ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<muld>:
  endbr64
  movsd  D19(%rip),%xmm4
  comisd %xmm4,%xmm0
  jb     L99
  movsd  D18(%rip),%xmm5
  mov    $0x1,%eax
  comisd %xmm0,%xmm5
  jae    L43
L99:
  comisd %xmm4,%xmm1
  jb     L66
  movsd  D18(%rip),%xmm4
  mov    $0x1,%eax
  comisd %xmm1,%xmm4
  jae    L43
L66:
  pxor   %xmm4,%xmm4
  comisd %xmm0,%xmm4
  ja     L98
L65:
  comisd %xmm4,%xmm0
  jbe    L0
  comisd %xmm1,%xmm4
  ja     L79
L94:
  comisd %xmm4,%xmm1
  jbe    L0
  divsd  %xmm1,%xmm3
  mov    $0x1,%eax
  comisd %xmm0,%xmm3
  jae    L43
  nopl   (%rax)
L0:
  test   %edi,%edi
  je     L13
  sub    $0x8,%rsp
  lea    D44(%rip),%rdi
  call   <warning@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
  xchg   %ax,%ax
L98:
  comisd %xmm1,%xmm4
  ja     L35
L113:
  comisd %xmm4,%xmm1
  jbe    L65
  movapd %xmm2,%xmm5
  mov    $0x1,%eax
  divsd  %xmm1,%xmm5
  comisd %xmm5,%xmm0
  jb     L65
L43:
  ret
L79:
  divsd  %xmm0,%xmm2
  mov    $0x1,%eax
  comisd %xmm2,%xmm1
  jb     L94
  ret
L35:
  movq   D29(%rip),%xmm5
  movapd %xmm0,%xmm6
  mov    $0x1,%eax
  xorpd  %xmm5,%xmm6
  xorpd  %xmm3,%xmm5
  divsd  %xmm1,%xmm5
  comisd %xmm6,%xmm5
  jb     L113
  ret
  nopl   0x0(%rax)
L13:
  xor    %eax,%eax
  ret
```

tool_calls: name: unpack_data
arguments: tag: D19
type_infered: byte[8]
name: unpack_data
arguments: tag: D18
type_infered: byte[8]
name: unpack_data
arguments: tag: D44
type_infered: string
name: unpack_data
arguments: tag: D29
type_infered: byte[8]

tool: {"D19": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0xbf]}

tool: {"D18": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: D44: overflow in constant expression


tool: {"D29": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
int muld(double x, double y, double min, double max, int needconst) {
  int cond = ((x >= -1 && x <= 1) || (y >= -1 && y <= 1) || (x < 0 && y < 0 && -x <= max / -y) || (x < 0 && y > 0 && x >= min / y) || (x > 0 && y < 0 && y >= min / x) || (x > 0 && y > 0 && x <= max / y));
  if (!cond && needconst) {
    warning("overflow in constant expression\n");
    cond = 1;
  }
  return cond;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  jle    L44
  mov    -0x18(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jge    L47
  mov    -0x20(%rbp),%rax
  sub    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L42
L47:
  mov    -0x18(%rbp),%rax
  and    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L42
L44:
  cmpq   $0xffffffffffffffff,-0x20(%rbp)
  jge    L42
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
L42:
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long test(long x, long y) {
  long val = 8 * x;
  if (y > 0) {
    if (x < y)
      val = y - x;
    else
      val = x & y;
  } else if (y <= -2)
    val = x + y;
  return val;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<test>:
  endbr64
  test   %rsi,%rsi
  jle    L6
  mov    %rsi,%rdx
  sub    %rdi,%rdx
  mov    %rdi,%rax
  and    %rsi,%rax
  cmp    %rsi,%rdi
  cmovl  %rdx,%rax
  ret
L6:
  lea    0x0(,%rdi,8),%rax
  add    %rsi,%rdi
  cmp    $0xffffffffffffffff,%rsi
  cmovl  %rdi,%rax
  ret
```

assistant: ```c
long test(long x, long y) {
  long val = 8 * x;
  if (y > 0) {
    if (x < y)
      val = y - x;
    else
      val = x & y;
  } else if (y <= -2)
    val = x + y;
  return val;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<test>:
  endbr64
  test   %rsi,%rsi
  jle    L1
  mov    %rsi,%rdx
  mov    %rdi,%rax
  sub    %rdi,%rdx
  and    %rsi,%rax
  cmp    %rsi,%rdi
  cmovl  %rdx,%rax
  ret
  nopl   (%rax)
L1:
  lea    0x0(,%rdi,8),%rax
  add    %rsi,%rdi
  cmp    $0xffffffffffffffff,%rsi
  cmovl  %rdi,%rax
  ret
```

assistant: ```c
long test(long x, long y) {
  long val = 8 * x;
  if (y > 0) {
    if (x < y)
      val = y - x;
    else
      val = x & y;
  } else if (y <= -2)
    val = x + y;
  return val;
}
```
user: Translate the below assembly code into C programming language:

```asm
<test>:
  endbr64
  test   %rsi,%rsi
  jle    L6
  mov    %rsi,%rdx
  mov    %rdi,%rax
  sub    %rdi,%rdx
  and    %rsi,%rax
  cmp    %rsi,%rdi
  cmovl  %rdx,%rax
  ret
  nopl   (%rax)
L6:
  lea    0x0(,%rdi,8),%rax
  add    %rsi,%rdi
  cmp    $0xffffffffffffffff,%rsi
  cmovl  %rdi,%rax
  ret
```

assistant: ```c
long test(long x, long y) {
  long val = 8 * x;
  if (y > 0) {
    if (x < y)
      val = y - x;
    else
      val = x & y;
  } else if (y <= -2)
    val = x + y;
  return val;
}
```
user: What is the c source code of the assembly code below:

```asm
<bar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    %r8d,-0x24(%rbp)
  mov    %r9d,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rdi
  call   <toto@plt>
  mov    0x18(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
int bar(int a, int b, int c, int d, int e, int f, int g, int h) {
  int res;
  toto(&res);
  return h;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<bar>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <toto@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    0x28(%rsp),%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bar(int a, int b, int c, int d, int e, int f, int g, int h) {
  int res;
  toto(&res);
  return h;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<bar>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <toto@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    0x28(%rsp),%eax
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bar(int a, int b, int c, int d, int e, int f, int g, int h) {
  int res;
  toto(&res);
  return h;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<bar>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <toto@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    0x28(%rsp),%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bar(int a, int b, int c, int d, int e, int f, int g, int h) {
  int res;
  toto(&res);
  return h;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<transpose>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %rdx,-0x30(%rbp)
  mov    %rcx,-0x38(%rbp)
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x10(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  mov    -0x24(%rbp),%edx
  movslq %edx,%rcx
  sub    $0x1,%rcx
  mov    %rcx,-0x8(%rbp)
  movslq %edx,%rcx
  mov    %rcx,%r8
  mov    $0x0,%r9d
  movl   $0x0,-0x18(%rbp)
  jmp    L4
L19:
  movl   $0x0,-0x14(%rbp)
  jmp    L14
L10:
  mov    -0x18(%rbp),%ecx
  movslq %ecx,%rsi
  movslq %eax,%rcx
  imul   %rsi,%rcx
  lea    0x0(,%rcx,8),%rsi
  mov    -0x30(%rbp),%rcx
  lea    (%rsi,%rcx,1),%rdi
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rsi
  movslq %edx,%rcx
  imul   %rsi,%rcx
  lea    0x0(,%rcx,8),%rsi
  mov    -0x38(%rbp),%rcx
  add    %rcx,%rsi
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rcx
  movsd  (%rdi,%rcx,8),%xmm0
  mov    -0x18(%rbp),%ecx
  movslq %ecx,%rcx
  movsd  %xmm0,(%rsi,%rcx,8)
  addl   $0x1,-0x14(%rbp)
L14:
  mov    -0x14(%rbp),%ecx
  cmp    -0x28(%rbp),%ecx
  jl     L10
  addl   $0x1,-0x18(%rbp)
L4:
  mov    -0x18(%rbp),%ecx
  cmp    -0x24(%rbp),%ecx
  jl     L19
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void transpose(int rows, int cols, double m[rows][cols], double m_trans[cols][rows]) {
  int i, j;
  for (i = 0; i < rows; i++)
    for (j = 0; j < cols; j++)
      m_trans[j][i] = m[i][j];
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<transpose>:
  endbr64
  mov    %edi,%r11d
  mov    %esi,%r9d
  movslq %esi,%r10
  movslq %edi,%rax
  test   %edi,%edi
  jle    L31
  mov    %rcx,%rdi
  shl    $0x3,%r10
  lea    -0x1(%rsi),%r8d
  lea    0x8(%rdx,%r8,8),%rsi
  lea    -0x1(%r11),%edx
  lea    0x8(%rcx,%rdx,8),%r11
  lea    0x0(,%rax,8),%rcx
  not    %r8
  shl    $0x3,%r8
  jmp    L11
L33:
  movsd  (%rax),%xmm0
  movsd  %xmm0,(%rdx)
  add    $0x8,%rax
  add    %rcx,%rdx
  cmp    %rsi,%rax
  jne    L33
L48:
  add    $0x8,%rdi
  add    %r10,%rsi
  cmp    %r11,%rdi
  je     L31
L11:
  lea    (%r8,%rsi,1),%rax
  mov    %rdi,%rdx
  test   %r9d,%r9d
  jg     L33
  jmp    L48
L31:
  ret
```

assistant: ```c
void transpose(int rows, int cols, double m[rows][cols], double m_trans[cols][rows]) {
  int i, j;
  for (i = 0; i < rows; i++)
    for (j = 0; j < cols; j++)
      m_trans[j][i] = m[i][j];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<transpose>:
  endbr64
  movslq %esi,%r9
  mov    %edi,%r10d
  movslq %edi,%rsi
  mov    %r9,%r8
  test   %edi,%edi
  jle    L17
  lea    -0x1(%r8),%eax
  shl    $0x3,%r9
  mov    %rcx,%rdi
  shl    $0x3,%rsi
  shl    $0x3,%rax
  lea    0x8(%rdx,%rax,1),%rcx
  lea    -0x1(%r10),%edx
  mov    $0xfffffffffffffff8,%r10
  lea    0x8(%rdi,%rdx,8),%r11
  sub    %rax,%r10
  nop
L26:
  lea    (%r10,%rcx,1),%rax
  mov    %rdi,%rdx
  test   %r8d,%r8d
  jle    L18
  nopl   0x0(%rax)
L28:
  movsd  (%rax),%xmm0
  add    $0x8,%rax
  movsd  %xmm0,(%rdx)
  add    %rsi,%rdx
  cmp    %rcx,%rax
  jne    L28
L18:
  add    $0x8,%rdi
  add    %r9,%rcx
  cmp    %r11,%rdi
  jne    L26
L17:
  ret
```

assistant: ```c
void transpose(int rows, int cols, double m[rows][cols], double m_trans[cols][rows]) {
  int i, j;
  for (i = 0; i < rows; i++)
    for (j = 0; j < cols; j++)
      m_trans[j][i] = m[i][j];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<transpose>:
  endbr64
  movslq %edi,%rax
  movslq %esi,%r8
  test   %eax,%eax
  jle    L39
  test   %esi,%esi
  jle    L39
  lea    -0x1(%rsi),%r11d
  mov    %rcx,%rdi
  shl    $0x3,%r8
  mov    $0xfffffffffffffff8,%r9
  shl    $0x3,%r11
  lea    0x0(,%rax,8),%rsi
  lea    0x8(%rdx,%r11,1),%rcx
  lea    -0x1(%rax),%edx
  sub    %r11,%r9
  lea    0x8(%rdi,%rdx,8),%r10
L19:
  lea    (%r9,%rcx,1),%rax
  mov    %rdi,%rdx
  nopw   0x0(%rax,%rax,1)
L24:
  movsd  (%rax),%xmm0
  add    $0x8,%rax
  movsd  %xmm0,(%rdx)
  add    %rsi,%rdx
  cmp    %rcx,%rax
  jne    L24
  add    $0x8,%rdi
  add    %r8,%rcx
  cmp    %r10,%rdi
  jne    L19
L39:
  ret
```

assistant: ```c
void transpose(int rows, int cols, double m[rows][cols], double m_trans[cols][rows]) {
  int i, j;
  for (i = 0; i < rows; i++)
    for (j = 0; j < cols; j++)
      m_trans[j][i] = m[i][j];
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<_calc_tokenThreshold>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x4(%rbp)
  mov    %al,-0x8(%rbp)
  cmpb   $0x5f,-0x4(%rbp)
  je     L17
  cmpb   $0x2f,-0x4(%rbp)
  jle    L39
  cmpb   $0x39,-0x4(%rbp)
  jg     L39
L17:
  cmpb   $0x2f,-0x8(%rbp)
  jle    L39
  cmpb   $0x39,-0x8(%rbp)
  jg     L39
  mov    $0x0,%eax
  jmp    L56
L39:
  mov    $0x1,%eax
L56:
  pop    %rbp
  ret
```

assistant: ```c
bool _calc_tokenThreshold(char c1, char c2) {
  if ((c1 == '_' || (c1 >= '0' && c1 <= '9')) && (c2 >= '0' && c2 <= '9'))
    return false;
  return true;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<_calc_tokenThreshold>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L6
  mov    $0x1,%eax
  cmp    $0x5f,%dil
  jne    L8
L6:
  sub    $0x30,%esi
  cmp    $0x9,%sil
  seta   %al
  movzbl %al,%eax
L8:
  ret
```

assistant: ```c
bool _calc_tokenThreshold(char c1, char c2) {
  if ((c1 == '_' || (c1 >= '0' && c1 <= '9')) && (c2 >= '0' && c2 <= '9'))
    return false;
  return true;
}
```
user: Translate the below assembly code into C programming language:

```asm
<_calc_tokenThreshold>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L5
  mov    $0x1,%eax
  cmp    $0x5f,%dil
  je     L5
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  sub    $0x30,%esi
  xor    %eax,%eax
  cmp    $0x9,%sil
  seta   %al
  ret
```

assistant: ```c
bool _calc_tokenThreshold(char c1, char c2) {
  if ((c1 == '_' || (c1 >= '0' && c1 <= '9')) && (c2 >= '0' && c2 <= '9'))
    return false;
  return true;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<_calc_tokenThreshold>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L16
  mov    $0x1,%eax
  cmp    $0x5f,%dil
  je     L16
  ret
  nopw   0x0(%rax,%rax,1)
L16:
  sub    $0x30,%esi
  xor    %eax,%eax
  cmp    $0x9,%sil
  seta   %al
  ret
```

assistant: ```c
bool _calc_tokenThreshold(char c1, char c2) {
  if ((c1 == '_' || (c1 >= '0' && c1 <= '9')) && (c2 >= '0' && c2 <= '9'))
    return false;
  return true;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xc9d,-0x8(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0x8(%rbp),%xmm0
  movss  D15(%rip),%xmm1
  divss  %xmm1,%xmm0
  movss  %xmm0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movd   %eax,%xmm0
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D15
type_infered: byte[4]
name: disassemble_data
arguments: label: D10
type_infered: string

tool: {"D15": [0x00,0x00,0x7a,0x44]}

tool: D10: %f


assistant: ```c
int main() {
  float a;
  int b = 3229;
  a = (float)(b) / 1000;
  printf("%f\n", a);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  movss  D17(%rip),%xmm0
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D17
datatype: byte[4]
name: unpack_data
arguments: data_tag: D16
datatype: string

tool: {"D17": [0xf0,0xa7,0x4e,0x40]}

tool: D16: %f


assistant: ```c
int main() {
  float a;
  int b = 3229;
  a = (float)(b) / 1000;
  printf("%f\n", a);
  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
<main>:
  endbr64
  sub    $0x8,%rsp
  movss  D1(%rip),%xmm0
  lea    D13(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D1
type_infered: byte[4]
name: parse_data
arguments: data_tag: D13
type_infered: string

tool: {"D1": [0xf0,0xa7,0x4e,0x40]}

tool: D13: %f


assistant: ```c
int main() {
  float a;
  int b = 3229;
  a = (float)(b) / 1000;
  printf("%f\n", a);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  movss  D7(%rip),%xmm0
  lea    D15(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D7
datatype: byte[4]
name: disassemble_data
arguments: data_label: D15
datatype: string

tool: {"D7": [0xf0,0xa7,0x4e,0x40]}

tool: D15: %f


assistant: ```c
int main() {
  float a;
  int b = 3229;
  a = (float)(b) / 1000;
  printf("%f\n", a);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<rop3_253_0_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  not    %eax
  mov    %al,-0x3(%rbp)
  movzbl -0x1c(%rbp),%eax
  or     -0x3(%rbp),%al
  mov    %al,-0x2(%rbp)
  movzbl -0x20(%rbp),%eax
  or     -0x2(%rbp),%al
  mov    %al,-0x1(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl -0x1(%rbp),%edx
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rop3_253_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = ~*D;
  stk2 = S | stk3;
  stk1 = T | stk2;
  *D = stk1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<rop3_253_0_0>:
  endbr64
  movzbl (%rdi),%eax
  not    %eax
  or     %esi,%edx
  or     %edx,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rop3_253_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = ~*D;
  stk2 = S | stk3;
  stk1 = T | stk2;
  *D = stk1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<rop3_253_0_0>:
  endbr64
  movzbl (%rdi),%eax
  or     %esi,%edx
  not    %eax
  or     %edx,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rop3_253_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = ~*D;
  stk2 = S | stk3;
  stk1 = T | stk2;
  *D = stk1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rop3_253_0_0>:
  endbr64
  movzbl (%rdi),%eax
  or     %esi,%edx
  not    %eax
  or     %edx,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rop3_253_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = ~*D;
  stk2 = S | stk3;
  stk1 = T | stk2;
  *D = stk1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<decrypt>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    %edx,%ebp
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %eax,%r13d
  mov    %r12,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L97
  mov    %eax,%edi
  mov    %r12,%rax
  mov    $0x0,%esi
  mov    $0x0,%edx
  jmp    L85
L79:
  add    %r13d,%esi
  jmp    L74
L18:
  mov    %edx,%ecx
  sub    %esi,%ecx
  movslq %ecx,%rcx
  movsbl (%rbx,%rcx,1),%r8d
  sub    $0x41,%r8d
  movslq %r8d,%rcx
  imul   $0x4ec4ec4f,%rcx,%rcx
  sar    $0x23,%rcx
  mov    %r8d,%r9d
  sar    $0x1f,%r9d
  sub    %r9d,%ecx
  imul   $0x1a,%ecx,%ecx
  sub    %ecx,%r8d
  add    %r8b,(%rax)
  jmp    L73
L93:
  cmp    $0x40,%cl
  jg     L2
  add    $0x1a,%ecx
  mov    %cl,(%rax)
L2:
  add    $0x1,%edx
  add    $0x1,%rax
  cmp    %edx,%edi
  je     L97
L85:
  mov    %edx,%ecx
  sub    %esi,%ecx
  movslq %ecx,%rcx
  cmpb   $0xa,(%rbx,%rcx,1)
  je     L79
L74:
  test   %ebp,%ebp
  je     L18
  mov    %edx,%ecx
  sub    %esi,%ecx
  movslq %ecx,%rcx
  movsbl (%rbx,%rcx,1),%r8d
  sub    $0x41,%r8d
  movslq %r8d,%rcx
  imul   $0x4ec4ec4f,%rcx,%rcx
  sar    $0x23,%rcx
  mov    %r8d,%r9d
  sar    $0x1f,%r9d
  sub    %r9d,%ecx
  imul   $0x1a,%ecx,%ecx
  sub    %ecx,%r8d
  sub    %r8b,(%rax)
L73:
  movzbl (%rax),%ecx
  cmp    $0x5a,%cl
  jle    L93
  sub    $0x1a,%ecx
  mov    %cl,(%rax)
  jmp    L2
L97:
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *decrypt(char *result, char *key, int deen) {
  int i = 0, j = 0;
  int ln = strlen(key), ln2 = strlen(result);
  while (i < ln2) {
    if (key[i - j] == '\n')
      j += ln;
    if (deen)
      result[i] -= (key[i - j] - 65) % 26;
    else
      result[i] += (key[i - j] - 65) % 26;
    if (result[i] > 'Z')
      result[i] -= 26;
    else if (result[i] < 'A')
      result[i] += 26;
    i++;
  }
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<decrypt>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  mov    %rsi,%rdi
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%r13d
  call   <strlen@plt>
  test   %eax,%eax
  jle    L15
  movslq %eax,%rdi
  xor    %ecx,%ecx
  xor    %esi,%esi
  jmp    L0
  nopw   0x0(%rax,%rax,1)
L25:
  sub    $0x1a,%eax
  mov    %al,(%r12,%rcx,1)
  add    $0x1,%rcx
  cmp    %rdi,%rcx
  je     L15
L0:
  mov    %ecx,%eax
  sub    %esi,%eax
  cltq
  movsbl (%rbx,%rax,1),%eax
  cmp    $0xa,%al
  jne    L24
  add    %r13d,%esi
  mov    %ecx,%eax
  sub    %esi,%eax
  cltq
  movsbl (%rbx,%rax,1),%eax
L24:
  sub    $0x41,%eax
  movzbl (%r12,%rcx,1),%edx
  movslq %eax,%r8
  mov    %eax,%r9d
  imul   $0x4ec4ec4f,%r8,%r8
  sar    $0x1f,%r9d
  sar    $0x23,%r8
  sub    %r9d,%r8d
  imul   $0x1a,%r8d,%r8d
  sub    %r8d,%eax
  mov    %edx,%r8d
  sub    %eax,%r8d
  add    %edx,%eax
  test   %ebp,%ebp
  cmovne %r8d,%eax
  cmp    $0x5a,%al
  jg     L25
  cmp    $0x40,%al
  lea    0x1a(%rax),%edx
  cmovle %edx,%eax
  mov    %al,(%r12,%rcx,1)
  add    $0x1,%rcx
  cmp    %rdi,%rcx
  jne    L0
L15:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *decrypt(char *result, char *key, int deen) {
  int i = 0, j = 0;
  int ln = strlen(key), ln2 = strlen(result);
  while (i < ln2) {
    if (key[i - j] == '\n')
      j += ln;
    if (deen)
      result[i] -= (key[i - j] - 65) % 26;
    else
      result[i] += (key[i - j] - 65) % 26;
    if (result[i] > 'Z')
      result[i] -= 26;
    else if (result[i] < 'A')
      result[i] += 26;
    i++;
  }
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<decrypt>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rdi,%r12
  mov    %rsi,%rdi
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%ebp
  call   <strlen@plt>
  test   %eax,%eax
  jle    L64
  movslq %eax,%rsi
  xor    %ecx,%ecx
  test   %r13d,%r13d
  je     L73
  xor    %edi,%edi
  jmp    L35
  nop
L28:
  sub    $0x1a,%edx
  mov    %dl,(%r12,%rcx,1)
  add    $0x1,%rcx
  cmp    %rcx,%rsi
  je     L64
L35:
  mov    %ecx,%eax
  sub    %edi,%eax
  cltq
  movsbl (%rbx,%rax,1),%eax
  cmp    $0xa,%al
  jne    L78
  add    %ebp,%edi
  mov    %ecx,%eax
  sub    %edi,%eax
  cltq
  movsbl (%rbx,%rax,1),%eax
L78:
  sub    $0x41,%eax
  movslq %eax,%rdx
  mov    %eax,%r8d
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x1f,%r8d
  sar    $0x23,%rdx
  sub    %r8d,%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  movzbl (%r12,%rcx,1),%edx
  sub    %eax,%edx
  cmp    $0x5a,%dl
  jg     L28
  cmp    $0x40,%dl
  lea    0x1a(%rdx),%eax
  cmovle %eax,%edx
  mov    %dl,(%r12,%rcx,1)
  add    $0x1,%rcx
  cmp    %rcx,%rsi
  jne    L35
L64:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L33:
  lea    0x1a(%rax),%edx
  cmp    $0x40,%al
  cmovle %edx,%eax
  mov    %al,(%r12,%rcx,1)
L109:
  add    $0x1,%rcx
  cmp    %rsi,%rcx
  je     L64
L73:
  mov    %ecx,%eax
  sub    %r13d,%eax
  cltq
  movsbl (%rbx,%rax,1),%eax
  cmp    $0xa,%al
  jne    L16
  add    %ebp,%r13d
  mov    %ecx,%eax
  sub    %r13d,%eax
  cltq
  movsbl (%rbx,%rax,1),%eax
L16:
  sub    $0x41,%eax
  movslq %eax,%rdx
  mov    %eax,%edi
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x1f,%edi
  sar    $0x23,%rdx
  sub    %edi,%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  add    (%r12,%rcx,1),%al
  cmp    $0x5a,%al
  jle    L33
  sub    $0x1a,%eax
  mov    %al,(%r12,%rcx,1)
  jmp    L109
```

assistant: ```c
char *decrypt(char *result, char *key, int deen) {
  int i = 0, j = 0;
  int ln = strlen(key), ln2 = strlen(result);
  while (i < ln2) {
    if (key[i - j] == '\n')
      j += ln;
    if (deen)
      result[i] -= (key[i - j] - 65) % 26;
    else
      result[i] += (key[i - j] - 65) % 26;
    if (result[i] > 'Z')
      result[i] -= 26;
    else if (result[i] < 'A')
      result[i] += 26;
    i++;
  }
  return result;
}
```