ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<invbinocdf>:
  endbr64
  push   %r12
  mov    %edi,%ecx
  pxor   %xmm2,%xmm2
  movapd %xmm0,%xmm4
  push   %rbx
  shr    $0x1f,%ecx
  mov    %edi,%ebx
  sub    $0x28,%rsp
  comisd %xmm1,%xmm2
  movsd  %xmm0,0x18(%rsp)
  movsd  D4(%rip),%xmm0
  seta   %al
  comisd %xmm0,%xmm1
  seta   %dl
  ucomisd %xmm0,%xmm4
  jp     L102
  jne    L102
  or     %al,%dl
  jne    L63
  test   %cl,%cl
  jne    L63
  test   %edi,%edi
  je     L72
  mov    %edi,%r12d
L49:
  add    $0x28,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax)
L102:
  movsd  0x18(%rsp),%xmm5
  or     %ecx,%eax
  or     %edx,%eax
  comisd %xmm5,%xmm2
  seta   %dl
  or     %dl,%al
  jne    L63
  comisd %xmm0,%xmm5
  ja     L63
  movsd  0x18(%rsp),%xmm6
  mov    $0xffffffff,%r12d
  comisd %xmm2,%xmm6
  jbe    L49
  nopl   (%rax)
L9:
  add    $0x1,%r12d
  movapd %xmm1,%xmm0
  mov    %ebx,%edi
  movsd  %xmm2,0x10(%rsp)
  mov    %r12d,%esi
  movsd  %xmm1,0x8(%rsp)
  call   <binopdf_mat@plt>
  movsd  0x10(%rsp),%xmm2
  movsd  0x18(%rsp),%xmm3
  movsd  0x8(%rsp),%xmm1
  addsd  %xmm0,%xmm2
  comisd %xmm2,%xmm3
  ja     L9
  add    $0x28,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
L63:
  mov    $0xffffffff,%r12d
  jmp    L49
L72:
  or     $0xffffffff,%r12d
  jmp    L9
```

tool_calls: name: parse_data
arguments: data_tag: D4
type_guessed: f64

tool: D4: 1

assistant: ```c
int invbinocdf(double y, int n, double p) {
  int x = 0;
  double cumdist = 0;
  double currpdf;
  int count;
  if (y == 1)
    x = n;
  if ((n < 0) | (p < 0) | (p > 1) | (y < 0.0) | (y > 1.0))
    x = -1;
  if (x == 0) {
    x = -1;
    while (cumdist < y) {
      x++;
      currpdf = binopdf_mat(n, x, p);
      cumdist += currpdf;
    }
  }
  return x;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<invbinocdf>:
  endbr64
  push   %r12
  mov    %edi,%ecx
  pxor   %xmm2,%xmm2
  movapd %xmm0,%xmm4
  push   %rbx
  shr    $0x1f,%ecx
  mov    %edi,%ebx
  sub    $0x28,%rsp
  comisd %xmm1,%xmm2
  movsd  %xmm0,0x18(%rsp)
  movsd  D6(%rip),%xmm0
  seta   %al
  comisd %xmm0,%xmm1
  seta   %dl
  ucomisd %xmm0,%xmm4
  jp     L109
  jne    L109
  or     %dl,%al
  jne    L16
  test   %cl,%cl
  jne    L16
  test   %edi,%edi
  je     L27
  mov    %edi,%r12d
L68:
  add    $0x28,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax)
L109:
  movsd  0x18(%rsp),%xmm5
  or     %ecx,%eax
  or     %edx,%eax
  comisd %xmm5,%xmm2
  seta   %dl
  or     %dl,%al
  jne    L16
  comisd %xmm0,%xmm5
  ja     L16
  comisd %xmm2,%xmm5
  mov    $0xffffffff,%r12d
  jbe    L68
  nop
L51:
  add    $0x1,%r12d
  movapd %xmm1,%xmm0
  mov    %ebx,%edi
  movsd  %xmm2,0x10(%rsp)
  mov    %r12d,%esi
  movsd  %xmm1,0x8(%rsp)
  call   <binopdf_mat@plt>
  movsd  0x10(%rsp),%xmm2
  movsd  0x18(%rsp),%xmm3
  movsd  0x8(%rsp),%xmm1
  addsd  %xmm0,%xmm2
  comisd %xmm2,%xmm3
  ja     L51
  add    $0x28,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
L16:
  mov    $0xffffffff,%r12d
  jmp    L68
L27:
  or     $0xffffffff,%r12d
  jmp    L51
```

tool_calls: name: read_data
arguments: data_tag: D6
data_type: f64

tool: D6: 1

assistant: ```c
int invbinocdf(double y, int n, double p) {
  int x = 0;
  double cumdist = 0;
  double currpdf;
  int count;
  if (y == 1)
    x = n;
  if ((n < 0) | (p < 0) | (p > 1) | (y < 0.0) | (y > 1.0))
    x = -1;
  if (x == 0) {
    x = -1;
    while (cumdist < y) {
      x++;
      currpdf = binopdf_mat(n, x, p);
      cumdist += currpdf;
    }
  }
  return x;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<be64get>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    $0x38,%edx
  mov    %edx,%ecx
  shl    %cl,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  mov    $0x30,%ecx
  shl    %cl,%eax
  or     %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%eax
  mov    $0x28,%ecx
  shl    %cl,%eax
  or     %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  add    $0xc,%rax
  mov    (%rax),%eax
  mov    $0x20,%ecx
  shl    %cl,%eax
  or     %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%eax
  shl    $0x18,%eax
  or     %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  add    $0x14,%rax
  mov    (%rax),%eax
  shl    $0x10,%eax
  or     %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  add    $0x18,%rax
  mov    (%rax),%eax
  shl    $0x8,%eax
  or     %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  add    $0x1c,%rax
  mov    (%rax),%eax
  or     %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x8,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int be64get(void *p, uint64_t *v) {
  uint8_t *a;
  a = p;
  *v = (uint64_t)a[0] << 56;
  *v |= (uint64_t)a[1] << 48;
  *v |= (uint64_t)a[2] << 40;
  *v |= (uint64_t)a[3] << 32;
  *v |= (uint64_t)a[4] << 24;
  *v |= a[5] << 16;
  *v |= a[6] << 8;
  *v |= a[7];
  return 8;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<be64get>:
  endbr64
  movl   $0x0,(%rsi)
  mov    0x10(%rdi),%eax
  shl    $0x18,%eax
  mov    %eax,(%rsi)
  mov    0x14(%rdi),%edx
  shl    $0x10,%edx
  or     %eax,%edx
  mov    %edx,(%rsi)
  mov    0x18(%rdi),%eax
  shl    $0x8,%eax
  or     %edx,%eax
  mov    %eax,(%rsi)
  or     0x1c(%rdi),%eax
  mov    %eax,(%rsi)
  mov    $0x8,%eax
  ret
```

assistant: ```c
unsigned int be64get(void *p, uint64_t *v) {
  uint8_t *a;
  a = p;
  *v = (uint64_t)a[0] << 56;
  *v |= (uint64_t)a[1] << 48;
  *v |= (uint64_t)a[2] << 40;
  *v |= (uint64_t)a[3] << 32;
  *v |= (uint64_t)a[4] << 24;
  *v |= a[5] << 16;
  *v |= a[6] << 8;
  *v |= a[7];
  return 8;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<be64get>:
  endbr64
  movl   $0x0,(%rsi)
  mov    0x10(%rdi),%eax
  shl    $0x18,%eax
  mov    %eax,(%rsi)
  mov    0x14(%rdi),%edx
  shl    $0x10,%edx
  or     %eax,%edx
  mov    %edx,(%rsi)
  mov    0x18(%rdi),%eax
  shl    $0x8,%eax
  or     %edx,%eax
  mov    %eax,(%rsi)
  or     0x1c(%rdi),%eax
  mov    %eax,(%rsi)
  mov    $0x8,%eax
  ret
```

assistant: ```c
unsigned int be64get(void *p, uint64_t *v) {
  uint8_t *a;
  a = p;
  *v = (uint64_t)a[0] << 56;
  *v |= (uint64_t)a[1] << 48;
  *v |= (uint64_t)a[2] << 40;
  *v |= (uint64_t)a[3] << 32;
  *v |= (uint64_t)a[4] << 24;
  *v |= a[5] << 16;
  *v |= a[6] << 8;
  *v |= a[7];
  return 8;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<be64get>:
  endbr64
  movl   $0x0,(%rsi)
  mov    0x10(%rdi),%eax
  shl    $0x18,%eax
  mov    %eax,(%rsi)
  mov    0x14(%rdi),%edx
  shl    $0x10,%edx
  or     %eax,%edx
  mov    %edx,(%rsi)
  mov    0x18(%rdi),%eax
  shl    $0x8,%eax
  or     %edx,%eax
  mov    %eax,(%rsi)
  or     0x1c(%rdi),%eax
  mov    %eax,(%rsi)
  mov    $0x8,%eax
  ret
```

assistant: ```c
unsigned int be64get(void *p, uint64_t *v) {
  uint8_t *a;
  a = p;
  *v = (uint64_t)a[0] << 56;
  *v |= (uint64_t)a[1] << 48;
  *v |= (uint64_t)a[2] << 40;
  *v |= (uint64_t)a[3] << 32;
  *v |= (uint64_t)a[4] << 24;
  *v |= a[5] << 16;
  *v |= a[6] << 8;
  *v |= a[7];
  return 8;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<memchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L4
L8:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  mov    -0x1c(%rbp),%edx
  cmp    %dl,%al
  jne    L13
  mov    -0x8(%rbp),%rax
  jmp    L29
L13:
  addq   $0x1,-0x8(%rbp)
L4:
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  test   %rax,%rax
  jne    L8
  mov    $0x0,%eax
L29:
  pop    %rbp
  ret
```

assistant: ```c
void *memchr(const void *buf, int ch, size_t n) {
  const unsigned char *src = buf;
  while (n--) {
    if (*src == (unsigned char)ch) {
      return (void *)src;
    }
    src++;
  }
  return NULL;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<memchr>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L10
  add    %rdi,%rdx
L29:
  cmp    %sil,(%rax)
  je     L28
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L29
  mov    $0x0,%eax
  ret
L10:
  mov    $0x0,%eax
L28:
  ret
```

assistant: ```c
void *memchr(const void *buf, int ch, size_t n) {
  const unsigned char *src = buf;
  while (n--) {
    if (*src == (unsigned char)ch) {
      return (void *)src;
    }
    src++;
  }
  return NULL;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<memchr>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L24
  add    %rdi,%rdx
  jmp    L35
  nopl   0x0(%rax)
L9:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L24
L35:
  cmp    %sil,(%rax)
  jne    L9
  ret
  nopw   0x0(%rax,%rax,1)
L24:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *memchr(const void *buf, int ch, size_t n) {
  const unsigned char *src = buf;
  while (n--) {
    if (*src == (unsigned char)ch) {
      return (void *)src;
    }
    src++;
  }
  return NULL;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<memchr>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L3
  add    %rdi,%rdx
  jmp    L28
  nopl   0x0(%rax)
L31:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L3
L28:
  cmp    %sil,(%rax)
  jne    L31
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *memchr(const void *buf, int ch, size_t n) {
  const unsigned char *src = buf;
  while (n--) {
    if (*src == (unsigned char)ch) {
      return (void *)src;
    }
    src++;
  }
  return NULL;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cmp_gt_ll>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jge    L9
  mov    $0x1,%eax
  jmp    L25
L9:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jge    L12
  mov    $0xffffffff,%eax
  jmp    L25
L12:
  mov    $0x0,%eax
L25:
  pop    %rbp
  ret
```

assistant: ```c
int cmp_gt_ll(const void *p, const void *q) {
  const long long a = *(const long long *)p;
  const long long b = *(const long long *)q;
  do {
    if (a < b)
      return +1;
    if (b < a)
      return -1;
    return 0;
  } while (0);
}
```
user: What is the c source code of the assembly code below:

```asm
<cmp_gt_ll>:
  endbr64
  mov    (%rdi),%rcx
  mov    (%rsi),%rdx
  mov    $0x1,%eax
  cmp    %rdx,%rcx
  jl     L6
  setg   %al
  movzbl %al,%eax
  neg    %eax
L6:
  ret
```

assistant: ```c
int cmp_gt_ll(const void *p, const void *q) {
  const long long a = *(const long long *)p;
  const long long b = *(const long long *)q;
  do {
    if (a < b)
      return +1;
    if (b < a)
      return -1;
    return 0;
  } while (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cmp_gt_ll>:
  endbr64
  mov    (%rsi),%rdx
  mov    $0x1,%eax
  cmp    %rdx,(%rdi)
  jl     L3
  setg   %al
  movzbl %al,%eax
  neg    %eax
L3:
  ret
```

assistant: ```c
int cmp_gt_ll(const void *p, const void *q) {
  const long long a = *(const long long *)p;
  const long long b = *(const long long *)q;
  do {
    if (a < b)
      return +1;
    if (b < a)
      return -1;
    return 0;
  } while (0);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<cmp_gt_ll>:
  endbr64
  mov    (%rsi),%rdx
  mov    $0x1,%eax
  cmp    %rdx,(%rdi)
  jl     L7
  setg   %al
  movzbl %al,%eax
  neg    %eax
L7:
  ret
```

assistant: ```c
int cmp_gt_ll(const void *p, const void *q) {
  const long long a = *(const long long *)p;
  const long long b = *(const long long *)q;
  do {
    if (a < b)
      return +1;
    if (b < a)
      return -1;
    return 0;
  } while (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rsi
  mov    %rsi,-0x18(%rbp)
  xor    %esi,%esi
  mov    %rsp,%rsi
  mov    %rsi,%r9
  movl   $0x3,-0x38(%rbp)
  movl   $0x4,-0x34(%rbp)
  mov    -0x38(%rbp),%esi
  movslq %esi,%rdi
  sub    $0x1,%rdi
  mov    %rdi,-0x30(%rbp)
  movslq %esi,%rdi
  mov    %rdi,%rax
  mov    $0x0,%edx
  imul   $0xc0,%rdx,%r8
  imul   $0x0,%rax,%rdi
  add    %r8,%rdi
  mov    $0xc0,%r8d
  mul    %r8
  add    %rdx,%rdi
  mov    %rdi,%rdx
  movslq %esi,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x3,%rax
  movslq %esi,%rax
  mov    %rax,%rcx
  mov    $0x0,%ebx
  imul   $0xc0,%rbx,%rdx
  imul   $0x0,%rcx,%rax
  lea    (%rdx,%rax,1),%rdi
  mov    $0xc0,%eax
  mul    %rcx
  lea    (%rdi,%rdx,1),%rcx
  mov    %rcx,%rdx
  movslq %esi,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x3,%rax
  mov    %rax,%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%ebx
  mov    $0x0,%edx
  div    %rbx
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L28:
  cmp    %rdx,%rsp
  je     L20
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L28
L20:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L6
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L6:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    $0x0,%eax
  mov    %r9,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int main(void) {
  int n = 3;
  int m = 4;
  int a[n][6];
  int(*p)[6];
  p = a;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
  int n = 3;
  int m = 4;
  int a[n][6];
  int(*p)[6];
  p = a;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  int n = 3;
  int m = 4;
  int a[n][6];
  int(*p)[6];
  p = a;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  jmp    L3
L1:
  cmpl   $0x20,-0x4(%rbp)
  je     L24
  cmpl   $0x9,-0x4(%rbp)
  je     L24
  cmpl   $0xa,-0x4(%rbp)
  jne    L53
L24:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L3
L53:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
L3:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x4(%rbp)
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jne    L1
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
main() {
  int c;
  while ((c = getchar()) != EOF) {
    if (c == ' ' || c == '\t' || c == '\n')
      putchar('\n');
    else
      putchar(c);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  jmp    L5
L9:
  mov    $0xa,%edi
  call   <putchar@plt>
L5:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  mov    D6(%rip),%rax
  cmp    %edi,(%rax)
  je     L0
  lea    -0x9(%rdi),%eax
  cmp    $0x1,%eax
  jbe    L9
  cmp    $0x20,%edi
  je     L9
  call   <putchar@plt>
  jmp    L5
L0:
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
main() {
  int c;
  while ((c = getchar()) != EOF) {
    if (c == ' ' || c == '\t' || c == '\n')
      putchar('\n');
    else
      putchar(c);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  mov    D9(%rip),%rbx
  nopl   0x0(%rax)
L47:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  je     L38
L39:
  lea    -0x9(%rdi),%eax
  cmp    $0x1,%eax
  jbe    L30
  cmp    $0x20,%edi
  je     L30
  call   <putchar@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  jne    L39
L38:
  xor    %eax,%eax
  pop    %rbx
  ret
L30:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L47
```

assistant: ```c
main() {
  int c;
  while ((c = getchar()) != EOF) {
    if (c == ' ' || c == '\t' || c == '\n')
      putchar('\n');
    else
      putchar(c);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbx
  mov    D6(%rip),%rbx
  nopl   0x0(%rax)
L22:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  je     L8
L32:
  lea    -0x9(%rdi),%eax
  cmp    $0x1,%eax
  jbe    L45
  cmp    $0x20,%edi
  je     L45
  call   <putchar@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  jne    L32
L8:
  xor    %eax,%eax
  pop    %rbx
  ret
L45:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L22
```

assistant: ```c
main() {
  int c;
  while ((c = getchar()) != EOF) {
    if (c == ' ' || c == '\t' || c == '\n')
      putchar('\n');
    else
      putchar(c);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D5
datatype: string

tool: D5: WEP SPE THREAD!!


assistant: ```c
int main(unsigned long long spe, unsigned long long argp, unsigned long long envp) {
  printf("WEP SPE THREAD!!\n");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D9
type_infered: string

tool: D9: WEP SPE THREAD!!


assistant: ```c
int main(unsigned long long spe, unsigned long long argp, unsigned long long envp) {
  printf("WEP SPE THREAD!!\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

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

tool_calls: name: parse_data
arguments: label: D9
datatype: string

tool: D9: WEP SPE THREAD!!


assistant: ```c
int main(unsigned long long spe, unsigned long long argp, unsigned long long envp) {
  printf("WEP SPE THREAD!!\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

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

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

tool: D1: WEP SPE THREAD!!


assistant: ```c
int main(unsigned long long spe, unsigned long long argp, unsigned long long envp) {
  printf("WEP SPE THREAD!!\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<remove_trailing_mascuerade>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  movq   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  lea    D5(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strstr@plt>
  mov    %rax,-0x8(%rbp)
  jmp    L12
L15:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x11(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
L12:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x11(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L15
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D5
data_type: string

tool: D5:  (MasC=uerade)..."

assistant: ```c
void remove_trailing_mascuerade(char *text) {
  int i = 0;
  char *sub = NULL;
  sub = strstr(text, " (MasC=uerade)...\"");
  while (sub[i + 17] != 0) {
    sub[i] = sub[i + 17];
    i++;
  }
  sub[i] = 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<remove_trailing_mascuerade>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rsi
  call   <strstr@plt>
  movzbl 0x11(%rax),%ecx
  test   %cl,%cl
  je     L11
  mov    $0x0,%edx
L3:
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  movzbl 0x11(%rax,%rdx,1),%ecx
  test   %cl,%cl
  jne    L3
L4:
  movb   $0x0,(%rax,%rdx,1)
  add    $0x8,%rsp
  ret
L11:
  mov    $0x0,%edx
  jmp    L4
```

tool_calls: name: read_data
arguments: data_tag: D0
type_guessed: string

tool: D0:  (MasC=uerade)..."

assistant: ```c
void remove_trailing_mascuerade(char *text) {
  int i = 0;
  char *sub = NULL;
  sub = strstr(text, " (MasC=uerade)...\"");
  while (sub[i + 17] != 0) {
    sub[i] = sub[i + 17];
    i++;
  }
  sub[i] = 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<remove_trailing_mascuerade>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rsi
  call   <strstr@plt>
  movzbl 0x11(%rax),%ecx
  test   %cl,%cl
  je     L4
  xor    %edx,%edx
  xchg   %ax,%ax
L8:
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  movzbl 0x11(%rax,%rdx,1),%ecx
  test   %cl,%cl
  jne    L8
  add    %rdx,%rax
L4:
  movb   $0x0,(%rax)
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D7
data_type: string

tool: D7:  (MasC=uerade)..."

assistant: ```c
void remove_trailing_mascuerade(char *text) {
  int i = 0;
  char *sub = NULL;
  sub = strstr(text, " (MasC=uerade)...\"");
  while (sub[i + 17] != 0) {
    sub[i] = sub[i + 17];
    i++;
  }
  sub[i] = 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<remove_trailing_mascuerade>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rsi
  call   <strstr@plt>
  movzbl 0x11(%rax),%ecx
  test   %cl,%cl
  je     L8
  xor    %edx,%edx
  xchg   %ax,%ax
L14:
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  movzbl 0x11(%rax,%rdx,1),%ecx
  test   %cl,%cl
  jne    L14
  add    %rdx,%rax
L8:
  movb   $0x0,(%rax)
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D0
type_guessed: string

tool: D0:  (MasC=uerade)..."

assistant: ```c
void remove_trailing_mascuerade(char *text) {
  int i = 0;
  char *sub = NULL;
  sub = strstr(text, " (MasC=uerade)...\"");
  while (sub[i + 17] != 0) {
    sub[i] = sub[i + 17];
    i++;
  }
  sub[i] = 0;
}
```