 << -1;
  printf("a=%d, b=%d, c=%d, d=%d:%u, e=%d:%u, f=%d, g=%d\n", a, b, c, (int)(d >> 32), (unsigned int)d, (int)(e >> 32), (unsigned int)e, f, g);
  aa = ((unsigned long long)0x200000000LL) >> 33;
  bb = ((unsigned long long)0x200000000LL) >> 65;
  cc = 3 >> -1;
  printf("aa=%d, bb=%d, cc=%d\n", aa, bb, cc);
  if (a != 512) {
    printf("Error %d\n", (int)2);
    exit(2);
  };
  if (b != 0) {
    printf("Error %d\n", (int)3);
    exit(3);
  };
  if (c != 2) {
    printf("Error %d\n", (int)4);
    exit(4);
  };
  if (d != 0) {
    printf("Error %d\n", (int)5);
    exit(5);
  };
  if (e != ((unsigned long long)0x200000000)) {
    printf("Error %d\n", (int)6);
    exit(6);
  };
  if (f != 0) {
    printf("Error %d\n", (int)7);
    exit(7);
  };
  if (g != 0) {
    printf("Error %d\n", (int)8);
    exit(8);
  };
  if (aa != 1) {
    printf("Error %d\n", (int)9);
    exit(9);
  };
  if (bb != 0) {
    printf("Error %d\n", (int)10);
    exit(10);
  };
  if (cc != 6) {
    printf("Error %d\n", (int)11);
    exit(11);
  };
  {
    printf("Success\n");
    exit(0);
  };
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0xffffffff,%ecx
  mov    $0x1,%ebx
  xor    %edx,%edx
  shl    %cl,%ebx
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  mov    $0x2,%ecx
  push   %rbx
  mov    $0x200,%esi
  lea    D37(%rip),%rdi
  xor    %eax,%eax
  push   $0x0
  push   $0x0
  push   $0x2
  call   <printf@plt>
  mov    $0xffffffff,%ecx
  add    $0x20,%rsp
  xor    %edx,%edx
  mov    $0x3,%eax
  mov    $0x1,%esi
  lea    D33(%rip),%rdi
  sar    %cl,%eax
  mov    %eax,%ecx
  xor    %eax,%eax
  call   <printf@plt>
  test   %ebx,%ebx
  jne    L6
L10:
  mov    $0xb,%esi
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xb,%edi
  call   <exit@plt>
  lea    D2(%rip),%rdi
  call   <puts@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
L6:
  mov    $0x8,%esi
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x8,%edi
  call   <exit@plt>
  jmp    L10
```

tool_calls: name: disassemble_data
arguments: tag: D37
type: string
name: disassemble_data
arguments: tag: D33
type: string
name: disassemble_data
arguments: tag: D6
type: string
name: disassemble_data
arguments: tag: D2
type: string

tool: D37: a=%d, b=%d, c=%d, d=%d:%u, e=%d:%u, f=%d, g=%d


tool: D33: aa=%d, bb=%d, cc=%d


tool: D6: Error %d


tool: D2: Success

assistant: ```c
int main() {
  int x = 257;
  int aa, bb, cc;
  int a = ((char)1) << 9;
  char b = ((char)1) << 9;
  int c = 1 << ((char)257);
  unsigned long long d = 1 << 33;
  unsigned long long e = ((unsigned long long)1) << 33;
  int f = 1 << 64;
  int g = 1 << -1;
  printf("a=%d, b=%d, c=%d, d=%d:%u, e=%d:%u, f=%d, g=%d\n", a, b, c, (int)(d >> 32), (unsigned int)d, (int)(e >> 32), (unsigned int)e, f, g);
  aa = ((unsigned long long)0x200000000LL) >> 33;
  bb = ((unsigned long long)0x200000000LL) >> 65;
  cc = 3 >> -1;
  printf("aa=%d, bb=%d, cc=%d\n", aa, bb, cc);
  if (a != 512) {
    printf("Error %d\n", (int)2);
    exit(2);
  };
  if (b != 0) {
    printf("Error %d\n", (int)3);
    exit(3);
  };
  if (c != 2) {
    printf("Error %d\n", (int)4);
    exit(4);
  };
  if (d != 0) {
    printf("Error %d\n", (int)5);
    exit(5);
  };
  if (e != ((unsigned long long)0x200000000)) {
    printf("Error %d\n", (int)6);
    exit(6);
  };
  if (f != 0) {
    printf("Error %d\n", (int)7);
    exit(7);
  };
  if (g != 0) {
    printf("Error %d\n", (int)8);
    exit(8);
  };
  if (aa != 1) {
    printf("Error %d\n", (int)9);
    exit(9);
  };
  if (bb != 0) {
    printf("Error %d\n", (int)10);
    exit(10);
  };
  if (cc != 6) {
    printf("Error %d\n", (int)11);
    exit(11);
  };
  {
    printf("Success\n");
    exit(0);
  };
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ue_linfo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L12
L2:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L12:
  cmpl   $0xf,-0x8(%rbp)
  jg     L22
  cmpl   $0x0,-0x4(%rbp)
  jne    L2
L22:
  mov    -0x8(%rbp),%eax
  add    %eax,%eax
  lea    0x1(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x8(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  sub    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void ue_linfo(int ue, int dummy, int *len, int *info) {
  int i, nn;
  nn = (ue + 1) / 2;
  for (i = 0; i < 16 && nn != 0; i++) {
    nn /= 2;
  }
  *len = 2 * i + 1;
  *info = ue + 1 - (1 << i);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ue_linfo>:
  endbr64
  mov    %rcx,%r8
  add    $0x1,%edi
  mov    %edi,%ecx
  shr    $0x1f,%ecx
  add    %edi,%ecx
  sar    %ecx
  mov    %ecx,%esi
  je     L2
  mov    $0x0,%ecx
L23:
  mov    %esi,%eax
  shr    $0x1f,%eax
  add    %esi,%eax
  sar    %eax
  mov    %eax,%esi
  add    $0x1,%ecx
  cmp    $0xf,%ecx
  jg     L2
  test   %eax,%eax
  jne    L23
L2:
  lea    0x1(%rcx,%rcx,1),%eax
  mov    %eax,(%rdx)
  mov    $0x1,%eax
  shl    %cl,%eax
  sub    %eax,%edi
  mov    %edi,(%r8)
  ret
```

assistant: ```c
void ue_linfo(int ue, int dummy, int *len, int *info) {
  int i, nn;
  nn = (ue + 1) / 2;
  for (i = 0; i < 16 && nn != 0; i++) {
    nn /= 2;
  }
  *len = 2 * i + 1;
  *info = ue + 1 - (1 << i);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ue_linfo>:
  endbr64
  lea    0x1(%rdi),%r9d
  mov    %rcx,%r8
  mov    %r9d,%eax
  shr    $0x1f,%eax
  add    %r9d,%eax
  sar    %eax
  je     L7
  xor    %ecx,%ecx
  nopw   0x0(%rax,%rax,1)
L19:
  mov    %eax,%esi
  add    $0x1,%ecx
  shr    $0x1f,%esi
  add    %esi,%eax
  sar    %eax
  cmp    $0xf,%ecx
  jg     L5
  test   %eax,%eax
  jne    L19
L5:
  mov    $0x1,%eax
  mov    %r9d,%edi
  lea    0x1(%rcx,%rcx,1),%esi
  shl    %cl,%eax
  mov    %esi,(%rdx)
  sub    %eax,%edi
  mov    %edi,(%r8)
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  mov    $0x1,%esi
  mov    %esi,(%rdx)
  mov    %edi,(%r8)
  ret
```

assistant: ```c
void ue_linfo(int ue, int dummy, int *len, int *info) {
  int i, nn;
  nn = (ue + 1) / 2;
  for (i = 0; i < 16 && nn != 0; i++) {
    nn /= 2;
  }
  *len = 2 * i + 1;
  *info = ue + 1 - (1 << i);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ue_linfo>:
  endbr64
  mov    %edi,%esi
  mov    %rdx,%r8
  mov    %rcx,%rdi
  lea    0x1(%rsi),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %eax,%edx
  shr    %edx
  je     L56
  test   %eax,%eax
  lea    0x4(%rsi),%edx
  cmovns %eax,%edx
  shr    $0x2,%edx
  je     L65
  test   %eax,%eax
  lea    0x8(%rsi),%edx
  cmovns %eax,%edx
  shr    $0x3,%edx
  je     L133
  test   %eax,%eax
  lea    0x10(%rsi),%edx
  cmovns %eax,%edx
  shr    $0x4,%edx
  je     L128
  test   %eax,%eax
  lea    0x20(%rsi),%edx
  cmovns %eax,%edx
  shr    $0x5,%edx
  je     L230
  test   %eax,%eax
  lea    0x40(%rsi),%edx
  cmovns %eax,%edx
  shr    $0x6,%edx
  je     L76
  test   %eax,%eax
  lea    0x80(%rsi),%edx
  cmovns %eax,%edx
  shr    $0x7,%edx
  je     L264
  test   %eax,%eax
  lea    0x100(%rsi),%edx
  cmovns %eax,%edx
  shr    $0x8,%edx
  je     L54
  test   %eax,%eax
  lea    0x200(%rsi),%edx
  cmovns %eax,%edx
  shr    $0x9,%edx
  je     L219
  test   %eax,%eax
  lea    0x400(%rsi),%edx
  cmovns %eax,%edx
  shr    $0xa,%edx
  je     L181
  test   %eax,%eax
  lea    0x7ff(%rax),%edx
  cmovns %eax,%edx
  shr    $0xb,%edx
  je     L240
  test   %eax,%eax
  lea    0xfff(%rax),%edx
  cmovns %eax,%edx
  shr    $0xc,%edx
  je     L180
  test   %eax,%eax
  lea    0x1fff(%rax),%edx
  cmovns %eax,%edx
  shr    $0xd,%edx
  je     L275
  test   %eax,%eax
  lea    0x3fff(%rax),%edx
  cmovns %eax,%edx
  shr    $0xe,%edx
  je     L87
  test   %eax,%eax
  lea    0x7fff(%rax),%edx
  cmovns %eax,%edx
  shr    $0xf,%edx
  je     L261
  add    $0x10000,%esi
  xor    %ecx,%ecx
  cmp    $0x1fffe,%esi
  seta   %cl
  add    $0xf,%ecx
L201:
  mov    $0x1,%edx
  lea    0x1(%rcx,%rcx,1),%r9d
  shl    %cl,%edx
  mov    %r9d,(%r8)
  sub    %edx,%eax
  mov    %eax,%esi
  mov    %esi,(%rdi)
  ret
L76:
  mov    $0x5,%ecx
  jmp    L201
  nopl   0x0(%rax)
L56:
  mov    $0x1,%r9d
  mov    %r9d,(%r8)
  mov    %esi,(%rdi)
  ret
  nopl   0x0(%rax)
L65:
  mov    $0x1,%ecx
  jmp    L201
L133:
  mov    $0x2,%ecx
  jmp    L201
L128:
  mov    $0x3,%ecx
  jmp    L201
L230:
  mov    $0x4,%ecx
  jmp    L201
L264:
  mov    $0x6,%ecx
  jmp    L201
L87:
  mov    $0xd,%ecx
  jmp    L201
L54:
  mov    $0x7,%ecx
  jmp    L201
L219:
  mov    $0x8,%ecx
  jmp    L201
L181:
  mov    $0x9,%ecx
  jmp    L201
L240:
  mov    $0xa,%ecx
  jmp    L201
L180:
  mov    $0xb,%ecx
  jmp    L201
L275:
  mov    $0xc,%ecx
  jmp    L201
L261:
  mov    $0xe,%ecx
  jmp    L201
```

assistant: ```c
void ue_linfo(int ue, int dummy, int *len, int *info) {
  int i, nn;
  nn = (ue + 1) / 2;
  for (i = 0; i < 16 && nn != 0; i++) {
    nn /= 2;
  }
  *len = 2 * i + 1;
  *info = ue + 1 - (1 << i);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<NthBucketSpiral>:
  endbr64
  push   %rbx
  mov    %esi,%r11d
  mov    %edx,%r10d
  mov    %rcx,%rsi
  cmp    %r11d,%edx
  mov    %r11d,%eax
  cmovle %edx,%eax
  sub    $0x1,%eax
  mov    %eax,%r9d
  shr    $0x1f,%r9d
  add    %eax,%r9d
  sar    %r9d
  mov    %edx,%ecx
  imul   %r11d,%ecx
  cmp    %ecx,%edi
  jge    L74
  mov    %r11d,%eax
  sub    %r11d,%r10d
L24:
  sub    $0x1,%eax
  lea    (%r10,%rax,1),%edx
  mov    %eax,%ecx
  imul   %edx,%ecx
  cmp    %edi,%ecx
  jg     L24
L56:
  cmp    %edx,%eax
  mov    %edx,%r11d
  cmovle %eax,%r11d
  mov    %r11d,%ebx
  shr    $0x1f,%ebx
  lea    (%r11,%rbx,1),%r10d
  and    $0x1,%r10d
  sub    %ebx,%r10d
  cmp    $0x1,%r10d
  je     L22
  lea    (%rcx,%rdx,1),%r10d
  cmp    %edi,%r10d
  jl     L89
  mov    %r11d,%r10d
  shr    $0x1f,%r10d
  add    %r11d,%r10d
  sar    %r10d
  neg    %r10d
  mov    %r10d,%eax
  add    %r10d,%edx
  sub    %ecx,%edi
  sub    %edi,%edx
L87:
  add    %r9d,%eax
  mov    %eax,(%rsi)
  lea    (%rdx,%r9,1),%eax
  mov    %eax,(%r8)
  pop    %rbx
  ret
L74:
  mov    %r11d,%eax
  jmp    L56
L22:
  lea    (%rcx,%rdx,1),%r10d
  cmp    %edi,%r10d
  jl     L53
  mov    %ebx,%edx
  add    %r11d,%edx
  sar    %edx
  neg    %edx
  add    %edx,%eax
  add    %edi,%edx
  sub    %ecx,%edx
  jmp    L87
L53:
  mov    %r11d,%ecx
  shr    $0x1f,%ecx
  add    %r11d,%ecx
  sar    %ecx
  neg    %ecx
  add    %ecx,%eax
  sub    %r10d,%edi
  sub    %edi,%eax
  add    %ecx,%edx
  jmp    L87
L89:
  mov    %r11d,%eax
  shr    $0x1f,%eax
  add    %r11d,%eax
  sar    %eax
  neg    %eax
  mov    %eax,%edx
  sub    %r10d,%edi
  add    %edi,%eax
  jmp    L87
```

assistant: ```c
void NthBucketSpiral(int n, int nxbuckets, int nybuckets, unsigned int *bucketx, unsigned int *buckety) {
  int nx, ny, nxny, minnxny, x, y;
  int minnbuckets;
  int center;
  minnbuckets = (nxbuckets < nybuckets) ? nxbuckets : nybuckets;
  center = (minnbuckets - 1) / 2;
  nx = nxbuckets;
  ny = nybuckets;
  while (n < nx * ny) {
    nx = nx - 1;
    ny = ny - 1;
  }
  nxny = nx * ny;
  minnxny = (nx < ny) ? nx : ny;
  if (minnxny % 2 == 1) {
    if (n <= (nxny + ny)) {
      x = nx - minnxny / 2;
      y = -minnxny / 2 + n - nxny;
    } else {
      x = nx - minnxny / 2 - (n - (nxny + ny));
      y = ny - minnxny / 2;
    }
  } else {
    if (n <= (nxny + ny)) {
      x = -minnxny / 2;
      y = ny - minnxny / 2 - (n - nxny);
    } else {
      x = -minnxny / 2 + (n - (nxny + ny));
      y = -minnxny / 2;
    }
  }
  *bucketx = x + center;
  *buckety = y + center;
}
```