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

```asm
<bcrypt>:
  endbr64
  push   %rbx
  lea    D4(%rip),%rbx
  mov    $0x3d,%ecx
  mov    %rbx,%rdx
  call   <bcrypt_hashpass@plt>
  test   %rax,%rax
  mov    $0x0,%eax
  cmove  %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *bcrypt(const char *pass, const uint8_t *salt) {
  static char gencrypted[61];
  if (bcrypt_hashpass(pass, salt, gencrypted, sizeof(gencrypted)) != 0)
    return NULL;
  return gencrypted;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<bcrypt>:
  endbr64
  push   %rbx
  lea    D9(%rip),%rbx
  mov    $0x3d,%ecx
  mov    %rbx,%rdx
  call   <bcrypt_hashpass@plt>
  test   %rax,%rax
  mov    $0x0,%eax
  cmove  %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *bcrypt(const char *pass, const uint8_t *salt) {
  static char gencrypted[61];
  if (bcrypt_hashpass(pass, salt, gencrypted, sizeof(gencrypted)) != 0)
    return NULL;
  return gencrypted;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<merge>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x28(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L80
L10:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jne    L74
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  jmp    L94
L74:
  mov    -0xc(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jne    L54
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  jmp    L94
L54:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jge    L57
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  jmp    L94
L57:
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
L94:
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  mov    %eax,(%rdx)
  addl   $0x1,-0x4(%rbp)
L80:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L10
  movl   $0x0,-0xc(%rbp)
  jmp    L37
L78:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L37:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L78
  nop
  nop
  leave
  ret
```

assistant: ```c
void merge(int *numbersA, int size, int mid, int *numbersB) {
  int i, j, k;
  numbersB = (int *)malloc(size * sizeof(int));
  for (i = 0, j = mid, k = 0; k < size; k++) {
    numbersB[k] = (j == size) ? numbersA[i++] : (i == mid) ? numbersA[j++] : (numbersA[j] < numbersA[i]) ? numbersA[j++] : numbersA[i++];
  }
  for (i = 0; i < size; i++) {
    numbersA[i] = numbersB[i];
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<merge>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    %edx,%r12d
  lea    0x0(,%rsi,4),%edi
  call   <malloc@plt>
  test   %ebp,%ebp
  jle    L99
  mov    %rax,%r9
  mov    %ebp,%eax
  lea    (%r9,%rax,4),%rdi
  mov    %r9,%rax
  mov    %r12d,%esi
  mov    $0x0,%edx
  jmp    L66
L79:
  movslq %edx,%rcx
  mov    (%rbx,%rcx,4),%ecx
  lea    0x1(%rdx),%edx
L90:
  mov    %ecx,(%rax)
  add    $0x4,%rax
  cmp    %rdi,%rax
  je     L93
L66:
  cmp    %esi,%ebp
  je     L79
  cmp    %edx,%r12d
  je     L58
  movslq %esi,%rcx
  mov    (%rbx,%rcx,4),%r8d
  movslq %edx,%rcx
  mov    (%rbx,%rcx,4),%ecx
  cmp    %ecx,%r8d
  jge    L27
  add    $0x1,%esi
  mov    %r8d,%ecx
  jmp    L90
L58:
  movslq %esi,%rcx
  mov    (%rbx,%rcx,4),%ecx
  lea    0x1(%rsi),%esi
  jmp    L90
L27:
  add    $0x1,%edx
  jmp    L90
L93:
  mov    %r9,%rax
  sub    %r9,%rbx
L87:
  mov    (%rax),%edx
  mov    %edx,(%rbx,%rax,1)
  add    $0x4,%rax
  cmp    %rdi,%rax
  jne    L87
L99:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void merge(int *numbersA, int size, int mid, int *numbersB) {
  int i, j, k;
  numbersB = (int *)malloc(size * sizeof(int));
  for (i = 0, j = mid, k = 0; k < size; k++) {
    numbersB[k] = (j == size) ? numbersA[i++] : (i == mid) ? numbersA[j++] : (numbersA[j] < numbersA[i]) ? numbersA[j++] : numbersA[i++];
  }
  for (i = 0; i < size; i++) {
    numbersA[i] = numbersB[i];
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<merge>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%rsi,4),%edi
  call   <malloc@plt>
  test   %ebp,%ebp
  jle    L14
  mov    %ebp,%edx
  mov    %r12d,%edi
  xor    %ecx,%ecx
  lea    (%rax,%rdx,4),%r8
  mov    %rax,%rdx
  jmp    L62
L31:
  movslq %edi,%rsi
  mov    (%rbx,%rsi,4),%esi
  cmp    %ecx,%r12d
  je     L49
  movslq %ecx,%r9
  mov    (%rbx,%r9,4),%r9d
  cmp    %esi,%r9d
  jg     L49
  add    $0x1,%ecx
  mov    %r9d,%esi
L7:
  mov    %esi,(%rdx)
  add    $0x4,%rdx
  cmp    %r8,%rdx
  je     L12
L62:
  cmp    %edi,%ebp
  jne    L31
  movslq %ecx,%rsi
  add    $0x4,%rdx
  add    $0x1,%ecx
  mov    (%rbx,%rsi,4),%esi
  mov    %esi,-0x4(%rdx)
  cmp    %r8,%rdx
  jne    L62
L12:
  sub    %rax,%rbx
  nopl   0x0(%rax)
L8:
  mov    (%rax),%edx
  mov    %edx,(%rbx,%rax,1)
  add    $0x4,%rax
  cmp    %r8,%rax
  jne    L8
L14:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L49:
  add    $0x1,%edi
  jmp    L7
```

assistant: ```c
void merge(int *numbersA, int size, int mid, int *numbersB) {
  int i, j, k;
  numbersB = (int *)malloc(size * sizeof(int));
  for (i = 0, j = mid, k = 0; k < size; k++) {
    numbersB[k] = (j == size) ? numbersA[i++] : (i == mid) ? numbersA[j++] : (numbersA[j] < numbersA[i]) ? numbersA[j++] : numbersA[i++];
  }
  for (i = 0; i < size; i++) {
    numbersA[i] = numbersB[i];
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<merge>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  lea    0x0(,%rsi,4),%edi
  push   %rbx
  mov    %esi,%ebx
  call   <malloc@plt>
  test   %ebx,%ebx
  jle    L71
  mov    %r12d,%edx
  mov    %rax,%rdi
  mov    %rax,%rsi
  xor    %ecx,%ecx
  xor    %eax,%eax
  cmp    %edx,%ebx
  je     L15
  nopl   (%rax)
L96:
  movslq %edx,%r8
  mov    0x0(%rbp,%r8,4),%r8d
  cmp    %eax,%r12d
  je     L8
  movslq %eax,%r9
  mov    0x0(%rbp,%r9,4),%r9d
  cmp    %r8d,%r9d
  jg     L8
  add    $0x1,%eax
  mov    %r9d,%r8d
L122:
  add    $0x1,%ecx
  mov    %r8d,(%rsi)
  cmp    %ecx,%ebx
  je     L7
  add    $0x4,%rsi
  cmp    %edx,%ebx
  jne    L96
L15:
  lea    0x1(%rax),%edx
  cltq
  add    $0x1,%ecx
  mov    0x0(%rbp,%rax,4),%eax
  mov    %eax,(%rsi)
  cmp    %ecx,%ebx
  je     L7
  movslq %ecx,%rax
  movslq %edx,%r8
  mov    %ebx,%r11d
  shl    $0x2,%rax
  lea    0x4(,%r8,4),%r8
  sub    %ecx,%r11d
  lea    (%rdi,%rax,1),%rsi
  lea    0x0(%rbp,%r8,1),%r10
  mov    %rsi,%r9
  sub    %r10,%r9
  cmp    $0x8,%r9
  jbe    L51
  lea    -0x1(%r11),%r9d
  cmp    $0x2,%r9d
  jbe    L51
  cmp    %ecx,%ebx
  mov    $0x1,%eax
  lea    -0x4(%rbp,%r8,1),%r9
  cmovle %eax,%r11d
  xor    %eax,%eax
  mov    %r11d,%r10d
  shr    $0x2,%r10d
  jmp    1060 <merge.cold>
  nopl   (%rax)
L8:
  add    $0x1,%edx
  jmp    L122
L7:
  lea    0x4(%rdi),%rdx
  mov    %rbp,%rax
  sub    %rdx,%rax
  cmp    $0x8,%rax
  jbe    L62
  lea    -0x1(%rbx),%eax
  cmp    $0x2,%eax
  jbe    L62
  mov    %ebx,%edx
  mov    %rbp,%rcx
  mov    %rdi,%rax
  shr    $0x2,%edx
  sub    %rdi,%rcx
  shl    $0x4,%rdx
  add    %rdi,%rdx
  nopl   (%rax)
L105:
  movdqu (%rax),%xmm0
  movups %xmm0,(%rcx,%rax,1)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L105
  mov    %ebx,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%bl
  je     L71
  mov    %eax,%ecx
  mov    (%rdi,%rcx,4),%esi
  lea    0x0(,%rcx,4),%rdx
  mov    %esi,0x0(%rbp,%rcx,4)
  lea    0x1(%rax),%ecx
  cmp    %ecx,%ebx
  jle    L71
  mov    0x4(%rdi,%rdx,1),%ecx
  add    $0x2,%eax
  mov    %ecx,0x4(%rbp,%rdx,1)
  cmp    %eax,%ebx
  jle    L71
  mov    0x8(%rdi,%rdx,1),%eax
  mov    %eax,0x8(%rbp,%rdx,1)
L71:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nop
L62:
  mov    %ebx,%ebx
  sub    %rdi,%rbp
  lea    (%rdi,%rbx,4),%rdx
  nopl   0x0(%rax)
L85:
  mov    (%rdi),%eax
  mov    %eax,0x0(%rbp,%rdi,1)
  add    $0x4,%rdi
  cmp    %rdx,%rdi
  jne    L85
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L51:
  neg    %rax
  sub    %rdi,%rax
  lea    -0x4(%r8,%rax,1),%rax
  add    %rbp,%rax
  jmp    L162
```

assistant: ```c
void merge(int *numbersA, int size, int mid, int *numbersB) {
  int i, j, k;
  numbersB = (int *)malloc(size * sizeof(int));
  for (i = 0, j = mid, k = 0; k < size; k++) {
    numbersB[k] = (j == size) ? numbersA[i++] : (i == mid) ? numbersA[j++] : (numbersA[j] < numbersA[i]) ? numbersA[j++] : numbersA[i++];
  }
  for (i = 0; i < size; i++) {
    numbersA[i] = numbersB[i];
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<ParseIoPacketAGS>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  and    $0xffffffc0,%eax
  cmp    $0xc0,%al
  je     L3
  mov    $0x1,%eax
  jmp    L20
L3:
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  and    $0xffffffc0,%eax
  cmp    $0x80,%al
  je     L34
  mov    $0x1,%eax
  jmp    L20
L34:
  mov    -0x8(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  and    $0xffffffc0,%eax
  cmp    $0x40,%al
  je     L41
  mov    $0x1,%eax
  jmp    L20
L41:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0xc,%eax
  and    $0x3f000,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x6,%eax
  and    $0xfc0,%eax
  or     %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  and    $0x3f,%eax
  or     %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
L20:
  pop    %rbp
  ret
```

assistant: ```c
int ParseIoPacketAGS(unsigned char *Packet, int *Type, int *Data) {
  if (0xc0 != (0xc0 & Packet[1]))
    return (1);
  if (0x80 != (0xc0 & Packet[2]))
    return (1);
  if (0x40 != (0xc0 & Packet[3]))
    return (1);
  *Type = Packet[0];
  *Data = ((Packet[1] & 077) << 12) | ((Packet[2] & 077) << 6) | (Packet[3] & 077);
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ParseIoPacketAGS>:
  endbr64
  mov    %rdi,%rax
  mov    %rdx,%rdi
  movzbl 0x1(%rax),%edx
  and    $0xffffffc0,%edx
  mov    $0x1,%ecx
  cmp    $0xc0,%dl
  jne    L18
  movzbl 0x2(%rax),%edx
  and    $0xffffffc0,%edx
  cmp    $0x80,%dl
  jne    L18
  movzbl 0x3(%rax),%edx
  and    $0xffffffc0,%edx
  cmp    $0x40,%dl
  jne    L18
  movzbl (%rax),%edx
  mov    %edx,(%rsi)
  movzbl 0x1(%rax),%edx
  shl    $0xc,%edx
  and    $0x3f000,%edx
  movzbl 0x2(%rax),%ecx
  shl    $0x6,%ecx
  and    $0xfc0,%ecx
  or     %ecx,%edx
  movzbl 0x3(%rax),%eax
  and    $0x3f,%eax
  or     %edx,%eax
  mov    %eax,(%rdi)
  mov    $0x0,%ecx
L18:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int ParseIoPacketAGS(unsigned char *Packet, int *Type, int *Data) {
  if (0xc0 != (0xc0 & Packet[1]))
    return (1);
  if (0x80 != (0xc0 & Packet[2]))
    return (1);
  if (0x40 != (0xc0 & Packet[3]))
    return (1);
  *Type = Packet[0];
  *Data = ((Packet[1] & 077) << 12) | ((Packet[2] & 077) << 6) | (Packet[3] & 077);
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ParseIoPacketAGS>:
  endbr64
  mov    %rdx,%rcx
  movzbl 0x1(%rdi),%edx
  mov    $0x1,%r8d
  and    $0xffffffc0,%edx
  cmp    $0xc0,%dl
  jne    L20
  movzbl 0x2(%rdi),%edx
  and    $0xffffffc0,%edx
  cmp    $0x80,%dl
  je     L19
L20:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L19:
  movzbl 0x3(%rdi),%edx
  and    $0xffffffc0,%edx
  cmp    $0x40,%dl
  jne    L20
  movzbl (%rdi),%edx
  xor    %r8d,%r8d
  mov    %edx,(%rsi)
  movzbl 0x1(%rdi),%edx
  movzbl 0x2(%rdi),%esi
  movzbl 0x3(%rdi),%eax
  shl    $0xc,%edx
  shl    $0x6,%esi
  and    $0x3f000,%edx
  and    $0x3f,%eax
  and    $0xfc0,%esi
  or     %esi,%edx
  or     %edx,%eax
  mov    %eax,(%rcx)
  mov    %r8d,%eax
  ret
```

assistant: ```c
int ParseIoPacketAGS(unsigned char *Packet, int *Type, int *Data) {
  if (0xc0 != (0xc0 & Packet[1]))
    return (1);
  if (0x80 != (0xc0 & Packet[2]))
    return (1);
  if (0x40 != (0xc0 & Packet[3]))
    return (1);
  *Type = Packet[0];
  *Data = ((Packet[1] & 077) << 12) | ((Packet[2] & 077) << 6) | (Packet[3] & 077);
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ParseIoPacketAGS>:
  endbr64
  mov    %rdx,%rcx
  movzbl 0x1(%rdi),%edx
  mov    $0x1,%r8d
  and    $0xffffffc0,%edx
  cmp    $0xc0,%dl
  jne    L1
  movzbl 0x2(%rdi),%edx
  and    $0xffffffc0,%edx
  cmp    $0x80,%dl
  je     L27
L1:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L27:
  movzbl 0x3(%rdi),%edx
  and    $0xffffffc0,%edx
  cmp    $0x40,%dl
  jne    L1
  movzbl (%rdi),%edx
  xor    %r8d,%r8d
  mov    %edx,(%rsi)
  movzbl 0x1(%rdi),%edx
  movzbl 0x2(%rdi),%esi
  movzbl 0x3(%rdi),%eax
  shl    $0xc,%edx
  shl    $0x6,%esi
  and    $0x3f000,%edx
  and    $0x3f,%eax
  and    $0xfc0,%esi
  or     %esi,%edx
  or     %edx,%eax
  mov    %eax,(%rcx)
  mov    %r8d,%eax
  ret
```

assistant: ```c
int ParseIoPacketAGS(unsigned char *Packet, int *Type, int *Data) {
  if (0xc0 != (0xc0 & Packet[1]))
    return (1);
  if (0x80 != (0xc0 & Packet[2]))
    return (1);
  if (0x40 != (0xc0 & Packet[3]))
    return (1);
  *Type = Packet[0];
  *Data = ((Packet[1] & 077) << 12) | ((Packet[2] & 077) << 6) | (Packet[3] & 077);
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<arm_int2str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L59
L61:
  addl   $0x1,-0x8(%rbp)
  mov    -0xc(%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,-0xc(%rbp)
L59:
  cmpl   $0x0,-0xc(%rbp)
  jne    L61
  cmpl   $0x0,-0x1c(%rbp)
  jns    L39
  addl   $0x1,-0x8(%rbp)
L39:
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L30
L48:
  mov    -0xc(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    -0x8(%rbp),%eax
  sub    -0x4(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rcx
  mov    -0x18(%rbp),%rax
  add    %rax,%rcx
  movslq %edx,%rax
  lea    D0(%rip),%rdx
  movzbl (%rax,%rdx,1),%eax
  mov    %al,(%rcx)
  mov    -0xc(%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,-0xc(%rbp)
  addl   $0x1,-0x4(%rbp)
L30:
  cmpl   $0x0,-0xc(%rbp)
  jne    L48
  cmpl   $0x0,-0x1c(%rbp)
  jns    L67
  mov    -0x18(%rbp),%rax
  movb   $0x2d,(%rax)
L67:
  cmpl   $0x0,-0x8(%rbp)
  jne    L17
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x30,(%rax)
  addl   $0x1,-0x8(%rbp)
L17:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
data_type: string

tool: D0: 0123456789

assistant: ```c
void arm_int2str(char *dst, int src) {
  int val, count = 0, pos = 0;
  static const char intchars[] = "0123456789";
  val = src;
  while (val) {
    count++;
    val = val / 10;
  }
  if (src < 0) {
    count++;
  }
  val = (src < 0) ? -src : src;
  while (val) {
    dst[count - pos - 1] = intchars[val % 10];
    val = val / 10;
    pos++;
  }
  if (src < 0) {
    dst[0] = '-';
  }
  if (count == 0) {
    dst[count] = '0';
    count++;
  }
  dst[count] = '\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
<arm_int2str>:
  endbr64
  test   %esi,%esi
  je     L28
  mov    %esi,%edx
  mov    $0x0,%ecx
L7:
  mov    %ecx,%r8d
  add    $0x1,%ecx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,%edx
  jne    L7
  mov    %esi,%edx
  neg    %edx
  cmovs  %esi,%edx
  test   %esi,%esi
  js     L53
L13:
  movslq %ecx,%rax
  lea    -0x1(%rdi,%rax,1),%r9
  lea    D7(%rip),%r10
L9:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%r8d
  sar    $0x1f,%r8d
  sub    %r8d,%eax
  lea    (%rax,%rax,4),%r8d
  add    %r8d,%r8d
  sub    %r8d,%edx
  movslq %edx,%rdx
  movzbl (%r10,%rdx,1),%edx
  mov    %dl,(%r9)
  mov    %eax,%edx
  sub    $0x1,%r9
  test   %eax,%eax
  jne    L9
  test   %esi,%esi
  js     L17
L14:
  test   %ecx,%ecx
  jne    L60
  movb   $0x30,(%rdi)
  mov    $0x1,%ecx
L60:
  movslq %ecx,%rcx
  movb   $0x0,(%rdi,%rcx,1)
  ret
L53:
  lea    0x2(%r8),%ecx
  mov    %esi,%edx
  neg    %edx
  cmovs  %esi,%edx
  jmp    L13
L17:
  movb   $0x2d,(%rdi)
  jmp    L14
L28:
  mov    %esi,%ecx
  jmp    L14
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
type_guessed: string

tool: D7: 0123456789

assistant: ```c
void arm_int2str(char *dst, int src) {
  int val, count = 0, pos = 0;
  static const char intchars[] = "0123456789";
  val = src;
  while (val) {
    count++;
    val = val / 10;
  }
  if (src < 0) {
    count++;
  }
  val = (src < 0) ? -src : src;
  while (val) {
    dst[count - pos - 1] = intchars[val % 10];
    val = val / 10;
    pos++;
  }
  if (src < 0) {
    dst[0] = '-';
  }
  if (count == 0) {
    dst[count] = '0';
    count++;
  }
  dst[count] = '\0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<arm_int2str>:
  endbr64
  mov    %esi,%r8d
  test   %esi,%esi
  je     L42
  mov    %esi,%ecx
  xor    %edx,%edx
  nop
L61:
  movslq %ecx,%rax
  mov    %ecx,%esi
  mov    %edx,%r9d
  add    $0x1,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  mov    %eax,%ecx
  jne    L61
  mov    %r8d,%eax
  test   %r8d,%r8d
  js     L62
L49:
  movslq %edx,%rsi
  lea    D5(%rip),%r11
  mov    $0xcccccccd,%r10d
  lea    -0x1(%rdi,%rsi,1),%r9
  nopl   0x0(%rax)
L15:
  mov    %eax,%edx
  sub    $0x1,%r9
  imul   %r10,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  cltq
  movzbl (%r11,%rax,1),%eax
  mov    %al,0x1(%r9)
  mov    %edx,%eax
  test   %edx,%edx
  jne    L15
  test   %r8d,%r8d
  js     L67
L29:
  mov    %rsi,%rdx
  movb   $0x0,(%rdi,%rdx,1)
  ret
  nopl   0x0(%rax,%rax,1)
L42:
  mov    $0x1,%edx
  movb   $0x30,(%rdi)
  movb   $0x0,(%rdi,%rdx,1)
  ret
  nopl   (%rax)
L67:
  movb   $0x2d,(%rdi)
  jmp    L29
  nopl   (%rax)
L62:
  neg    %eax
  lea    0x2(%r9),%edx
  jmp    L49
```

tool_calls: name: unpack_data
arguments: data_label: D5
datatype: string

tool: D5: 0123456789

assistant: ```c
void arm_int2str(char *dst, int src) {
  int val, count = 0, pos = 0;
  static const char intchars[] = "0123456789";
  val = src;
  while (val) {
    count++;
    val = val / 10;
  }
  if (src < 0) {
    count++;
  }
  val = (src < 0) ? -src : src;
  while (val) {
    dst[count - pos - 1] = intchars[val % 10];
    val = val / 10;
    pos++;
  }
  if (src < 0) {
    dst[0] = '-';
  }
  if (count == 0) {
    dst[count] = '0';
    count++;
  }
  dst[count] = '\0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<arm_int2str>:
  endbr64
  mov    %esi,%r8d
  test   %esi,%esi
  je     L65
  mov    %esi,%ecx
  xor    %edx,%edx
  nop
L33:
  movslq %ecx,%rax
  mov    %ecx,%esi
  mov    %edx,%r9d
  add    $0x1,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  mov    %eax,%ecx
  jne    L33
  mov    %r8d,%eax
  test   %r8d,%r8d
  js     L67
L5:
  movslq %edx,%rsi
  lea    D5(%rip),%r11
  mov    $0xcccccccd,%r10d
  lea    -0x1(%rdi,%rsi,1),%r9
  nopl   0x0(%rax)
L37:
  mov    %eax,%edx
  sub    $0x1,%r9
  imul   %r10,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  cltq
  movzbl (%r11,%rax,1),%eax
  mov    %al,0x1(%r9)
  mov    %edx,%eax
  test   %edx,%edx
  jne    L37
  test   %r8d,%r8d
  js     L16
L38:
  mov    %rsi,%rdx
  movb   $0x0,(%rdi,%rdx,1)
  ret
  nopl   0x0(%rax,%rax,1)
L65:
  mov    $0x1,%edx
  movb   $0x30,(%rdi)
  movb   $0x0,(%rdi,%rdx,1)
  ret
  nopl   (%rax)
L16:
  movb   $0x2d,(%rdi)
  jmp    L38
  nopl   (%rax)
L67:
  neg    %eax
  lea    0x2(%r9),%edx
  jmp    L5
```

tool_calls: name: disassemble_data
arguments: data_tag: D5
datatype: string

tool: D5: 0123456789

assistant: ```c
void arm_int2str(char *dst, int src) {
  int val, count = 0, pos = 0;
  static const char intchars[] = "0123456789";
  val = src;
  while (val) {
    count++;
    val = val / 10;
  }
  if (src < 0) {
    count++;
  }
  val = (src < 0) ? -src : src;
  while (val) {
    dst[count - pos - 1] = intchars[val % 10];
    val = val / 10;
    pos++;
  }
  if (src < 0) {
    dst[0] = '-';
  }
  if (count == 0) {
    dst[count] = '0';
    count++;
  }
  dst[count] = '\0';
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L8:
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  test   %eax,%eax
  jle    L5
  mov    -0xc(%rbp),%eax
  cmp    $0xa,%eax
  jle    L37
L5:
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L37:
  mov    -0xc(%rbp),%eax
  test   %eax,%eax
  jle    L8
  mov    -0xc(%rbp),%eax
  cmp    $0xa,%eax
  jg     L8
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L35
  call   <__stack_chk_fail@plt>
L35:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D29
type_guessed: string
name: retrieve_data
arguments: data_tag: D11
type_guessed: string
name: retrieve_data
arguments: data_tag: D12
type_guessed: string
name: retrieve_data
arguments: data_tag: D40
type_guessed: string
name: retrieve_data
arguments: data_tag: D45
type_guessed: string

tool: D29: do-while Loop Example 2


tool: D11: Enter a number from 1 to 10: 

tool: D12: %d

tool: D40: Number is not in proper range!

tool: D45: 
You entered %d.



assistant: ```c
int main() {
  int num;
  printf("do-while Loop Example 2\n\n");
  do {
    printf("Enter a number from 1 to 10: ");
    scanf("%d", &num);
    if (num < 1 || num > 10)
      printf("Number is not in proper range!\n");
  } while (num < 1 || num > 10);
  printf("\nYou entered %d.\n\n", num);
  return 0;
}
```