ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  movq   $0x0,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L120
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x30(%rbp),%rdx
  add    %rdx,%rax
  movb   $0x30,(%rax)
  mov    -0x30(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x30(%rbp),%rax
  jmp    L11
L120:
  cmpq   $0x0,-0x28(%rbp)
  jns    L45
  cmpl   $0xa,-0x34(%rbp)
  je     L73
  cmpl   $0x64,-0x34(%rbp)
  je     L73
  cmpl   $0x69,-0x34(%rbp)
  jne    L45
L73:
  movl   $0xa,-0x34(%rbp)
  movl   $0x1,-0x14(%rbp)
  negq   -0x28(%rbp)
  jmp    L122
L45:
  cmpl   $0x78,-0x34(%rbp)
  jne    L63
  movl   $0x10,-0x34(%rbp)
  jmp    L8
L63:
  cmpl   $0x62,-0x34(%rbp)
  jne    L68
  movl   $0x2,-0x34(%rbp)
  jmp    L8
L68:
  movl   $0xa,-0x34(%rbp)
L122:
  jmp    L8
L46:
  mov    -0x34(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x28(%rbp),%rax
  cqto
  idiv   %rcx
  mov    %rdx,-0x8(%rbp)
  cmpq   $0x9,-0x8(%rbp)
  jle    L51
  mov    -0x8(%rbp),%rax
  add    $0x57,%eax
  mov    %eax,%ecx
  jmp    L7
L51:
  mov    -0x8(%rbp),%rax
  add    $0x30,%eax
  mov    %eax,%ecx
L7:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x30(%rbp),%rdx
  add    %rdx,%rax
  mov    %cl,(%rax)
  mov    -0x34(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x28(%rbp),%rax
  cqto
  idiv   %rsi
  mov    %rax,-0x28(%rbp)
L8:
  cmpq   $0x0,-0x28(%rbp)
  jne    L46
  cmpl   $0x0,-0x14(%rbp)
  je     L118
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x30(%rbp),%rdx
  add    %rdx,%rax
  movb   $0x2d,(%rax)
L118:
  mov    -0x30(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <reverse@plt>
  mov    -0x30(%rbp),%rax
L11:
  leave
  ret
```

assistant: ```c
char *itoa(long long num, char *str, int base) {
  size_t i = 0;
  bool is_negative = false;
  if (num == 0) {
    str[i++] = '0';
    str[i] = '\0';
    return str;
  }
  if (num < 0 && (base == 10 || base == 'd' || base == 'i')) {
    base = 10;
    is_negative = true;
    num = -num;
  } else if (base == 'x') {
    base = 16;
  } else if (base == 'b') {
    base = 2;
  } else {
    base = 10;
  }
  while (num != 0) {
    long long rem = num % base;
    str[i++] = (rem > 9) ? (rem - 10) + 'a' : rem + '0';
    num = num / base;
  }
  if (is_negative)
    str[i++] = '-';
  str[i] = '\0';
  reverse(str, i);
  return str;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<itoa>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  test   %rdi,%rdi
  je     L105
  mov    %rdi,%rcx
  js     L82
L55:
  cmp    $0x78,%edx
  je     L90
  mov    $0x0,%r9d
  cmp    $0x62,%edx
  setne  %sil
  movzbl %sil,%esi
  lea    0x2(,%rsi,8),%esi
L93:
  mov    $0x0,%r8d
  movslq %esi,%rsi
  jmp    L19
L105:
  movb   $0x30,(%rsi)
  movb   $0x0,0x1(%rsi)
  jmp    L69
L82:
  cmp    $0xa,%edx
  sete   %al
  cmp    $0x64,%edx
  sete   %sil
  or     %sil,%al
  jne    L104
  cmp    $0x69,%edx
  jne    L55
L104:
  neg    %rcx
  mov    $0x1,%r9d
  mov    $0xa,%esi
  jmp    L93
L68:
  mov    %rdi,%r8
L19:
  mov    %rcx,%rax
  cqto
  idiv   %rsi
  lea    0x57(%rdx),%edi
  lea    0x30(%rdx),%eax
  cmp    $0xa,%rdx
  cmovge %edi,%eax
  lea    0x1(%r8),%rdi
  mov    %al,-0x1(%rbx,%rdi,1)
  mov    %rcx,%rax
  cqto
  idiv   %rsi
  mov    %rax,%rcx
  test   %rax,%rax
  jne    L68
  test   %r9d,%r9d
  je     L50
  movb   $0x2d,(%rbx,%rdi,1)
  lea    0x2(%r8),%rdi
L50:
  movb   $0x0,(%rbx,%rdi,1)
  mov    %rdi,%rsi
  mov    %rbx,%rdi
  call   <reverse@plt>
L69:
  mov    %rbx,%rax
  pop    %rbx
  ret
L90:
  mov    $0x0,%r9d
  mov    $0x10,%esi
  jmp    L93
```

assistant: ```c
char *itoa(long long num, char *str, int base) {
  size_t i = 0;
  bool is_negative = false;
  if (num == 0) {
    str[i++] = '0';
    str[i] = '\0';
    return str;
  }
  if (num < 0 && (base == 10 || base == 'd' || base == 'i')) {
    base = 10;
    is_negative = true;
    num = -num;
  } else if (base == 'x') {
    base = 16;
  } else if (base == 'b') {
    base = 2;
  } else {
    base = 10;
  }
  while (num != 0) {
    long long rem = num % base;
    str[i++] = (rem > 9) ? (rem - 10) + 'a' : rem + '0';
    num = num / base;
  }
  if (is_negative)
    str[i++] = '-';
  str[i] = '\0';
  reverse(str, i);
  return str;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<itoa>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  test   %rdi,%rdi
  je     L37
  mov    %rdi,%rax
  js     L46
L33:
  cmp    $0x78,%edx
  je     L40
  xor    %edi,%edi
  cmp    $0x62,%edx
  setne  %dil
  xor    %r9d,%r9d
  lea    0x2(,%rdi,8),%rdi
L58:
  xor    %r8d,%r8d
  jmp    L64
  nopl   0x0(%rax)
L9:
  mov    %rsi,%r8
L64:
  cqto
  idiv   %rdi
  lea    0x57(%rdx),%esi
  lea    0x30(%rdx),%ecx
  cmp    $0xa,%rdx
  cmovge %esi,%ecx
  lea    0x1(%r8),%rsi
  mov    %cl,-0x1(%r12,%rsi,1)
  test   %rax,%rax
  jne    L9
  test   %r9d,%r9d
  je     L88
  movb   $0x2d,(%r12,%rsi,1)
  lea    0x2(%r8),%rsi
L88:
  movb   $0x0,(%r12,%rsi,1)
  mov    %r12,%rdi
  call   <reverse@plt>
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L46:
  cmp    $0xa,%edx
  sete   %cl
  cmp    $0x64,%edx
  sete   %sil
  or     %sil,%cl
  jne    L41
  cmp    $0x69,%edx
  jne    L33
L41:
  neg    %rax
  mov    $0xa,%edi
  mov    $0x1,%r9d
  jmp    L58
  nopl   0x0(%rax,%rax,1)
L37:
  mov    $0x30,%eax
  mov    %ax,(%rsi)
  mov    %r12,%rax
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L40:
  mov    $0x10,%edi
  xor    %r9d,%r9d
  jmp    L58
```

assistant: ```c
char *itoa(long long num, char *str, int base) {
  size_t i = 0;
  bool is_negative = false;
  if (num == 0) {
    str[i++] = '0';
    str[i] = '\0';
    return str;
  }
  if (num < 0 && (base == 10 || base == 'd' || base == 'i')) {
    base = 10;
    is_negative = true;
    num = -num;
  } else if (base == 'x') {
    base = 16;
  } else if (base == 'b') {
    base = 2;
  } else {
    base = 10;
  }
  while (num != 0) {
    long long rem = num % base;
    str[i++] = (rem > 9) ? (rem - 10) + 'a' : rem + '0';
    num = num / base;
  }
  if (is_negative)
    str[i++] = '-';
  str[i] = '\0';
  reverse(str, i);
  return str;
}
```
user: What is the c source code of the assembly code below:

```asm
<itoa>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  test   %rdi,%rdi
  je     L89
  mov    %rdi,%rax
  js     L10
L37:
  cmp    $0x78,%edx
  je     L100
  xor    %ecx,%ecx
  cmp    $0x62,%edx
  setne  %cl
  xor    %r8d,%r8d
  lea    0x2(,%rcx,8),%rcx
L116:
  xor    %edi,%edi
  jmp    L18
  nopw   0x0(%rax,%rax,1)
L43:
  add    $0x57,%edx
  mov    %dl,-0x1(%r12,%rsi,1)
  test   %rax,%rax
  je     L23
L69:
  mov    %rsi,%rdi
L18:
  cqto
  lea    0x1(%rdi),%rsi
  idiv   %rcx
  cmp    $0x9,%rdx
  jg     L43
  add    $0x30,%edx
  mov    %dl,-0x1(%r12,%rsi,1)
  test   %rax,%rax
  jne    L69
L23:
  test   %r8d,%r8d
  je     L48
  movb   $0x2d,(%r12,%rsi,1)
  lea    0x2(%rdi),%rsi
L48:
  movb   $0x0,(%r12,%rsi,1)
  mov    %r12,%rdi
  call   <reverse@plt>
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   (%rax)
L10:
  cmp    $0xa,%edx
  sete   %cl
  cmp    $0x64,%edx
  sete   %sil
  or     %sil,%cl
  jne    L30
  cmp    $0x69,%edx
  jne    L37
L30:
  neg    %rax
  mov    $0xa,%ecx
  mov    $0x1,%r8d
  jmp    L116
  xchg   %ax,%ax
L89:
  mov    $0x30,%eax
  mov    %ax,(%rsi)
  mov    %r12,%rax
  pop    %r12
  ret
  xchg   %ax,%ax
L100:
  mov    $0x10,%ecx
  xor    %r8d,%r8d
  jmp    L116
```

assistant: ```c
char *itoa(long long num, char *str, int base) {
  size_t i = 0;
  bool is_negative = false;
  if (num == 0) {
    str[i++] = '0';
    str[i] = '\0';
    return str;
  }
  if (num < 0 && (base == 10 || base == 'd' || base == 'i')) {
    base = 10;
    is_negative = true;
    num = -num;
  } else if (base == 'x') {
    base = 16;
  } else if (base == 'b') {
    base = 2;
  } else {
    base = 10;
  }
  while (num != 0) {
    long long rem = num % base;
    str[i++] = (rem > 9) ? (rem - 10) + 'a' : rem + '0';
    num = num / base;
  }
  if (is_negative)
    str[i++] = '-';
  str[i] = '\0';
  reverse(str, i);
  return str;
}
```
user: Translate the below assembly code into C programming language:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  sub    $0x99f4bb,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L49
L57:
  movl   $0x0,-0x14(%rbp)
  jmp    L62
L61:
  mov    -0xc(%rbp),%eax
  mov    -0x14(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  add    %ecx,%edx
  cmp    %edx,%eax
  jbe    L6
  lea    -0xc(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  shl    $0x2,%rax
  add    %rdx,%rax
  movzwl (%rax),%eax
  mov    %ax,-0x16(%rbp)
  lea    -0xc(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    $0x2,%rax
  lea    -0xc(%rbp),%rcx
  mov    -0x14(%rbp),%edx
  shl    $0x2,%rdx
  add    %rcx,%rdx
  movzwl (%rax),%eax
  mov    %ax,(%rdx)
  lea    -0xc(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x2(%rax),%rdx
  movzwl -0x16(%rbp),%eax
  mov    %ax,(%rdx)
  lea    -0xc(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    $0x2,%rax
  movzwl (%rax),%eax
  mov    %ax,-0x16(%rbp)
  lea    -0xc(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    -0xc(%rbp),%rcx
  mov    -0x14(%rbp),%edx
  shl    $0x2,%rdx
  add    %rcx,%rdx
  add    $0x2,%rdx
  movzwl (%rax),%eax
  mov    %ax,(%rdx)
  lea    -0xc(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  shl    $0x2,%rax
  add    %rax,%rdx
  movzwl -0x16(%rbp),%eax
  mov    %ax,(%rdx)
  jmp    L35
L6:
  mov    -0xc(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0xc(%rbp)
L35:
  addl   $0x2,-0x14(%rbp)
L62:
  cmpl   $0x0,-0x14(%rbp)
  je     L61
  addl   $0x2,-0x10(%rbp)
L49:
  cmpl   $0x0,-0x10(%rbp)
  je     L57
  mov    -0xc(%rbp),%eax
  imul   $0x6be98b82,%eax,%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L69
  call   <__stack_chk_fail@plt>
L69:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local2;
  unsigned int local1;
  unsigned short copy11;
  {
    state[0UL] = input[0UL] + 4284877637U;
    local1 = 0UL;
    while (local1 < 1UL) {
      local2 = 0UL;
      while (local2 < 1UL) {
        if (state[0UL] > local2 + local1) {
          copy11 = *((unsigned short *)(&state[local2]) + 0);
          *((unsigned short *)(&state[local2]) + 0) = *((unsigned short *)(&state[local2]) + 1);
          *((unsigned short *)(&state[local2]) + 1) = copy11;
          copy11 = *((unsigned short *)(&state[local2]) + 1);
          *((unsigned short *)(&state[local2]) + 1) = *((unsigned short *)(&state[local2]) + 0);
          *((unsigned short *)(&state[local2]) + 0) = copy11;
        } else {
          state[0UL] += state[0UL];
        }
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] * 1810467714U;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  sub    $0x99f4bb,%eax
  imul   $0x6be98b82,%eax,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local2;
  unsigned int local1;
  unsigned short copy11;
  {
    state[0UL] = input[0UL] + 4284877637U;
    local1 = 0UL;
    while (local1 < 1UL) {
      local2 = 0UL;
      while (local2 < 1UL) {
        if (state[0UL] > local2 + local1) {
          copy11 = *((unsigned short *)(&state[local2]) + 0);
          *((unsigned short *)(&state[local2]) + 0) = *((unsigned short *)(&state[local2]) + 1);
          *((unsigned short *)(&state[local2]) + 1) = copy11;
          copy11 = *((unsigned short *)(&state[local2]) + 1);
          *((unsigned short *)(&state[local2]) + 1) = *((unsigned short *)(&state[local2]) + 0);
          *((unsigned short *)(&state[local2]) + 0) = copy11;
        } else {
          state[0UL] += state[0UL];
        }
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] * 1810467714U;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  sub    $0x99f4bb,%eax
  mov    %eax,%edx
  je     L5
  mov    %ax,%dx
  mov    %edx,%eax
L5:
  imul   $0x6be98b82,%eax,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local2;
  unsigned int local1;
  unsigned short copy11;
  {
    state[0UL] = input[0UL] + 4284877637U;
    local1 = 0UL;
    while (local1 < 1UL) {
      local2 = 0UL;
      while (local2 < 1UL) {
        if (state[0UL] > local2 + local1) {
          copy11 = *((unsigned short *)(&state[local2]) + 0);
          *((unsigned short *)(&state[local2]) + 0) = *((unsigned short *)(&state[local2]) + 1);
          *((unsigned short *)(&state[local2]) + 1) = copy11;
          copy11 = *((unsigned short *)(&state[local2]) + 1);
          *((unsigned short *)(&state[local2]) + 1) = *((unsigned short *)(&state[local2]) + 0);
          *((unsigned short *)(&state[local2]) + 0) = copy11;
        } else {
          state[0UL] += state[0UL];
        }
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] * 1810467714U;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  sub    $0x99f4bb,%eax
  mov    %eax,%edx
  je     L2
  mov    %ax,%dx
  mov    %edx,%eax
L2:
  imul   $0x6be98b82,%eax,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local2;
  unsigned int local1;
  unsigned short copy11;
  {
    state[0UL] = input[0UL] + 4284877637U;
    local1 = 0UL;
    while (local1 < 1UL) {
      local2 = 0UL;
      while (local2 < 1UL) {
        if (state[0UL] > local2 + local1) {
          copy11 = *((unsigned short *)(&state[local2]) + 0);
          *((unsigned short *)(&state[local2]) + 0) = *((unsigned short *)(&state[local2]) + 1);
          *((unsigned short *)(&state[local2]) + 1) = copy11;
          copy11 = *((unsigned short *)(&state[local2]) + 1);
          *((unsigned short *)(&state[local2]) + 1) = *((unsigned short *)(&state[local2]) + 0);
          *((unsigned short *)(&state[local2]) + 0) = copy11;
        } else {
          state[0UL] += state[0UL];
        }
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] * 1810467714U;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_putnbr_fd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  lea    D2(%rip),%rdx
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <ft_itoa_base@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <ft_putstr_fd@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
type_guessed: string

tool: D2: 0123456789

assistant: ```c
void ft_putnbr_fd(long n, int fd) {
  char *number;
  number = ft_itoa_base(n, 10, "0123456789");
  ft_putstr_fd(number, fd);
  free(number);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<ft_putnbr_fd>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%ebp
  lea    D4(%rip),%rdx
  mov    $0xa,%esi
  call   <ft_itoa_base@plt>
  mov    %rax,%rbx
  mov    %ebp,%esi
  mov    %rax,%rdi
  call   <ft_putstr_fd@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

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

tool: D4: 0123456789

assistant: ```c
void ft_putnbr_fd(long n, int fd) {
  char *number;
  number = ft_itoa_base(n, 10, "0123456789");
  ft_putstr_fd(number, fd);
  free(number);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_putnbr_fd>:
  endbr64
  push   %r12
  lea    D8(%rip),%rdx
  mov    %esi,%r12d
  mov    $0xa,%esi
  push   %rbp
  sub    $0x8,%rsp
  call   <ft_itoa_base@plt>
  mov    %r12d,%esi
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <ft_putstr_fd@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1070 <free@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D8
type: string

tool: D8: 0123456789

assistant: ```c
void ft_putnbr_fd(long n, int fd) {
  char *number;
  number = ft_itoa_base(n, 10, "0123456789");
  ft_putstr_fd(number, fd);
  free(number);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_putnbr_fd>:
  endbr64
  push   %r12
  lea    D5(%rip),%rdx
  mov    %esi,%r12d
  mov    $0xa,%esi
  push   %rbp
  sub    $0x8,%rsp
  call   <ft_itoa_base@plt>
  mov    %r12d,%esi
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <ft_putstr_fd@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1070 <free@plt>
```

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

tool: D5: 0123456789

assistant: ```c
void ft_putnbr_fd(long n, int fd) {
  char *number;
  number = ft_itoa_base(n, 10, "0123456789");
  ft_putstr_fd(number, fd);
  free(number);
}
```
user: What is the c source code of the assembly code below:

```asm
<simulateWar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    %r8d,-0x2c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L25
L2:
  mov    -0x28(%rbp),%ecx
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <simulateWarOneDay@plt>
  mov    -0x28(%rbp),%ecx
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <areTheSameGrid@plt>
  test   %rax,%rax
  jne    L16
  mov    -0x28(%rbp),%ecx
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <copyGrid@plt>
  addl   $0x1,-0x4(%rbp)
L25:
  mov    -0x4(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L2
  jmp    L12
L16:
  nop
L12:
  nop
  leave
  ret
```

assistant: ```c
void simulateWar(char current[][100 + 1], char result[][100 + 1], int numRow, int numCol, int numDay) {
  int day;
  for (day = 0; day < numDay; day++) {
    simulateWarOneDay(current, result, numRow, numCol);
    if (areTheSameGrid(current, result, numRow, numCol))
      break;
    copyGrid(current, result, numRow, numCol);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<simulateWar>:
  endbr64
  test   %r8d,%r8d
  jle    L7
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %edx,%r12d
  mov    %ecx,%r13d
  mov    %r8d,%r15d
  mov    $0x0,%r14d
L28:
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <simulateWarOneDay@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <areTheSameGrid@plt>
  test   %rax,%rax
  jne    L13
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <copyGrid@plt>
  add    $0x1,%r14d
  cmp    %r14d,%r15d
  jne    L28
L13:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L7:
  ret
```

assistant: ```c
void simulateWar(char current[][100 + 1], char result[][100 + 1], int numRow, int numCol, int numDay) {
  int day;
  for (day = 0; day < numDay; day++) {
    simulateWarOneDay(current, result, numRow, numCol);
    if (areTheSameGrid(current, result, numRow, numCol))
      break;
    copyGrid(current, result, numRow, numCol);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<simulateWar>:
  endbr64
  test   %r8d,%r8d
  jle    L11
  push   %r15
  mov    %r8d,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L12
  nopl   0x0(%rax,%rax,1)
L24:
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <copyGrid@plt>
  add    $0x1,%r14d
  cmp    %r14d,%r15d
  je     L8
L12:
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <simulateWarOneDay@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <areTheSameGrid@plt>
  test   %rax,%rax
  je     L24
L8:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L11:
  ret
```

assistant: ```c
void simulateWar(char current[][100 + 1], char result[][100 + 1], int numRow, int numCol, int numDay) {
  int day;
  for (day = 0; day < numDay; day++) {
    simulateWarOneDay(current, result, numRow, numCol);
    if (areTheSameGrid(current, result, numRow, numCol))
      break;
    copyGrid(current, result, numRow, numCol);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<simulateWar>:
  endbr64
  test   %r8d,%r8d
  jle    L0
  push   %r15
  mov    %r8d,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L4
  nopl   0x0(%rax,%rax,1)
L39:
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <copyGrid@plt>
  add    $0x1,%r14d
  cmp    %r14d,%r15d
  je     L2
L4:
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <simulateWarOneDay@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <areTheSameGrid@plt>
  test   %rax,%rax
  je     L39
L2:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L0:
  ret
```

assistant: ```c
void simulateWar(char current[][100 + 1], char result[][100 + 1], int numRow, int numCol, int numDay) {
  int day;
  for (day = 0; day < numDay; day++) {
    simulateWarOneDay(current, result, numRow, numCol);
    if (areTheSameGrid(current, result, numRow, numCol))
      break;
    copyGrid(current, result, numRow, numCol);
  }
}
```