ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<broken_longjmp>:
  endbr64
  push   %rbp
  pxor   %xmm0,%xmm0
  mov    %rsp,%rbp
  sub    $0x110,%rsp
  movdqu (%rdi),%xmm1
  movdqu 0x10(%rdi),%xmm2
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    0x20(%rdi),%rax
  movups %xmm0,(%rdi)
  lea    -0x110(%rbp),%rcx
  movq   $0x0,0x20(%rdi)
  movaps %xmm1,-0x110(%rbp)
  movups %xmm0,0x10(%rdi)
  mov    %rax,-0xf0(%rbp)
  xor    %eax,%eax
  movaps %xmm2,-0x100(%rbp)
  mov    -0x108(%rbp),%rdx
  mov    -0x110(%rbp),%rdi
  rdsspq %rax
  sub    -0x100(%rbp),%rax
  je     L28
  neg    %rax
  mov    %rax,%rsi
  shr    $0x3,%rax
  cmp    $0x7ff,%rsi
  jbe    L27
  mov    $0xff,%esi
L1:
  incsspq %rsi
  sub    $0xff,%rax
  cmp    $0xff,%rax
  ja     L1
L27:
  incsspq %rax
  mov    0x18(%rcx),%rsp
  mov    %rdi,%rbp
  jmp    *%rdx
  nopl   0x0(%rax)
L28:
  mov    0x18(%rcx),%rsp
  mov    %rdi,%rbp
  jmp    *%rdx
```

assistant: ```c
__attribute((no)) void broken_longjmp(void *p) {
  void *buf[32];
  __builtin_memcpy(buf, p, 5 * sizeof(void *));
  __builtin_memset(p, 0, 5 * sizeof(void *));
  __builtin_longjmp(buf, 1);
}
```
user: What is the c source code of the assembly code below?

```asm
<broken_longjmp>:
  endbr64
  push   %rbp
  pxor   %xmm0,%xmm0
  mov    %rsp,%rbp
  sub    $0x110,%rsp
  movdqu (%rdi),%xmm1
  movdqu 0x10(%rdi),%xmm2
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    0x20(%rdi),%rax
  movups %xmm0,(%rdi)
  lea    -0x110(%rbp),%rcx
  movq   $0x0,0x20(%rdi)
  movaps %xmm1,-0x110(%rbp)
  movups %xmm0,0x10(%rdi)
  mov    %rax,-0xf0(%rbp)
  xor    %eax,%eax
  movaps %xmm2,-0x100(%rbp)
  mov    -0x108(%rbp),%rdx
  mov    -0x110(%rbp),%rdi
  rdsspq %rax
  sub    -0x100(%rbp),%rax
  je     L9
  neg    %rax
  mov    %rax,%rsi
  shr    $0x3,%rax
  cmp    $0x7ff,%rsi
  jbe    L2
  mov    $0xff,%esi
L25:
  incsspq %rsi
  sub    $0xff,%rax
  cmp    $0xff,%rax
  ja     L25
L2:
  incsspq %rax
  mov    0x18(%rcx),%rsp
  mov    %rdi,%rbp
  jmp    *%rdx
  nopl   0x0(%rax)
L9:
  mov    0x18(%rcx),%rsp
  mov    %rdi,%rbp
  jmp    *%rdx
```

assistant: ```c
__attribute((no)) void broken_longjmp(void *p) {
  void *buf[32];
  __builtin_memcpy(buf, p, 5 * sizeof(void *));
  __builtin_memset(p, 0, 5 * sizeof(void *));
  __builtin_longjmp(buf, 1);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<mp2vpicx_set_fcode11>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x18,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  add    $0x18,%rax
  and    $0xf,%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x18,%rax
  mov    (%rax),%ecx
  mov    -0xc(%rbp),%eax
  shl    $0x4,%eax
  movzbl %al,%edx
  mov    -0x8(%rbp),%rax
  add    $0x18,%rax
  or     %ecx,%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void mp2vpicx_set_fcode11(uint8_t *p_mp2vpicx, uint8_t i_fcode11) {
  p_mp2vpicx[6] &= 0x0f;
  p_mp2vpicx[6] |= (i_fcode11 << 4) & 0xf0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<mp2vpicx_set_fcode11>:
  endbr64
  mov    0x18(%rdi),%eax
  and    $0xf,%eax
  shl    $0x4,%esi
  movzbl %sil,%esi
  or     %esi,%eax
  mov    %eax,0x18(%rdi)
  ret
```

assistant: ```c
void mp2vpicx_set_fcode11(uint8_t *p_mp2vpicx, uint8_t i_fcode11) {
  p_mp2vpicx[6] &= 0x0f;
  p_mp2vpicx[6] |= (i_fcode11 << 4) & 0xf0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mp2vpicx_set_fcode11>:
  endbr64
  mov    0x18(%rdi),%eax
  shl    $0x4,%esi
  movzbl %sil,%esi
  and    $0xf,%eax
  or     %esi,%eax
  mov    %eax,0x18(%rdi)
  ret
```

assistant: ```c
void mp2vpicx_set_fcode11(uint8_t *p_mp2vpicx, uint8_t i_fcode11) {
  p_mp2vpicx[6] &= 0x0f;
  p_mp2vpicx[6] |= (i_fcode11 << 4) & 0xf0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <daemon@plt>
  mov    $0x40,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  daemon(0, 0);
  return 64;
}
```
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
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <daemon@plt>
  mov    $0x40,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  daemon(0, 0);
  return 64;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  xor    %edi,%edi
  call   <daemon@plt>
  mov    $0x40,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  daemon(0, 0);
  return 64;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  xor    %edi,%edi
  call   <daemon@plt>
  mov    $0x40,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  daemon(0, 0);
  return 64;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<i_to_a>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  lea    D8(%rip),%rax
  mov    %rax,-0x8(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  js     L11
L28:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %ecx,%eax
  add    $0x30,%eax
  subq   $0x1,-0x8(%rbp)
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L28
  mov    -0x8(%rbp),%rax
  jmp    L5
L11:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    $0x30,%eax
  sub    %edx,%eax
  subq   $0x1,-0x8(%rbp)
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L11
  subq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movb   $0x2d,(%rax)
  mov    -0x8(%rbp),%rax
L5:
  pop    %rbp
  ret
```

assistant: ```c
char *i_to_a(int i) {
  static char buf[19 + 2];
  char *p = buf + 19 + 1;
  if (i >= 0) {
    do {
      *--p = '0' + (i % 10);
      i /= 10;
    } while (i != 0);
    return p;
  } else {
    do {
      *--p = '0' - (i % 10);
      i /= 10;
    } while (i != 0);
    *--p = '-';
  }
  return p;
}
```
user: What is the c source code of the assembly code below:

```asm
<i_to_a>:
  endbr64
  mov    %edi,%ecx
  lea    D5(%rip),%rdi
  mov    $0x30,%r8d
  test   %ecx,%ecx
  jns    L39
L23:
  mov    %rdi,%rax
  sub    $0x1,%rdi
  movslq %ecx,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%esi
  lea    (%r8,%rsi,2),%esi
  sub    %ecx,%esi
  mov    %sil,(%rdi)
  mov    %edx,%ecx
  test   %edx,%edx
  jne    L23
  lea    -0x2(%rax),%rdx
  movb   $0x2d,-0x1(%rdi)
L19:
  mov    %rdx,%rax
  ret
L39:
  mov    %rdi,%rdx
L9:
  sub    $0x1,%rdx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%ecx
  add    $0x30,%ecx
  mov    %cl,(%rdx)
  mov    %eax,%ecx
  test   %eax,%eax
  jne    L9
  jmp    L19
```

assistant: ```c
char *i_to_a(int i) {
  static char buf[19 + 2];
  char *p = buf + 19 + 1;
  if (i >= 0) {
    do {
      *--p = '0' + (i % 10);
      i /= 10;
    } while (i != 0);
    return p;
  } else {
    do {
      *--p = '0' - (i % 10);
      i /= 10;
    } while (i != 0);
    *--p = '-';
  }
  return p;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<i_to_a>:
  endbr64
  mov    %edi,%edx
  mov    $0x30,%r9d
  lea    D7(%rip),%rdi
  test   %edx,%edx
  jns    L29
  nopw   0x0(%rax,%rax,1)
L4:
  movslq %edx,%rcx
  mov    %edx,%eax
  mov    %rdi,%r8
  sub    $0x1,%rdi
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%eax
  sar    $0x22,%rcx
  sub    %eax,%ecx
  lea    (%rcx,%rcx,4),%esi
  lea    (%r9,%rsi,2),%esi
  sub    %edx,%esi
  mov    %ecx,%edx
  mov    %sil,(%rdi)
  test   %ecx,%ecx
  jne    L4
  sub    $0x2,%r8
  movb   $0x2d,-0x1(%rdi)
  mov    %r8,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L29:
  mov    %rdi,%r8
  mov    $0xcccccccd,%esi
  nopl   0x0(%rax,%rax,1)
L13:
  mov    %edx,%eax
  sub    $0x1,%r8
  imul   %rsi,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  add    $0x30,%edx
  mov    %dl,(%r8)
  mov    %eax,%edx
  test   %eax,%eax
  jne    L13
  mov    %r8,%rax
  ret
```

assistant: ```c
char *i_to_a(int i) {
  static char buf[19 + 2];
  char *p = buf + 19 + 1;
  if (i >= 0) {
    do {
      *--p = '0' + (i % 10);
      i /= 10;
    } while (i != 0);
    return p;
  } else {
    do {
      *--p = '0' - (i % 10);
      i /= 10;
    } while (i != 0);
    *--p = '-';
  }
  return p;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<i_to_a>:
  endbr64
  mov    %edi,%edx
  mov    $0x30,%r9d
  lea    D8(%rip),%rdi
  test   %edx,%edx
  jns    L21
  nopw   0x0(%rax,%rax,1)
L4:
  movslq %edx,%rcx
  mov    %edx,%eax
  mov    %rdi,%r8
  sub    $0x1,%rdi
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%eax
  sar    $0x22,%rcx
  sub    %eax,%ecx
  lea    (%rcx,%rcx,4),%esi
  lea    (%r9,%rsi,2),%esi
  sub    %edx,%esi
  mov    %ecx,%edx
  mov    %sil,(%rdi)
  test   %ecx,%ecx
  jne    L4
  sub    $0x2,%r8
  movb   $0x2d,-0x1(%rdi)
  mov    %r8,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L21:
  mov    %rdi,%r8
  mov    $0xcccccccd,%esi
  nopl   0x0(%rax,%rax,1)
L19:
  mov    %edx,%eax
  sub    $0x1,%r8
  imul   %rsi,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  add    $0x30,%edx
  mov    %dl,(%r8)
  mov    %eax,%edx
  test   %eax,%eax
  jne    L19
  mov    %r8,%rax
  ret
```

assistant: ```c
char *i_to_a(int i) {
  static char buf[19 + 2];
  char *p = buf + 19 + 1;
  if (i >= 0) {
    do {
      *--p = '0' + (i % 10);
      i /= 10;
    } while (i != 0);
    return p;
  } else {
    do {
      *--p = '0' - (i % 10);
      i /= 10;
    } while (i != 0);
    *--p = '-';
  }
  return p;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<isCopyOfHolder>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x2,%esi
  mov    $0x3,%edi
  call   <__CrestCall@plt>
  mov    $0x4,%edi
  call   <__CrestReturn@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void isCopyOfHolder(void) {
  {
    __CrestCall(3, 2);
    {
      __CrestReturn(4);
      return;
    }
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<isCopyOfHolder>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%esi
  mov    $0x3,%edi
  call   <__CrestCall@plt>
  mov    $0x4,%edi
  call   <__CrestReturn@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void isCopyOfHolder(void) {
  {
    __CrestCall(3, 2);
    {
      __CrestReturn(4);
      return;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<isCopyOfHolder>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%edi
  mov    $0x2,%esi
  call   <__CrestCall@plt>
  mov    $0x4,%edi
  add    $0x8,%rsp
  jmp    1060 <__CrestReturn@plt>
```

assistant: ```c
void isCopyOfHolder(void) {
  {
    __CrestCall(3, 2);
    {
      __CrestReturn(4);
      return;
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<isCopyOfHolder>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%edi
  mov    $0x2,%esi
  call   <__CrestCall@plt>
  mov    $0x4,%edi
  add    $0x8,%rsp
  jmp    1060 <__CrestReturn@plt>
```

assistant: ```c
void isCopyOfHolder(void) {
  {
    __CrestCall(3, 2);
    {
      __CrestReturn(4);
      return;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getSum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    %ecx,-0x14(%rbp)
  cmpl   $0x0,-0x10(%rbp)
  js     L11
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L11
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L32
L11:
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L23
L32:
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%esi
  mov    -0x14(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    $0x0,%r9d
  mov    %ecx,%r8d
  mov    %edx,%ecx
  mov    %esi,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <getSumUtil@plt>
L23:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
type_infered: string

tool: D1: Invalid Input

assistant: ```c
int getSum(int *st, int n, int qs, int qe) {
  if (qs < 0 || qe > n - 1 || qs > qe) {
    printf("Invalid Input");
    return -1;
  }
  return getSumUtil(st, 0, n - 1, qs, qe, 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getSum>:
  endbr64
  sub    $0x8,%rsp
  mov    %edx,%eax
  mov    %ecx,%r8d
  mov    %edx,%ecx
  shr    $0x1f,%ecx
  cmp    %r8d,%edx
  setg   %dl
  or     %cl,%dl
  jne    L23
  cmp    %r8d,%esi
  jle    L23
  lea    -0x1(%rsi),%edx
  mov    $0x0,%r9d
  mov    %eax,%ecx
  mov    $0x0,%esi
  call   <getSumUtil@plt>
L0:
  add    $0x8,%rsp
  ret
L23:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L0
```

tool_calls: name: unpack_data
arguments: tag: D4
type_infered: string

tool: D4: Invalid Input

assistant: ```c
int getSum(int *st, int n, int qs, int qe) {
  if (qs < 0 || qe > n - 1 || qs > qe) {
    printf("Invalid Input");
    return -1;
  }
  return getSumUtil(st, 0, n - 1, qs, qe, 0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getSum>:
  endbr64
  mov    %edx,%r10d
  shr    $0x1f,%edx
  cmp    %ecx,%r10d
  setg   %al
  or     %dl,%al
  jne    L3
  mov    %ecx,%r8d
  cmp    %ecx,%esi
  jle    L3
  lea    -0x1(%rsi),%edx
  xor    %r9d,%r9d
  mov    %r10d,%ecx
  xor    %esi,%esi
  jmp    1060 <getSumUtil@plt>
  nopl   0x0(%rax,%rax,1)
L3:
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D0: Invalid Input

assistant: ```c
int getSum(int *st, int n, int qs, int qe) {
  if (qs < 0 || qe > n - 1 || qs > qe) {
    printf("Invalid Input");
    return -1;
  }
  return getSumUtil(st, 0, n - 1, qs, qe, 0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getSum>:
  endbr64
  mov    %edx,%r10d
  shr    $0x1f,%edx
  cmp    %ecx,%r10d
  setg   %al
  or     %dl,%al
  jne    L0
  mov    %ecx,%r8d
  cmp    %ecx,%esi
  jle    L0
  lea    -0x1(%rsi),%edx
  xor    %r9d,%r9d
  mov    %r10d,%ecx
  xor    %esi,%esi
  jmp    1060 <getSumUtil@plt>
  nopl   0x0(%rax,%rax,1)
L0:
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
data_type: string

tool: D6: Invalid Input

assistant: ```c
int getSum(int *st, int n, int qs, int qe) {
  if (qs < 0 || qe > n - 1 || qs > qe) {
    printf("Invalid Input");
    return -1;
  }
  return getSumUtil(st, 0, n - 1, qs, qe, 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<compute>:
  endbr64
  sub    $0x218,%rsp
  mov    %edi,%eax
  mov    %rdx,%rdi
  mov    %fs:0x28,%rdx
  mov    %rdx,0x208(%rsp)
  xor    %edx,%edx
  test   %eax,%eax
  je     L29
  mov    %rsi,%r8
  mov    %rcx,%r9
  mov    %rsp,%r10
  lea    0x100(%rsp),%rcx
  mov    %r10,%rax
L31:
  movq   $0x0,(%rax)
  movq   $0x0,0x100(%rax)
  add    $0x8,%rax
  cmp    %rcx,%rax
  jne    L31
  mov    %r8,%rsi
  add    $0x2000,%r8
  jmp    L30
L61:
  add    $0x100,%rdi
  add    $0x8,%rsi
  cmp    %r8,%rsi
  je     L68
L30:
  movsd  (%rsi),%xmm3
  movsd  0x2000(%rsi),%xmm2
  mov    %r10,%rax
  mov    %rdi,%rdx
L86:
  movsd  (%rdx),%xmm0
  movapd %xmm0,%xmm1
  mulsd  %xmm3,%xmm1
  addsd  (%rax),%xmm1
  movsd  %xmm1,(%rax)
  mulsd  %xmm2,%xmm0
  addsd  0x100(%rax),%xmm0
  movsd  %xmm0,0x100(%rax)
  add    $0x8,%rdx
  add    $0x8,%rax
  cmp    %rcx,%rax
  jne    L86
  jmp    L61
L68:
  mov    $0x0,%eax
L20:
  movsd  (%rsp,%rax,1),%xmm0
  movsd  %xmm0,(%r9,%rax,1)
  add    $0x8,%rax
  cmp    $0x100,%rax
  jne    L20
  add    $0x2000,%r9
  mov    $0x0,%eax
  lea    0x100(%rsp),%rdx
L49:
  movsd  (%rdx,%rax,1),%xmm0
  movsd  %xmm0,(%r9,%rax,1)
  add    $0x8,%rax
  cmp    $0x100,%rax
  jne    L49
L29:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  add    $0x218,%rsp
  ret
L53:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void compute(int flag, double rows[2][1024], double col[1024][32], double *prod, int num_rows) {
#pragma HLS inline off
  if (flag) {
    double local_prod[2][32];
#pragma HLS array_partition variable = local_prod dim = 1 complete
    int i, j, k;
    for (j = 0; j < 32; j++) {
#pragma HLS pipeline
      for (i = 0; i < 2; i++) {
#pragma HLS unroll
        local_prod[i][j] = 0.0;
      }
    }
    for (j = 0; j < 1024; j++) {
      for (k = 0; k < 32; k++) {
#pragma HLS pipeline
#pragma HLS dependence variable = local_prod inter false
        for (i = 0; i < 2; i++) {
#pragma HLS unroll
          local_prod[i][k] += rows[i][j] * col[j][k];
        }
      }
    }
    for (i = 0; i < 2; i++) {
      for (j = 0; j < 32; j++) {
#pragma HLS pipeline
        prod[i * 1024 + j] = local_prod[i][j];
      }
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<TA_GetVersionString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D5(%rip),%rax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D5
type_infered: string

tool: D5: 0.6.0-dev (Mar 15 2022 19:27:03)

assistant: ```c
const char *TA_GetVersionString(void) {
  if (sizeof("dev") > 1) {
    return "0"
           "."
           "6"
           "."
           "0"
           "-"
           "dev"
           " "
           "("
           "Mar 15 2022"
           " "
           "19:27:03"
           ")";
  } else {
    return "0"
           "."
           "6"
           "."
           "0"
           " "
           "("
           "Mar 15 2022"
           " "
           "19:27:03"
           ")";
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<TA_GetVersionString>:
  endbr64
  lea    D2(%rip),%rax
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
type: string

tool: D2: 0.6.0-dev (Mar 15 2022 19:27:03)

assistant: ```c
const char *TA_GetVersionString(void) {
  if (sizeof("dev") > 1) {
    return "0"
           "."
           "6"
           "."
           "0"
           "-"
           "dev"
           " "
           "("
           "Mar 15 2022"
           " "
           "19:27:03"
           ")";
  } else {
    return "0"
           "."
           "6"
           "."
           "0"
           " "
           "("
           "Mar 15 2022"
           " "
           "19:27:03"
           ")";
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<TA_GetVersionString>:
  endbr64
  lea    D4(%rip),%rax
  ret
```

tool_calls: name: read_data
arguments: data_label: D4
datatype: string

tool: D4: 0.6.0-dev (Mar 15 2022 19:27:03)

assistant: ```c
const char *TA_GetVersionString(void) {
  if (sizeof("dev") > 1) {
    return "0"
           "."
           "6"
           "."
           "0"
           "-"
           "dev"
           " "
           "("
           "Mar 15 2022"
           " "
           "19:27:03"
           ")";
  } else {
    return "0"
           "."
           "6"
           "."
           "0"
           " "
           "("
           "Mar 15 2022"
           " "
           "19:27:03"
           ")";
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<TA_GetVersionString>:
  endbr64
  lea    D8(%rip),%rax
  ret
```

tool_calls: name: disassemble_data
arguments: label: D8
type_guessed: string

tool: D8: 0.6.0-dev (Mar 15 2022 19:27:03)

assistant: ```c
const char *TA_GetVersionString(void) {
  if (sizeof("dev") > 1) {
    return "0"
           "."
           "6"
           "."
           "0"
           "-"
           "dev"
           " "
           "("
           "Mar 15 2022"
           " "
           "19:27:03"
           ")";
  } else {
    return "0"
           "."
           "6"
           "."
           "0"
           " "
           "("
           "Mar 15 2022"
           " "
           "19:27:03"
           ")";
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<__strcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    -0x20(%rbp),%rdx
  mov    %rax,%rsi
  mov    %rdx,%rdi
L4:
  lods   %ds:(%rsi),%al
  scas   %es:(%rdi),%al
  jne    L16
  test   %al,%al
  jne    L4
  xor    %eax,%eax
  jmp    L28
L16:
  sbb    %eax,%eax
  or     $0x1,%al
L28:
  mov    %edi,%edx
  mov    %esi,%ecx
  mov    %eax,-0xc(%rbp)
  mov    %ecx,-0x8(%rbp)
  mov    %edx,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int __strcmp(const char *s1, const char *s2) {
  int d0, d1, ret;
  asm volatile("1: lodsb;"
               "scasb;"
               "jne 2f;"
               "testb %%al, %%al;"
               "jne 1b;"
               "xorl %%eax, %%eax;"
               "jmp 3f;"
               "2: sbbl %%eax, %%eax;"
               "orb $1, %%al;"
               "3:"
               : "=a"(ret), "=&S"(d0), "=&D"(d1)
               : "1"(s1), "2"(s2)
               : "memory");
  return ret;
}
```
user: What is the c source code of the assembly code below?

```asm
<__strcmp>:
  endbr64
  mov    %rdi,%rax
  mov    %rsi,%rdi
  mov    %rax,%rsi
L27:
  lods   %ds:(%rsi),%al
  scas   %es:(%rdi),%al
  jne    L24
  test   %al,%al
  jne    L27
  xor    %eax,%eax
  jmp    L15
L24:
  sbb    %eax,%eax
  or     $0x1,%al
L15:
  ret
```

assistant: ```c
int __strcmp(const char *s1, const char *s2) {
  int d0, d1, ret;
  asm volatile("1: lodsb;"
               "scasb;"
               "jne 2f;"
               "testb %%al, %%al;"
               "jne 1b;"
               "xorl %%eax, %%eax;"
               "jmp 3f;"
               "2: sbbl %%eax, %%eax;"
               "orb $1, %%al;"
               "3:"
               : "=a"(ret), "=&S"(d0), "=&D"(d1)
               : "1"(s1), "2"(s2)
               : "memory");
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<__strcmp>:
  endbr64
  mov    %rdi,%r8
  mov    %rsi,%rdi
  mov    %r8,%rsi
L11:
  lods   %ds:(%rsi),%al
  scas   %es:(%rdi),%al
  jne    L29
  test   %al,%al
  jne    L11
  xor    %eax,%eax
  jmp    L28
L29:
  sbb    %eax,%eax
  or     $0x1,%al
L28:
  ret
```

assistant: ```c
int __strcmp(const char *s1, const char *s2) {
  int d0, d1, ret;
  asm volatile("1: lodsb;"
               "scasb;"
               "jne 2f;"
               "testb %%al, %%al;"
               "jne 1b;"
               "xorl %%eax, %%eax;"
               "jmp 3f;"
               "2: sbbl %%eax, %%eax;"
               "orb $1, %%al;"
               "3:"
               : "=a"(ret), "=&S"(d0), "=&D"(d1)
               : "1"(s1), "2"(s2)
               : "memory");
  return ret;
}
```