ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<parse_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L67
L48:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5c,%al
  jne    L1
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x78,%al
  je     L25
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x58,%al
  jne    L1
L25:
  mov    -0x10(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <hexval@plt>
  mov    %eax,-0x18(%rbp)
  cmpl   $0xffffffff,-0x18(%rbp)
  je     L1
  mov    -0x10(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <hexval@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0xffffffff,-0x14(%rbp)
  je     L1
  mov    -0x18(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addq   $0x4,-0x10(%rbp)
  jmp    L67
L1:
  mov    -0x10(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L67:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L48
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  sub    -0x30(%rbp),%rax
  leave
  ret
```

assistant: ```c
int parse_string(char *input, char *output) {
  char *c = input;
  char *o = output;
  int h1, h2;
  while (*c) {
    if (c[0] == '\\' && (c[1] == 'x' || c[1] == 'X') && (h2 = hexval(c[2])) != -1 && (h1 = hexval(c[3])) != -1) {
      *o++ = h2 * 16 + h1;
      c += 4;
      continue;
    }
    *o++ = *c++;
  }
  *o = 0;
  return (o - output);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<parse_string>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%r13
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L46
  mov    %rdi,%rbx
  lea    0x1(%rsi),%rbp
  jmp    L69
L76:
  mov    %rbp,%rax
  movzbl (%rbx),%edx
  lea    0x1(%rbx),%rbx
L75:
  mov    %dl,-0x1(%rbp)
  movzbl (%rbx),%edx
  add    $0x1,%rbp
  test   %dl,%dl
  je     L36
L69:
  cmp    $0x5c,%dl
  jne    L76
  movzbl 0x1(%rbx),%eax
  and    $0xffffffdf,%eax
  cmp    $0x58,%al
  jne    L76
  movsbl 0x2(%rbx),%edi
  call   <hexval@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L76
  movsbl 0x3(%rbx),%edi
  call   <hexval@plt>
  mov    %eax,%edx
  cmp    $0xffffffff,%eax
  je     L76
  mov    %rbp,%rax
  shl    $0x4,%r12d
  add    %r12d,%edx
  add    $0x4,%rbx
  jmp    L75
L46:
  mov    %rsi,%rax
L36:
  movb   $0x0,(%rax)
  sub    %r13d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int parse_string(char *input, char *output) {
  char *c = input;
  char *o = output;
  int h1, h2;
  while (*c) {
    if (c[0] == '\\' && (c[1] == 'x' || c[1] == 'X') && (h2 = hexval(c[2])) != -1 && (h1 = hexval(c[3])) != -1) {
      *o++ = h2 * 16 + h1;
      c += 4;
      continue;
    }
    *o++ = *c++;
  }
  *o = 0;
  return (o - output);
}
```
user: What is the c source code of the assembly code below?

```asm
<parse_string>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%eax
  test   %al,%al
  je     L33
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  jmp    L32
L42:
  movzbl (%rbx),%eax
  add    $0x1,%rbx
  mov    %al,-0x1(%rbp)
  movzbl (%rbx),%eax
  test   %al,%al
  je     L55
L32:
  add    $0x1,%rbp
  cmp    $0x5c,%al
  jne    L42
  movzbl 0x1(%rbx),%eax
  and    $0xffffffdf,%eax
  cmp    $0x58,%al
  jne    L42
  movsbl 0x2(%rbx),%edi
  call   <hexval@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L42
  movsbl 0x3(%rbx),%edi
  call   <hexval@plt>
  cmp    $0xffffffff,%eax
  je     L42
  shl    $0x4,%r12d
  add    $0x4,%rbx
  add    %r12d,%eax
  mov    %al,-0x1(%rbp)
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L32
L55:
  movb   $0x0,0x0(%rbp)
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  sub    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L33:
  mov    %rsi,%rbp
  xor    %eax,%eax
  movb   $0x0,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int parse_string(char *input, char *output) {
  char *c = input;
  char *o = output;
  int h1, h2;
  while (*c) {
    if (c[0] == '\\' && (c[1] == 'x' || c[1] == 'X') && (h2 = hexval(c[2])) != -1 && (h1 = hexval(c[3])) != -1) {
      *o++ = h2 * 16 + h1;
      c += 4;
      continue;
    }
    *o++ = *c++;
  }
  *o = 0;
  return (o - output);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<parse_string>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%eax
  test   %al,%al
  je     L83
  mov    %rdi,%rbp
  mov    %rsi,%r12
  jmp    L53
  nopl   0x0(%rax)
L25:
  add    $0x1,%rbp
L89:
  mov    %al,-0x1(%r12)
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L26
L53:
  add    $0x1,%r12
  cmp    $0x5c,%al
  jne    L25
  movzbl 0x1(%rbp),%edx
  and    $0xffffffdf,%edx
  cmp    $0x58,%dl
  jne    L25
  movsbl 0x2(%rbp),%edi
  call   <hexval@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  jne    L75
L23:
  movzbl 0x0(%rbp),%eax
  jmp    L25
  nopl   0x0(%rax,%rax,1)
L26:
  movb   $0x0,(%r12)
  mov    %r12d,%eax
  add    $0x8,%rsp
  pop    %rbx
  sub    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L75:
  movsbl 0x3(%rbp),%edi
  call   <hexval@plt>
  cmp    $0xffffffff,%eax
  je     L23
  shl    $0x4,%ebx
  add    $0x4,%rbp
  add    %ebx,%eax
  jmp    L89
  nopl   0x0(%rax)
L83:
  mov    %rsi,%r12
  xor    %eax,%eax
  movb   $0x0,(%r12)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int parse_string(char *input, char *output) {
  char *c = input;
  char *o = output;
  int h1, h2;
  while (*c) {
    if (c[0] == '\\' && (c[1] == 'x' || c[1] == 'X') && (h2 = hexval(c[2])) != -1 && (h1 = hexval(c[3])) != -1) {
      *o++ = h2 * 16 + h1;
      c += 4;
      continue;
    }
    *o++ = *c++;
  }
  *o = 0;
  return (o - output);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ML_clear_vram>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%eax
  call   <ML_vram_adress@plt>
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  neg    %eax
  and    $0x3,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  cltq
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L32
L25:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  addl   $0x1,-0x1c(%rbp)
L32:
  mov    -0x1c(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L25
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  cltq
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L14
L22:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  addl   $0x1,-0x1c(%rbp)
L14:
  cmpl   $0xfe,-0x1c(%rbp)
  jle    L22
  mov    -0x14(%rbp),%eax
  add    $0x3fc,%eax
  cltq
  add    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  and    $0x3,%eax
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L23
L13:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  addl   $0x1,-0x1c(%rbp)
L23:
  mov    -0x1c(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L13
  nop
  nop
  leave
  ret
```

assistant: ```c
void ML_clear_vram() {
  int i, end, *pointer_long, vram;
  char *pointer_byte;
  vram = (int)ML_vram_adress();
  end = 4 - vram & 3;
  pointer_byte = (char *)vram;
  for (i = 0; i < end; i++)
    pointer_byte[i] = 0;
  pointer_long = (int *)(vram + end);
  for (i = 0; i < 255; i++)
    pointer_long[i] = 0;
  pointer_byte += 1020 + end;
  end = vram & 3;
  for (i = 0; i < end; i++)
    pointer_byte[i] = 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ML_clear_vram>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <ML_vram_adress@plt>
  mov    %eax,%edi
  mov    %eax,%esi
  neg    %esi
  and    $0x3,%esi
  je     L33
  movslq %eax,%rdx
  lea    -0x1(%rsi),%ecx
  lea    0x1(%rdx,%rcx,1),%rcx
L30:
  movb   $0x0,(%rdx)
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L30
L33:
  lea    (%rsi,%rax,1),%edx
  movslq %edx,%rdx
  lea    0x3fc(%rdx),%rcx
L4:
  movl   $0x0,(%rdx)
  add    $0x4,%rdx
  cmp    %rcx,%rdx
  jne    L4
  and    $0x3,%eax
  je     L26
  mov    $0x0,%edx
  movslq %edi,%rdi
  movslq %esi,%rsi
L31:
  lea    (%rdx,%rdi,1),%rcx
  movb   $0x0,0x3fc(%rsi,%rcx,1)
  add    $0x1,%rdx
  cmp    %edx,%eax
  jg     L31
L26:
  add    $0x8,%rsp
  ret
```

assistant: ```c
void ML_clear_vram() {
  int i, end, *pointer_long, vram;
  char *pointer_byte;
  vram = (int)ML_vram_adress();
  end = 4 - vram & 3;
  pointer_byte = (char *)vram;
  for (i = 0; i < end; i++)
    pointer_byte[i] = 0;
  pointer_long = (int *)(vram + end);
  for (i = 0; i < 255; i++)
    pointer_long[i] = 0;
  pointer_byte += 1020 + end;
  end = vram & 3;
  for (i = 0; i < end; i++)
    pointer_byte[i] = 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
<ML_clear_vram>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ML_vram_adress@plt>
  mov    %eax,%edx
  mov    %rax,%rsi
  movslq %eax,%r8
  neg    %edx
  and    $0x3,%edx
  je     L37
  xor    %eax,%eax
L18:
  mov    %eax,%ecx
  add    $0x1,%eax
  movb   $0x0,(%r8,%rcx,1)
  cmp    %edx,%eax
  jb     L18
L37:
  lea    (%rdx,%rsi,1),%eax
  add    $0x3fc,%edx
  cltq
  movslq %edx,%rdx
  lea    0x8(%rax),%rdi
  movq   $0x0,(%rax)
  movq   $0x0,0x3f4(%rax)
  and    $0xfffffffffffffff8,%rdi
  sub    %rdi,%rax
  lea    0x3fc(%rax),%ecx
  xor    %eax,%eax
  shr    $0x3,%ecx
  rep stos %rax,%es:(%rdi)
  mov    %esi,%eax
  and    $0x3,%eax
  je     L35
  add    %rdx,%r8
  xor    %edx,%edx
L9:
  mov    %edx,%ecx
  add    $0x1,%edx
  movb   $0x0,(%r8,%rcx,1)
  cmp    %eax,%edx
  jb     L9
L35:
  add    $0x8,%rsp
  ret
```

assistant: ```c
void ML_clear_vram() {
  int i, end, *pointer_long, vram;
  char *pointer_byte;
  vram = (int)ML_vram_adress();
  end = 4 - vram & 3;
  pointer_byte = (char *)vram;
  for (i = 0; i < end; i++)
    pointer_byte[i] = 0;
  pointer_long = (int *)(vram + end);
  for (i = 0; i < 255; i++)
    pointer_long[i] = 0;
  pointer_byte += 1020 + end;
  end = vram & 3;
  for (i = 0; i < end; i++)
    pointer_byte[i] = 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ML_clear_vram>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ML_vram_adress@plt>
  mov    %eax,%edx
  mov    %rax,%rsi
  movslq %eax,%r8
  neg    %edx
  and    $0x3,%edx
  je     L18
  xor    %eax,%eax
L9:
  mov    %eax,%ecx
  add    $0x1,%eax
  movb   $0x0,(%r8,%rcx,1)
  cmp    %edx,%eax
  jb     L9
L18:
  lea    (%rdx,%rsi,1),%eax
  add    $0x3fc,%edx
  cltq
  movslq %edx,%rdx
  lea    0x8(%rax),%rdi
  movq   $0x0,(%rax)
  movq   $0x0,0x3f4(%rax)
  and    $0xfffffffffffffff8,%rdi
  sub    %rdi,%rax
  lea    0x3fc(%rax),%ecx
  xor    %eax,%eax
  shr    $0x3,%ecx
  rep stos %rax,%es:(%rdi)
  mov    %esi,%eax
  and    $0x3,%eax
  je     L32
  add    %rdx,%r8
  xor    %edx,%edx
L0:
  mov    %edx,%ecx
  add    $0x1,%edx
  movb   $0x0,(%r8,%rcx,1)
  cmp    %eax,%edx
  jb     L0
L32:
  add    $0x8,%rsp
  ret
```

assistant: ```c
void ML_clear_vram() {
  int i, end, *pointer_long, vram;
  char *pointer_byte;
  vram = (int)ML_vram_adress();
  end = 4 - vram & 3;
  pointer_byte = (char *)vram;
  for (i = 0; i < end; i++)
    pointer_byte[i] = 0;
  pointer_long = (int *)(vram + end);
  for (i = 0; i < 255; i++)
    pointer_long[i] = 0;
  pointer_byte += 1020 + end;
  end = vram & 3;
  for (i = 0; i < end; i++)
    pointer_byte[i] = 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<PopCnt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  xor    %rcx,%rcx
  test   %rax,%rax
  je     L4
L17:
  lea    -0x1(%rax),%rdx
  inc    %rcx
  and    %rdx,%rax
  jne    L17
L4:
  mov    %rcx,-0x18(%rbp)
  mov    %rax,-0x10(%rbp)
  mov    %rdx,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
int PopCnt(long word) {
  long dummy, dummy2, dummy3;
  asm("          xorq    %0, %0    "
      "\n\t"
      "          testq   %1, %1    "
      "\n\t"
      "          jz      2f        "
      "\n\t"
      "1:        leaq    -1(%1),%2 "
      "\n\t"
      "          incq    %0        "
      "\n\t"
      "          andq    %2, %1    "
      "\n\t"
      "          jnz     1b        "
      "\n\t"
      "2:                          "
      "\n\t"
      : "=&r"(dummy), "=&r"(dummy2), "=&r"(dummy3)
      : "1"((long)(word))
      : "cc");
  return (dummy);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<PopCnt>:
  endbr64
  xor    %rax,%rax
  test   %rdi,%rdi
  je     L5
L9:
  lea    -0x1(%rdi),%rdx
  inc    %rax
  and    %rdx,%rdi
  jne    L9
L5:
  ret
```

assistant: ```c
int PopCnt(long word) {
  long dummy, dummy2, dummy3;
  asm("          xorq    %0, %0    "
      "\n\t"
      "          testq   %1, %1    "
      "\n\t"
      "          jz      2f        "
      "\n\t"
      "1:        leaq    -1(%1),%2 "
      "\n\t"
      "          incq    %0        "
      "\n\t"
      "          andq    %2, %1    "
      "\n\t"
      "          jnz     1b        "
      "\n\t"
      "2:                          "
      "\n\t"
      : "=&r"(dummy), "=&r"(dummy2), "=&r"(dummy3)
      : "1"((long)(word))
      : "cc");
  return (dummy);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<PopCnt>:
  endbr64
  xor    %rax,%rax
  test   %rdi,%rdi
  je     L5
L0:
  lea    -0x1(%rdi),%rdx
  inc    %rax
  and    %rdx,%rdi
  jne    L0
L5:
  ret
```

assistant: ```c
int PopCnt(long word) {
  long dummy, dummy2, dummy3;
  asm("          xorq    %0, %0    "
      "\n\t"
      "          testq   %1, %1    "
      "\n\t"
      "          jz      2f        "
      "\n\t"
      "1:        leaq    -1(%1),%2 "
      "\n\t"
      "          incq    %0        "
      "\n\t"
      "          andq    %2, %1    "
      "\n\t"
      "          jnz     1b        "
      "\n\t"
      "2:                          "
      "\n\t"
      : "=&r"(dummy), "=&r"(dummy2), "=&r"(dummy3)
      : "1"((long)(word))
      : "cc");
  return (dummy);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<PopCnt>:
  endbr64
  xor    %rax,%rax
  test   %rdi,%rdi
  je     L16
L8:
  lea    -0x1(%rdi),%rdx
  inc    %rax
  and    %rdx,%rdi
  jne    L8
L16:
  ret
```

assistant: ```c
int PopCnt(long word) {
  long dummy, dummy2, dummy3;
  asm("          xorq    %0, %0    "
      "\n\t"
      "          testq   %1, %1    "
      "\n\t"
      "          jz      2f        "
      "\n\t"
      "1:        leaq    -1(%1),%2 "
      "\n\t"
      "          incq    %0        "
      "\n\t"
      "          andq    %2, %1    "
      "\n\t"
      "          jnz     1b        "
      "\n\t"
      "2:                          "
      "\n\t"
      : "=&r"(dummy), "=&r"(dummy2), "=&r"(dummy3)
      : "1"((long)(word))
      : "cc");
  return (dummy);
}
```
user: What is the c source code of the assembly code below:

```asm
<SetPixel>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %rcx,-0x28(%rbp)
  cmpl   $0x7f,-0x14(%rbp)
  jg     L25
  cmpl   $0x3f,-0x18(%rbp)
  jg     L25
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  lea    0x7(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x3,%eax
  shl    $0x7,%eax
  cltq
  shl    $0x2,%rax
  add    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  cltd
  shr    $0x1d,%edx
  add    %edx,%eax
  and    $0x7,%eax
  sub    %edx,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  je     L4
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  or     %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L35
L4:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  not    %eax
  and    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L35
L25:
  nop
L35:
  pop    %rbp
  ret
```

assistant: ```c
void SetPixel(uint32_t xpos, uint32_t ypos, uint32_t val, uint8_t *buffer) {
  uint8_t *ptr;
  uint32_t bit_number;
  if ((xpos >= 128) || (ypos >= 64)) {
    return;
  }
  ptr = buffer + xpos;
  ptr += 128 * (ypos / 8);
  bit_number = ypos % 8;
  if (val) {
    *ptr |= (1 << bit_number);
  } else {
    *ptr &= ~(1 << bit_number);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<SetPixel>:
  endbr64
  cmp    $0x7f,%edi
  jg     L2
  cmp    $0x3f,%esi
  jg     L2
  lea    0x7(%rsi),%eax
  test   %esi,%esi
  cmovns %esi,%eax
  sar    $0x3,%eax
  shl    $0x7,%eax
  cltq
  movslq %edi,%rdi
  add    %rdi,%rax
  lea    (%rcx,%rax,4),%rax
  test   %edx,%edx
  je     L1
  mov    %esi,%ecx
  and    $0x7,%ecx
  mov    $0x1,%edx
  shl    %cl,%edx
  or     %edx,(%rax)
  ret
L1:
  mov    %esi,%ecx
  and    $0x7,%ecx
  mov    $0xfffffffe,%edx
  rol    %cl,%edx
  and    %edx,(%rax)
L2:
  ret
```

assistant: ```c
void SetPixel(uint32_t xpos, uint32_t ypos, uint32_t val, uint8_t *buffer) {
  uint8_t *ptr;
  uint32_t bit_number;
  if ((xpos >= 128) || (ypos >= 64)) {
    return;
  }
  ptr = buffer + xpos;
  ptr += 128 * (ypos / 8);
  bit_number = ypos % 8;
  if (val) {
    *ptr |= (1 << bit_number);
  } else {
    *ptr &= ~(1 << bit_number);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<SetPixel>:
  endbr64
  cmp    $0x7f,%edi
  jg     L15
  cmp    $0x3f,%esi
  jg     L15
  test   %esi,%esi
  lea    0x7(%rsi),%eax
  movslq %edi,%rdi
  cmovns %esi,%eax
  and    $0x7,%esi
  sar    $0x3,%eax
  shl    $0x7,%eax
  cltq
  add    %rdi,%rax
  lea    (%rcx,%rax,4),%rdi
  mov    %esi,%ecx
  mov    $0x1,%eax
  shl    %cl,%eax
  mov    %eax,%ecx
  or     (%rdi),%eax
  not    %ecx
  and    (%rdi),%ecx
  test   %edx,%edx
  cmove  %ecx,%eax
  mov    %eax,(%rdi)
L15:
  ret
```

assistant: ```c
void SetPixel(uint32_t xpos, uint32_t ypos, uint32_t val, uint8_t *buffer) {
  uint8_t *ptr;
  uint32_t bit_number;
  if ((xpos >= 128) || (ypos >= 64)) {
    return;
  }
  ptr = buffer + xpos;
  ptr += 128 * (ypos / 8);
  bit_number = ypos % 8;
  if (val) {
    *ptr |= (1 << bit_number);
  } else {
    *ptr &= ~(1 << bit_number);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<SetPixel>:
  endbr64
  cmp    $0x7f,%edi
  jg     L15
  cmp    $0x3f,%esi
  jg     L15
  test   %esi,%esi
  lea    0x7(%rsi),%eax
  movslq %edi,%rdi
  cmovns %esi,%eax
  and    $0x7,%esi
  sar    $0x3,%eax
  shl    $0x7,%eax
  cltq
  add    %rdi,%rax
  lea    (%rcx,%rax,4),%rdi
  mov    %esi,%ecx
  mov    $0x1,%eax
  shl    %cl,%eax
  mov    %eax,%ecx
  or     (%rdi),%eax
  not    %ecx
  and    (%rdi),%ecx
  test   %edx,%edx
  cmove  %ecx,%eax
  mov    %eax,(%rdi)
L15:
  ret
```

assistant: ```c
void SetPixel(uint32_t xpos, uint32_t ypos, uint32_t val, uint8_t *buffer) {
  uint8_t *ptr;
  uint32_t bit_number;
  if ((xpos >= 128) || (ypos >= 64)) {
    return;
  }
  ptr = buffer + xpos;
  ptr += 128 * (ypos / 8);
  bit_number = ypos % 8;
  if (val) {
    *ptr |= (1 << bit_number);
  } else {
    *ptr &= ~(1 << bit_number);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x6a72666b6a6d7374,%rax
  movabs $0x6f706a666b6c6473,%rdx
  mov    %rax,-0x40(%rbp)
  mov    %rdx,-0x38(%rbp)
  movw   $0x7265,-0x30(%rbp)
  movb   $0x0,-0x2e(%rbp)
  movabs $0x6b73636b666c6473,%rax
  movabs $0x6f7a696f72666a64,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  movw   $0x6f72,-0x10(%rbp)
  movb   $0x0,-0xe(%rbp)
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <memcpy@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x48(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x20(%rbp),%rcx
  mov    -0x48(%rbp),%rax
  mov    $0x12,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %rax,-0x48(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D4
datatype: string
name: disassemble_data
arguments: data_label: D9
datatype: string

tool: D4: str => %s | 

tool: D9: str -> %s


assistant: ```c
int main(int argc, char *argv[]) {
  char tmp[] = "tsmjkfrjsdlkfjpoer";
  char tmp2[] = "sdlfkcskdjfroizoro";
  char *str;
  memcpy(NULL, NULL, 0);
  str = strdup(tmp);
  printf("str => %s | ", str);
  str = memcpy(str, tmp2, sizeof(tmp) - 1);
  printf("str -> %s\n", str);
  free(str);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movabs $0x6a72666b6a6d7374,%rax
  movabs $0x6f706a666b6c6473,%rdx
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  movw   $0x7265,0x10(%rsp)
  movb   $0x0,0x12(%rsp)
  movabs $0x6b73636b666c6473,%rax
  movabs $0x6f7a696f72666a64,%rdx
  mov    %rax,0x20(%rsp)
  mov    %rdx,0x28(%rsp)
  movw   $0x6f72,0x30(%rsp)
  movb   $0x0,0x32(%rsp)
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <memcpy@plt>
  mov    %rsp,%rdi
  call   <strdup@plt>
  mov    %rax,%rbx
  mov    %rax,%rsi
  lea    D17(%rip),%rdi
  call   <printf@plt>
  lea    0x20(%rsp),%rsi
  mov    $0x12,%edx
  mov    %rbx,%rdi
  call   <memcpy@plt>
  mov    %rax,%rbx
  mov    %rax,%rsi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D17
datatype: string
name: extract_symbolic_data
arguments: tag: D6
datatype: string

tool: D17: str => %s | 

tool: D6: str -> %s


assistant: ```c
int main(int argc, char *argv[]) {
  char tmp[] = "tsmjkfrjsdlkfjpoer";
  char tmp2[] = "sdlfkcskdjfroizoro";
  char *str;
  memcpy(NULL, NULL, 0);
  str = strdup(tmp);
  printf("str => %s | ", str);
  str = memcpy(str, tmp2, sizeof(tmp) - 1);
  printf("str -> %s\n", str);
  free(str);
  return 0;
}
```