ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<UI_ConcatArgs>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq %edx,%rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %ebp,%ebp
  jle    L61
  xor    %eax,%eax
  mov    %edi,%r15d
  call   <trap_Argc@plt>
  mov    %eax,%r13d
  cmp    %r15d,%eax
  jle    L87
  lea    -0x2(%r12,%rbp,1),%r14
  mov    %r12,%rbx
  sub    $0x1,%rbp
  nopw   0x0(%rax,%rax,1)
L55:
  mov    %r15d,%edi
  call   <UI_Argv@plt>
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L12
  lea    (%r12,%rbp,1),%rcx
  cmp    %rbx,%rcx
  ja     L11
  jmp    L12
  nopw   0x0(%rax,%rax,1)
L81:
  cmp    %rcx,%rbx
  je     L58
L11:
  add    $0x1,%rax
  mov    %dl,(%rbx)
  add    $0x1,%rbx
  movzbl (%rax),%edx
  test   %dl,%dl
  jne    L81
L12:
  mov    %rbx,%rcx
L58:
  cmp    %rcx,%r14
  jbe    L50
  add    $0x1,%r15d
  movb   $0x20,(%rcx)
  lea    0x1(%rcx),%rbx
  cmp    %r15d,%r13d
  jne    L55
  mov    %rbx,%rcx
L50:
  movb   $0x0,(%rcx)
L61:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L87:
  mov    %r12,%rcx
  jmp    L50
```

assistant: ```c
char *UI_ConcatArgs(int arg, char *buf, int len) {
  char *p;
  int c;
  if (len <= 0)
    return buf;
  p = buf;
  c = trap_Argc();
  for (; arg < c; arg++) {
    char *argp = UI_Argv(arg);
    while (*argp && p < &buf[len - 1])
      *p++ = *argp++;
    if (p < &buf[len - 2])
      *p++ = ' ';
    else
      break;
  }
  *p = '\0';
  return buf;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<UI_ConcatArgs>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq %edx,%rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %ebp,%ebp
  jle    L90
  xor    %eax,%eax
  mov    %edi,%r15d
  call   <trap_Argc@plt>
  mov    %eax,%r13d
  cmp    %r15d,%eax
  jle    L88
  lea    -0x2(%r12,%rbp,1),%r14
  mov    %r12,%rbx
  sub    $0x1,%rbp
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %r15d,%edi
  call   <UI_Argv@plt>
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L33
  lea    (%r12,%rbp,1),%rcx
  cmp    %rbx,%rcx
  ja     L81
  jmp    L33
  nopw   0x0(%rax,%rax,1)
L93:
  cmp    %rcx,%rbx
  je     L72
L81:
  add    $0x1,%rax
  mov    %dl,(%rbx)
  add    $0x1,%rbx
  movzbl (%rax),%edx
  test   %dl,%dl
  jne    L93
L33:
  mov    %rbx,%rcx
L72:
  cmp    %rcx,%r14
  jbe    L92
  add    $0x1,%r15d
  movb   $0x20,(%rcx)
  lea    0x1(%rcx),%rbx
  cmp    %r15d,%r13d
  jne    L6
  mov    %rbx,%rcx
L92:
  movb   $0x0,(%rcx)
L90:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L88:
  mov    %r12,%rcx
  jmp    L92
```

assistant: ```c
char *UI_ConcatArgs(int arg, char *buf, int len) {
  char *p;
  int c;
  if (len <= 0)
    return buf;
  p = buf;
  c = trap_Argc();
  for (; arg < c; arg++) {
    char *argp = UI_Argv(arg);
    while (*argp && p < &buf[len - 1])
      *p++ = *argp++;
    if (p < &buf[len - 2])
      *p++ = ' ';
    else
      break;
  }
  *p = '\0';
  return buf;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ezxml_ent_ok>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  jmp    L82
L116:
  addq   $0x1,-0x20(%rbp)
L82:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L45
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x26,%al
  jne    L116
L45:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L128
  mov    $0x1,%eax
  jmp    L138
L128:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L42
  mov    $0x0,%eax
  jmp    L138
L42:
  movl   $0x0,-0x4(%rbp)
  jmp    L62
L81:
  addl   $0x2,-0x4(%rbp)
L62:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L22
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rsi
  mov    -0x28(%rbp),%rax
  add    %rsi,%rax
  mov    (%rax),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L81
L22:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L108
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rcx
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <ezxml_ent_ok>
  test   %eax,%eax
  jne    L108
  mov    $0x0,%eax
  jmp    L138
L108:
  addq   $0x1,-0x20(%rbp)
  jmp    L82
L138:
  leave
  ret
```

assistant: ```c
int ezxml_ent_ok(char *name, char *s, char **ent) {
  int i;
  for (;; s++) {
    while (*s && *s != '&')
      s++;
    if (!*s)
      return 1;
    if (!strncmp(s + 1, name, strlen(name)))
      return 0;
    for (i = 0; ent[i] && strncmp(ent[i], s + 1, strlen(ent[i])); i += 2)
      ;
    if (ent[i] && !ezxml_ent_ok(name, ent[i + 1], ent))
      return 0;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<ezxml_ent_ok>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r14
  mov    %rsi,%rbx
  mov    %rdx,%r13
L41:
  movzbl (%rbx),%eax
  test   %al,%al
  je     L31
  cmp    $0x26,%al
  je     L31
L5:
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  test   %al,%al
  je     L31
  cmp    $0x26,%al
  jne    L5
L31:
  test   %al,%al
  je     L84
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  add    $0x1,%rbx
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  je     L73
  mov    0x0(%r13),%rdi
  mov    %r13,%rbp
  mov    $0x0,%r12d
  test   %rdi,%rdi
  je     L41
L25:
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %rbx,%rsi
  mov    0x0(%rbp),%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  je     L17
  add    $0x10,%rbp
  mov    0x0(%rbp),%rdi
  add    $0x2,%r12
  test   %rdi,%rdi
  jne    L25
  jmp    L41
L17:
  cmpq   $0x0,0x0(%rbp)
  je     L41
  mov    0x8(%r13,%r12,8),%rsi
  mov    %r13,%rdx
  mov    %r14,%rdi
  call   <ezxml_ent_ok>
  test   %eax,%eax
  jne    L41
  jmp    L49
L84:
  mov    $0x1,%eax
L49:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L73:
  mov    $0x0,%eax
  jmp    L49
```

assistant: ```c
int ezxml_ent_ok(char *name, char *s, char **ent) {
  int i;
  for (;; s++) {
    while (*s && *s != '&')
      s++;
    if (!*s)
      return 1;
    if (!strncmp(s + 1, name, strlen(name)))
      return 0;
    for (i = 0; ent[i] && strncmp(ent[i], s + 1, strlen(ent[i])); i += 2)
      ;
    if (ent[i] && !ezxml_ent_ok(name, ent[i + 1], ent))
      return 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ezxml_ent_ok>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L46
  nopw   0x0(%rax,%rax,1)
L32:
  add    $0x1,%rbx
L46:
  movzbl (%rbx),%eax
  test   %al,%al
  je     L54
  cmp    $0x26,%al
  jne    L32
  test   %al,%al
  je     L54
  mov    %r14,%rdi
  add    $0x1,%rbx
  call   <strlen@plt>
  mov    %r14,%rsi
  mov    %rbx,%rdi
  mov    %eax,%edx
  call   <strncmp@plt>
  test   %rax,%rax
  je     L77
  mov    0x0(%r13),%rdi
  mov    %r13,%rbp
  xor    %r12d,%r12d
  test   %rdi,%rdi
  jne    L30
  jmp    L46
  nopw   0x0(%rax,%rax,1)
L17:
  mov    0x10(%rbp),%rdi
  add    $0x10,%rbp
  add    $0x2,%r12
  test   %rdi,%rdi
  je     L46
L30:
  call   <strlen@plt>
  mov    0x0(%rbp),%rdi
  mov    %rbx,%rsi
  mov    %eax,%edx
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L17
  cmpq   $0x0,0x0(%rbp)
  je     L46
  mov    0x8(%r13,%r12,8),%rsi
  mov    %r13,%rdx
  mov    %r14,%rdi
  call   <ezxml_ent_ok>
  test   %eax,%eax
  jne    L46
L77:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L54:
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int ezxml_ent_ok(char *name, char *s, char **ent) {
  int i;
  for (;; s++) {
    while (*s && *s != '&')
      s++;
    if (!*s)
      return 1;
    if (!strncmp(s + 1, name, strlen(name)))
      return 0;
    for (i = 0; ent[i] && strncmp(ent[i], s + 1, strlen(ent[i])); i += 2)
      ;
    if (ent[i] && !ezxml_ent_ok(name, ent[i + 1], ent))
      return 0;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ezxml_ent_ok>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L72
  nopw   0x0(%rax,%rax,1)
L52:
  add    $0x1,%rbx
L72:
  movzbl (%rbx),%eax
  test   %al,%al
  je     L35
  cmp    $0x26,%al
  jne    L52
  test   %al,%al
  je     L35
  mov    %r14,%rdi
  add    $0x1,%rbx
  call   <strlen@plt>
  mov    %r14,%rsi
  mov    %rbx,%rdi
  mov    %eax,%edx
  call   <strncmp@plt>
  test   %rax,%rax
  je     L51
  mov    0x0(%r13),%rdi
  mov    %r13,%rbp
  xor    %r12d,%r12d
  test   %rdi,%rdi
  jne    L42
  jmp    L72
  nopw   0x0(%rax,%rax,1)
L64:
  mov    0x10(%rbp),%rdi
  add    $0x10,%rbp
  add    $0x2,%r12
  test   %rdi,%rdi
  je     L72
L42:
  call   <strlen@plt>
  mov    0x0(%rbp),%rdi
  mov    %rbx,%rsi
  mov    %eax,%edx
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L64
  cmpq   $0x0,0x0(%rbp)
  je     L72
  mov    0x8(%r13,%r12,8),%rsi
  mov    %r13,%rdx
  mov    %r14,%rdi
  call   <ezxml_ent_ok>
  test   %eax,%eax
  jne    L72
L51:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L35:
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int ezxml_ent_ok(char *name, char *s, char **ent) {
  int i;
  for (;; s++) {
    while (*s && *s != '&')
      s++;
    if (!*s)
      return 1;
    if (!strncmp(s + 1, name, strlen(name)))
      return 0;
    for (i = 0; ent[i] && strncmp(ent[i], s + 1, strlen(ent[i])); i += 2)
      ;
    if (ent[i] && !ezxml_ent_ok(name, ent[i + 1], ent))
      return 0;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<vec3_addk>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  movss  %xmm0,-0x14(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    -0x8(%rbp),%rax
  movss  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  movss  (%rax),%xmm0
  mulss  -0x14(%rbp),%xmm0
  addss  %xmm1,%xmm0
  mov    -0x20(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mulss  -0x14(%rbp),%xmm0
  mov    -0x20(%rbp),%rax
  add    $0x4,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm0
  mulss  -0x14(%rbp),%xmm0
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void vec3_addk(const float a[3], const float b[3], float k, float out[3]) {
  out[0] = a[0] + b[0] * k;
  out[1] = a[1] + b[1] * k;
  out[2] = a[2] + b[2] * k;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<vec3_addk>:
  endbr64
  movaps %xmm0,%xmm1
  mulss  (%rsi),%xmm1
  addss  (%rdi),%xmm1
  movss  %xmm1,(%rdx)
  movaps %xmm0,%xmm1
  mulss  0x4(%rsi),%xmm1
  addss  0x4(%rdi),%xmm1
  movss  %xmm1,0x4(%rdx)
  mulss  0x8(%rsi),%xmm0
  addss  0x8(%rdi),%xmm0
  movss  %xmm0,0x8(%rdx)
  ret
```

assistant: ```c
void vec3_addk(const float a[3], const float b[3], float k, float out[3]) {
  out[0] = a[0] + b[0] * k;
  out[1] = a[1] + b[1] * k;
  out[2] = a[2] + b[2] * k;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<vec3_addk>:
  endbr64
  movss  (%rsi),%xmm1
  mulss  %xmm0,%xmm1
  addss  (%rdi),%xmm1
  movss  %xmm1,(%rdx)
  movss  0x4(%rsi),%xmm1
  mulss  %xmm0,%xmm1
  addss  0x4(%rdi),%xmm1
  movss  %xmm1,0x4(%rdx)
  mulss  0x8(%rsi),%xmm0
  addss  0x8(%rdi),%xmm0
  movss  %xmm0,0x8(%rdx)
  ret
```

assistant: ```c
void vec3_addk(const float a[3], const float b[3], float k, float out[3]) {
  out[0] = a[0] + b[0] * k;
  out[1] = a[1] + b[1] * k;
  out[2] = a[2] + b[2] * k;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<vec3_addk>:
  endbr64
  movss  (%rsi),%xmm1
  mulss  %xmm0,%xmm1
  addss  (%rdi),%xmm1
  movss  %xmm1,(%rdx)
  movss  0x4(%rsi),%xmm1
  mulss  %xmm0,%xmm1
  addss  0x4(%rdi),%xmm1
  movss  %xmm1,0x4(%rdx)
  mulss  0x8(%rsi),%xmm0
  addss  0x8(%rdi),%xmm0
  movss  %xmm0,0x8(%rdx)
  ret
```

assistant: ```c
void vec3_addk(const float a[3], const float b[3], float k, float out[3]) {
  out[0] = a[0] + b[0] * k;
  out[1] = a[1] + b[1] * k;
  out[2] = a[2] + b[2] * k;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<thirt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x200,%rsp
  mov    %rdi,-0x1f8(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x1f8(%rbp),%rax
  mov    %rax,-0x1e8(%rbp)
  mov    -0x1f8(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x1e0(%rbp)
  movq   $0x1,-0x1d0(%rbp)
  movq   $0xa,-0x1c8(%rbp)
  movq   $0x9,-0x1c0(%rbp)
  movq   $0xc,-0x1b8(%rbp)
  movq   $0x3,-0x1b0(%rbp)
  movq   $0x4,-0x1a8(%rbp)
  jmp    L7
L55:
  mov    -0x1e8(%rbp),%rax
  mov    %rax,-0x1e0(%rbp)
  mov    -0x1e0(%rbp),%rax
  mov    %rax,-0x1d8(%rbp)
  movl   $0x0,-0x1f0(%rbp)
L23:
  mov    -0x1d8(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  mov    %rcx,%rsi
  sar    $0x3f,%rsi
  sub    %rsi,%rax
  mov    %rax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  mov    -0x1f0(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    %ecx,-0x1f0(%rbp)
  cltq
  mov    %edx,-0x1a0(%rbp,%rax,4)
  mov    -0x1d8(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x1d8(%rbp)
  cmpq   $0x0,-0x1d8(%rbp)
  jg     L23
  movq   $0x0,-0x1e8(%rbp)
  movl   $0x0,-0x1ec(%rbp)
  jmp    L27
L3:
  mov    -0x1ec(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  movslq %eax,%rsi
  mov    -0x1ec(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x2aaaaaab,%rax,%rax
  shr    $0x20,%rax
  mov    %ecx,%edi
  sar    $0x1f,%edi
  mov    %eax,%edx
  sub    %edi,%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  movslq %edx,%rax
  mov    -0x1d0(%rbp,%rax,8),%rax
  imul   %rsi,%rax
  add    %rax,-0x1e8(%rbp)
  addl   $0x1,-0x1ec(%rbp)
L27:
  mov    -0x1ec(%rbp),%eax
  cmp    -0x1f0(%rbp),%eax
  jl     L3
L7:
  mov    -0x1e8(%rbp),%rax
  cmp    -0x1e0(%rbp),%rax
  jne    L55
  mov    -0x1e8(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

assistant: ```c
long long thirt(long long n) {
  long long new_num = n;
  long long old_num = n - 1;
  long long rems[] = {1, 10, 9, 12, 3, 4};
  while (new_num != old_num) {
    old_num = new_num;
    long long tmp = old_num;
    int arr_cnt = 0;
    int arr[100];
    do {
      arr[arr_cnt++] = tmp % 10;
    } while ((tmp /= 10) > 0);
    new_num = 0;
    for (int i = 0; i < arr_cnt; ++i)
      new_num += arr[i] * rems[i % 6];
  }
  return new_num;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<thirt>:
  endbr64
  sub    $0x1d8,%rsp
  mov    %rdi,%r8
  mov    %fs:0x28,%rax
  mov    %rax,0x1c8(%rsp)
  xor    %eax,%eax
  movq   $0x1,(%rsp)
  movq   $0xa,0x8(%rsp)
  movq   $0x9,0x10(%rsp)
  movq   $0xc,0x18(%rsp)
  movq   $0x3,0x20(%rsp)
  movq   $0x4,0x28(%rsp)
  movabs $0x6666666666666667,%rdi
  jmp    L63
L27:
  mov    %r9,%r8
L63:
  lea    0x30(%rsp),%r10
  mov    %r8,%rcx
  mov    $0x0,%r9d
L4:
  mov    %r9d,%esi
  add    $0x1,%r9d
  mov    %rcx,%rax
  imul   %rdi
  sar    $0x2,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  mov    %rcx,%r11
  sub    %rax,%r11
  mov    %r11d,(%r10)
  mov    %rcx,%rax
  mov    %rdx,%rcx
  add    $0x4,%r10
  cmp    $0x9,%rax
  jg     L4
  test   %r9d,%r9d
  jle    L50
  mov    $0x0,%edx
  mov    $0x0,%r9d
L33:
  movslq %edx,%rax
  imul   $0x2aaaaaab,%rax,%rax
  shr    $0x20,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,2),%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  movslq %ecx,%rcx
  movslq 0x30(%rsp,%rdx,4),%rax
  imul   (%rsp,%rcx,8),%rax
  add    %rax,%r9
  add    $0x1,%rdx
  lea    -0x1(%rdx),%eax
  cmp    %esi,%eax
  jl     L33
L49:
  cmp    %r8,%r9
  jne    L27
  mov    0x1c8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  mov    %r8,%rax
  add    $0x1d8,%rsp
  ret
L50:
  mov    $0x0,%r9d
  jmp    L49
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long thirt(long long n) {
  long long new_num = n;
  long long old_num = n - 1;
  long long rems[] = {1, 10, 9, 12, 3, 4};
  while (new_num != old_num) {
    old_num = new_num;
    long long tmp = old_num;
    int arr_cnt = 0;
    int arr[100];
    do {
      arr[arr_cnt++] = tmp % 10;
    } while ((tmp /= 10) > 0);
    new_num = 0;
    for (int i = 0; i < arr_cnt; ++i)
      new_num += arr[i] * rems[i % 6];
  }
  return new_num;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<thirt>:
  endbr64
  movabs $0x6666666666666667,%rsi
  push   %rbx
  mov    %rdi,%r8
  mov    $0xaaaaaaab,%edi
  sub    $0x1d0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1c8(%rsp)
  xor    %eax,%eax
  movq   $0x1,(%rsp)
  lea    0x30(%rsp),%rcx
  movq   $0xa,0x8(%rsp)
  movq   $0x9,0x10(%rsp)
  movq   $0xc,0x18(%rsp)
  movq   $0x3,0x20(%rsp)
  movq   $0x4,0x28(%rsp)
  nopl   0x0(%rax,%rax,1)
L7:
  mov    %r8,%r9
  xor    %r10d,%r10d
  cs nopw 0x0(%rax,%rax,1)
L21:
  mov    %r9,%rax
  mov    %r9,%rbx
  mov    %r10d,%r11d
  imul   %rsi
  mov    %r9,%rax
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rbx
  mov    %r9,%rax
  mov    %rdx,%r9
  mov    %ebx,(%rcx,%r10,4)
  add    $0x1,%r10
  cmp    $0x9,%rax
  jg     L21
  add    $0x2,%r11d
  mov    $0x1,%r9d
  mov    $0x1,%edx
  xor    %r10d,%r10d
  jmp    L51
  nopl   0x0(%rax)
L50:
  mov    %eax,%edx
  imul   %rdi,%rdx
  shr    $0x22,%rdx
  lea    (%rdx,%rdx,2),%edx
  add    %edx,%edx
  sub    %edx,%eax
  cltq
  mov    (%rsp,%rax,8),%rdx
L51:
  movslq -0x4(%rcx,%r9,4),%rax
  imul   %rdx,%rax
  add    %rax,%r10
  mov    %r9d,%eax
  add    $0x1,%r9
  cmp    %r9,%r11
  jne    L50
  cmp    %r10,%r8
  je     L54
  mov    %r10,%r8
  jmp    L7
  nopl   0x0(%rax)
L54:
  mov    0x1c8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x1d0,%rsp
  mov    %r8,%rax
  pop    %rbx
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long thirt(long long n) {
  long long new_num = n;
  long long old_num = n - 1;
  long long rems[] = {1, 10, 9, 12, 3, 4};
  while (new_num != old_num) {
    old_num = new_num;
    long long tmp = old_num;
    int arr_cnt = 0;
    int arr[100];
    do {
      arr[arr_cnt++] = tmp % 10;
    } while ((tmp /= 10) > 0);
    new_num = 0;
    for (int i = 0; i < arr_cnt; ++i)
      new_num += arr[i] * rems[i % 6];
  }
  return new_num;
}
```
user: What is the c source code of the assembly code below:

```asm
<thirt>:
  endbr64
  push   %rbx
  mov    %rdi,%r10
  mov    $0xaaaaaaab,%r9d
  movabs $0x6666666666666667,%r8
  sub    $0x1d0,%rsp
  movdqa D5(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x1c8(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rdi
  movaps %xmm0,(%rsp)
  movdqa D20(%rip),%xmm0
  movaps %xmm0,0x10(%rsp)
  movdqa D14(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  nopl   (%rax)
L44:
  mov    %r10,%rcx
  xor    %esi,%esi
  nopl   (%rax)
L14:
  mov    %rcx,%rax
  mov    %rcx,%rbx
  mov    %esi,%r11d
  imul   %r8
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rbx
  mov    %rcx,%rax
  mov    %rdx,%rcx
  mov    %ebx,(%rdi,%rsi,4)
  add    $0x1,%rsi
  cmp    $0x9,%rax
  jg     L14
  movslq 0x30(%rsp),%rsi
  test   %r11d,%r11d
  je     L15
  add    $0x1,%r11d
  mov    $0x1,%edx
  nopw   0x0(%rax,%rax,1)
L20:
  mov    %edx,%eax
  mov    %edx,%ecx
  imul   %r9,%rax
  shr    $0x22,%rax
  lea    (%rax,%rax,2),%eax
  add    %eax,%eax
  sub    %eax,%ecx
  movslq (%rdi,%rdx,4),%rax
  add    $0x1,%rdx
  movslq %ecx,%rcx
  imul   (%rsp,%rcx,8),%rax
  add    %rax,%rsi
  cmp    %rdx,%r11
  jne    L20
L15:
  cmp    %rsi,%r10
  je     L4
  mov    %rsi,%r10
  jmp    L44
L4:
  mov    0x1c8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x1d0,%rsp
  mov    %r10,%rax
  pop    %rbx
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D5
data_type: byte[16]
name: retrieve_data
arguments: data_label: D20
data_type: byte[16]
name: retrieve_data
arguments: data_label: D14
data_type: byte[16]

tool: {"D5": [0x01,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x0a,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: {"D20": [0x09,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x0c,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: {"D14": [0x03,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
long long thirt(long long n) {
  long long new_num = n;
  long long old_num = n - 1;
  long long rems[] = {1, 10, 9, 12, 3, 4};
  while (new_num != old_num) {
    old_num = new_num;
    long long tmp = old_num;
    int arr_cnt = 0;
    int arr[100];
    do {
      arr[arr_cnt++] = tmp % 10;
    } while ((tmp /= 10) > 0);
    new_num = 0;
    for (int i = 0; i < arr_cnt; ++i)
      new_num += arr[i] * rems[i % 6];
  }
  return new_num;
}
```
user: What is the c source code of the assembly code below:

```asm
<emit_rotclsr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <emit_lsrs_imm@plt>
  mov    -0x4(%rbp),%edx
  mov    -0x4(%rbp),%ecx
  mov    -0x4(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <emit_adcs@plt>
  mov    -0x8(%rbp),%edx
  mov    -0x8(%rbp),%ecx
  mov    -0x8(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <emit_adc@plt>
  nop
  leave
  ret
```

assistant: ```c
void emit_rotclsr(int t, int sr) {
  emit_lsrs_imm(sr, 1, sr);
  emit_adcs(t, t, t);
  emit_adc(sr, sr, sr);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<emit_rotclsr>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    %esi,%edx
  mov    $0x1,%esi
  mov    %ebx,%edi
  call   <emit_lsrs_imm@plt>
  mov    %ebp,%edx
  mov    %ebp,%esi
  mov    %ebp,%edi
  call   <emit_adcs@plt>
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %ebx,%edi
  call   <emit_adc@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void emit_rotclsr(int t, int sr) {
  emit_lsrs_imm(sr, 1, sr);
  emit_adcs(t, t, t);
  emit_adc(sr, sr, sr);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<emit_rotclsr>:
  endbr64
  push   %r12
  mov    %esi,%edx
  mov    %edi,%r12d
  push   %rbp
  mov    %esi,%ebp
  mov    $0x1,%esi
  mov    %ebp,%edi
  sub    $0x8,%rsp
  call   <emit_lsrs_imm@plt>
  mov    %r12d,%edx
  mov    %r12d,%esi
  mov    %r12d,%edi
  call   <emit_adcs@plt>
  add    $0x8,%rsp
  mov    %ebp,%edx
  mov    %ebp,%esi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    1080 <emit_adc@plt>
```

assistant: ```c
void emit_rotclsr(int t, int sr) {
  emit_lsrs_imm(sr, 1, sr);
  emit_adcs(t, t, t);
  emit_adc(sr, sr, sr);
}
```