ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x10,%edi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <malloc@plt>
  mov    $0x8000,%edi
  mov    %rax,%r12
  call   <malloc@plt>
  mov    $0x8000,%edx
  xor    %esi,%esi
  mov    %rax,%r14
  mov    %rax,%rdi
  call   <memset@plt>
  mov    %r12,%rsi
  mov    %r14,%rdi
  mov    %r14,%rbx
  call   <write_block@plt>
  lea    0x8000(%r14),%r13
  mov    %r14,%rbp
  nopl   (%rax)
L29:
  mov    %rbp,%rdi
  mov    $0x10,%edx
  mov    %r12,%rsi
  add    $0x10,%rbp
  call   <memcpy@plt>
  cmp    %r13,%rbp
  jne    L29
  mov    %r12,%rsi
  mov    %r14,%rdi
  mov    %r14,%rbp
  call   <write_block@plt>
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %rbp,%rdi
  mov    $0x10,%edx
  mov    %r12,%rsi
  add    $0x10,%rbp
  call   <memcpy@plt>
  cmp    %r13,%rbp
  jne    L1
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <write_block@plt>
  nopl   0x0(%rax)
L22:
  mov    %rbx,%rdi
  mov    $0x10,%edx
  mov    %r12,%rsi
  add    $0x10,%rbx
  call   <memcpy@plt>
  cmp    %r13,%rbx
  jne    L22
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <write_block@plt>
  mov    %r14,%rdi
  call   <free@plt>
  mov    %r12,%rdi
  call   <free@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned char *hash = malloc(16);
  void *block = malloc(32768);
  memset(block, 0, 32768);
  write_block(block, hash);
  for (int i = 0; i < 32768; i += 16)
    memcpy(block + i, hash, 16);
  write_block(block, hash);
  for (int i = 0; i < 32768; i += 16)
    memcpy(block + i, hash, 16);
  write_block(block, hash);
  for (int i = 0; i < 32768; i += 16)
    memcpy(block + i, hash, 16);
  write_block(block, hash);
  free(block);
  free(hash);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x10,%edi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <malloc@plt>
  mov    $0x8000,%edi
  mov    %rax,%r12
  call   <malloc@plt>
  mov    $0x8000,%edx
  xor    %esi,%esi
  mov    %rax,%r14
  mov    %rax,%rdi
  call   <memset@plt>
  mov    %r12,%rsi
  mov    %r14,%rdi
  mov    %r14,%rbx
  call   <write_block@plt>
  lea    0x8000(%r14),%r13
  mov    %r14,%rbp
  nopl   (%rax)
L14:
  mov    %rbp,%rdi
  mov    $0x10,%edx
  mov    %r12,%rsi
  add    $0x10,%rbp
  call   <memcpy@plt>
  cmp    %r13,%rbp
  jne    L14
  mov    %r12,%rsi
  mov    %r14,%rdi
  mov    %r14,%rbp
  call   <write_block@plt>
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %rbp,%rdi
  mov    $0x10,%edx
  mov    %r12,%rsi
  add    $0x10,%rbp
  call   <memcpy@plt>
  cmp    %r13,%rbp
  jne    L1
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <write_block@plt>
  nopl   0x0(%rax)
L21:
  mov    %rbx,%rdi
  mov    $0x10,%edx
  mov    %r12,%rsi
  add    $0x10,%rbx
  call   <memcpy@plt>
  cmp    %r13,%rbx
  jne    L21
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <write_block@plt>
  mov    %r14,%rdi
  call   <free@plt>
  mov    %r12,%rdi
  call   <free@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned char *hash = malloc(16);
  void *block = malloc(32768);
  memset(block, 0, 32768);
  write_block(block, hash);
  for (int i = 0; i < 32768; i += 16)
    memcpy(block + i, hash, 16);
  write_block(block, hash);
  for (int i = 0; i < 32768; i += 16)
    memcpy(block + i, hash, 16);
  write_block(block, hash);
  for (int i = 0; i < 32768; i += 16)
    memcpy(block + i, hash, 16);
  write_block(block, hash);
  free(block);
  free(hash);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x32,-0x8(%rbp)
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x8(%rbp),%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  mov    %eax,%edi
  call   <ceil@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x2,-0xc(%rbp)
  jmp    L1
L26:
  mov    -0x4(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <judge@plt>
  test   %rax,%rax
  je     L4
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L7
L4:
  nop
L7:
  addl   $0x1,-0xc(%rbp)
L1:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L26
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D1
datatype: string

tool: D1: %d 

assistant: ```c
int main() {
  int n = 50;
  int endNum = (int)(ceil(sqrt((double)n)));
  for (int i = 2; i <= n; i++) {
    if (judge(i, endNum) != 0) {
      printf("%d ", i);
    } else {
      continue;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  movsd  D16(%rip),%xmm0
  call   <sqrt@plt>
  mov    %eax,%edi
  call   <ceil@plt>
  mov    %eax,%ebp
  mov    $0x2,%ebx
  lea    D6(%rip),%r12
  jmp    L15
L23:
  add    $0x1,%ebx
  cmp    $0x33,%ebx
  je     L24
L15:
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <judge@plt>
  test   %rax,%rax
  je     L23
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  jmp    L23
L24:
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D16
type_guessed: byte[8]
name: extract_symbolic_data
arguments: data_tag: D6
type_guessed: string

tool: {"D16": [0x00,0x00,0x00,0x00,0x00,0x00,0x49,0x40]}

tool: D6: %d 

assistant: ```c
int main() {
  int n = 50;
  int endNum = (int)(ceil(sqrt((double)n)));
  for (int i = 2; i <= n; i++) {
    if (judge(i, endNum) != 0) {
      printf("%d ", i);
    } else {
      continue;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r12
  movsd  D13(%rip),%xmm0
  lea    D5(%rip),%r12
  push   %rbp
  push   %rbx
  mov    $0x2,%ebx
  call   <sqrt@plt>
  mov    %eax,%edi
  call   <ceil@plt>
  mov    %eax,%ebp
  nopw   0x0(%rax,%rax,1)
L11:
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <judge@plt>
  test   %rax,%rax
  je     L6
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
L6:
  add    $0x1,%ebx
  cmp    $0x33,%ebx
  jne    L11
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D13
data_type: byte[8]
name: parse_data
arguments: data_tag: D5
data_type: string

tool: {"D13": [0x00,0x00,0x00,0x00,0x00,0x00,0x49,0x40]}

tool: D5: %d 

assistant: ```c
int main() {
  int n = 50;
  int endNum = (int)(ceil(sqrt((double)n)));
  for (int i = 2; i <= n; i++) {
    if (judge(i, endNum) != 0) {
      printf("%d ", i);
    } else {
      continue;
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  movsd  D4(%rip),%xmm0
  lea    D14(%rip),%r12
  push   %rbp
  push   %rbx
  mov    $0x2,%ebx
  call   <sqrt@plt>
  mov    %eax,%edi
  call   <ceil@plt>
  mov    %eax,%ebp
  nopw   0x0(%rax,%rax,1)
L14:
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <judge@plt>
  test   %rax,%rax
  je     L2
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
L2:
  add    $0x1,%ebx
  cmp    $0x33,%ebx
  jne    L14
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
type_infered: byte[8]
name: disassemble_data
arguments: data_tag: D14
type_infered: string

tool: {"D4": [0x00,0x00,0x00,0x00,0x00,0x00,0x49,0x40]}

tool: D14: %d 

assistant: ```c
int main() {
  int n = 50;
  int endNum = (int)(ceil(sqrt((double)n)));
  for (int i = 2; i <= n; i++) {
    if (judge(i, endNum) != 0) {
      printf("%d ", i);
    } else {
      continue;
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fibonacci>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  cmpl   $0x1,-0x24(%rbp)
  jg     L8
  mov    -0x24(%rbp),%eax
  cltq
  mov    %rax,-0x18(%rbp)
  jmp    L18
L8:
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <fibonacci>
  mov    %rax,%rbx
  mov    -0x24(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,%edi
  call   <fibonacci>
  add    %rbx,%rax
  mov    %rax,-0x18(%rbp)
L18:
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
long fibonacci(int n) {
  long r;
  if (n <= 1) {
    r = n;
  } else {
    r = fibonacci(n - 1) + fibonacci(n - 2);
  }
  return r;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<fibonacci>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  movslq %edi,%rax
  cmp    $0x1,%edi
  jg     L4
L18:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L4:
  lea    -0x1(%rdi),%edi
  call   <fibonacci>
  mov    %rax,%rbp
  lea    -0x2(%rbx),%edi
  call   <fibonacci>
  add    %rbp,%rax
  jmp    L18
```

assistant: ```c
long fibonacci(int n) {
  long r;
  if (n <= 1) {
    r = n;
  } else {
    r = fibonacci(n - 1) + fibonacci(n - 2);
  }
  return r;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<putAlignmentPattern>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  sub    $0x2,%eax
  imul   -0x1c(%rbp),%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  cltq
  sub    $0x2,%rax
  add    %rax,-0x18(%rbp)
  lea    D6(%rip),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L16
L22:
  movl   $0x0,-0x10(%rbp)
  jmp    L33
L32:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x10(%rbp)
L33:
  cmpl   $0x4,-0x10(%rbp)
  jle    L32
  mov    -0x1c(%rbp),%eax
  cltq
  add    %rax,-0x18(%rbp)
  addq   $0x5,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L16:
  cmpl   $0x4,-0xc(%rbp)
  jle    L22
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void putAlignmentPattern(unsigned char *frame, int width, int ox, int oy) {
  static const unsigned char finder[] = {
      0xa1, 0xa1, 0xa1, 0xa1, 0xa1, 0xa1, 0xa0, 0xa0, 0xa0, 0xa1, 0xa1, 0xa0, 0xa1, 0xa0, 0xa1, 0xa1, 0xa0, 0xa0, 0xa0, 0xa1, 0xa1, 0xa1, 0xa1, 0xa1, 0xa1,
  };
  int x, y;
  const unsigned char *s;
  frame += (oy - 2) * width + ox - 2;
  s = finder;
  for (y = 0; y < 5; y++) {
    for (x = 0; x < 5; x++) {
      frame[x] = s[x];
    }
    frame += width;
    s += 5;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<putAlignmentPattern>:
  endbr64
  sub    $0x2,%ecx
  imul   %esi,%ecx
  add    %edx,%ecx
  movslq %ecx,%rcx
  lea    -0x2(%rcx,%rdi,1),%rdi
  lea    D6(%rip),%rcx
  movslq %esi,%rsi
  lea    0x19(%rcx),%r8
L8:
  mov    $0x0,%eax
L5:
  movzbl (%rcx,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    $0x5,%rax
  jne    L5
  add    %rsi,%rdi
  add    $0x5,%rcx
  cmp    %r8,%rcx
  jne    L8
  ret
```

assistant: ```c
void putAlignmentPattern(unsigned char *frame, int width, int ox, int oy) {
  static const unsigned char finder[] = {
      0xa1, 0xa1, 0xa1, 0xa1, 0xa1, 0xa1, 0xa0, 0xa0, 0xa0, 0xa1, 0xa1, 0xa0, 0xa1, 0xa0, 0xa1, 0xa1, 0xa0, 0xa0, 0xa0, 0xa1, 0xa1, 0xa1, 0xa1, 0xa1, 0xa1,
  };
  int x, y;
  const unsigned char *s;
  frame += (oy - 2) * width + ox - 2;
  s = finder;
  for (y = 0; y < 5; y++) {
    for (x = 0; x < 5; x++) {
      frame[x] = s[x];
    }
    frame += width;
    s += 5;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<putAlignmentPattern>:
  endbr64
  sub    $0x2,%ecx
  imul   %esi,%ecx
  movslq %esi,%rsi
  add    %edx,%ecx
  mov    $0xffffffa1,%edx
  movslq %ecx,%rcx
  lea    -0x2(%rcx,%rdi,1),%rdi
  lea    D2(%rip),%rcx
  lea    0x19(%rcx),%r8
L32:
  xor    %eax,%eax
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    $0x5,%rax
  je     L33
L12:
  movzbl (%rcx,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    $0x5,%rax
  jne    L12
L33:
  add    $0x5,%rcx
  add    %rsi,%rdi
  cmp    %r8,%rcx
  je     L10
  movzbl (%rcx),%edx
  jmp    L32
  nopl   0x0(%rax,%rax,1)
L10:
  ret
```

assistant: ```c
void putAlignmentPattern(unsigned char *frame, int width, int ox, int oy) {
  static const unsigned char finder[] = {
      0xa1, 0xa1, 0xa1, 0xa1, 0xa1, 0xa1, 0xa0, 0xa0, 0xa0, 0xa1, 0xa1, 0xa0, 0xa1, 0xa0, 0xa1, 0xa1, 0xa0, 0xa0, 0xa0, 0xa1, 0xa1, 0xa1, 0xa1, 0xa1, 0xa1,
  };
  int x, y;
  const unsigned char *s;
  frame += (oy - 2) * width + ox - 2;
  s = finder;
  for (y = 0; y < 5; y++) {
    for (x = 0; x < 5; x++) {
      frame[x] = s[x];
    }
    frame += width;
    s += 5;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<putAlignmentPattern>:
  endbr64
  sub    $0x2,%ecx
  imul   %esi,%ecx
  add    %edx,%ecx
  movslq %esi,%rdx
  movslq %ecx,%rcx
  lea    -0x2(%rcx,%rdi,1),%rax
  movl   $0xa1a1a1a1,(%rax)
  movb   $0xa1,0x4(%rax)
  add    %rdx,%rax
  movl   $0xa0a0a0a1,(%rax)
  movb   $0xa1,0x4(%rax)
  add    %rdx,%rax
  movl   $0xa0a1a0a1,(%rax)
  movb   $0xa1,0x4(%rax)
  add    %rdx,%rax
  movl   $0xa0a0a0a1,(%rax)
  movb   $0xa1,0x4(%rax)
  add    %rdx,%rax
  movl   $0xa1a1a1a1,(%rax)
  movb   $0xa1,0x4(%rax)
  ret
```

assistant: ```c
void putAlignmentPattern(unsigned char *frame, int width, int ox, int oy) {
  static const unsigned char finder[] = {
      0xa1, 0xa1, 0xa1, 0xa1, 0xa1, 0xa1, 0xa0, 0xa0, 0xa0, 0xa1, 0xa1, 0xa0, 0xa1, 0xa0, 0xa1, 0xa1, 0xa0, 0xa0, 0xa0, 0xa1, 0xa1, 0xa1, 0xa1, 0xa1, 0xa1,
  };
  int x, y;
  const unsigned char *s;
  frame += (oy - 2) * width + ox - 2;
  s = finder;
  for (y = 0; y < 5; y++) {
    for (x = 0; x < 5; x++) {
      frame[x] = s[x];
    }
    frame += width;
    s += 5;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rsi
  lea    -0x10(%rbp),%rcx
  lea    -0x14(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rsi,%r8
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <__get_cpuid@plt>
  test   %eax,%eax
  jne    L33
  mov    $0x0,%eax
  jmp    L25
L33:
  mov    -0x10(%rbp),%eax
  and    $0x10000000,%eax
  test   %eax,%eax
  je     L29
  mov    $0x0,%eax
  call   <vector_1_x@plt>
L29:
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
L25:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

assistant: ```c
int main() {
  unsigned int eax, ebx, ecx, edx;
  if (!__get_cpuid(1, &eax, &ebx, &ecx, &edx))
    return 0;
  if (ecx & (1 << 28))
    vector_1_x();
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  mov    $0x1,%edi
  call   <__get_cpuid@plt>
  test   %eax,%eax
  je     L30
  testb  $0x10,0x13(%rsp)
  jne    L36
L15:
  mov    $0x0,%edi
  call   <exit@plt>
L30:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L36:
  mov    $0x0,%eax
  call   <vector_1_x@plt>
  jmp    L15
L33:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  unsigned int eax, ebx, ecx, edx;
  if (!__get_cpuid(1, &eax, &ebx, &ecx, &edx))
    return 0;
  if (ecx & (1 << 28))
    vector_1_x();
  exit(0);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    $0x1,%edi
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  call   <__get_cpuid@plt>
  test   %eax,%eax
  je     L33
  testb  $0x10,0x13(%rsp)
  jne    L3
L32:
  xor    %edi,%edi
  call   <exit@plt>
L33:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L3:
  xor    %eax,%eax
  call   <vector_1_x@plt>
  jmp    L32
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  unsigned int eax, ebx, ecx, edx;
  if (!__get_cpuid(1, &eax, &ebx, &ecx, &edx))
    return 0;
  if (ecx & (1 << 28))
    vector_1_x();
  exit(0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    $0x1,%edi
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  call   <__get_cpuid@plt>
  test   %eax,%eax
  je     L8
  testb  $0x10,0x13(%rsp)
  jne    L36
L4:
  xor    %edi,%edi
  call   <exit@plt>
L8:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L36:
  xor    %eax,%eax
  call   <vector_1_x@plt>
  jmp    L4
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  unsigned int eax, ebx, ecx, edx;
  if (!__get_cpuid(1, &eax, &ebx, &ecx, &edx))
    return 0;
  if (ecx & (1 << 28))
    vector_1_x();
  exit(0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<is_labelchar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <isalnum@plt>
  test   %rax,%rax
  jne    L57
  cmpl   $0x2d,-0x4(%rbp)
  jne    L32
L57:
  mov    $0x1,%eax
  jmp    L25
L32:
  cmpl   $0x5f,-0x4(%rbp)
  jne    L15
  mov    $0x3,%esi
  mov    $0x0,%edi
  call   <ds_get_boolean@plt>
  test   %rax,%rax
  je     L15
  mov    $0x1,%eax
  jmp    L25
L15:
  mov    $0x0,%eax
L25:
  leave
  ret
```

assistant: ```c
int is_labelchar(int ich) {
  if ((isalnum(ich)) || (ich == '-'))
    return 1;
  if (ich == '_' && ds_get_boolean(0, 3))
    return 1;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<is_labelchar>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <isalnum@plt>
  cmp    $0x2d,%ebx
  je     L22
  test   %rax,%rax
  jne    L22
  cmp    $0x5f,%ebx
  jne    L3
  mov    $0x3,%esi
  mov    $0x0,%edi
  call   <ds_get_boolean@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  jmp    L3
L22:
  mov    $0x1,%eax
L3:
  pop    %rbx
  ret
```

assistant: ```c
int is_labelchar(int ich) {
  if ((isalnum(ich)) || (ich == '-'))
    return 1;
  if (ich == '_' && ds_get_boolean(0, 3))
    return 1;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<is_labelchar>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <isalnum@plt>
  cmp    $0x2d,%ebx
  je     L11
  test   %rax,%rax
  jne    L11
  cmp    $0x5f,%ebx
  je     L12
  pop    %rbx
  ret
  nopl   (%rax)
L11:
  mov    $0x1,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  xor    %edi,%edi
  mov    $0x3,%esi
  call   <ds_get_boolean@plt>
  pop    %rbx
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int is_labelchar(int ich) {
  if ((isalnum(ich)) || (ich == '-'))
    return 1;
  if (ich == '_' && ds_get_boolean(0, 3))
    return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<is_labelchar>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <isalnum@plt>
  cmp    $0x2d,%ebx
  je     L9
  test   %rax,%rax
  jne    L9
  cmp    $0x5f,%ebx
  je     L16
  pop    %rbx
  ret
  nopl   (%rax)
L9:
  mov    $0x1,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L16:
  xor    %edi,%edi
  mov    $0x3,%esi
  call   <ds_get_boolean@plt>
  pop    %rbx
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int is_labelchar(int ich) {
  if ((isalnum(ich)) || (ich == '-'))
    return 1;
  if (ich == '_' && ds_get_boolean(0, 3))
    return 1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<generateSinusoidalAlea>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movsd  D23(%rip),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    D23(%rip),%rax
  movq   %rax,%xmm0
  call   <cos@plt>
  add    %eax,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  %xmm0,D1(%rip)
  movsd  D1(%rip),%xmm1
  pxor   %xmm0,%xmm0
  comisd %xmm1,%xmm0
  jbe    L9
  movsd  D1(%rip),%xmm0
  movq   D1(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  movsd  %xmm0,D40(%rip)
L9:
  movsd  -0x8(%rbp),%xmm1
  movsd  D40(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D1
type: byte[8]
name: disassemble_data
arguments: tag: D40
type: byte[8]

tool: {"D1": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

tool: {"D40": [0x00,0x00,0x00,0xe8,0x76,0x48,0x37,0x42]}

assistant: ```c
double generateSinusoidalAlea(void) {
  static double xn = 1;
  double value = xn;
  xn = 2 * cos(xn);
  if (xn < 0)
    xn = xn * -1;
  return value * 100000000000;
}
```
user: Translate the below assembly code into C programming language:

```asm
<show>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <showConst@plt>
  mov    $0x0,%eax
  call   <showId@plt>
  mov    $0x0,%eax
  call   <showPIF@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void show() {
  showConst();
  showId();
  showPIF();
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<show>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <showConst@plt>
  mov    $0x0,%eax
  call   <showId@plt>
  mov    $0x0,%eax
  call   <showPIF@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void show() {
  showConst();
  showId();
  showPIF();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<show>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <showConst@plt>
  xor    %eax,%eax
  call   <showId@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1090 <showPIF@plt>
```

assistant: ```c
void show() {
  showConst();
  showId();
  showPIF();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<show>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <showConst@plt>
  xor    %eax,%eax
  call   <showId@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1090 <showPIF@plt>
```

assistant: ```c
void show() {
  showConst();
  showId();
  showPIF();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<fun1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D0
type_infered: string

tool: D0: hello world.


assistant: ```c
int fun1(void) {
  printf("hello world.\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<fun1>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D1
type_guessed: string

tool: D1: hello world.


assistant: ```c
int fun1(void) {
  printf("hello world.\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fun1>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D7
type_guessed: string

tool: D7: hello world.


assistant: ```c
int fun1(void) {
  printf("hello world.\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fun1>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D9
datatype: string

tool: D9: hello world.


assistant: ```c
int fun1(void) {
  printf("hello world.\n");
  return 0;
}
```