ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<asn1_item_clear>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  add    $0xffffff80,%eax
  cmp    $0x5,%eax
  ja     L34
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D12(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D12(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    -0x20(%rbp),%rax
  mov    0x8(%rax),%rax
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L60
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L60
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rcx
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   *%rcx
  jmp    L34
L60:
  mov    -0x18(%rbp),%rax
  movq   $0x0,(%rax)
  jmp    L34
  mov    -0x20(%rbp),%rax
  mov    0x4(%rax),%eax
  test   %eax,%eax
  je     L26
  mov    -0x20(%rbp),%rax
  mov    0x4(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <asn1_template_clear@plt>
  jmp    L34
L26:
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <asn1_primitive_clear@plt>
  jmp    L34
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <asn1_primitive_clear@plt>
  jmp    L34
  mov    -0x18(%rbp),%rax
  movq   $0x0,(%rax)
  nop
L34:
  nop
  leave
  ret
```

assistant: ```c
void asn1_item_clear(ASN1_VALUE **pval, const ASN1_ITEM *it) {
  const ASN1_EXTERN_FUNCS *ef;
  switch (it->itype) {
  case ASN1_ITYPE_EXTERN:
    ef = it->funcs;
    if (ef && ef->asn1_ex_clear)
      ef->asn1_ex_clear(pval, it);
    else
      *pval = NULL;
    break;
  case ASN1_ITYPE_PRIMITIVE:
    if (it->templates)
      asn1_template_clear(pval, it->templates);
    else
      asn1_primitive_clear(pval, it);
    break;
  case ASN1_ITYPE_MSTRING:
    asn1_primitive_clear(pval, it);
    break;
  case ASN1_ITYPE_CHOICE:
  case ASN1_ITYPE_SEQUENCE:
  case ASN1_ITYPE_NDEF_SEQUENCE:
    *pval = NULL;
    break;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<asn1_item_clear>:
  endbr64
  mov    (%rsi),%eax
  add    $0xffffff80,%eax
  cmp    $0x5,%eax
  ja     L30
  sub    $0x8,%rsp
  mov    %eax,%eax
  lea    D1(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    0x8(%rsi),%rax
  test   %rax,%rax
  je     L18
  mov    (%rax),%rax
  test   %rax,%rax
  je     L18
  call   *%rax
  jmp    L38
L18:
  movq   $0x0,(%rdi)
  jmp    L38
  mov    0x4(%rsi),%eax
  test   %eax,%eax
  je     L79
  mov    %eax,%esi
  call   <asn1_template_clear@plt>
  jmp    L38
L79:
  call   <asn1_primitive_clear@plt>
  jmp    L38
  call   <asn1_primitive_clear@plt>
L38:
  add    $0x8,%rsp
  ret
  movq   $0x0,(%rdi)
  jmp    L38
L30:
  ret
```

assistant: ```c
void asn1_item_clear(ASN1_VALUE **pval, const ASN1_ITEM *it) {
  const ASN1_EXTERN_FUNCS *ef;
  switch (it->itype) {
  case ASN1_ITYPE_EXTERN:
    ef = it->funcs;
    if (ef && ef->asn1_ex_clear)
      ef->asn1_ex_clear(pval, it);
    else
      *pval = NULL;
    break;
  case ASN1_ITYPE_PRIMITIVE:
    if (it->templates)
      asn1_template_clear(pval, it->templates);
    else
      asn1_primitive_clear(pval, it);
    break;
  case ASN1_ITYPE_MSTRING:
    asn1_primitive_clear(pval, it);
    break;
  case ASN1_ITYPE_CHOICE:
  case ASN1_ITYPE_SEQUENCE:
  case ASN1_ITYPE_NDEF_SEQUENCE:
    *pval = NULL;
    break;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<asn1_item_clear>:
  endbr64
  mov    (%rsi),%eax
  add    $0xffffff80,%eax
  cmp    $0x5,%eax
  ja     L3
  lea    D0(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nop
  mov    0x8(%rsi),%rax
  test   %rax,%rax
  jne    L22
  nopl   0x0(%rax)
L31:
  movq   $0x0,(%rdi)
L3:
  ret
  nopl   0x0(%rax,%rax,1)
  mov    0x4(%rsi),%r8d
  test   %r8d,%r8d
  jne    L38
  jmp    1060 <asn1_primitive_clear@plt>
  xchg   %ax,%ax
L22:
  mov    (%rax),%rax
  test   %rax,%rax
  je     L31
  jmp    *%rax
  nopw   0x0(%rax,%rax,1)
L38:
  mov    %r8d,%esi
  jmp    1070 <asn1_template_clear@plt>
```

assistant: ```c
void asn1_item_clear(ASN1_VALUE **pval, const ASN1_ITEM *it) {
  const ASN1_EXTERN_FUNCS *ef;
  switch (it->itype) {
  case ASN1_ITYPE_EXTERN:
    ef = it->funcs;
    if (ef && ef->asn1_ex_clear)
      ef->asn1_ex_clear(pval, it);
    else
      *pval = NULL;
    break;
  case ASN1_ITYPE_PRIMITIVE:
    if (it->templates)
      asn1_template_clear(pval, it->templates);
    else
      asn1_primitive_clear(pval, it);
    break;
  case ASN1_ITYPE_MSTRING:
    asn1_primitive_clear(pval, it);
    break;
  case ASN1_ITYPE_CHOICE:
  case ASN1_ITYPE_SEQUENCE:
  case ASN1_ITYPE_NDEF_SEQUENCE:
    *pval = NULL;
    break;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<asn1_item_clear>:
  endbr64
  mov    (%rsi),%eax
  add    $0xffffff80,%eax
  cmp    $0x5,%eax
  ja     L32
  lea    D9(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nop
  mov    0x8(%rsi),%rax
  test   %rax,%rax
  jne    L29
  nopl   0x0(%rax)
L4:
  movq   $0x0,(%rdi)
L32:
  ret
  nopl   0x0(%rax,%rax,1)
  mov    0x4(%rsi),%r8d
  test   %r8d,%r8d
  jne    L12
  jmp    1060 <asn1_primitive_clear@plt>
  xchg   %ax,%ax
L29:
  mov    (%rax),%rax
  test   %rax,%rax
  je     L4
  jmp    *%rax
  nopw   0x0(%rax,%rax,1)
L12:
  mov    %r8d,%esi
  jmp    1070 <asn1_template_clear@plt>
```

assistant: ```c
void asn1_item_clear(ASN1_VALUE **pval, const ASN1_ITEM *it) {
  const ASN1_EXTERN_FUNCS *ef;
  switch (it->itype) {
  case ASN1_ITYPE_EXTERN:
    ef = it->funcs;
    if (ef && ef->asn1_ex_clear)
      ef->asn1_ex_clear(pval, it);
    else
      *pval = NULL;
    break;
  case ASN1_ITYPE_PRIMITIVE:
    if (it->templates)
      asn1_template_clear(pval, it->templates);
    else
      asn1_primitive_clear(pval, it);
    break;
  case ASN1_ITYPE_MSTRING:
    asn1_primitive_clear(pval, it);
    break;
  case ASN1_ITYPE_CHOICE:
  case ASN1_ITYPE_SEQUENCE:
  case ASN1_ITYPE_NDEF_SEQUENCE:
    *pval = NULL;
    break;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sqrtll>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  movq   $0x1,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L0
L35:
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdx
  shr    $0x3f,%rdx
  add    %rdx,%rax
  sar    %rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  imul   %rax,%rax
  cmp    %rax,-0x28(%rbp)
  jle    L45
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L0
L45:
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
L0:
  mov    -0x18(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jg     L26
  mov    -0x10(%rbp),%rax
  sub    -0x18(%rbp),%rax
  cmp    $0x1,%rax
  jg     L35
L26:
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long long sqrtll(long long n) {
  long long lo = 1;
  long long hi = n;
  long long pivot;
  while (lo <= hi && hi - lo > 1) {
    pivot = (hi + lo) / 2;
    if (pivot * pivot < n)
      lo = pivot;
    else
      hi = pivot;
  }
  return lo;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<sqrtll>:
  endbr64
  mov    %rdi,%rcx
  mov    $0x1,%edx
  jmp    L10
L4:
  lea    (%rdx,%rcx,1),%rsi
  mov    %rsi,%rax
  shr    $0x3f,%rax
  add    %rsi,%rax
  sar    %rax
  mov    %rax,%rsi
  imul   %rax,%rsi
  cmp    %rdi,%rsi
  cmovge %rax,%rcx
  cmovl  %rax,%rdx
L10:
  cmp    %rcx,%rdx
  jg     L22
  mov    %rcx,%rax
  sub    %rdx,%rax
  cmp    $0x1,%rax
  jg     L4
L22:
  mov    %rdx,%rax
  ret
```

assistant: ```c
long long sqrtll(long long n) {
  long long lo = 1;
  long long hi = n;
  long long pivot;
  while (lo <= hi && hi - lo > 1) {
    pivot = (hi + lo) / 2;
    if (pivot * pivot < n)
      lo = pivot;
    else
      hi = pivot;
  }
  return lo;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<sqrtll>:
  endbr64
  mov    %rdi,%rdx
  mov    $0x1,%r8d
  jmp    L23
  nop
L22:
  mov    %rdx,%rax
  sub    %r8,%rax
  cmp    $0x1,%rax
  jle    L29
  lea    (%r8,%rdx,1),%rcx
  mov    %rcx,%rax
  shr    $0x3f,%rax
  add    %rcx,%rax
  sar    %rax
  mov    %rax,%rcx
  imul   %rax,%rcx
  cmp    %rdi,%rcx
  cmovge %rax,%rdx
  cmovl  %rax,%r8
L23:
  cmp    %rdx,%r8
  jle    L22
L29:
  mov    %r8,%rax
  ret
```

assistant: ```c
long long sqrtll(long long n) {
  long long lo = 1;
  long long hi = n;
  long long pivot;
  while (lo <= hi && hi - lo > 1) {
    pivot = (hi + lo) / 2;
    if (pivot * pivot < n)
      lo = pivot;
    else
      hi = pivot;
  }
  return lo;
}
```
user: What is the c source code of the assembly code below:

```asm
<sqrtll>:
  endbr64
  mov    %rdi,%rdx
  mov    $0x1,%r8d
  jmp    L3
  nop
L28:
  mov    %rdx,%rax
  sub    %r8,%rax
  cmp    $0x1,%rax
  jle    L15
  lea    (%r8,%rdx,1),%rcx
  mov    %rcx,%rax
  shr    $0x3f,%rax
  add    %rcx,%rax
  sar    %rax
  mov    %rax,%rcx
  imul   %rax,%rcx
  cmp    %rdi,%rcx
  cmovge %rax,%rdx
  cmovl  %rax,%r8
L3:
  cmp    %rdx,%r8
  jle    L28
L15:
  mov    %r8,%rax
  ret
```

assistant: ```c
long long sqrtll(long long n) {
  long long lo = 1;
  long long hi = n;
  long long pivot;
  while (lo <= hi && hi - lo > 1) {
    pivot = (hi + lo) / 2;
    if (pivot * pivot < n)
      lo = pivot;
    else
      hi = pivot;
  }
  return lo;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L7
L2:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <compose@plt>
  addl   $0x1,-0x4(%rbp)
L7:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L2
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int i;
  for (i = 1; i < argc; ++i) {
    compose(argv[i], strlen(argv[i]));
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  cmp    $0x1,%edi
  jle    L2
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    0x8(%rsi),%rbx
  lea    -0x2(%rdi),%eax
  lea    0x10(%rsi,%rax,8),%rbp
L4:
  mov    (%rbx),%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  mov    (%rbx),%rdi
  call   <compose@plt>
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L4
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int i;
  for (i = 1; i < argc; ++i) {
    compose(argv[i], strlen(argv[i]));
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  cmp    $0x1,%edi
  jle    L15
  push   %rbp
  lea    -0x2(%rdi),%eax
  push   %rbx
  lea    0x10(%rsi,%rax,8),%rbp
  lea    0x8(%rsi),%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L3:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <strlen@plt>
  mov    -0x8(%rbx),%rdi
  mov    %eax,%esi
  call   <compose@plt>
  cmp    %rbp,%rbx
  jne    L3
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L15:
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int i;
  for (i = 1; i < argc; ++i) {
    compose(argv[i], strlen(argv[i]));
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  cmp    $0x1,%edi
  jle    L8
  push   %rbp
  lea    -0x2(%rdi),%eax
  push   %rbx
  lea    0x10(%rsi,%rax,8),%rbp
  lea    0x8(%rsi),%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L11:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <strlen@plt>
  mov    -0x8(%rbx),%rdi
  mov    %eax,%esi
  call   <compose@plt>
  cmp    %rbp,%rbx
  jne    L11
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L8:
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int i;
  for (i = 1; i < argc; ++i) {
    compose(argv[i], strlen(argv[i]));
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x2,%edi
  call   <play@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  play(2);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <play@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  play(2);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <play@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  play(2);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <play@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  play(2);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <unlink@plt>
  mov    $0x0,%eax
  call   <datagram_test@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D6
type_infered: string

tool: D6: /tmp/123

assistant: ```c
int main(int argc, char *argv[]) {
  (void)unlink("/tmp/123");
  datagram_test();
  if (0)
    stream_test();
  return (0);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <unlink@plt>
  mov    $0x0,%eax
  call   <datagram_test@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D2
datatype: string

tool: D2: /tmp/123

assistant: ```c
int main(int argc, char *argv[]) {
  (void)unlink("/tmp/123");
  datagram_test();
  if (0)
    stream_test();
  return (0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <unlink@plt>
  xor    %eax,%eax
  call   <datagram_test@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D0
type: string

tool: D0: /tmp/123

assistant: ```c
int main(int argc, char *argv[]) {
  (void)unlink("/tmp/123");
  datagram_test();
  if (0)
    stream_test();
  return (0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

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

tool_calls: name: unpack_data
arguments: data_label: D1
type_infered: string

tool: D1: /tmp/123

assistant: ```c
int main(int argc, char *argv[]) {
  (void)unlink("/tmp/123");
  datagram_test();
  if (0)
    stream_test();
  return (0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<exit_signal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    $0x0,%eax
  call   <reconos_cleanup@plt>
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D1
datatype: string

tool: D1: [reconos-core] aborted


assistant: ```c
void exit_signal(int sig) {
  reconos_cleanup();
  printf("[reconos-core] aborted\n");
  exit(0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

tool_calls: name: retrieve_data
arguments: data_tag: D9
type: string

tool: D9: [reconos-core] aborted


assistant: ```c
void exit_signal(int sig) {
  reconos_cleanup();
  printf("[reconos-core] aborted\n");
  exit(0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<exit_signal>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <reconos_cleanup@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1090 <exit@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D1
type: string

tool: D1: [reconos-core] aborted


assistant: ```c
void exit_signal(int sig) {
  reconos_cleanup();
  printf("[reconos-core] aborted\n");
  exit(0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<exit_signal>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <reconos_cleanup@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1090 <exit@plt>
```

tool_calls: name: read_data
arguments: tag: D0
datatype: string

tool: D0: [reconos-core] aborted


assistant: ```c
void exit_signal(int sig) {
  reconos_cleanup();
  printf("[reconos-core] aborted\n");
  exit(0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  add    $0x136ff696,%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x18(%rbp)
  nop
  mov    -0x10(%rbp),%rax
  mov    $0xab39b775,%edx
  add    %rax,%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local2;
  unsigned long local1;
  char copy11;
  unsigned short copy12;
  {
    state[0UL] = (input[0UL] + 51238316UL) + 274866410UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      local2 = 0UL;
      while (local2 < 0UL) {
        copy11 = *((char *)(&state[local1]) + 5);
        *((char *)(&state[local1]) + 5) = *((char *)(&state[local1]) + 3);
        *((char *)(&state[local1]) + 3) = copy11;
        copy11 = *((char *)(&state[local1]) + 3);
        *((char *)(&state[local1]) + 3) = *((char *)(&state[local1]) + 2);
        *((char *)(&state[local1]) + 2) = copy11;
        copy12 = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 3);
        *((unsigned short *)(&state[local1]) + 3) = copy12;
        local2++;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] + 736209367UL) + 2136477086UL;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<RandomFunc>:
  endbr64
  mov    $0xbea9ae0b,%eax
  add    (%rdi),%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local2;
  unsigned long local1;
  char copy11;
  unsigned short copy12;
  {
    state[0UL] = (input[0UL] + 51238316UL) + 274866410UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      local2 = 0UL;
      while (local2 < 0UL) {
        copy11 = *((char *)(&state[local1]) + 5);
        *((char *)(&state[local1]) + 5) = *((char *)(&state[local1]) + 3);
        *((char *)(&state[local1]) + 3) = copy11;
        copy11 = *((char *)(&state[local1]) + 3);
        *((char *)(&state[local1]) + 3) = *((char *)(&state[local1]) + 2);
        *((char *)(&state[local1]) + 2) = copy11;
        copy12 = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 3);
        *((unsigned short *)(&state[local1]) + 3) = copy12;
        local2++;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] + 736209367UL) + 2136477086UL;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<RandomFunc>:
  endbr64
  mov    $0xbea9ae0b,%eax
  add    (%rdi),%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local2;
  unsigned long local1;
  char copy11;
  unsigned short copy12;
  {
    state[0UL] = (input[0UL] + 51238316UL) + 274866410UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      local2 = 0UL;
      while (local2 < 0UL) {
        copy11 = *((char *)(&state[local1]) + 5);
        *((char *)(&state[local1]) + 5) = *((char *)(&state[local1]) + 3);
        *((char *)(&state[local1]) + 3) = copy11;
        copy11 = *((char *)(&state[local1]) + 3);
        *((char *)(&state[local1]) + 3) = *((char *)(&state[local1]) + 2);
        *((char *)(&state[local1]) + 2) = copy11;
        copy12 = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 3);
        *((unsigned short *)(&state[local1]) + 3) = copy12;
        local2++;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] + 736209367UL) + 2136477086UL;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<RandomFunc>:
  endbr64
  mov    $0xbea9ae0b,%eax
  add    (%rdi),%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local2;
  unsigned long local1;
  char copy11;
  unsigned short copy12;
  {
    state[0UL] = (input[0UL] + 51238316UL) + 274866410UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      local2 = 0UL;
      while (local2 < 0UL) {
        copy11 = *((char *)(&state[local1]) + 5);
        *((char *)(&state[local1]) + 5) = *((char *)(&state[local1]) + 3);
        *((char *)(&state[local1]) + 3) = copy11;
        copy11 = *((char *)(&state[local1]) + 3);
        *((char *)(&state[local1]) + 3) = *((char *)(&state[local1]) + 2);
        *((char *)(&state[local1]) + 2) = copy11;
        copy12 = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 3);
        *((unsigned short *)(&state[local1]) + 3) = copy12;
        local2++;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] + 736209367UL) + 2136477086UL;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<byte_copy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L40
  cmpq   $0x0,-0x20(%rbp)
  je     L40
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L24
L44:
  mov    -0x8(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x10(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L24:
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  test   %rax,%rax
  jne    L44
  jmp    L7
L40:
  nop
L7:
  pop    %rbp
  ret
```

assistant: ```c
void byte_copy(void *dst, const void *src, size_t size) {
  if (!dst || !src)
    return;
  char *restrict d = dst;
  const char *restrict s = src;
  while (size--)
    *d++ = *s++;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<byte_copy>:
  endbr64
  test   %rdi,%rdi
  je     L24
  test   %rsi,%rsi
  je     L24
  test   %rdx,%rdx
  je     L24
  mov    $0x0,%eax
L9:
  movzbl (%rsi,%rax,1),%ecx
  mov    %cl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L9
L24:
  ret
```

assistant: ```c
void byte_copy(void *dst, const void *src, size_t size) {
  if (!dst || !src)
    return;
  char *restrict d = dst;
  const char *restrict s = src;
  while (size--)
    *d++ = *s++;
}
```