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
<call_passwordbox>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r12
  mov    %rsi,%rdi
  call   <arg_rest@plt>
  mov    %eax,(%r12)
  mov    $0x0,%edx
  mov    $0x4,%esi
  mov    %rbx,%rdi
  call   <optional_str@plt>
  mov    %eax,%r13d
  mov    $0x3,%esi
  mov    %rbx,%rdi
  call   <numeric_arg@plt>
  mov    %eax,%r12d
  mov    $0x2,%esi
  mov    %rbx,%rdi
  call   <numeric_arg@plt>
  mov    %eax,%edx
  mov    0x8(%rbx),%rsi
  mov    $0x1,%r9d
  mov    %r13d,%r8d
  mov    %r12d,%ecx
  mov    %rbp,%rdi
  call   <dialog_inputbox@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int call_passwordbox(const char *t, char *av[], int *offset_add) {
  *offset_add = arg_rest(av);
  return dialog_inputbox(t, av[1], numeric_arg(av, 2), numeric_arg(av, 3), optional_str(av, 4, 0), 1);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<call_passwordbox>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <arg_rest@plt>
  xor    %edx,%edx
  mov    %rbx,%rdi
  mov    $0x4,%esi
  mov    %eax,(%r12)
  call   <optional_str@plt>
  mov    %rbx,%rdi
  mov    $0x3,%esi
  mov    %eax,%r13d
  call   <numeric_arg@plt>
  mov    %rbx,%rdi
  mov    $0x2,%esi
  mov    %eax,%r12d
  call   <numeric_arg@plt>
  mov    0x8(%rbx),%rsi
  add    $0x8,%rsp
  mov    %r13d,%r8d
  pop    %rbx
  mov    %r12d,%ecx
  mov    %rbp,%rdi
  mov    %eax,%edx
  pop    %rbp
  mov    $0x1,%r9d
  pop    %r12
  pop    %r13
  jmp    10a0 <dialog_inputbox@plt>
```

assistant: ```c
int call_passwordbox(const char *t, char *av[], int *offset_add) {
  *offset_add = arg_rest(av);
  return dialog_inputbox(t, av[1], numeric_arg(av, 2), numeric_arg(av, 3), optional_str(av, 4, 0), 1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<call_passwordbox>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <arg_rest@plt>
  xor    %edx,%edx
  mov    %rbx,%rdi
  mov    $0x4,%esi
  mov    %eax,(%r12)
  call   <optional_str@plt>
  mov    %rbx,%rdi
  mov    $0x3,%esi
  mov    %eax,%r13d
  call   <numeric_arg@plt>
  mov    %rbx,%rdi
  mov    $0x2,%esi
  mov    %eax,%r12d
  call   <numeric_arg@plt>
  mov    0x8(%rbx),%rsi
  add    $0x8,%rsp
  mov    %r13d,%r8d
  pop    %rbx
  mov    %r12d,%ecx
  mov    %rbp,%rdi
  mov    %eax,%edx
  pop    %rbp
  mov    $0x1,%r9d
  pop    %r12
  pop    %r13
  jmp    10a0 <dialog_inputbox@plt>
```

assistant: ```c
int call_passwordbox(const char *t, char *av[], int *offset_add) {
  *offset_add = arg_rest(av);
  return dialog_inputbox(t, av[1], numeric_arg(av, 2), numeric_arg(av, 3), optional_str(av, 4, 0), 1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<TestByteOrder>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movw   $0x1,-0x12(%rbp)
  lea    -0x12(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  setne  %al
  movzbl %al,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int TestByteOrder() {
  short int word = 0x0001;
  char *byte = (char *)&word;
  return (byte[0] ? 1 : 0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<TestByteOrder>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int TestByteOrder() {
  short int word = 0x0001;
  char *byte = (char *)&word;
  return (byte[0] ? 1 : 0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<TestByteOrder>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int TestByteOrder() {
  short int word = 0x0001;
  char *byte = (char *)&word;
  return (byte[0] ? 1 : 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<TestByteOrder>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int TestByteOrder() {
  short int word = 0x0001;
  char *byte = (char *)&word;
  return (byte[0] ? 1 : 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<suma>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x28(%rbp),%rax
  movl   $0x0,(%rax)
  movl   $0x0,-0x8(%rbp)
  jmp    L58
L8:
  movl   $0x0,-0x4(%rbp)
  jmp    L32
L26:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jle    L3
  mov    -0x28(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rcx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  add    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
L3:
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x1c(%rbp)
  jg     L4
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0xc(%rbp)
L4:
  addl   $0x1,-0x4(%rbp)
L32:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L26
  addl   $0x1,-0x8(%rbp)
L58:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L8
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int suma(int **a, int n, int m, int *sp) {
  int ss = 0, i, j;
  *sp = 0;
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) {
      if (i > j)
        *sp += a[i][j];
      if (i + j > n - 1)
        ss += a[i][j];
    }
  return ss;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<suma>:
  endbr64
  mov    %esi,%r10d
  movl   $0x0,(%rcx)
  test   %esi,%esi
  jle    L64
  mov    $0x0,%r8d
  mov    $0x0,%r11d
  jmp    L52
L82:
  add    $0x1,%eax
  add    $0x4,%rsi
  cmp    %eax,%edx
  je     L20
L57:
  cmp    %eax,%r8d
  jle    L58
  mov    (%rdi),%r9
  mov    (%r9,%rsi,1),%r9d
  add    %r9d,(%rcx)
L58:
  lea    (%rax,%r8,1),%r9d
  cmp    %r10d,%r9d
  jl     L82
  mov    (%rdi),%r9
  add    (%r9,%rsi,1),%r11d
  jmp    L82
L20:
  add    $0x1,%r8d
  add    $0x8,%rdi
  cmp    %r8d,%r10d
  je     L11
L52:
  mov    $0x0,%esi
  mov    $0x0,%eax
  test   %edx,%edx
  jg     L57
  jmp    L20
L64:
  mov    $0x0,%r11d
L11:
  mov    %r11d,%eax
  ret
```

assistant: ```c
int suma(int **a, int n, int m, int *sp) {
  int ss = 0, i, j;
  *sp = 0;
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) {
      if (i > j)
        *sp += a[i][j];
      if (i + j > n - 1)
        ss += a[i][j];
    }
  return ss;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<suma>:
  endbr64
  movl   $0x0,(%rcx)
  mov    %esi,%r10d
  test   %esi,%esi
  jle    L12
  xor    %r8d,%r8d
  xor    %r11d,%r11d
  nopw   0x0(%rax,%rax,1)
L40:
  xor    %esi,%esi
  xor    %eax,%eax
  test   %edx,%edx
  jle    L19
  nopl   0x0(%rax,%rax,1)
L25:
  cmp    %eax,%r8d
  jle    L3
  mov    (%rdi),%r9
  mov    (%r9,%rsi,1),%r9d
  add    %r9d,(%rcx)
L3:
  lea    (%rax,%r8,1),%r9d
  cmp    %r10d,%r9d
  jl     L49
  mov    (%rdi),%r9
  add    (%r9,%rsi,1),%r11d
L49:
  add    $0x1,%eax
  add    $0x4,%rsi
  cmp    %eax,%edx
  jne    L25
L19:
  add    $0x1,%r8d
  add    $0x8,%rdi
  cmp    %r8d,%r10d
  jne    L40
  mov    %r11d,%eax
  ret
L12:
  xor    %r11d,%r11d
  mov    %r11d,%eax
  ret
```

assistant: ```c
int suma(int **a, int n, int m, int *sp) {
  int ss = 0, i, j;
  *sp = 0;
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) {
      if (i > j)
        *sp += a[i][j];
      if (i + j > n - 1)
        ss += a[i][j];
    }
  return ss;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<suma>:
  endbr64
  movl   $0x0,(%rcx)
  test   %esi,%esi
  jle    L24
  push   %r13
  mov    %esi,%r8d
  mov    %edx,%r11d
  xor    %esi,%esi
  push   %r12
  xor    %r9d,%r9d
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %rcx,%rbx
  nopl   0x0(%rax)
L18:
  test   %r11d,%r11d
  jle    L99
  test   %esi,%esi
  je     L75
  cmp    %esi,%r11d
  mov    %esi,%r13d
  mov    (%rbx),%ecx
  mov    (%r12),%rbp
  cmovle %r11d,%r13d
  xor    %edx,%edx
  xor    %eax,%eax
  nopl   (%rax)
L77:
  lea    0x0(%rbp,%rdx,1),%rdi
  lea    (%rax,%rsi,1),%r10d
  add    (%rdi),%ecx
  mov    %ecx,(%rbx)
  cmp    %r10d,%r8d
  jg     L59
  add    (%rdi),%r9d
L59:
  add    $0x1,%eax
  add    $0x4,%rdx
  cmp    %r13d,%eax
  jl     L77
  cmp    %eax,%r11d
  jle    L99
L89:
  movslq %eax,%rdx
  shl    $0x2,%rdx
  nopl   (%rax)
L4:
  lea    (%rax,%rsi,1),%ecx
  cmp    %ecx,%r8d
  jg     L63
  mov    (%r12),%rcx
  add    (%rcx,%rdx,1),%r9d
L63:
  add    $0x1,%eax
  add    $0x4,%rdx
  cmp    %eax,%r11d
  jg     L4
L99:
  add    $0x1,%esi
  add    $0x8,%r12
  cmp    %esi,%r8d
  jne    L18
  pop    %rbx
  mov    %r9d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L75:
  xor    %eax,%eax
  jmp    L89
L24:
  xor    %eax,%eax
  ret
```

assistant: ```c
int suma(int **a, int n, int m, int *sp) {
  int ss = 0, i, j;
  *sp = 0;
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) {
      if (i > j)
        *sp += a[i][j];
      if (i + j > n - 1)
        ss += a[i][j];
    }
  return ss;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memset>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L19
L11:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x1c(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L19:
  mov    -0x4(%rbp),%eax
  cltq
  cmp    %rax,-0x28(%rbp)
  ja     L11
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
void *memset(void *ptr, int value, size_t num) {
  int i;
  for (i = 0; i < num; i++)
    ((unsigned char *)ptr)[i] = value;
  return ptr;
}
```
user: What is the c source code of the assembly code below?

```asm
<memset>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L7
  mov    %rdi,%rcx
  add    %rdi,%rdx
L15:
  mov    %sil,(%rcx)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L15
L7:
  ret
```

assistant: ```c
void *memset(void *ptr, int value, size_t num) {
  int i;
  for (i = 0; i < num; i++)
    ((unsigned char *)ptr)[i] = value;
  return ptr;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<memset>:
  endbr64
  test   %rdx,%rdx
  je     L2
  sub    $0x8,%rsp
  movzbl %sil,%esi
  call   <memset@plt>
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L2:
  mov    %rdi,%rax
  ret
```

assistant: ```c
void *memset(void *ptr, int value, size_t num) {
  int i;
  for (i = 0; i < num; i++)
    ((unsigned char *)ptr)[i] = value;
  return ptr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<memset>:
  endbr64
  test   %rdx,%rdx
  je     L9
  sub    $0x8,%rsp
  movzbl %sil,%esi
  call   <memset@plt>
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  mov    %rdi,%rax
  ret
```

assistant: ```c
void *memset(void *ptr, int value, size_t num) {
  int i;
  for (i = 0; i < num; i++)
    ((unsigned char *)ptr)[i] = value;
  return ptr;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xrop3_230_0_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x24(%rbp)
  mov    %al,-0x28(%rbp)
  movl   $0xff,-0x18(%rbp)
  movzbl -0x24(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x24(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x14(%rbp)
  movzbl -0x28(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x28(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  and    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  notl   -0xc(%rbp)
  mov    -0x18(%rbp),%eax
  and    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  xor    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned xrop3_230_0_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = T & S;
  stk3 = ~stk3;
  stk2 = D & stk3;
  stk1 = S ^ stk2;
  return stk1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<xrop3_230_0_0>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  movzbl %sil,%esi
  mov    %esi,%eax
  shl    $0x8,%eax
  or     %esi,%eax
  and    %edi,%eax
  not    %eax
  movzbl %al,%eax
  xor    %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_230_0_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = T & S;
  stk3 = ~stk3;
  stk2 = D & stk3;
  stk1 = S ^ stk2;
  return stk1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xrop3_230_0_0>:
  endbr64
  movzbl %dil,%eax
  movzbl %sil,%esi
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  mov    %esi,%eax
  shl    $0x8,%eax
  or     %esi,%eax
  and    %edi,%eax
  not    %eax
  movzbl %al,%eax
  xor    %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_230_0_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = T & S;
  stk3 = ~stk3;
  stk2 = D & stk3;
  stk1 = S ^ stk2;
  return stk1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<xrop3_230_0_0>:
  endbr64
  movzbl %dil,%eax
  movzbl %sil,%esi
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  mov    %esi,%eax
  shl    $0x8,%eax
  or     %esi,%eax
  and    %edi,%eax
  not    %eax
  movzbl %al,%eax
  xor    %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_230_0_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = T & S;
  stk3 = ~stk3;
  stk2 = D & stk3;
  stk1 = S ^ stk2;
  return stk1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<decode3to2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <codetovalue@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jns    L44
  mov    -0x8(%rbp),%rax
  jmp    L46
L44:
  mov    -0x8(%rbp),%rax
  mov    %eax,-0xc(%rbp)
  shll   $0x6,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <codetovalue@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jns    L18
  mov    -0x8(%rbp),%rax
  jmp    L46
L18:
  mov    -0x8(%rbp),%rax
  or     %eax,-0xc(%rbp)
  shll   $0x4,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <codetovalue@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jns    L3
  mov    -0x8(%rbp),%rax
  jmp    L46
L3:
  mov    -0x8(%rbp),%rax
  sar    $0x2,%eax
  or     %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  sar    $0x8,%eax
  movzbl %al,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    -0xc(%rbp),%eax
  movzbl %al,%eax
  mov    %eax,(%rdx)
  mov    $0x0,%eax
L46:
  leave
  ret
```

assistant: ```c
int decode3to2(const char *src, uint8_t *dest) {
  uint32_t b32 = 0;
  int32_t bits;
  bits = codetovalue(src[0]);
  if (bits < 0) {
    return bits;
  }
  b32 = (uint32_t)bits;
  b32 <<= 6;
  bits = codetovalue(src[1]);
  if (bits < 0) {
    return bits;
  }
  b32 |= (uint32_t)bits;
  b32 <<= 4;
  bits = codetovalue(src[2]);
  if (bits < 0) {
    return bits;
  }
  b32 |= ((uint32_t)bits) >> 2;
  dest[0] = (b32 >> 8) & 0xFF;
  dest[1] = b32 & 0xFF;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<decode3to2>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbp
  movsbl (%rdi),%edi
  call   <codetovalue@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  js     L20
  movsbl 0x1(%r13),%edi
  call   <codetovalue@plt>
  mov    %rax,%r12
  test   %rax,%rax
  js     L20
  movsbl 0x2(%r13),%edi
  call   <codetovalue@plt>
  test   %rax,%rax
  js     L20
  shl    $0x6,%ebx
  or     %r12d,%ebx
  shl    $0x4,%ebx
  sar    $0x2,%eax
  or     %ebx,%eax
  movzbl %ah,%edx
  mov    %edx,0x0(%rbp)
  and    $0xff,%eax
  mov    %eax,0x4(%rbp)
  mov    $0x0,%eax
L20:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int decode3to2(const char *src, uint8_t *dest) {
  uint32_t b32 = 0;
  int32_t bits;
  bits = codetovalue(src[0]);
  if (bits < 0) {
    return bits;
  }
  b32 = (uint32_t)bits;
  b32 <<= 6;
  bits = codetovalue(src[1]);
  if (bits < 0) {
    return bits;
  }
  b32 |= (uint32_t)bits;
  b32 <<= 4;
  bits = codetovalue(src[2]);
  if (bits < 0) {
    return bits;
  }
  b32 |= ((uint32_t)bits) >> 2;
  dest[0] = (b32 >> 8) & 0xFF;
  dest[1] = b32 & 0xFF;
  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
<decode3to2>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  call   <codetovalue@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  js     L5
  movsbl 0x1(%r12),%edi
  call   <codetovalue@plt>
  mov    %rax,%r13
  test   %rax,%rax
  js     L5
  movsbl 0x2(%r12),%edi
  call   <codetovalue@plt>
  test   %rax,%rax
  js     L5
  shl    $0x6,%ebx
  sar    $0x2,%eax
  or     %r13d,%ebx
  shl    $0x4,%ebx
  or     %ebx,%eax
  movzbl %ah,%edx
  and    $0xff,%eax
  mov    %edx,0x0(%rbp)
  mov    %eax,0x4(%rbp)
  xor    %eax,%eax
L5:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int decode3to2(const char *src, uint8_t *dest) {
  uint32_t b32 = 0;
  int32_t bits;
  bits = codetovalue(src[0]);
  if (bits < 0) {
    return bits;
  }
  b32 = (uint32_t)bits;
  b32 <<= 6;
  bits = codetovalue(src[1]);
  if (bits < 0) {
    return bits;
  }
  b32 |= (uint32_t)bits;
  b32 <<= 4;
  bits = codetovalue(src[2]);
  if (bits < 0) {
    return bits;
  }
  b32 |= ((uint32_t)bits) >> 2;
  dest[0] = (b32 >> 8) & 0xFF;
  dest[1] = b32 & 0xFF;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<decode3to2>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  call   <codetovalue@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  js     L24
  movsbl 0x1(%r12),%edi
  call   <codetovalue@plt>
  mov    %rax,%r13
  test   %rax,%rax
  js     L24
  movsbl 0x2(%r12),%edi
  call   <codetovalue@plt>
  test   %rax,%rax
  js     L24
  shl    $0x6,%ebx
  sar    $0x2,%eax
  or     %r13d,%ebx
  shl    $0x4,%ebx
  or     %ebx,%eax
  movzbl %ah,%edx
  and    $0xff,%eax
  mov    %edx,0x0(%rbp)
  mov    %eax,0x4(%rbp)
  xor    %eax,%eax
L24:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int decode3to2(const char *src, uint8_t *dest) {
  uint32_t b32 = 0;
  int32_t bits;
  bits = codetovalue(src[0]);
  if (bits < 0) {
    return bits;
  }
  b32 = (uint32_t)bits;
  b32 <<= 6;
  bits = codetovalue(src[1]);
  if (bits < 0) {
    return bits;
  }
  b32 |= (uint32_t)bits;
  b32 <<= 4;
  bits = codetovalue(src[2]);
  if (bits < 0) {
    return bits;
  }
  b32 |= ((uint32_t)bits) >> 2;
  dest[0] = (b32 >> 8) & 0xFF;
  dest[1] = b32 & 0xFF;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<is_directory>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <stat@plt>
  test   %rax,%rax
  je     L31
  mov    $0x0,%eax
  jmp    L11
L31:
  mov    -0xc(%rbp),%edx
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  and    %edx,%eax
  test   %eax,%eax
  jne    L12
  mov    $0x0,%eax
  jmp    L11
L12:
  mov    $0x1,%eax
L11:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L22
  call   <__stack_chk_fail@plt>
L22:
  leave
  ret
```

assistant: ```c
int is_directory(const char *filename) {
  struct stat s;
  if (stat(filename, &s) != 0) {
    return 0;
  } else if (!(s.st_mode & S_IFDIR)) {
    return 0;
  }
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<is_directory>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  jne    L3
  mov    D3(%rip),%rdx
  mov    0x4(%rsp),%eax
  and    (%rdx),%eax
  setne  %al
  movzbl %al,%eax
L3:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_directory(const char *filename) {
  struct stat s;
  if (stat(filename, &s) != 0) {
    return 0;
  } else if (!(s.st_mode & S_IFDIR)) {
    return 0;
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<is_directory>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L9
  mov    D7(%rip),%rdx
  mov    0x4(%rsp),%eax
  and    (%rdx),%eax
  setne  %al
  movzbl %al,%eax
L9:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_directory(const char *filename) {
  struct stat s;
  if (stat(filename, &s) != 0) {
    return 0;
  } else if (!(s.st_mode & S_IFDIR)) {
    return 0;
  }
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<is_directory>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L11
  mov    D2(%rip),%rdx
  mov    0x4(%rsp),%eax
  and    (%rdx),%eax
  setne  %al
  movzbl %al,%eax
L11:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_directory(const char *filename) {
  struct stat s;
  if (stat(filename, &s) != 0) {
    return 0;
  } else if (!(s.st_mode & S_IFDIR)) {
    return 0;
  }
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<atexit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <on_exit@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int atexit(void (*f)(void)) {
  on_exit(f, 0);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<atexit>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  call   <on_exit@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int atexit(void (*f)(void)) {
  on_exit(f, 0);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<atexit>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  call   <on_exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int atexit(void (*f)(void)) {
  on_exit(f, 0);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<atexit>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  call   <on_exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int atexit(void (*f)(void)) {
  on_exit(f, 0);
  return 0;
}
```