ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<xdbase_mkdir>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x420,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x418(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L170
  mov    %rdi,%rsi
  lea    0x10(%rsp),%rdi
  call   <strcpy@plt>
  movzbl 0x10(%rsp),%edx
  test   %dl,%dl
  je     L91
  mov    $0xffffffff,%ebp
  lea    0x10(%rsp),%rax
  lea    0xc(%rsp),%r12
  jmp    L51
L109:
  mov    %rax,%rbx
L19:
  test   %ebp,%ebp
  jg     L188
L103:
  lea    0x1(%rbx),%rax
  movzbl 0x1(%rbx),%edx
  test   %dl,%dl
  je     L84
L51:
  mov    %rax,%rbx
  cmp    $0x2f,%dl
  jne    L103
  add    $0x1,%ebp
  cmpb   $0x2f,(%rax)
  jne    L19
L169:
  cmpb   $0x2f,0x1(%rax)
  jne    L109
  add    $0x1,%rax
  cmpb   $0x2f,(%rax)
  je     L169
  mov    %rax,%rbx
  jmp    L19
L188:
  movb   $0x0,(%rbx)
  mov    $0x4,%edx
  mov    $0x0,%esi
  mov    %r12,%rdi
  call   <memset@plt>
  lea    0x10(%rsp),%rdi
  mov    %r12,%rsi
  call   <stat@plt>
  test   %eax,%eax
  je     L163
  lea    0x10(%rsp),%rdi
  mov    $0x1ed,%esi
  call   <mkdir@plt>
  test   %rax,%rax
  je     L184
  mov    $0xffffffff,%eax
  jmp    L158
L163:
  mov    0xc(%rsp),%edi
  call   <S_ISDIR@plt>
  test   %eax,%eax
  je     L175
L184:
  movb   $0x2f,(%rbx)
  jmp    L103
L84:
  mov    $0x0,%eax
L158:
  mov    0x418(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L147
  add    $0x420,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L170:
  mov    $0xffffffff,%eax
  jmp    L158
L91:
  mov    $0x0,%eax
  jmp    L158
L175:
  mov    $0xffffffff,%eax
  jmp    L158
L147:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int xdbase_mkdir(char *path) {
  struct stat st;
  char fullpath[1024];
  char *p = NULL;
  int level = -1, ret = -1;
  if (path) {
    strcpy(fullpath, path);
    p = fullpath;
    while (*p != '\0') {
      if (*p == '/') {
        level++;
        while (*p != '\0' && *p == '/' && *(p + 1) == '/')
          ++p;
        if (level > 0) {
          *p = '\0';
          memset(&st, 0, sizeof(struct stat));
          ret = stat(fullpath, &st);
          if (ret == 0 && !S_ISDIR(st.st_mode))
            return -1;
          if (ret != 0 && mkdir(fullpath, 0755) != 0)
            return -1;
          *p = '/';
        }
      }
      ++p;
    }
    return 0;
  }
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<xdbase_mkdir>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x420,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x418(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L120
  lea    0x10(%rsp),%r13
  mov    %rdi,%rsi
  mov    $0xffffffff,%r12d
  mov    %r13,%rdi
  mov    %r13,%rbx
  lea    0xc(%rsp),%r14
  call   <strcpy@plt>
  movzbl 0x10(%rsp),%eax
  test   %al,%al
  jne    L70
  jmp    L68
  nopl   0x0(%rax,%rax,1)
L89:
  movzbl 0x1(%rbp),%eax
  add    $0x1,%rbx
  test   %al,%al
  je     L68
L70:
  mov    %rbx,%rbp
  cmp    $0x2f,%al
  jne    L89
  add    $0x1,%r12d
  cmpb   $0x2f,(%rbx)
  je     L88
  jmp    L112
  xchg   %ax,%ax
L21:
  add    $0x1,%rbx
L88:
  cmpb   $0x2f,0x1(%rbx)
  je     L21
  mov    %rbx,%rbp
L112:
  test   %r12d,%r12d
  je     L89
  movb   $0x0,(%rbx)
  xor    %esi,%esi
  mov    %r14,%rdi
  mov    $0x4,%edx
  call   <memset@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <stat@plt>
  test   %eax,%eax
  je     L3
  mov    $0x1ed,%esi
  mov    %r13,%rdi
  call   <mkdir@plt>
  test   %rax,%rax
  jne    L120
L92:
  movb   $0x2f,(%rbx)
  movzbl 0x1(%rbp),%eax
  add    $0x1,%rbx
  test   %al,%al
  jne    L70
  nopl   0x0(%rax)
L68:
  xor    %eax,%eax
L25:
  mov    0x418(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L19
  add    $0x420,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L3:
  mov    0xc(%rsp),%edi
  call   <S_ISDIR@plt>
  test   %eax,%eax
  jne    L92
L120:
  mov    $0xffffffff,%eax
  jmp    L25
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int xdbase_mkdir(char *path) {
  struct stat st;
  char fullpath[1024];
  char *p = NULL;
  int level = -1, ret = -1;
  if (path) {
    strcpy(fullpath, path);
    p = fullpath;
    while (*p != '\0') {
      if (*p == '/') {
        level++;
        while (*p != '\0' && *p == '/' && *(p + 1) == '/')
          ++p;
        if (level > 0) {
          *p = '\0';
          memset(&st, 0, sizeof(struct stat));
          ret = stat(fullpath, &st);
          if (ret == 0 && !S_ISDIR(st.st_mode))
            return -1;
          if (ret != 0 && mkdir(fullpath, 0755) != 0)
            return -1;
          *p = '/';
        }
      }
      ++p;
    }
    return 0;
  }
  return -1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<xdbase_mkdir>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x420,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x418(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L56
  lea    0x10(%rsp),%r13
  mov    %rdi,%rsi
  mov    $0xffffffff,%r12d
  mov    %r13,%rdi
  mov    %r13,%rbx
  lea    0xc(%rsp),%r14
  call   <strcpy@plt>
  movzbl 0x10(%rsp),%eax
  test   %al,%al
  jne    L64
  jmp    L44
  nopl   0x0(%rax,%rax,1)
L5:
  movzbl 0x1(%rbp),%eax
  add    $0x1,%rbx
  test   %al,%al
  je     L44
L64:
  mov    %rbx,%rbp
  cmp    $0x2f,%al
  jne    L5
  add    $0x1,%r12d
  cmpb   $0x2f,0x1(%rbx)
  jne    L122
  nopl   (%rax)
L10:
  add    $0x1,%rbx
  cmpb   $0x2f,0x1(%rbx)
  je     L10
  mov    %rbx,%rbp
L122:
  test   %r12d,%r12d
  je     L5
  movb   $0x0,(%rbx)
  xor    %esi,%esi
  mov    %r14,%rdi
  mov    $0x4,%edx
  call   <memset@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <stat@plt>
  test   %eax,%eax
  je     L79
  mov    $0x1ed,%esi
  mov    %r13,%rdi
  call   <mkdir@plt>
  test   %rax,%rax
  jne    L56
L134:
  movb   $0x2f,(%rbx)
  movzbl 0x1(%rbp),%eax
  add    $0x1,%rbx
  test   %al,%al
  jne    L64
  nopl   0x0(%rax)
L44:
  xor    %eax,%eax
L94:
  mov    0x418(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x420,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L79:
  mov    0xc(%rsp),%edi
  call   <S_ISDIR@plt>
  test   %eax,%eax
  jne    L134
L56:
  mov    $0xffffffff,%eax
  jmp    L94
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int xdbase_mkdir(char *path) {
  struct stat st;
  char fullpath[1024];
  char *p = NULL;
  int level = -1, ret = -1;
  if (path) {
    strcpy(fullpath, path);
    p = fullpath;
    while (*p != '\0') {
      if (*p == '/') {
        level++;
        while (*p != '\0' && *p == '/' && *(p + 1) == '/')
          ++p;
        if (level > 0) {
          *p = '\0';
          memset(&st, 0, sizeof(struct stat));
          ret = stat(fullpath, &st);
          if (ret == 0 && !S_ISDIR(st.st_mode))
            return -1;
          if (ret != 0 && mkdir(fullpath, 0755) != 0)
            return -1;
          *p = '/';
        }
      }
      ++p;
    }
    return 0;
  }
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<dm_round>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L7:
  mov    -0x10(%rbp),%eax
  sub    $0x61c88647,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x4(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x4(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  xor    %eax,%ecx
  mov    %ecx,%edx
  mov    -0x4(%rbp),%eax
  sar    $0x5,%eax
  mov    %eax,%ecx
  mov    -0x20(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  add    %ecx,%eax
  xor    %edx,%eax
  add    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x8(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  xor    %eax,%ecx
  mov    %ecx,%edx
  mov    -0x8(%rbp),%eax
  sar    $0x5,%eax
  mov    %eax,%ecx
  mov    -0x20(%rbp),%rax
  add    $0xc,%rax
  mov    (%rax),%eax
  add    %ecx,%eax
  xor    %edx,%eax
  add    %eax,-0x4(%rbp)
  subl   $0x1,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jne    L7
  mov    -0x28(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  add    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  add    %eax,%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int dm_round(int rounds, uint32_t *array, uint32_t *h0, uint32_t *h1) {
  uint32_t sum = 0;
  int n = 0;
  uint32_t b0 = 0;
  uint32_t b1 = 0;
  b0 = *h0;
  b1 = *h1;
  n = rounds;
  do {
    sum += 0x9E3779B9;
    b0 += ((b1 << 4) + array[0]) ^ (b1 + sum) ^ ((b1 >> 5) + array[1]);
    b1 += ((b0 << 4) + array[2]) ^ (b0 + sum) ^ ((b0 >> 5) + array[3]);
  } while (--n);
  *h0 += b0;
  *h1 += b1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<dm_round>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbp
  mov    %rcx,%rbx
  mov    (%rdx),%r12d
  mov    (%rcx),%edx
  mov    (%rsi),%r11d
  mov    0x4(%rsi),%r10d
  mov    0x8(%rsi),%r9d
  mov    0xc(%rsi),%r8d
  imul   $0x9e3779b9,%edi,%edi
  mov    %r12d,%ecx
  mov    $0x0,%esi
L4:
  sub    $0x61c88647,%esi
  mov    %edx,%eax
  shl    $0x4,%eax
  add    %r11d,%eax
  mov    %edx,%r13d
  sar    $0x5,%r13d
  add    %r10d,%r13d
  xor    %r13d,%eax
  lea    (%rdx,%rsi,1),%r13d
  xor    %r13d,%eax
  add    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x4,%eax
  add    %r9d,%eax
  mov    %ecx,%r13d
  sar    $0x5,%r13d
  add    %r8d,%r13d
  xor    %r13d,%eax
  lea    (%rcx,%rsi,1),%r13d
  xor    %r13d,%eax
  add    %eax,%edx
  cmp    %edi,%esi
  jne    L4
  add    %ecx,%r12d
  mov    %r12d,0x0(%rbp)
  add    %edx,(%rbx)
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int dm_round(int rounds, uint32_t *array, uint32_t *h0, uint32_t *h1) {
  uint32_t sum = 0;
  int n = 0;
  uint32_t b0 = 0;
  uint32_t b1 = 0;
  b0 = *h0;
  b1 = *h1;
  n = rounds;
  do {
    sum += 0x9E3779B9;
    b0 += ((b1 << 4) + array[0]) ^ (b1 + sum) ^ ((b1 >> 5) + array[1]);
    b1 += ((b0 << 4) + array[2]) ^ (b0 + sum) ^ ((b0 >> 5) + array[3]);
  } while (--n);
  *h0 += b0;
  *h1 += b1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<dm_round>:
  endbr64
  push   %r13
  imul   $0x9e3779b9,%edi,%edi
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    (%rdx),%r12d
  mov    %rcx,%rbx
  mov    (%rcx),%edx
  mov    (%rsi),%r11d
  mov    0x4(%rsi),%r10d
  mov    0x8(%rsi),%r9d
  mov    %r12d,%ecx
  mov    0xc(%rsi),%r8d
  xor    %esi,%esi
  nop
L2:
  mov    %edx,%eax
  mov    %edx,%r13d
  sub    $0x61c88647,%esi
  shl    $0x4,%eax
  sar    $0x5,%r13d
  add    %r10d,%r13d
  add    %r11d,%eax
  xor    %r13d,%eax
  lea    (%rdx,%rsi,1),%r13d
  xor    %r13d,%eax
  add    %eax,%ecx
  mov    %ecx,%eax
  mov    %ecx,%r13d
  shl    $0x4,%eax
  sar    $0x5,%r13d
  add    %r8d,%r13d
  add    %r9d,%eax
  xor    %r13d,%eax
  lea    (%rcx,%rsi,1),%r13d
  xor    %r13d,%eax
  add    %eax,%edx
  cmp    %esi,%edi
  jne    L2
  add    %ecx,%r12d
  xor    %eax,%eax
  mov    %r12d,0x0(%rbp)
  add    %edx,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int dm_round(int rounds, uint32_t *array, uint32_t *h0, uint32_t *h1) {
  uint32_t sum = 0;
  int n = 0;
  uint32_t b0 = 0;
  uint32_t b1 = 0;
  b0 = *h0;
  b1 = *h1;
  n = rounds;
  do {
    sum += 0x9E3779B9;
    b0 += ((b1 << 4) + array[0]) ^ (b1 + sum) ^ ((b1 >> 5) + array[1]);
    b1 += ((b0 << 4) + array[2]) ^ (b0 + sum) ^ ((b0 >> 5) + array[3]);
  } while (--n);
  *h0 += b0;
  *h1 += b1;
  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
<dm_round>:
  endbr64
  push   %r13
  imul   $0x9e3779b9,%edi,%edi
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    (%rdx),%r12d
  mov    %rcx,%rbx
  mov    (%rcx),%edx
  mov    (%rsi),%r11d
  mov    0x4(%rsi),%r10d
  mov    0x8(%rsi),%r9d
  mov    %r12d,%ecx
  mov    0xc(%rsi),%r8d
  xor    %esi,%esi
  nop
L7:
  mov    %edx,%eax
  mov    %edx,%r13d
  sub    $0x61c88647,%esi
  shl    $0x4,%eax
  sar    $0x5,%r13d
  add    %r10d,%r13d
  add    %r11d,%eax
  xor    %r13d,%eax
  lea    (%rdx,%rsi,1),%r13d
  xor    %r13d,%eax
  add    %eax,%ecx
  mov    %ecx,%eax
  mov    %ecx,%r13d
  shl    $0x4,%eax
  sar    $0x5,%r13d
  add    %r8d,%r13d
  add    %r9d,%eax
  xor    %r13d,%eax
  lea    (%rcx,%rsi,1),%r13d
  xor    %r13d,%eax
  add    %eax,%edx
  cmp    %esi,%edi
  jne    L7
  add    %ecx,%r12d
  xor    %eax,%eax
  mov    %r12d,0x0(%rbp)
  add    %edx,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int dm_round(int rounds, uint32_t *array, uint32_t *h0, uint32_t *h1) {
  uint32_t sum = 0;
  int n = 0;
  uint32_t b0 = 0;
  uint32_t b1 = 0;
  b0 = *h0;
  b1 = *h1;
  n = rounds;
  do {
    sum += 0x9E3779B9;
    b0 += ((b1 << 4) + array[0]) ^ (b1 + sum) ^ ((b1 >> 5) + array[1]);
    b1 += ((b0 << 4) + array[2]) ^ (b0 + sum) ^ ((b0 >> 5) + array[3]);
  } while (--n);
  *h0 += b0;
  *h1 += b1;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<mySqrt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jg     L72
  mov    $0x0,%eax
  jmp    L29
L72:
  movq   $0x0,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  add    $0x1,%eax
  cltq
  mov    %rax,-0x18(%rbp)
  jmp    L22
L0:
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdx
  shr    $0x3f,%rdx
  add    %rdx,%rax
  sar    %rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  imul   %rax,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  cmp    %rax,-0x8(%rbp)
  jne    L18
  mov    -0x10(%rbp),%rax
  jmp    L29
L18:
  mov    -0x24(%rbp),%eax
  cltq
  cmp    %rax,-0x8(%rbp)
  jge    L55
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x20(%rbp)
  jmp    L22
L55:
  mov    -0x10(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x18(%rbp)
L22:
  mov    -0x20(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jle    L0
  mov    -0x18(%rbp),%rax
L29:
  pop    %rbp
  ret
```

assistant: ```c
int mySqrt(int x) {
  if (x <= 0)
    return (0);
  long long l, u, m, t;
  l = 0;
  u = x / 2 + 1;
  while (l <= u) {
    m = (l + u) / 2;
    t = m * m;
    if (t == x)
      return ((int)m);
    if (t < x)
      l = m + 1;
    else
      u = m - 1;
  }
  return ((int)u);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<mySqrt>:
  endbr64
  mov    $0x0,%eax
  test   %edi,%edi
  jle    L39
  mov    %edi,%eax
  shr    $0x1f,%eax
  add    %edi,%eax
  sar    %eax
  add    $0x1,%eax
  cltq
  mov    %rax,%rdx
  shr    $0x3f,%rdx
  add    %rax,%rdx
  sar    %rdx
  mov    %rdx,%rcx
  imul   %rdx,%rdx
  movslq %edi,%rdi
  cmp    %rdi,%rdx
  je     L52
  mov    $0x0,%esi
  jmp    L65
L52:
  mov    %ecx,%eax
  ret
L24:
  lea    -0x1(%rcx),%rax
L11:
  cmp    %rax,%rsi
  jg     L39
  lea    (%rsi,%rax,1),%rcx
  mov    %rcx,%rdx
  shr    $0x3f,%rdx
  add    %rcx,%rdx
  sar    %rdx
  mov    %rdx,%rcx
  imul   %rdx,%rdx
  cmp    %rdi,%rdx
  je     L52
L65:
  cmp    %rdx,%rdi
  jle    L24
  lea    0x1(%rcx),%rsi
  jmp    L11
L39:
  ret
```

assistant: ```c
int mySqrt(int x) {
  if (x <= 0)
    return (0);
  long long l, u, m, t;
  l = 0;
  u = x / 2 + 1;
  while (l <= u) {
    m = (l + u) / 2;
    t = m * m;
    if (t == x)
      return ((int)m);
    if (t < x)
      l = m + 1;
    else
      u = m - 1;
  }
  return ((int)u);
}
```
user: What is the c source code of the assembly code below:

```asm
<mySqrt>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  jle    L0
  mov    %edi,%eax
  xor    %ecx,%ecx
  movslq %edi,%rdi
  sar    %eax
  add    $0x1,%eax
  cltq
  jmp    L50
  nopw   0x0(%rax,%rax,1)
L52:
  lea    0x1(%rdx),%rcx
  cmp    %rax,%rcx
  jg     L0
L50:
  lea    (%rcx,%rax,1),%rdx
  sar    %rdx
  mov    %rdx,%rsi
  imul   %rdx,%rsi
  cmp    %rsi,%rdi
  je     L16
  jg     L52
  lea    -0x1(%rdx),%rax
  cmp    %rax,%rcx
  jle    L50
L0:
  ret
  nopl   0x0(%rax,%rax,1)
L16:
  mov    %edx,%eax
  ret
```

assistant: ```c
int mySqrt(int x) {
  if (x <= 0)
    return (0);
  long long l, u, m, t;
  l = 0;
  u = x / 2 + 1;
  while (l <= u) {
    m = (l + u) / 2;
    t = m * m;
    if (t == x)
      return ((int)m);
    if (t < x)
      l = m + 1;
    else
      u = m - 1;
  }
  return ((int)u);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<mySqrt>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  jle    L20
  mov    %edi,%eax
  xor    %ecx,%ecx
  movslq %edi,%rdi
  sar    %eax
  add    $0x1,%eax
  cltq
  nopl   0x0(%rax,%rax,1)
L27:
  lea    (%rcx,%rax,1),%rdx
  sar    %rdx
  mov    %rdx,%rsi
  imul   %rdx,%rsi
  cmp    %rsi,%rdi
  je     L28
  jle    L37
  lea    0x1(%rdx),%rcx
  cmp    %rax,%rcx
  jle    L27
  ret
  nop
L37:
  lea    -0x1(%rdx),%rax
  cmp    %rcx,%rax
  jge    L27
L20:
  ret
  nopw   0x0(%rax,%rax,1)
L28:
  mov    %edx,%eax
  ret
```

assistant: ```c
int mySqrt(int x) {
  if (x <= 0)
    return (0);
  long long l, u, m, t;
  l = 0;
  u = x / 2 + 1;
  while (l <= u) {
    m = (l + u) / 2;
    t = m * m;
    if (t == x)
      return ((int)m);
    if (t < x)
      l = m + 1;
    else
      u = m - 1;
  }
  return ((int)u);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sanitize_input>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L0
L1:
  addl   $0x1,-0x4(%rbp)
L0:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jae    L35
  cmpl   $0x3b,-0x4(%rbp)
  je     L35
  cmpl   $0x3d,-0x4(%rbp)
  jne    L1
L35:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int sanitize_input(const char *input, unsigned int input_len) {
  unsigned int i = 0;
  while (i < input_len && i != ';' && i != '=')
    i++;
  return i;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sanitize_input>:
  endbr64
  test   %esi,%esi
  je     L14
  mov    $0x0,%eax
L29:
  mov    %eax,%edx
  add    $0x1,%eax
  sub    $0x3a,%edx
  test   $0xfffffffd,%edx
  je     L27
  cmp    %eax,%esi
  ja     L29
  ret
L14:
  mov    %esi,%eax
L27:
  ret
```

assistant: ```c
int sanitize_input(const char *input, unsigned int input_len) {
  unsigned int i = 0;
  while (i < input_len && i != ';' && i != '=')
    i++;
  return i;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sanitize_input>:
  endbr64
  xor    %eax,%eax
  test   %esi,%esi
  je     L17
  nopw   0x0(%rax,%rax,1)
L29:
  add    $0x1,%eax
  cmp    %eax,%esi
  jbe    L17
  cmp    $0x3b,%eax
  jne    L29
  ret
  nopl   (%rax)
L17:
  ret
```

assistant: ```c
int sanitize_input(const char *input, unsigned int input_len) {
  unsigned int i = 0;
  while (i < input_len && i != ';' && i != '=')
    i++;
  return i;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<sanitize_input>:
  endbr64
  xor    %eax,%eax
  test   %esi,%esi
  je     L27
  nopw   0x0(%rax,%rax,1)
L19:
  add    $0x1,%eax
  cmp    %eax,%esi
  jbe    L27
  cmp    $0x3b,%eax
  jne    L19
  ret
  nopl   (%rax)
L27:
  ret
```

assistant: ```c
int sanitize_input(const char *input, unsigned int input_len) {
  unsigned int i = 0;
  while (i < input_len && i != ';' && i != '=')
    i++;
  return i;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<DoCountTest>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xffffffff,-0x10(%rbp)
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <printf@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    %eax,-0x1c(%rbp)
  je     L5
  mov    -0x10(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <Fail@plt>
L5:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0xc(%rbp)
  je     L12
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%ecx
  mov    -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    %ecx,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <Fail@plt>
L12:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D1
data_type: string
name: unpack_data
arguments: tag: D13
data_type: string

tool: D1: Expected count parameter to resolve to %d, got %d


tool: D13: Expected printf to return %d, got %d.


assistant: ```c
void DoCountTest(const char *formatstr, int param, const char *checkstr) {
  int ret;
  int n = -1;
  ret = printf(formatstr, &n);
  if (n != param) {
    Fail("Expected count parameter to resolve to %d, got %d\n", param, n);
  }
  if (ret != strlen(checkstr)) {
    Fail("Expected printf to return %d, got %d.\n", strlen(checkstr), ret);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<DoCountTest>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %esi,%r12d
  mov    %rdx,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0xffffffff,0x4(%rsp)
  lea    0x4(%rsp),%rsi
  call   <printf@plt>
  mov    %eax,%ebx
  mov    0x4(%rsp),%edx
  cmp    %r12d,%edx
  jne    L13
L39:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jne    L1
L8:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L13:
  mov    %r12d,%esi
  lea    D8(%rip),%rdi
  call   <Fail@plt>
  jmp    L39
L1:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  mov    %ebx,%edx
  lea    D18(%rip),%rdi
  call   <Fail@plt>
  jmp    L8
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D8
type: string
name: extract_symbolic_data
arguments: label: D18
type: string

tool: D8: Expected count parameter to resolve to %d, got %d


tool: D18: Expected printf to return %d, got %d.


assistant: ```c
void DoCountTest(const char *formatstr, int param, const char *checkstr) {
  int ret;
  int n = -1;
  ret = printf(formatstr, &n);
  if (n != param) {
    Fail("Expected count parameter to resolve to %d, got %d\n", param, n);
  }
  if (ret != strlen(checkstr)) {
    Fail("Expected printf to return %d, got %d.\n", strlen(checkstr), ret);
  }
}
```