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

```asm
<C>:
  endbr64
  movsd  D0(%rip),%xmm0
  test   %edi,%edi
  je     L3
  ret
  nopl   0x0(%rax)
L3:
  sub    $0x18,%rsp
  mov    $0x2,%edi
  movsd  %xmm0,0x8(%rsp)
  call   <sqrt@plt>
  pxor   %xmm1,%xmm1
  movsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  cvtsi2sd %eax,%xmm1
  divsd  %xmm1,%xmm0
  ret
```

tool_calls: name: read_data
arguments: data_label: D0
datatype: f64

tool: D0: 1

assistant: ```c
double C(int32_t xi) {
  if (!xi) {
    return 1. / sqrt(2.);
  } else {
    return 1.;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<C>:
  endbr64
  movsd  D2(%rip),%xmm0
  test   %edi,%edi
  je     L9
  ret
  nopl   0x0(%rax)
L9:
  sub    $0x18,%rsp
  mov    $0x2,%edi
  movsd  %xmm0,0x8(%rsp)
  call   <sqrt@plt>
  pxor   %xmm1,%xmm1
  movsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  cvtsi2sd %eax,%xmm1
  divsd  %xmm1,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D2
data_type: f64

tool: D2: 1

assistant: ```c
double C(int32_t xi) {
  if (!xi) {
    return 1. / sqrt(2.);
  } else {
    return 1.;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<length>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    -0x1c(%rbp),%ecx
  mov    %rax,%rdx
  mov    %rdx,%rdi
  xor    %al,%al
  repnz scas %es:(%rdi),%al
  mov    %rdi,%rdx
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  sub    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
int length(const char *src, int maxlen) {
  const char *s = src;

  __asm__ volatile("xor %%al,%%al; repne scasb" : "=D"(src) : "0"(src), "c"(maxlen) : "%eax");
  return src - s;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<length>:
  endbr64
  mov    %rdi,%rdx
  mov    %esi,%ecx
  xor    %al,%al
  repnz scas %es:(%rdi),%al
  mov    %edi,%eax
  sub    %edx,%eax
  ret
```

assistant: ```c
int length(const char *src, int maxlen) {
  const char *s = src;

  __asm__ volatile("xor %%al,%%al; repne scasb" : "=D"(src) : "0"(src), "c"(maxlen) : "%eax");
  return src - s;
}
```
user: Translate the below assembly code into C programming language:

```asm
<length>:
  endbr64
  mov    %rdi,%rdx
  mov    %esi,%ecx
  xor    %al,%al
  repnz scas %es:(%rdi),%al
  mov    %edi,%eax
  sub    %edx,%eax
  ret
```

assistant: ```c
int length(const char *src, int maxlen) {
  const char *s = src;

  __asm__ volatile("xor %%al,%%al; repne scasb" : "=D"(src) : "0"(src), "c"(maxlen) : "%eax");
  return src - s;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<length>:
  endbr64
  mov    %rdi,%rdx
  mov    %esi,%ecx
  xor    %al,%al
  repnz scas %es:(%rdi),%al
  mov    %edi,%eax
  sub    %edx,%eax
  ret
```

assistant: ```c
int length(const char *src, int maxlen) {
  const char *s = src;

  __asm__ volatile("xor %%al,%%al; repne scasb" : "=D"(src) : "0"(src), "c"(maxlen) : "%eax");
  return src - s;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L15
L11:
  mov    -0x8(%rbp),%rax
  shl    $0x5,%rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    $0x0,%edx
  divq   -0x28(%rbp)
  mov    %rdx,-0x8(%rbp)
L15:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jne    L11
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long hash(unsigned char *str, unsigned long seed, unsigned long p) {
  unsigned long hash = seed;
  int c;
  while ((c = *str++)) {
    hash = (((hash << 5) + hash) + c) % p;
  }
  return hash;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hash>:
  endbr64
  mov    %rdx,%r9
  lea    0x1(%rdi),%r8
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L15
L4:
  mov    %rsi,%rax
  shl    $0x5,%rax
  add    %rsi,%rax
  movzbl %cl,%ecx
  add    %rcx,%rax
  mov    $0x0,%edx
  div    %r9
  mov    %rdx,%rsi
  add    $0x1,%r8
  movzbl -0x1(%r8),%ecx
  test   %cl,%cl
  jne    L4
L15:
  mov    %rsi,%rax
  ret
```

assistant: ```c
unsigned long hash(unsigned char *str, unsigned long seed, unsigned long p) {
  unsigned long hash = seed;
  int c;
  while ((c = *str++)) {
    hash = (((hash << 5) + hash) + c) % p;
  }
  return hash;
}
```
user: What is the c source code of the assembly code below?

```asm
<hash>:
  endbr64
  movzbl (%rdi),%ecx
  mov    %rsi,%r8
  mov    %rdx,%r9
  lea    0x1(%rdi),%rsi
  test   %cl,%cl
  je     L13
  nopl   (%rax)
L2:
  mov    %r8,%rax
  xor    %edx,%edx
  add    $0x1,%rsi
  shl    $0x5,%rax
  add    %r8,%rax
  add    %rcx,%rax
  movzbl -0x1(%rsi),%ecx
  div    %r9
  mov    %rdx,%r8
  test   %cl,%cl
  jne    L2
L13:
  mov    %r8,%rax
  ret
```

assistant: ```c
unsigned long hash(unsigned char *str, unsigned long seed, unsigned long p) {
  unsigned long hash = seed;
  int c;
  while ((c = *str++)) {
    hash = (((hash << 5) + hash) + c) % p;
  }
  return hash;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hash>:
  endbr64
  movzbl (%rdi),%ecx
  mov    %rsi,%r8
  mov    %rdx,%r9
  lea    0x1(%rdi),%rsi
  test   %cl,%cl
  je     L19
  nopl   (%rax)
L0:
  mov    %r8,%rax
  xor    %edx,%edx
  add    $0x1,%rsi
  shl    $0x5,%rax
  add    %r8,%rax
  add    %rcx,%rax
  movzbl -0x1(%rsi),%ecx
  div    %r9
  mov    %rdx,%r8
  test   %cl,%cl
  jne    L0
L19:
  mov    %r8,%rax
  ret
```

assistant: ```c
unsigned long hash(unsigned char *str, unsigned long seed, unsigned long p) {
  unsigned long hash = seed;
  int c;
  while ((c = *str++)) {
    hash = (((hash << 5) + hash) + c) % p;
  }
  return hash;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<_sbrk>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    D24(%rip),%rax
  mov    %rax,-0x10(%rbp)
  lea    -0x14(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    D24(%rip),%rdx
  mov    -0x14(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    %rax,D14(%rip)
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jae    L4
  mov    D14(%rip),%rax
  cmp    %rax,-0x8(%rbp)
  jb     L22
L4:
  mov    D14(%rip),%rax
  cmp    %rax,-0x10(%rbp)
  jbe    L1
L22:
  mov    D14(%rip),%rax
  mov    (%rax),%edx
  mov    D67(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffffffffffff,%rax
  jmp    L38
L1:
  mov    -0x10(%rbp),%rax
L38:
  pop    %rbp
  ret
```

assistant: ```c
char *_sbrk(int adj) {
  extern char end;
  static char *heap = &end;
  char *rv = heap;
  char *heaptop = (char *)&adj;
  heap += adj;
  if (rv < heaptop && heap > heaptop || heap < rv) {
    errno = ENOMEM;
    return (char *)-1;
  }
  return rv;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<_sbrk>:
  endbr64
  mov    %edi,%eax
  mov    %edi,-0x4(%rsp)
  mov    D12(%rip),%rdx
  cltq
  add    %rdx,%rax
  mov    %rax,D19(%rip)
  lea    -0x4(%rsp),%rcx
  cmp    %rcx,%rax
  jbe    L31
  cmp    %rcx,%rdx
  jb     L25
L31:
  cmp    %rdx,%rax
  jb     L25
L18:
  mov    %rdx,%rax
  ret
L25:
  mov    D19(%rip),%rax
  mov    (%rax),%edx
  mov    D31(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffffffffffff,%rdx
  jmp    L18
```

assistant: ```c
char *_sbrk(int adj) {
  extern char end;
  static char *heap = &end;
  char *rv = heap;
  char *heaptop = (char *)&adj;
  heap += adj;
  if (rv < heaptop && heap > heaptop || heap < rv) {
    errno = ENOMEM;
    return (char *)-1;
  }
  return rv;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_sbrk>:
  endbr64
  mov    D8(%rip),%r8
  movslq %edi,%rax
  lea    -0x4(%rsp),%rdx
  mov    %edi,-0x4(%rsp)
  add    %r8,%rax
  mov    %rax,D33(%rip)
  cmp    %rdx,%rax
  jbe    L8
  cmp    %rdx,%r8
  jb     L12
L8:
  cmp    %r8,%rax
  jb     L12
  mov    %r8,%rax
  ret
  nopl   0x0(%rax)
L12:
  mov    D33(%rip),%rax
  mov    $0xffffffffffffffff,%r8
  mov    (%rax),%edx
  mov    D29(%rip),%rax
  mov    %edx,(%rax)
  mov    %r8,%rax
  ret
```

assistant: ```c
char *_sbrk(int adj) {
  extern char end;
  static char *heap = &end;
  char *rv = heap;
  char *heaptop = (char *)&adj;
  heap += adj;
  if (rv < heaptop && heap > heaptop || heap < rv) {
    errno = ENOMEM;
    return (char *)-1;
  }
  return rv;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<_sbrk>:
  endbr64
  mov    D39(%rip),%r8
  movslq %edi,%rax
  lea    -0x4(%rsp),%rdx
  mov    %edi,-0x4(%rsp)
  add    %r8,%rax
  mov    %rax,D19(%rip)
  cmp    %rdx,%rax
  jbe    L17
  cmp    %rdx,%r8
  jb     L23
L17:
  cmp    %r8,%rax
  jb     L23
  mov    %r8,%rax
  ret
  nopl   0x0(%rax)
L23:
  mov    D19(%rip),%rax
  mov    $0xffffffffffffffff,%r8
  mov    (%rax),%edx
  mov    D4(%rip),%rax
  mov    %edx,(%rax)
  mov    %r8,%rax
  ret
```

assistant: ```c
char *_sbrk(int adj) {
  extern char end;
  static char *heap = &end;
  char *rv = heap;
  char *heaptop = (char *)&adj;
  heap += adj;
  if (rv < heaptop && heap > heaptop || heap < rv) {
    errno = ENOMEM;
    return (char *)-1;
  }
  return rv;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<string_getnextword>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    %rcx,-0x38(%rbp)
  mov    -0x30(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L26
L44:
  addl   $0x1,-0x10(%rbp)
L26:
  mov    -0x10(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jge    L51
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <char_isspace@plt>
  test   %rax,%rax
  jne    L44
L51:
  mov    -0x10(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L46
  mov    $0x0,%eax
  jmp    L47
L46:
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  jmp    L40
L69:
  addl   $0x1,-0x14(%rbp)
L40:
  mov    -0x14(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jge    L74
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <char_isspace@plt>
  test   %rax,%rax
  je     L69
L74:
  mov    -0x14(%rbp),%eax
  sub    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <jdns_alloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L10
  mov    $0x0,%eax
  jmp    L47
L10:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %eax,%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
L47:
  leave
  ret
```

assistant: ```c
unsigned char *string_getnextword(unsigned char *in, int size, int pos, int *newpos) {
  int n;
  int at;
  int len;
  unsigned char *out;
  at = pos;
  while (at < size && char_isspace(in[at]))
    ++at;
  if (at >= size)
    return 0;
  n = at;
  while (n < size && !char_isspace(in[n]))
    ++n;
  len = n - at;
  out = (unsigned char *)jdns_alloc(len + 1);
  if (!out)
    return 0;
  memcpy(out, in + at, len);
  out[len] = 0;
  *newpos = at + len;
  return out;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<string_getnextword>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    %esi,%edx
  jge    L1
  mov    %rdi,%rbp
  mov    %esi,%r12d
  mov    %rcx,%r14
  movslq %edx,%rbx
  lea    -0x1(%rsi),%eax
  sub    %edx,%eax
  lea    0x1(%rbx,%rax,1),%r15
L40:
  mov    %ebx,%r13d
  movzbl 0x0(%rbp,%rbx,1),%edi
  call   <char_isspace@plt>
  test   %rax,%rax
  je     L50
  add    $0x1,%rbx
  cmp    %r15,%rbx
  jne    L40
L1:
  mov    $0x0,%r13d
  jmp    L67
L50:
  cmp    %ebx,%r12d
  jle    L81
  movslq %ebx,%r15
  add    %rbp,%r15
  mov    %r15,%rbp
L28:
  movzbl 0x0(%rbp),%edi
  call   <char_isspace@plt>
  test   %rax,%rax
  jne    L75
  add    $0x1,%ebx
  add    $0x1,%rbp
  cmp    %ebx,%r12d
  jne    L28
  mov    %r12d,%ebx
L75:
  mov    %ebx,%r12d
  sub    %r13d,%r12d
  lea    0x1(%r12),%edi
  call   <jdns_alloc@plt>
  mov    %rax,%rbp
  mov    $0x0,%r13d
  test   %rax,%rax
  je     L67
  mov    %rax,%r13
  mov    %r12d,%edx
  mov    %r15,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movslq %r12d,%r12
  movb   $0x0,0x0(%rbp,%r12,1)
  mov    %ebx,(%r14)
L67:
  mov    %r13,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L81:
  mov    $0x0,%r13d
  jmp    L67
```

assistant: ```c
unsigned char *string_getnextword(unsigned char *in, int size, int pos, int *newpos) {
  int n;
  int at;
  int len;
  unsigned char *out;
  at = pos;
  while (at < size && char_isspace(in[at]))
    ++at;
  if (at >= size)
    return 0;
  n = at;
  while (n < size && !char_isspace(in[n]))
    ++n;
  len = n - at;
  out = (unsigned char *)jdns_alloc(len + 1);
  if (!out)
    return 0;
  memcpy(out, in + at, len);
  out[len] = 0;
  *newpos = at + len;
  return out;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<string_getnextword>:
  endbr64
  push   %r15
  push   %r14
  mov    %rcx,%r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  movslq %edx,%rbp
  push   %rbx
  mov    %rbp,%rbx
  add    %rdi,%rbp
  sub    $0x8,%rsp
  cmp    %esi,%ebx
  jl     L46
  jmp    L70
  nopl   (%rax)
L62:
  add    $0x1,%ebx
  add    $0x1,%rbp
  cmp    %ebx,%r13d
  je     L70
L46:
  movzbl 0x0(%rbp),%edi
  call   <char_isspace@plt>
  test   %rax,%rax
  jne    L62
  movslq %ebx,%rax
  mov    %ebx,%r15d
  add    %rax,%r12
  nopl   0x0(%rax,%rax,1)
L19:
  movzbl (%r12),%edi
  call   <char_isspace@plt>
  test   %rax,%rax
  jne    L79
  add    $0x1,%r15d
  add    $0x1,%r12
  cmp    %r15d,%r13d
  jg     L19
L79:
  mov    %r15d,%r13d
  sub    %ebx,%r13d
  lea    0x1(%r13),%edi
  call   <jdns_alloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L70
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  movslq %r13d,%r13
  call   <memcpy@plt>
  movb   $0x0,(%r12,%r13,1)
  mov    %r15d,(%r14)
  jmp    L68
  cs nopw 0x0(%rax,%rax,1)
L70:
  xor    %r12d,%r12d
L68:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
unsigned char *string_getnextword(unsigned char *in, int size, int pos, int *newpos) {
  int n;
  int at;
  int len;
  unsigned char *out;
  at = pos;
  while (at < size && char_isspace(in[at]))
    ++at;
  if (at >= size)
    return 0;
  n = at;
  while (n < size && !char_isspace(in[n]))
    ++n;
  len = n - at;
  out = (unsigned char *)jdns_alloc(len + 1);
  if (!out)
    return 0;
  memcpy(out, in + at, len);
  out[len] = 0;
  *newpos = at + len;
  return out;
}
```
user: Translate the below assembly code into C programming language:

```asm
<string_getnextword>:
  endbr64
  push   %r15
  push   %r14
  mov    %rcx,%r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  movslq %edx,%rbp
  push   %rbx
  mov    %rbp,%rbx
  add    %rdi,%rbp
  sub    $0x8,%rsp
  cmp    %esi,%ebx
  jl     L36
  jmp    L45
  nopl   (%rax)
L20:
  add    $0x1,%ebx
  add    $0x1,%rbp
  cmp    %ebx,%r13d
  je     L45
L36:
  movzbl 0x0(%rbp),%edi
  call   <char_isspace@plt>
  test   %rax,%rax
  jne    L20
  movslq %ebx,%rax
  mov    %ebx,%r15d
  add    %rax,%r12
  nopl   0x0(%rax,%rax,1)
L2:
  movzbl (%r12),%edi
  call   <char_isspace@plt>
  test   %rax,%rax
  jne    L18
  add    $0x1,%r15d
  add    $0x1,%r12
  cmp    %r15d,%r13d
  jg     L2
L18:
  mov    %r15d,%r13d
  sub    %ebx,%r13d
  lea    0x1(%r13),%edi
  call   <jdns_alloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L45
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  movslq %r13d,%r13
  call   <memcpy@plt>
  movb   $0x0,(%r12,%r13,1)
  mov    %r15d,(%r14)
  jmp    L7
  cs nopw 0x0(%rax,%rax,1)
L45:
  xor    %r12d,%r12d
L7:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
unsigned char *string_getnextword(unsigned char *in, int size, int pos, int *newpos) {
  int n;
  int at;
  int len;
  unsigned char *out;
  at = pos;
  while (at < size && char_isspace(in[at]))
    ++at;
  if (at >= size)
    return 0;
  n = at;
  while (n < size && !char_isspace(in[n]))
    ++n;
  len = n - at;
  out = (unsigned char *)jdns_alloc(len + 1);
  if (!out)
    return 0;
  memcpy(out, in + at, len);
  out[len] = 0;
  *newpos = at + len;
  return out;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Bpda_file_error>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%esi
  mov    $0x1,%edi
  mov    $0x0,%eax
  call   <Kill_bplatform_and_exit@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D7
datatype: string

tool: D7: 
  BMotif: can't open %s for %s



assistant: ```c
void Bpda_file_error(file, mode) char *file;
char *mode;
{
  printf("\n  BMotif: can't open %s for %s\n\n", file, mode);
  Kill_bplatform_and_exit(1, 0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Bpda_file_error>:
  endbr64
  sub    $0x8,%rsp
  mov    %rsi,%rdx
  mov    %rdi,%rsi
  lea    D0(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%esi
  mov    $0x1,%edi
  mov    $0x0,%eax
  call   <Kill_bplatform_and_exit@plt>
  add    $0x8,%rsp
  ret
```

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

tool: D0: 
  BMotif: can't open %s for %s



assistant: ```c
void Bpda_file_error(file, mode) char *file;
char *mode;
{
  printf("\n  BMotif: can't open %s for %s\n\n", file, mode);
  Kill_bplatform_and_exit(1, 0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Bpda_file_error>:
  endbr64
  sub    $0x8,%rsp
  mov    %rsi,%rdx
  xor    %eax,%eax
  mov    %rdi,%rsi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %esi,%esi
  mov    $0x1,%edi
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <Kill_bplatform_and_exit@plt>
```

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

tool: D6: 
  BMotif: can't open %s for %s



assistant: ```c
void Bpda_file_error(file, mode) char *file;
char *mode;
{
  printf("\n  BMotif: can't open %s for %s\n\n", file, mode);
  Kill_bplatform_and_exit(1, 0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Bpda_file_error>:
  endbr64
  sub    $0x8,%rsp
  mov    %rsi,%rdx
  xor    %eax,%eax
  mov    %rdi,%rsi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %esi,%esi
  mov    $0x1,%edi
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <Kill_bplatform_and_exit@plt>
```

tool_calls: name: parse_data
arguments: label: D7
datatype: string

tool: D7: 
  BMotif: can't open %s for %s



assistant: ```c
void Bpda_file_error(file, mode) char *file;
char *mode;
{
  printf("\n  BMotif: can't open %s for %s\n\n", file, mode);
  Kill_bplatform_and_exit(1, 0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<test_add_sub>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%ecx
  mov    -0x4(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <test_add@plt>
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%ecx
  mov    -0x4(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <test_sub@plt>
  mov    -0xc(%rbp),%edx
  mov    -0x4(%rbp),%ecx
  mov    -0x8(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <test_sub@plt>
  nop
  leave
  ret
```

assistant: ```c
void test_add_sub(int64_t x, int64_t y, uint32_t n) {
  test_add(x, y, n);
  test_sub(x, y, n);
  test_sub(y, x, n);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<test_add_sub>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    %edx,%r12d
  call   <test_add@plt>
  mov    %r12d,%edx
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <test_sub@plt>
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <test_sub@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void test_add_sub(int64_t x, int64_t y, uint32_t n) {
  test_add(x, y, n);
  test_sub(x, y, n);
  test_sub(y, x, n);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<test_add_sub>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %esi,%ebp
  call   <test_add@plt>
  mov    %r13d,%edx
  mov    %ebp,%esi
  mov    %r12d,%edi
  call   <test_sub@plt>
  mov    %r13d,%edx
  mov    %r12d,%esi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <test_sub@plt>
```

assistant: ```c
void test_add_sub(int64_t x, int64_t y, uint32_t n) {
  test_add(x, y, n);
  test_sub(x, y, n);
  test_sub(y, x, n);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<test_add_sub>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %esi,%ebp
  call   <test_add@plt>
  mov    %r13d,%edx
  mov    %ebp,%esi
  mov    %r12d,%edi
  call   <test_sub@plt>
  mov    %r13d,%edx
  mov    %r12d,%esi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <test_sub@plt>
```

assistant: ```c
void test_add_sub(int64_t x, int64_t y, uint32_t n) {
  test_add(x, y, n);
  test_sub(x, y, n);
  test_sub(y, x, n);
}
```
user: What is the c source code of the assembly code below?

```asm
<voptwave>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  movsd  %xmm0,-0x38(%rbp)
  mov    %edi,-0x3c(%rbp)
  mov    %esi,-0x40(%rbp)
  mov    %edx,-0x44(%rbp)
  mov    %rcx,-0x50(%rbp)
  mov    %r8,-0x58(%rbp)
  mov    %r9,-0x60(%rbp)
  movsd  -0x38(%rbp),%xmm0
  movsd  D3(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x28(%rbp)
  mov    -0x50(%rbp),%rbx
  mov    -0x58(%rbp),%r12
  mov    -0x60(%rbp),%r14
  mov    $0x0,%r13d
  jmp    L14
L19:
  movsd  (%rbx),%xmm1
  movsd  D3(%rip),%xmm0
  addsd  %xmm1,%xmm0
  mulsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,(%r12)
  mov    %r14,%rax
  lea    0x4(%rax),%r14
  movl   $0x0,(%rax)
  mov    -0x40(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rax,%rbx
  mov    -0x44(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rax,%r12
  add    $0x1,%r13d
L14:
  cmp    -0x3c(%rbp),%r13d
  jl     L19
  mov    $0x0,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
type_infered: f64

tool: D3: 299792458

assistant: ```c
int voptwave(restwav, nvopt, svopt, swave, vopt, wave, stat)
double restwav;
int nvopt, svopt, swave;
const double vopt[];
double wave[];
int stat[];
{
  double r;
  register int ivopt, *statp;
  register const double *voptp;
  register double *wavep;
  r = restwav / 2.99792458e8;
  voptp = vopt;
  wavep = wave;
  statp = stat;
  for (ivopt = 0; ivopt < nvopt; ivopt++) {
    *wavep = r * (2.99792458e8 + *voptp);
    *(statp++) = 0;
    voptp += svopt;
    wavep += swave;
  }
  return 0;
}
```