ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<input>:
  endbr64
  add    $0x5a,%rdi
  mov    %rsi,%r8
  lea    0x51(%rsi),%r9
  jmp    L1
L42:
  add    $0x1,%rax
  cmp    $0xa,%rax
  je     L67
L48:
  cmp    %al,(%rcx)
  je     L42
  movb   $0x0,(%rcx,%rax,1)
  jmp    L42
L67:
  add    $0x1,%rsi
  add    $0xa,%rdx
  cmp    %rdi,%rdx
  je     L12
L77:
  movzbl (%rsi),%eax
  sub    $0x30,%eax
  mov    %rdx,%rcx
  mov    %al,(%rdx)
  test   %al,%al
  je     L67
  mov    $0x1,%eax
  jmp    L48
L12:
  add    $0x5a,%rdi
  add    $0x9,%r8
  cmp    %r9,%r8
  je     L66
L1:
  lea    -0x5a(%rdi),%rdx
  mov    %r8,%rsi
  jmp    L77
L66:
  ret
```

assistant: ```c
void input(char grid[9][9][10], char string[81]) {
  char i, j, k;
  for (i = 0; i < 9; ++i)
    for (j = 0; j < 9; ++j) {
      grid[i][j][0] = string[9 * i + j] - '0';
      if (grid[i][j][0])
        for (k = 1; k < 10; ++k)
          if (k != grid[i][j][0])
            grid[i][j][k] = 0;
    }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<input>:
  endbr64
  lea    0x5a(%rdi),%r8
  lea    0x51(%rsi),%r9
  nopl   0x0(%rax)
L74:
  lea    -0x5a(%r8),%rcx
  mov    %rsi,%rdi
  jmp    L77
  nopl   0x0(%rax)
L60:
  add    $0xa,%rcx
  add    $0x1,%rdi
  cmp    %rcx,%r8
  je     L1
L77:
  movzbl (%rdi),%eax
  lea    -0x30(%rax),%edx
  mov    %dl,(%rcx)
  test   %dl,%dl
  je     L60
  mov    $0x1,%eax
  jmp    L35
L62:
  movzbl (%rcx),%edx
L35:
  cmp    %al,%dl
  je     L16
  movb   $0x0,(%rcx,%rax,1)
L16:
  add    $0x1,%rax
  cmp    $0xa,%rax
  jne    L62
  add    $0xa,%rcx
  add    $0x1,%rdi
  cmp    %rcx,%r8
  jne    L77
L1:
  add    $0x9,%rsi
  add    $0x5a,%r8
  cmp    %r9,%rsi
  jne    L74
  ret
```

assistant: ```c
void input(char grid[9][9][10], char string[81]) {
  char i, j, k;
  for (i = 0; i < 9; ++i)
    for (j = 0; j < 9; ++j) {
      grid[i][j][0] = string[9 * i + j] - '0';
      if (grid[i][j][0])
        for (k = 1; k < 10; ++k)
          if (k != grid[i][j][0])
            grid[i][j][k] = 0;
    }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<input>:
  endbr64
  add    $0x5a,%rdi
  lea    0x51(%rsi),%r8
  nopl   0x0(%rax)
L98:
  lea    -0x5a(%rdi),%rdx
  mov    %rsi,%rcx
  jmp    L60
  nopl   0x0(%rax)
L159:
  add    $0xa,%rdx
  add    $0x1,%rcx
  cmp    %rdx,%rdi
  je     L73
L60:
  movzbl (%rcx),%eax
  sub    $0x30,%eax
  mov    %al,(%rdx)
  test   %al,%al
  je     L159
  cmp    $0x1,%al
  je     L65
  movb   $0x0,0x1(%rdx)
  cmp    $0x2,%al
  je     L46
  movb   $0x0,0x2(%rdx)
  cmp    $0x3,%al
  jne    L54
  movb   $0x0,0x4(%rdx)
L92:
  movb   $0x0,0x5(%rdx)
  cmp    $0x6,%al
  jne    L20
  movb   $0x0,0x7(%rdx)
L78:
  movb   $0x0,0x8(%rdx)
  cmp    $0x9,%al
  je     L159
  movb   $0x0,0x9(%rdx)
  nopl   0x0(%rax)
L53:
  add    $0xa,%rdx
  add    $0x1,%rcx
  cmp    %rdx,%rdi
  jne    L60
  nopl   (%rax)
L73:
  add    $0x9,%rsi
  add    $0x5a,%rdi
  cmp    %r8,%rsi
  jne    L98
  ret
  xchg   %ax,%ax
L65:
  movb   $0x0,0x2(%rdx)
L54:
  movb   $0x0,0x3(%rdx)
  cmp    $0x4,%al
  jne    L129
  movb   $0x0,0x5(%rdx)
L20:
  movb   $0x0,0x6(%rdx)
  cmp    $0x7,%al
  jne    L15
  movb   $0x0,0x8(%rdx)
  movb   $0x0,0x9(%rdx)
  jmp    L53
  nopw   0x0(%rax,%rax,1)
L46:
  movb   $0x0,0x3(%rdx)
L129:
  movb   $0x0,0x4(%rdx)
  cmp    $0x5,%al
  jne    L92
  movb   $0x0,0x6(%rdx)
L15:
  movb   $0x0,0x7(%rdx)
  cmp    $0x8,%al
  jne    L78
  movb   $0x0,0x9(%rdx)
  jmp    L53
```

assistant: ```c
void input(char grid[9][9][10], char string[81]) {
  char i, j, k;
  for (i = 0; i < 9; ++i)
    for (j = 0; j < 9; ++j) {
      grid[i][j][0] = string[9 * i + j] - '0';
      if (grid[i][j][0])
        for (k = 1; k < 10; ++k)
          if (k != grid[i][j][0])
            grid[i][j][k] = 0;
    }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ungz>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %rdi,-0x68(%rbp)
  mov    %esi,-0x6c(%rbp)
  mov    %rdx,-0x78(%rbp)
  mov    %ecx,-0x70(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xffffffff,-0x58(%rbp)
  movl   $0xffffffff,-0x54(%rbp)
  movq   $0x0,-0x50(%rbp)
  movq   $0x0,-0x48(%rbp)
  movq   $0x0,-0x40(%rbp)
  movq   $0x0,-0x38(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  mov    -0x70(%rbp),%eax
  mov    %eax,-0x4c(%rbp)
  mov    -0x4c(%rbp),%eax
  mov    %eax,-0x50(%rbp)
  mov    -0x6c(%rbp),%eax
  mov    %eax,-0x44(%rbp)
  mov    -0x44(%rbp),%eax
  mov    %eax,-0x48(%rbp)
  mov    -0x78(%rbp),%rax
  mov    %rax,-0x40(%rbp)
  mov    -0x68(%rbp),%rax
  mov    %rax,-0x38(%rbp)
  mov    D32(%rip),%rax
  mov    (%rax),%rax
  mov    %rax,-0x20(%rbp)
  mov    D32(%rip),%rax
  mov    (%rax),%rax
  mov    %rax,-0x28(%rbp)
  mov    D32(%rip),%rax
  mov    (%rax),%rax
  mov    %rax,-0x30(%rbp)
  lea    -0x50(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <inflateInit2@plt>
  mov    %eax,-0x58(%rbp)
  mov    D52(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x58(%rbp)
  jne    L27
  mov    D29(%rip),%rax
  mov    (%rax),%edx
  lea    -0x50(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <inflate@plt>
  mov    %eax,-0x58(%rbp)
  mov    D37(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x58(%rbp)
  jne    L4
  mov    -0x48(%rbp),%eax
  mov    %eax,-0x54(%rbp)
  lea    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <inflateEnd@plt>
  mov    -0x54(%rbp),%eax
  jmp    L13
L4:
  lea    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <inflateEnd@plt>
  mov    -0x58(%rbp),%eax
  jmp    L13
L27:
  lea    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <inflateEnd@plt>
  mov    -0x58(%rbp),%eax
L13:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

assistant: ```c
int ungz(char *dst, const uint32_t dstLen, char *src, const uint32_t srcLen) {
  int err = -1;
  int ret = -1;
  z_stream strm = {0};
  strm.total_in = strm.avail_in = srcLen;
  strm.total_out = strm.avail_out = dstLen;
  strm.next_in = (unsigned char *)src;
  strm.next_out = (unsigned char *)dst;
  strm.zalloc = Z_NULL;
  strm.zfree = Z_NULL;
  strm.opaque = Z_NULL;
  err = inflateInit2(&strm, 47);
  if (err == Z_OK) {
    err = inflate(&strm, Z_FINISH);
    if (err == Z_STREAM_END)
      ret = strm.total_out;
    else {
      inflateEnd(&strm);
      return err;
    }
  } else {
    inflateEnd(&strm);
    return err;
  }
  inflateEnd(&strm);
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ungz>:
  endbr64
  push   %rbx
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x38(%rsp)
  mov    %ecx,0x4(%rsp)
  mov    %ecx,(%rsp)
  mov    %esi,0xc(%rsp)
  mov    %esi,0x8(%rsp)
  mov    %rdx,0x10(%rsp)
  mov    %rdi,0x18(%rsp)
  mov    D2(%rip),%rax
  mov    (%rax),%rax
  mov    %rax,0x30(%rsp)
  mov    %rax,0x28(%rsp)
  mov    %rax,0x20(%rsp)
  mov    %rsp,%rdi
  mov    $0x2f,%esi
  call   <inflateInit2@plt>
  mov    %eax,%ebx
  mov    D22(%rip),%rax
  cmp    %ebx,(%rax)
  jne    L49
  mov    %rsp,%rdi
  mov    D14(%rip),%rax
  mov    (%rax),%esi
  call   <inflate@plt>
  mov    %eax,%ebx
  mov    D7(%rip),%rax
  cmp    %ebx,(%rax)
  jne    L20
  mov    0x8(%rsp),%ebx
  mov    %rsp,%rdi
  call   <inflateEnd@plt>
  jmp    L22
L20:
  mov    %rsp,%rdi
  call   <inflateEnd@plt>
  jmp    L22
L49:
  mov    %rsp,%rdi
  call   <inflateEnd@plt>
L22:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    %ebx,%eax
  add    $0x50,%rsp
  pop    %rbx
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ungz(char *dst, const uint32_t dstLen, char *src, const uint32_t srcLen) {
  int err = -1;
  int ret = -1;
  z_stream strm = {0};
  strm.total_in = strm.avail_in = srcLen;
  strm.total_out = strm.avail_out = dstLen;
  strm.next_in = (unsigned char *)src;
  strm.next_out = (unsigned char *)dst;
  strm.zalloc = Z_NULL;
  strm.zfree = Z_NULL;
  strm.opaque = Z_NULL;
  err = inflateInit2(&strm, 47);
  if (err == Z_OK) {
    err = inflate(&strm, Z_FINISH);
    if (err == Z_STREAM_END)
      ret = strm.total_out;
    else {
      inflateEnd(&strm);
      return err;
    }
  } else {
    inflateEnd(&strm);
    return err;
  }
  inflateEnd(&strm);
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ungz>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    D20(%rip),%rax
  mov    %rsp,%rbp
  mov    %esi,0xc(%rsp)
  mov    %esi,0x8(%rsp)
  mov    $0x2f,%esi
  mov    (%rax),%rax
  mov    %rdi,0x18(%rsp)
  mov    %rbp,%rdi
  movq   $0x0,0x38(%rsp)
  mov    %ecx,0x4(%rsp)
  mov    %ecx,(%rsp)
  mov    %rdx,0x10(%rsp)
  mov    %rax,0x30(%rsp)
  mov    %rax,0x28(%rsp)
  mov    %rax,0x20(%rsp)
  call   <inflateInit2@plt>
  mov    %eax,%r12d
  mov    D10(%rip),%rax
  cmp    %r12d,(%rax)
  jne    L22
  mov    D31(%rip),%rax
  mov    %rbp,%rdi
  mov    (%rax),%esi
  call   <inflate@plt>
  mov    %eax,%r12d
  mov    D11(%rip),%rax
  cmp    %r12d,(%rax)
  je     L1
L22:
  mov    %rbp,%rdi
  call   <inflateEnd@plt>
L34:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x58,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L1:
  mov    %rbp,%rdi
  mov    0x8(%rsp),%r12d
  call   <inflateEnd@plt>
  jmp    L34
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ungz(char *dst, const uint32_t dstLen, char *src, const uint32_t srcLen) {
  int err = -1;
  int ret = -1;
  z_stream strm = {0};
  strm.total_in = strm.avail_in = srcLen;
  strm.total_out = strm.avail_out = dstLen;
  strm.next_in = (unsigned char *)src;
  strm.next_out = (unsigned char *)dst;
  strm.zalloc = Z_NULL;
  strm.zfree = Z_NULL;
  strm.opaque = Z_NULL;
  err = inflateInit2(&strm, 47);
  if (err == Z_OK) {
    err = inflate(&strm, Z_FINISH);
    if (err == Z_STREAM_END)
      ret = strm.total_out;
    else {
      inflateEnd(&strm);
      return err;
    }
  } else {
    inflateEnd(&strm);
    return err;
  }
  inflateEnd(&strm);
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ungz>:
  endbr64
  push   %r12
  movd   %esi,%xmm1
  movd   %ecx,%xmm0
  movq   %rdi,%xmm2
  push   %rbp
  punpckldq %xmm1,%xmm1
  punpckldq %xmm0,%xmm0
  mov    $0x2f,%esi
  punpcklqdq %xmm1,%xmm0
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    D15(%rip),%rax
  movaps %xmm0,(%rsp)
  mov    %rsp,%rbp
  movq   %rdx,%xmm0
  mov    %rbp,%rdi
  movq   $0x0,0x38(%rsp)
  mov    (%rax),%rax
  punpcklqdq %xmm2,%xmm0
  movaps %xmm0,0x10(%rsp)
  movq   %rax,%xmm0
  mov    %rax,0x30(%rsp)
  punpcklqdq %xmm0,%xmm0
  movaps %xmm0,0x20(%rsp)
  call   <inflateInit2@plt>
  mov    %eax,%r12d
  mov    D20(%rip),%rax
  cmp    %r12d,(%rax)
  jne    L8
  mov    D26(%rip),%rax
  mov    %rbp,%rdi
  mov    (%rax),%esi
  call   <inflate@plt>
  mov    %eax,%r12d
  mov    D0(%rip),%rax
  cmp    %r12d,(%rax)
  je     L13
L8:
  mov    %rbp,%rdi
  call   <inflateEnd@plt>
L22:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x58,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L13:
  mov    %rbp,%rdi
  mov    0x8(%rsp),%r12d
  call   <inflateEnd@plt>
  jmp    L22
L31:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ungz(char *dst, const uint32_t dstLen, char *src, const uint32_t srcLen) {
  int err = -1;
  int ret = -1;
  z_stream strm = {0};
  strm.total_in = strm.avail_in = srcLen;
  strm.total_out = strm.avail_out = dstLen;
  strm.next_in = (unsigned char *)src;
  strm.next_out = (unsigned char *)dst;
  strm.zalloc = Z_NULL;
  strm.zfree = Z_NULL;
  strm.opaque = Z_NULL;
  err = inflateInit2(&strm, 47);
  if (err == Z_OK) {
    err = inflate(&strm, Z_FINISH);
    if (err == Z_STREAM_END)
      ret = strm.total_out;
    else {
      inflateEnd(&strm);
      return err;
    }
  } else {
    inflateEnd(&strm);
    return err;
  }
  inflateEnd(&strm);
  return ret;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<max>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L17
L1:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  comisd -0x8(%rbp),%xmm0
  jbe    L24
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
L24:
  addl   $0x1,-0xc(%rbp)
L17:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L1
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double max(double value[], int n) {
  int i;
  double max = value[0];
  for (i = 0; i < n; i++)
    if (max < value[i])
      max = value[i];
  return (max);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<max>:
  endbr64
  movsd  (%rdi),%xmm0
  test   %esi,%esi
  jle    L14
  add    $0x8,%rdi
  lea    -0x1(%rsi),%eax
  lea    (%rdi,%rax,8),%rax
  jmp    L28
L4:
  movsd  (%rdi),%xmm1
  maxsd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
  add    $0x8,%rdi
L28:
  cmp    %rax,%rdi
  jne    L4
L14:
  ret
```

assistant: ```c
double max(double value[], int n) {
  int i;
  double max = value[0];
  for (i = 0; i < n; i++)
    if (max < value[i])
      max = value[i];
  return (max);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<max>:
  endbr64
  movsd  (%rdi),%xmm0
  test   %esi,%esi
  jle    L6
  add    $0x8,%rdi
  lea    -0x1(%rsi),%eax
  lea    (%rdi,%rax,8),%rax
  jmp    L19
  nopl   0x0(%rax)
L3:
  movsd  (%rdi),%xmm1
  add    $0x8,%rdi
  maxsd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
L19:
  cmp    %rax,%rdi
  jne    L3
L6:
  ret
```

assistant: ```c
double max(double value[], int n) {
  int i;
  double max = value[0];
  for (i = 0; i < n; i++)
    if (max < value[i])
      max = value[i];
  return (max);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<max>:
  endbr64
  movsd  (%rdi),%xmm0
  test   %esi,%esi
  jle    L13
  cmp    $0x1,%esi
  je     L13
  lea    -0x2(%rsi),%edx
  lea    0x8(%rdi),%rax
  lea    0x10(%rdi,%rdx,8),%rdx
  nopl   (%rax)
L12:
  movsd  (%rax),%xmm1
  add    $0x8,%rax
  maxsd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
  cmp    %rdx,%rax
  jne    L12
L13:
  ret
```

assistant: ```c
double max(double value[], int n) {
  int i;
  double max = value[0];
  for (i = 0; i < n; i++)
    if (max < value[i])
      max = value[i];
  return (max);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ADCIntRegister>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  cmpq   $0x40038000,-0x18(%rbp)
  je     L0
  cmpq   $0x40039000,-0x18(%rbp)
  jne    L24
L0:
  mov    $0x1,%eax
  jmp    L30
L24:
  mov    $0x0,%eax
L30:
  mov    %eax,%edi
  call   <DEBUG_ASSERT@plt>
  cmpq   $0x3,-0x20(%rbp)
  setbe  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <DEBUG_ASSERT@plt>
  cmpq   $0x40038000,-0x18(%rbp)
  jne    L33
  mov    -0x20(%rbp),%rax
  add    $0x1e,%rax
  jmp    L42
L33:
  mov    -0x20(%rbp),%rax
  add    $0x40,%rax
L42:
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <IntRegister@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <IntEnable@plt>
  nop
  leave
  ret
```

assistant: ```c
void ADCIntRegister(unsigned long ulBase, unsigned long ulSequenceNum, void (*pfnHandler)(void)) {
  unsigned long ulInt;
  DEBUG_ASSERT((ulBase == 0x40038000) || (ulBase == 0x40039000));
  DEBUG_ASSERT(ulSequenceNum < 4);
  ulInt = ((ulBase == 0x40038000) ? (30 + ulSequenceNum) : (64 + ulSequenceNum));
  IntRegister(ulInt, pfnHandler);
  IntEnable(ulInt);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ADCIntRegister>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r12
  mov    %rdi,%rax
  and    $0xef,%ah
  cmp    $0x40038000,%rax
  sete   %dil
  movzbl %dil,%edi
  call   <DEBUG_ASSERT@plt>
  cmp    $0x3,%rbx
  setbe  %dil
  movzbl %dil,%edi
  call   <DEBUG_ASSERT@plt>
  lea    0x1e(%rbx),%rax
  add    $0x40,%rbx
  cmp    $0x40038000,%rbp
  cmove  %rax,%rbx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <IntRegister@plt>
  mov    %rbx,%rdi
  call   <IntEnable@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void ADCIntRegister(unsigned long ulBase, unsigned long ulSequenceNum, void (*pfnHandler)(void)) {
  unsigned long ulInt;
  DEBUG_ASSERT((ulBase == 0x40038000) || (ulBase == 0x40039000));
  DEBUG_ASSERT(ulSequenceNum < 4);
  ulInt = ((ulBase == 0x40038000) ? (30 + ulSequenceNum) : (64 + ulSequenceNum));
  IntRegister(ulInt, pfnHandler);
  IntEnable(ulInt);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ADCIntRegister>:
  endbr64
  mov    %rdi,%rax
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  and    $0xef,%ah
  mov    %rdi,%rbp
  xor    %edi,%edi
  push   %rbx
  cmp    $0x40038000,%rax
  mov    %rsi,%rbx
  sete   %dil
  call   <DEBUG_ASSERT@plt>
  xor    %edi,%edi
  cmp    $0x3,%rbx
  setbe  %dil
  call   <DEBUG_ASSERT@plt>
  lea    0x1e(%rbx),%rax
  add    $0x40,%rbx
  mov    %r12,%rsi
  cmp    $0x40038000,%rbp
  cmovne %rbx,%rax
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <IntRegister@plt>
  pop    %rbx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1090 <IntEnable@plt>
```

assistant: ```c
void ADCIntRegister(unsigned long ulBase, unsigned long ulSequenceNum, void (*pfnHandler)(void)) {
  unsigned long ulInt;
  DEBUG_ASSERT((ulBase == 0x40038000) || (ulBase == 0x40039000));
  DEBUG_ASSERT(ulSequenceNum < 4);
  ulInt = ((ulBase == 0x40038000) ? (30 + ulSequenceNum) : (64 + ulSequenceNum));
  IntRegister(ulInt, pfnHandler);
  IntEnable(ulInt);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ADCIntRegister>:
  endbr64
  mov    %rdi,%rax
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  and    $0xef,%ah
  mov    %rdi,%rbp
  xor    %edi,%edi
  push   %rbx
  cmp    $0x40038000,%rax
  mov    %rsi,%rbx
  sete   %dil
  call   <DEBUG_ASSERT@plt>
  xor    %edi,%edi
  cmp    $0x3,%rbx
  setbe  %dil
  call   <DEBUG_ASSERT@plt>
  lea    0x1e(%rbx),%rax
  add    $0x40,%rbx
  mov    %r12,%rsi
  cmp    $0x40038000,%rbp
  cmovne %rbx,%rax
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <IntRegister@plt>
  pop    %rbx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1090 <IntEnable@plt>
```

assistant: ```c
void ADCIntRegister(unsigned long ulBase, unsigned long ulSequenceNum, void (*pfnHandler)(void)) {
  unsigned long ulInt;
  DEBUG_ASSERT((ulBase == 0x40038000) || (ulBase == 0x40039000));
  DEBUG_ASSERT(ulSequenceNum < 4);
  ulInt = ((ulBase == 0x40038000) ? (30 + ulSequenceNum) : (64 + ulSequenceNum));
  IntRegister(ulInt, pfnHandler);
  IntEnable(ulInt);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x2,%edi
  call   <fat@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

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

tool: D7: %d


assistant: ```c
void main(void) {
  int m = fat(2);
  printf("%d\n", m);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <fat@plt>
  mov    %eax,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D3
data_type: string

tool: D3: %d


assistant: ```c
void main(void) {
  int m = fat(2);
  printf("%d\n", m);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <fat@plt>
  lea    D3(%rip),%rdi
  add    $0x8,%rsp
  mov    %eax,%esi
  jmp    1060 <printf@plt>
```

tool_calls: name: read_data
arguments: label: D3
type_guessed: string

tool: D3: %d


assistant: ```c
void main(void) {
  int m = fat(2);
  printf("%d\n", m);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <fat@plt>
  lea    D1(%rip),%rdi
  add    $0x8,%rsp
  mov    %eax,%esi
  jmp    1060 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: label: D1
type: string

tool: D1: %d


assistant: ```c
void main(void) {
  int m = fat(2);
  printf("%d\n", m);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<PWMIntEnable>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x14,%rax
  mov    (%rax),%rax
  mov    -0x8(%rbp),%rdx
  add    $0x14,%rdx
  or     -0x10(%rbp),%rax
  mov    %rax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void PWMIntEnable(unsigned long ulBase, unsigned long ulGenFault) {
  ;

  ;
  (*((volatile unsigned long *)(ulBase + 0x00000014))) |= ulGenFault;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<PWMIntEnable>:
  endbr64
  mov    0x14(%rdi),%rax
  or     %rsi,%rax
  mov    %rax,0x14(%rdi)
  ret
```

assistant: ```c
void PWMIntEnable(unsigned long ulBase, unsigned long ulGenFault) {
  ;

  ;
  (*((volatile unsigned long *)(ulBase + 0x00000014))) |= ulGenFault;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<PWMIntEnable>:
  endbr64
  mov    0x14(%rdi),%rax
  or     %rsi,%rax
  mov    %rax,0x14(%rdi)
  ret
```

assistant: ```c
void PWMIntEnable(unsigned long ulBase, unsigned long ulGenFault) {
  ;

  ;
  (*((volatile unsigned long *)(ulBase + 0x00000014))) |= ulGenFault;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<PWMIntEnable>:
  endbr64
  mov    0x14(%rdi),%rax
  or     %rsi,%rax
  mov    %rax,0x14(%rdi)
  ret
```

assistant: ```c
void PWMIntEnable(unsigned long ulBase, unsigned long ulGenFault) {
  ;

  ;
  (*((volatile unsigned long *)(ulBase + 0x00000014))) |= ulGenFault;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x1,-0x4(%rbp)
  movl   $0x2b,-0x4(%rbp)
  addl   $0x1,-0x4(%rbp)
  addl   $0x2,-0x4(%rbp)
  nop
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int a = 1;
  switch (1) {
  case 0:
    a += a + 3;
    a += 1;
    return 0;
  case 1:
    a = 43;
    a += 1;
    a += 2;
    break;
  case 2:
    a = 4;
    break;
  case 3:
    a = 5;
    break;
  default:
    return 4;
  }
  return a;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  mov    $0x2e,%eax
  ret
```

assistant: ```c
int main() {
  int a = 1;
  switch (1) {
  case 0:
    a += a + 3;
    a += 1;
    return 0;
  case 1:
    a = 43;
    a += 1;
    a += 2;
    break;
  case 2:
    a = 4;
    break;
  case 3:
    a = 5;
    break;
  default:
    return 4;
  }
  return a;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  mov    $0x2e,%eax
  ret
```

assistant: ```c
int main() {
  int a = 1;
  switch (1) {
  case 0:
    a += a + 3;
    a += 1;
    return 0;
  case 1:
    a = 43;
    a += 1;
    a += 2;
    break;
  case 2:
    a = 4;
    break;
  case 3:
    a = 5;
    break;
  default:
    return 4;
  }
  return a;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  mov    $0x2e,%eax
  ret
```

assistant: ```c
int main() {
  int a = 1;
  switch (1) {
  case 0:
    a += a + 3;
    a += 1;
    return 0;
  case 1:
    a = 43;
    a += 1;
    a += 2;
    break;
  case 2:
    a = 4;
    break;
  case 3:
    a = 5;
    break;
  default:
    return 4;
  }
  return a;
}
```