ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %edx,-0x38(%rbp)
  movl   $0x0,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x34(%rbp)
  jns    L73
  cmpl   $0xa,-0x38(%rbp)
  jne    L73
  mov    -0x20(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x20(%rbp)
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2d,(%rax)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
L73:
  cmpl   $0x10,-0x38(%rbp)
  jne    L15
  mov    -0x20(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x20(%rbp)
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movb   $0x30,(%rax)
  mov    -0x20(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x20(%rbp)
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movb   $0x78,(%rax)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
L15:
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x18(%rbp)
L68:
  mov    -0x18(%rbp),%eax
  cltd
  idivl  -0x38(%rbp)
  mov    %edx,-0x14(%rbp)
  cmpl   $0x9,-0x14(%rbp)
  jg     L47
  mov    -0x14(%rbp),%eax
  lea    0x30(%rax),%ecx
  mov    -0x20(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x20(%rbp)
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L37
L47:
  mov    -0x14(%rbp),%eax
  lea    0x57(%rax),%ecx
  mov    -0x20(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x20(%rbp)
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L37:
  mov    -0x18(%rbp),%eax
  cltd
  idivl  -0x38(%rbp)
  mov    %eax,-0x18(%rbp)
  cmpl   $0x0,-0x18(%rbp)
  jne    L68
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L8
L69:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x21(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  movzbl -0x21(%rbp),%edx
  mov    %dl,(%rax)
  addq   $0x1,-0x10(%rbp)
  subq   $0x1,-0x8(%rbp)
L8:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L69
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x40(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *itoa(int val, char *buf, int radix) {
  uint32_t i = 0;
  uint32_t start = i;
  if (val < 0 && radix == 10) {
    buf[i++] = '-';
    start = i;
  }
  if (radix == 16) {
    buf[i++] = '0';
    buf[i++] = 'x';
    start = i;
  }
  int x = val;
  do {
    int a = x % radix;
    if (a < 10)
      buf[i++] = a + '0';
    else
      buf[i++] = a + 'a' - 10;
  } while (x /= radix);
  char *s = buf + start;
  char *e = buf + (i - 1);
  while (s < e) {
    char t = *s;
    *s = *e;
    *e = t;
    s++;
    e--;
  }
  buf[i] = 0;
  return buf;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<itoa>:
  endbr64
  mov    %edi,%ecx
  mov    %rsi,%r10
  mov    %edx,%edi
  test   %ecx,%ecx
  jns    L53
  cmp    $0xa,%edx
  jne    L53
  movb   $0x2d,(%rsi)
  mov    $0x1,%esi
L86:
  lea    0x1(%rsi),%r9d
  movslq %esi,%rsi
  add    %r10,%rsi
  mov    %rsi,%r8
  jmp    L84
L53:
  mov    $0x0,%esi
  cmp    $0x10,%edi
  jne    L86
  movb   $0x30,(%r10)
  movb   $0x78,0x1(%r10)
  mov    $0x2,%esi
  jmp    L86
L16:
  mov    %r9d,%r11d
  add    $0x30,%edx
  mov    %dl,(%r8)
L79:
  mov    %ecx,%eax
  cltd
  idiv   %edi
  mov    %eax,%ecx
  add    $0x1,%r9d
  add    $0x1,%r8
  test   %eax,%eax
  je     L40
L84:
  mov    %ecx,%eax
  cltd
  idiv   %edi
  cmp    $0x9,%edx
  jle    L16
  mov    %r9d,%r11d
  add    $0x57,%edx
  mov    %dl,(%r8)
  jmp    L79
L40:
  movslq %r11d,%r11
  lea    -0x1(%r10,%r11,1),%rax
  cmp    %rax,%rsi
  jae    L67
L35:
  movzbl (%rsi),%edx
  movzbl (%rax),%ecx
  mov    %cl,(%rsi)
  mov    %dl,(%rax)
  add    $0x1,%rsi
  sub    $0x1,%rax
  cmp    %rax,%rsi
  jb     L35
L67:
  movb   $0x0,(%r10,%r11,1)
  mov    %r10,%rax
  ret
```

assistant: ```c
char *itoa(int val, char *buf, int radix) {
  uint32_t i = 0;
  uint32_t start = i;
  if (val < 0 && radix == 10) {
    buf[i++] = '-';
    start = i;
  }
  if (radix == 16) {
    buf[i++] = '0';
    buf[i++] = 'x';
    start = i;
  }
  int x = val;
  do {
    int a = x % radix;
    if (a < 10)
      buf[i++] = a + '0';
    else
      buf[i++] = a + 'a' - 10;
  } while (x /= radix);
  char *s = buf + start;
  char *e = buf + (i - 1);
  while (s < e) {
    char t = *s;
    *s = *e;
    *e = t;
    s++;
    e--;
  }
  buf[i] = 0;
  return buf;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<itoa>:
  endbr64
  mov    %edi,%eax
  mov    %rsi,%r10
  mov    %edx,%r9d
  test   %edi,%edi
  jns    L64
  cmp    $0xa,%edx
  jne    L64
  movb   $0x2d,(%rsi)
  lea    0x1(%rsi),%rcx
  mov    $0x1,%r8d
L79:
  movslq %r8d,%rdi
  add    %r10,%rdi
  nopl   0x0(%rax,%rax,1)
L23:
  cltd
  add    $0x1,%r8d
  idiv   %r9d
  cmp    $0x9,%edx
  lea    0x30(%rdx),%r11d
  lea    0x57(%rdx),%esi
  cmovle %r11d,%esi
  add    $0x1,%rdi
  mov    %sil,-0x1(%rdi)
  test   %eax,%eax
  jne    L23
  movslq %r8d,%r8
  lea    -0x1(%r10,%r8,1),%rax
  cmp    %rcx,%rax
  jbe    L69
  nop
L59:
  movzbl (%rcx),%edx
  movzbl (%rax),%esi
  add    $0x1,%rcx
  sub    $0x1,%rax
  mov    %sil,-0x1(%rcx)
  mov    %dl,0x1(%rax)
  cmp    %rax,%rcx
  jb     L59
L69:
  movb   $0x0,(%r10,%r8,1)
  mov    %r10,%rax
  ret
  nopl   0x0(%rax,%rax,1)
L64:
  cmp    $0x10,%r9d
  jne    L45
  mov    $0x7830,%edx
  lea    0x2(%r10),%rcx
  mov    $0x2,%r8d
  mov    %dx,(%r10)
  jmp    L79
  cs nopw 0x0(%rax,%rax,1)
L45:
  mov    %r10,%rcx
  xor    %r8d,%r8d
  jmp    L79
```

assistant: ```c
char *itoa(int val, char *buf, int radix) {
  uint32_t i = 0;
  uint32_t start = i;
  if (val < 0 && radix == 10) {
    buf[i++] = '-';
    start = i;
  }
  if (radix == 16) {
    buf[i++] = '0';
    buf[i++] = 'x';
    start = i;
  }
  int x = val;
  do {
    int a = x % radix;
    if (a < 10)
      buf[i++] = a + '0';
    else
      buf[i++] = a + 'a' - 10;
  } while (x /= radix);
  char *s = buf + start;
  char *e = buf + (i - 1);
  while (s < e) {
    char t = *s;
    *s = *e;
    *e = t;
    s++;
    e--;
  }
  buf[i] = 0;
  return buf;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<itoa>:
  endbr64
  mov    %edi,%eax
  mov    %rsi,%r9
  mov    %edx,%r8d
  test   %edi,%edi
  jns    L47
  cmp    $0xa,%edx
  jne    L47
  movb   $0x2d,(%rsi)
  lea    0x1(%rsi),%rcx
  mov    $0x1,%edi
L109:
  movslq %edi,%rsi
  add    %r9,%rsi
  nop
L1:
  cltd
  add    $0x1,%edi
  idiv   %r8d
  cmp    $0x9,%edx
  jg     L25
  add    $0x30,%edx
  add    $0x1,%rsi
  mov    %dl,-0x1(%rsi)
  test   %eax,%eax
  jne    L1
L34:
  movslq %edi,%rdi
  lea    -0x1(%r9,%rdi,1),%rax
  cmp    %rcx,%rax
  jbe    L78
  nopw   0x0(%rax,%rax,1)
L8:
  movzbl (%rcx),%edx
  movzbl (%rax),%esi
  add    $0x1,%rcx
  sub    $0x1,%rax
  mov    %sil,-0x1(%rcx)
  mov    %dl,0x1(%rax)
  cmp    %rax,%rcx
  jb     L8
L78:
  movb   $0x0,(%r9,%rdi,1)
  mov    %r9,%rax
  ret
  nopl   0x0(%rax,%rax,1)
L25:
  add    $0x57,%edx
  add    $0x1,%rsi
  mov    %dl,-0x1(%rsi)
  test   %eax,%eax
  jne    L1
  jmp    L34
  nopl   0x0(%rax,%rax,1)
L47:
  cmp    $0x10,%r8d
  jne    L104
  mov    $0x7830,%edx
  lea    0x2(%r9),%rcx
  mov    $0x2,%edi
  mov    %dx,(%r9)
  jmp    L109
  nopl   (%rax)
L104:
  mov    %r9,%rcx
  xor    %edi,%edi
  jmp    L109
```

assistant: ```c
char *itoa(int val, char *buf, int radix) {
  uint32_t i = 0;
  uint32_t start = i;
  if (val < 0 && radix == 10) {
    buf[i++] = '-';
    start = i;
  }
  if (radix == 16) {
    buf[i++] = '0';
    buf[i++] = 'x';
    start = i;
  }
  int x = val;
  do {
    int a = x % radix;
    if (a < 10)
      buf[i++] = a + '0';
    else
      buf[i++] = a + 'a' - 10;
  } while (x /= radix);
  char *s = buf + start;
  char *e = buf + (i - 1);
  while (s < e) {
    char t = *s;
    *s = *e;
    *e = t;
    s++;
    e--;
  }
  buf[i] = 0;
  return buf;
}
```
user: What is the c source code of the assembly code below:

```asm
<fun_stub_base_ll_call>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <fun_stub_base_ll_stub@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D18
type: string
name: unpack_data
arguments: label: D14
type: string

tool: D18: fun_stub_base_v_call:a = %d 


tool: D14: fun_stub_base_v_stub return %lld


assistant: ```c
long long fun_stub_base_ll_call(int a) {
  printf("fun_stub_base_v_call:a = %d \n", a);
  long long x = fun_stub_base_ll_stub(a);
  printf("fun_stub_base_v_stub return %lld\n", x);
  return x;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<fun_stub_base_ll_call>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  movslq %edi,%rsi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,%edi
  call   <fun_stub_base_ll_stub@plt>
  mov    %rax,%rbx
  mov    %rax,%rsi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: label: D6
type: string
name: unpack_data
arguments: label: D10
type: string

tool: D6: fun_stub_base_v_call:a = %d 


tool: D10: fun_stub_base_v_stub return %lld


assistant: ```c
long long fun_stub_base_ll_call(int a) {
  printf("fun_stub_base_v_call:a = %d \n", a);
  long long x = fun_stub_base_ll_stub(a);
  printf("fun_stub_base_v_stub return %lld\n", x);
  return x;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fun_stub_base_ll_call>:
  endbr64
  push   %r12
  movslq %edi,%rsi
  lea    D19(%rip),%rdi
  mov    %rsi,%r12
  call   <printf@plt>
  mov    %r12d,%edi
  call   <fun_stub_base_ll_stub@plt>
  lea    D6(%rip),%rdi
  mov    %rax,%r12
  mov    %rax,%rsi
  call   <printf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D19
data_type: string
name: retrieve_data
arguments: data_label: D6
data_type: string

tool: D19: fun_stub_base_v_call:a = %d 


tool: D6: fun_stub_base_v_stub return %lld


assistant: ```c
long long fun_stub_base_ll_call(int a) {
  printf("fun_stub_base_v_call:a = %d \n", a);
  long long x = fun_stub_base_ll_stub(a);
  printf("fun_stub_base_v_stub return %lld\n", x);
  return x;
}
```
user: What is the c source code of the assembly code below:

```asm
<fun_stub_base_ll_call>:
  endbr64
  push   %r12
  movslq %edi,%rsi
  lea    D8(%rip),%rdi
  mov    %rsi,%r12
  call   <printf@plt>
  mov    %r12d,%edi
  call   <fun_stub_base_ll_stub@plt>
  lea    D5(%rip),%rdi
  mov    %rax,%r12
  mov    %rax,%rsi
  call   <printf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: label: D8
type_guessed: string
name: parse_data
arguments: label: D5
type_guessed: string

tool: D8: fun_stub_base_v_call:a = %d 


tool: D5: fun_stub_base_v_stub return %lld


assistant: ```c
long long fun_stub_base_ll_call(int a) {
  printf("fun_stub_base_v_call:a = %d \n", a);
  long long x = fun_stub_base_ll_stub(a);
  printf("fun_stub_base_v_stub return %lld\n", x);
  return x;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<trim_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L82
L51:
  addq   $0x1,-0x8(%rbp)
L82:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L72
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isalnum@plt>
  test   %rax,%rax
  je     L51
L72:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L21
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  je     L21
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memmove@plt>
L21:
  mov    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  jmp    L85
L88:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isalnum@plt>
  test   %rax,%rax
  jne    L52
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  subq   $0x1,-0x8(%rbp)
L85:
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jne    L88
  jmp    L3
L52:
  nop
L3:
  nop
  leave
  ret
```

assistant: ```c
void trim_string(char *string) {
  char *p;
  for (p = string; *p && !isalnum(*p); ++p)
    ;
  if (*p && p != string) {
    (void)memmove(string, p, strlen(p) + 1);
  }
  for (p = strrchr(string, '\0'); p != string; --p) {
    if (isalnum(*p)) {
      break;
    }
    *p = '\0';
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<trim_string>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  movzbl (%rdi),%edi
  mov    %rbp,%rbx
  test   %dil,%dil
  je     L20
L36:
  movsbl %dil,%edi
  call   <isalnum@plt>
  test   %rax,%rax
  jne    L64
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L36
L20:
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <strrchr@plt>
  mov    %rax,%rbx
  cmp    %rax,%rbp
  je     L33
L37:
  movsbl (%rbx),%edi
  call   <isalnum@plt>
  test   %rax,%rax
  jne    L33
  movb   $0x0,(%rbx)
  sub    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L37
L33:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L64:
  cmpb   $0x0,(%rbx)
  je     L20
  cmp    %rbx,%rbp
  je     L20
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <memmove@plt>
  jmp    L20
```

assistant: ```c
void trim_string(char *string) {
  char *p;
  for (p = string; *p && !isalnum(*p); ++p)
    ;
  if (*p && p != string) {
    (void)memmove(string, p, strlen(p) + 1);
  }
  for (p = strrchr(string, '\0'); p != string; --p) {
    if (isalnum(*p)) {
      break;
    }
    *p = '\0';
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<trim_string>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %r12,%rbp
  push   %rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L47
  jmp    L37
  nopl   0x0(%rax,%rax,1)
L53:
  movsbl 0x1(%rbp),%edi
  add    $0x1,%rbp
  test   %dil,%dil
  je     L37
L47:
  call   <isalnum@plt>
  test   %rax,%rax
  je     L53
  cmpb   $0x0,0x0(%rbp)
  je     L37
  cmp    %rbp,%r12
  je     L37
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  lea    0x1(%rax),%rdx
  call   <memmove@plt>
L37:
  xor    %esi,%esi
  mov    %r12,%rdi
  call   <strrchr@plt>
  mov    %rax,%rbx
  cmp    %rax,%r12
  jne    L26
  jmp    L10
  nopl   (%rax)
L5:
  movb   $0x0,(%rbx)
  sub    $0x1,%rbx
  cmp    %rbx,%r12
  je     L10
L26:
  movsbl (%rbx),%edi
  call   <isalnum@plt>
  test   %rax,%rax
  je     L5
L10:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void trim_string(char *string) {
  char *p;
  for (p = string; *p && !isalnum(*p); ++p)
    ;
  if (*p && p != string) {
    (void)memmove(string, p, strlen(p) + 1);
  }
  for (p = strrchr(string, '\0'); p != string; --p) {
    if (isalnum(*p)) {
      break;
    }
    *p = '\0';
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<trim_string>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %r12,%rbp
  push   %rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L74
  jmp    L96
  nopl   0x0(%rax,%rax,1)
L23:
  movsbl 0x1(%rbp),%edi
  add    $0x1,%rbp
  test   %dil,%dil
  je     L96
L74:
  call   <isalnum@plt>
  test   %rax,%rax
  je     L23
  cmpb   $0x0,0x0(%rbp)
  je     L96
  cmp    %rbp,%r12
  je     L96
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  lea    0x1(%rax),%rdx
  call   <memmove@plt>
L96:
  xor    %esi,%esi
  mov    %r12,%rdi
  call   <strrchr@plt>
  mov    %rax,%rbx
  cmp    %rax,%r12
  jne    L28
  jmp    L77
  nopl   (%rax)
L16:
  movb   $0x0,(%rbx)
  sub    $0x1,%rbx
  cmp    %rbx,%r12
  je     L77
L28:
  movsbl (%rbx),%edi
  call   <isalnum@plt>
  test   %rax,%rax
  je     L16
L77:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void trim_string(char *string) {
  char *p;
  for (p = string; *p && !isalnum(*p); ++p)
    ;
  if (*p && p != string) {
    (void)memmove(string, p, strlen(p) + 1);
  }
  for (p = strrchr(string, '\0'); p != string; --p) {
    if (isalnum(*p)) {
      break;
    }
    *p = '\0';
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  sub    $0x99f4bb,%eax
  je     L12
  lea    (%rax,%rax,1),%edx
L5:
  imul   $0x6185fdd6,%edx,%edx
  add    $0x6e28b14a,%edx
  mov    %edx,(%rsi)
  ret
L12:
  rol    $0x10,%eax
  mov    %eax,%edx
  mov    %eax,%ecx
  shl    $0x8,%ecx
  sar    $0x18,%ecx
  mov    %cl,%dl
  movzbl %al,%eax
  shl    $0x10,%eax
  and    $0xff00ffff,%edx
  or     %eax,%edx
  jmp    L5
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy11;
  unsigned short copy12;
  unsigned short copy13;
  char copy14;
  {
    state[0UL] = input[0UL] + 4284877637U;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] > local1) {
        if (state[0UL] > local1) {
          copy11 = *((unsigned short *)(&state[local1]) + 1);
          *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
          *((unsigned short *)(&state[local1]) + 0) = copy11;
          copy11 = *((unsigned short *)(&state[local1]) + 0);
          *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 1);
          *((unsigned short *)(&state[local1]) + 1) = copy11;
          state[0UL] += state[0UL];
        } else {
          copy12 = *((unsigned short *)(&state[0UL]) + 0);
          *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = copy12;
          copy12 = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 0);
          *((unsigned short *)(&state[0UL]) + 0) = copy12;
          state[local1] = state[0UL] * state[0UL];
        }
      } else {
        copy13 = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 1);
        *((unsigned short *)(&state[local1]) + 1) = copy13;
        copy14 = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 2);
        *((char *)(&state[0UL]) + 2) = copy14;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] + 323928751UL) * 1636171222U;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%edx
  sub    $0x99f4bb,%edx
  lea    (%rdx,%rdx,1),%eax
  mov    $0x0,%edx
  cmove  %edx,%eax
  imul   $0x6185fdd6,%eax,%eax
  add    $0x6e28b14a,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy11;
  unsigned short copy12;
  unsigned short copy13;
  char copy14;
  {
    state[0UL] = input[0UL] + 4284877637U;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] > local1) {
        if (state[0UL] > local1) {
          copy11 = *((unsigned short *)(&state[local1]) + 1);
          *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
          *((unsigned short *)(&state[local1]) + 0) = copy11;
          copy11 = *((unsigned short *)(&state[local1]) + 0);
          *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 1);
          *((unsigned short *)(&state[local1]) + 1) = copy11;
          state[0UL] += state[0UL];
        } else {
          copy12 = *((unsigned short *)(&state[0UL]) + 0);
          *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = copy12;
          copy12 = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 0);
          *((unsigned short *)(&state[0UL]) + 0) = copy12;
          state[local1] = state[0UL] * state[0UL];
        }
      } else {
        copy13 = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 1);
        *((unsigned short *)(&state[local1]) + 1) = copy13;
        copy14 = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 2);
        *((char *)(&state[0UL]) + 2) = copy14;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] + 323928751UL) * 1636171222U;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%edx
  sub    $0x99f4bb,%edx
  lea    (%rdx,%rdx,1),%eax
  mov    $0x0,%edx
  cmove  %edx,%eax
  imul   $0x6185fdd6,%eax,%eax
  add    $0x6e28b14a,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy11;
  unsigned short copy12;
  unsigned short copy13;
  char copy14;
  {
    state[0UL] = input[0UL] + 4284877637U;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] > local1) {
        if (state[0UL] > local1) {
          copy11 = *((unsigned short *)(&state[local1]) + 1);
          *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
          *((unsigned short *)(&state[local1]) + 0) = copy11;
          copy11 = *((unsigned short *)(&state[local1]) + 0);
          *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 1);
          *((unsigned short *)(&state[local1]) + 1) = copy11;
          state[0UL] += state[0UL];
        } else {
          copy12 = *((unsigned short *)(&state[0UL]) + 0);
          *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = copy12;
          copy12 = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 0);
          *((unsigned short *)(&state[0UL]) + 0) = copy12;
          state[local1] = state[0UL] * state[0UL];
        }
      } else {
        copy13 = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 1);
        *((unsigned short *)(&state[local1]) + 1) = copy13;
        copy14 = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 2);
        *((char *)(&state[0UL]) + 2) = copy14;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] + 323928751UL) * 1636171222U;
  }
}
```