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

```asm
<invSqrt>:
  endbr64
  movq   %xmm0,%rdx
  mov    $0x5f3759df,%eax
  movsd  D9(%rip),%xmm1
  mulsd  D14(%rip),%xmm0
  sar    %rdx
  sub    %rdx,%rax
  movq   %rax,%xmm2
  mulsd  %xmm2,%xmm0
  mulsd  %xmm2,%xmm0
  subsd  %xmm0,%xmm1
  mulsd  %xmm2,%xmm1
  movapd %xmm1,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: label: D9
type_infered: f64
name: parse_data
arguments: label: D14
type_infered: f64

tool: D9: 1.5

tool: D14: 0.5

assistant: ```c
double invSqrt(double x) {
  double halfx = 0.5 * x;
  double y = x;
  long i = (*(long *)&y);
  i = 0x5f3759df - (i >> 1);
  y = (*(double *)&i);
  y = y * (1.5 - (halfx * y * y));
  return y;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<is_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x3,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  imul   %rax,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L23
L22:
  addq   $0x1,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  shl    $0x2,%rax
  add    %rax,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L23:
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jae    L13
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x10(%rbp)
  mov    %rdx,%rax
  test   %rax,%rax
  jne    L22
L13:
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x10(%rbp)
  mov    %rdx,%rax
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
int is_prime(unsigned long int candidate) {
  unsigned long int divn = 3;
  unsigned long int sq = divn * divn;
  while (sq < candidate && candidate % divn != 0) {
    ++divn;
    sq += 4 * divn;
    ++divn;
  }
  return candidate % divn != 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<is_prime>:
  endbr64
  cmp    $0x9,%rdi
  jbe    L26
  movabs $0xaaaaaaaaaaaaaaab,%rdx
  mov    %rdi,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    %rax
  and    $0xfffffffffffffffe,%rdx
  add    %rax,%rdx
  cmp    %rdx,%rdi
  je     L20
  mov    $0x10,%r8d
  mov    $0x9,%esi
  mov    $0x3,%ecx
L3:
  add    %r8,%rsi
  add    $0x2,%rcx
  cmp    %rsi,%rdi
  jbe    L6
  add    $0x8,%r8
  mov    %rdi,%rax
  mov    $0x0,%edx
  div    %rcx
  test   %rdx,%rdx
  jne    L3
L6:
  mov    %rdi,%rax
  mov    $0x0,%edx
  div    %rcx
  test   %rdx,%rdx
  setne  %al
  movzbl %al,%eax
  ret
L26:
  mov    $0x3,%ecx
  jmp    L6
L20:
  mov    $0x3,%ecx
  jmp    L6
```

assistant: ```c
int is_prime(unsigned long int candidate) {
  unsigned long int divn = 3;
  unsigned long int sq = divn * divn;
  while (sq < candidate && candidate % divn != 0) {
    ++divn;
    sq += 4 * divn;
    ++divn;
  }
  return candidate % divn != 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<is_prime>:
  endbr64
  movabs $0xaaaaaaaaaaaaaaab,%rcx
  mov    %rdi,%rax
  mul    %rcx
  mov    %rdx,%rcx
  shr    %rdx
  and    $0xfffffffffffffffe,%rcx
  add    %rdx,%rcx
  mov    %rdi,%rdx
  sub    %rcx,%rdx
  cmp    $0x9,%rdi
  jbe    L23
  test   %rdx,%rdx
  je     L23
  mov    $0x10,%r8d
  mov    $0x9,%esi
  mov    $0x3,%ecx
  jmp    L0
  nopl   0x0(%rax)
L39:
  add    $0x8,%r8
  test   %rdx,%rdx
  je     L23
L0:
  add    $0x2,%rcx
  mov    %rdi,%rax
  xor    %edx,%edx
  add    %r8,%rsi
  div    %rcx
  cmp    %rsi,%rdi
  ja     L39
L23:
  xor    %eax,%eax
  test   %rdx,%rdx
  setne  %al
  ret
```

assistant: ```c
int is_prime(unsigned long int candidate) {
  unsigned long int divn = 3;
  unsigned long int sq = divn * divn;
  while (sq < candidate && candidate % divn != 0) {
    ++divn;
    sq += 4 * divn;
    ++divn;
  }
  return candidate % divn != 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<is_prime>:
  endbr64
  movabs $0xaaaaaaaaaaaaaaab,%rcx
  mov    %rdi,%rax
  mul    %rcx
  mov    %rdx,%rcx
  shr    %rdx
  and    $0xfffffffffffffffe,%rcx
  add    %rdx,%rcx
  mov    %rdi,%rdx
  sub    %rcx,%rdx
  cmp    $0x9,%rdi
  jbe    L18
  test   %rdx,%rdx
  je     L18
  mov    $0x10,%r8d
  mov    $0x9,%esi
  mov    $0x3,%ecx
  jmp    L3
  nopl   0x0(%rax)
L2:
  add    $0x8,%r8
  test   %rdx,%rdx
  je     L18
L3:
  add    $0x2,%rcx
  mov    %rdi,%rax
  xor    %edx,%edx
  add    %r8,%rsi
  div    %rcx
  cmp    %rsi,%rdi
  ja     L2
L18:
  xor    %eax,%eax
  test   %rdx,%rdx
  setne  %al
  ret
```

assistant: ```c
int is_prime(unsigned long int candidate) {
  unsigned long int divn = 3;
  unsigned long int sq = divn * divn;
  while (sq < candidate && candidate % divn != 0) {
    ++divn;
    sq += 4 * divn;
    ++divn;
  }
  return candidate % divn != 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<write_data>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  movq   $0x0,-0x10(%rbp)
L55:
  mov    -0x38(%rbp),%rax
  sub    -0x10(%rbp),%rax
  mov    %rax,%rdx
  mov    D28(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jbe    L18
  mov    D28(%rip),%rax
  mov    (%rax),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L67
L18:
  mov    -0x38(%rbp),%rax
  sub    -0x10(%rbp),%rax
  mov    %rax,-0x18(%rbp)
L67:
  mov    -0x30(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  js     L48
  mov    -0x8(%rbp),%rax
  add    %rax,-0x10(%rbp)
  jmp    L69
L48:
  mov    D18(%rip),%rax
  mov    (%rax),%rdx
  mov    D16(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L69
  mov    $0x0,%eax
  jmp    L42
L69:
  mov    -0x10(%rbp),%rax
  cmp    -0x38(%rbp),%rax
  jb     L55
  mov    $0x1,%eax
L42:
  leave
  ret
```

assistant: ```c
int write_data(int fd, const void *buf, size_t nbytes) {
  size_t towrite, written = 0;
  ssize_t result;
  do {
    if (nbytes - written > SSIZE_MAX)
      towrite = SSIZE_MAX;
    else
      towrite = nbytes - written;
    if ((result = write(fd, (const char *)buf + written, towrite)) >= 0)
      written += result;
    else if (errno != EINTR)
      return 0;
  } while (written < nbytes);
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<write_data>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r13d
  mov    %rsi,%r12
  mov    %rdx,%rbp
  mov    $0x0,%ebx
  mov    D24(%rip),%r14
  jmp    L21
L39:
  add    %rax,%rbx
L2:
  cmp    %rbp,%rbx
  jae    L43
L21:
  mov    %rbp,%rdx
  sub    %rbx,%rdx
  mov    (%r14),%rax
  cmp    %rax,%rdx
  cmova  %rax,%rdx
  lea    (%r12,%rbx,1),%rsi
  mov    %r13d,%edi
  call   <write@plt>
  test   %rax,%rax
  jns    L39
  mov    D1(%rip),%rdx
  mov    D3(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L2
  mov    $0x0,%eax
  jmp    L12
L43:
  mov    $0x1,%eax
L12:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int write_data(int fd, const void *buf, size_t nbytes) {
  size_t towrite, written = 0;
  ssize_t result;
  do {
    if (nbytes - written > SSIZE_MAX)
      towrite = SSIZE_MAX;
    else
      towrite = nbytes - written;
    if ((result = write(fd, (const char *)buf + written, towrite)) >= 0)
      written += result;
    else if (errno != EINTR)
      return 0;
  } while (written < nbytes);
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<write_data>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    D3(%rip),%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  jmp    L21
  nopl   0x0(%rax)
L0:
  add    %rax,%r14
L26:
  cmp    %rbx,%r14
  jae    L14
L21:
  mov    0x0(%r13),%rax
  mov    %rbx,%rdx
  lea    0x0(%rbp,%r14,1),%rsi
  mov    %r12d,%edi
  sub    %r14,%rdx
  cmp    %rax,%rdx
  cmova  %rax,%rdx
  call   <write@plt>
  test   %rax,%rax
  jns    L0
  mov    D27(%rip),%rax
  mov    D15(%rip),%rdx
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L26
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L14:
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int write_data(int fd, const void *buf, size_t nbytes) {
  size_t towrite, written = 0;
  ssize_t result;
  do {
    if (nbytes - written > SSIZE_MAX)
      towrite = SSIZE_MAX;
    else
      towrite = nbytes - written;
    if ((result = write(fd, (const char *)buf + written, towrite)) >= 0)
      written += result;
    else if (errno != EINTR)
      return 0;
  } while (written < nbytes);
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<write_data>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    D14(%rip),%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  jmp    L38
  nopl   0x0(%rax)
L20:
  add    %rax,%r14
L17:
  cmp    %rbx,%r14
  jae    L23
L38:
  mov    0x0(%r13),%rax
  mov    %rbx,%rdx
  lea    0x0(%rbp,%r14,1),%rsi
  mov    %r12d,%edi
  sub    %r14,%rdx
  cmp    %rax,%rdx
  cmova  %rax,%rdx
  call   <write@plt>
  test   %rax,%rax
  jns    L20
  mov    D19(%rip),%rax
  mov    D7(%rip),%rdx
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L17
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L23:
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int write_data(int fd, const void *buf, size_t nbytes) {
  size_t towrite, written = 0;
  ssize_t result;
  do {
    if (nbytes - written > SSIZE_MAX)
      towrite = SSIZE_MAX;
    else
      towrite = nbytes - written;
    if ((result = write(fd, (const char *)buf + written, towrite)) >= 0)
      written += result;
    else if (errno != EINTR)
      return 0;
  } while (written < nbytes);
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ADCSequenceUnderflow>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x40038000,-0x4(%rbp)
  je     L29
  cmpl   $0x40039000,-0x4(%rbp)
  jne    L4
L29:
  mov    $0x1,%eax
  jmp    L12
L4:
  mov    $0x0,%eax
L12:
  mov    %eax,%edi
  call   <ASSERT@plt>
  cmpl   $0x3,-0x8(%rbp)
  setle  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <ASSERT@plt>
  mov    -0x4(%rbp),%eax
  add    $0x18,%eax
  cltq
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  and    %edx,%eax
  leave
  ret
```

assistant: ```c
int32_t ADCSequenceUnderflow(uint32_t ui32Base, uint32_t ui32SequenceNum) {
  ASSERT((ui32Base == 0x40038000) || (ui32Base == 0x40039000));
  ASSERT(ui32SequenceNum < 4);
  return ((*((volatile uint32_t *)(ui32Base + 0x00000018))) & (1 << ui32SequenceNum));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ADCSequenceUnderflow>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    %edi,%eax
  and    $0xef,%ah
  cmp    $0x40038000,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  cmp    $0x3,%ebp
  setle  %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  add    $0x18,%ebx
  movslq %ebx,%rbx
  mov    (%rbx),%edx
  mov    $0x1,%eax
  mov    %ebp,%ecx
  shl    %cl,%eax
  and    %edx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int32_t ADCSequenceUnderflow(uint32_t ui32Base, uint32_t ui32SequenceNum) {
  ASSERT((ui32Base == 0x40038000) || (ui32Base == 0x40039000));
  ASSERT(ui32SequenceNum < 4);
  return ((*((volatile uint32_t *)(ui32Base + 0x00000018))) & (1 << ui32SequenceNum));
}
```
user: What is the c source code of the assembly code below?

```asm
<ADCSequenceUnderflow>:
  endbr64
  push   %rbp
  mov    %edi,%eax
  mov    %esi,%ebp
  push   %rbx
  and    $0xef,%ah
  mov    %edi,%ebx
  xor    %edi,%edi
  sub    $0x8,%rsp
  cmp    $0x40038000,%eax
  sete   %dil
  call   <ASSERT@plt>
  xor    %edi,%edi
  cmp    $0x3,%ebp
  setle  %dil
  add    $0x18,%ebx
  call   <ASSERT@plt>
  movslq %ebx,%rbx
  mov    %ebp,%ecx
  mov    $0x1,%eax
  mov    (%rbx),%edx
  shl    %cl,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  and    %edx,%eax
  ret
```

assistant: ```c
int32_t ADCSequenceUnderflow(uint32_t ui32Base, uint32_t ui32SequenceNum) {
  ASSERT((ui32Base == 0x40038000) || (ui32Base == 0x40039000));
  ASSERT(ui32SequenceNum < 4);
  return ((*((volatile uint32_t *)(ui32Base + 0x00000018))) & (1 << ui32SequenceNum));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ADCSequenceUnderflow>:
  endbr64
  push   %rbp
  mov    %edi,%eax
  mov    %esi,%ebp
  push   %rbx
  and    $0xef,%ah
  mov    %edi,%ebx
  xor    %edi,%edi
  sub    $0x8,%rsp
  cmp    $0x40038000,%eax
  sete   %dil
  call   <ASSERT@plt>
  xor    %edi,%edi
  cmp    $0x3,%ebp
  setle  %dil
  add    $0x18,%ebx
  call   <ASSERT@plt>
  movslq %ebx,%rbx
  mov    %ebp,%ecx
  mov    $0x1,%eax
  mov    (%rbx),%edx
  shl    %cl,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  and    %edx,%eax
  ret
```

assistant: ```c
int32_t ADCSequenceUnderflow(uint32_t ui32Base, uint32_t ui32SequenceNum) {
  ASSERT((ui32Base == 0x40038000) || (ui32Base == 0x40039000));
  ASSERT(ui32SequenceNum < 4);
  return ((*((volatile uint32_t *)(ui32Base + 0x00000018))) & (1 << ui32SequenceNum));
}
```
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)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D2
type_guessed: string

tool: D2: the duplicate %p, the original %p


assistant: ```c
main(int argc, char *argv[]) {
  char *str;
  str = strdup(argv[0]);
  printf("the duplicate %p, the original %p\n", str, argv);
  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
  push   %rbx
  mov    %rsi,%rbx
  mov    (%rsi),%rdi
  call   <strdup@plt>
  mov    %rax,%rsi
  mov    %rbx,%rdx
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

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

tool: D0: the duplicate %p, the original %p


assistant: ```c
main(int argc, char *argv[]) {
  char *str;
  str = strdup(argv[0]);
  printf("the duplicate %p, the original %p\n", str, argv);
  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
  push   %r12
  mov    (%rsi),%rdi
  mov    %rsi,%r12
  call   <strdup@plt>
  mov    %r12,%rdx
  lea    D8(%rip),%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: tag: D8
datatype: string

tool: D8: the duplicate %p, the original %p


assistant: ```c
main(int argc, char *argv[]) {
  char *str;
  str = strdup(argv[0]);
  printf("the duplicate %p, the original %p\n", str, argv);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  mov    (%rsi),%rdi
  mov    %rsi,%r12
  call   <strdup@plt>
  mov    %r12,%rdx
  lea    D8(%rip),%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D8
type_infered: string

tool: D8: the duplicate %p, the original %p


assistant: ```c
main(int argc, char *argv[]) {
  char *str;
  str = strdup(argv[0]);
  printf("the duplicate %p, the original %p\n", str, argv);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<getDirectory>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L9
  mov    -0x18(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  sub    -0x18(%rbp),%rax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L10
L9:
  mov    -0x20(%rbp),%rax
  lea    D6(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L10:
  nop
  leave
  ret
```

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

tool: D6: .

assistant: ```c
void getDirectory(const char *filename, char *directorySpec) {
  int lengthOfDirectory;
  if (strrchr(filename, '/') != 0) {
    lengthOfDirectory = strrchr(filename, '/') - filename;
    strncpy(directorySpec, filename, lengthOfDirectory);
    directorySpec[lengthOfDirectory] = '\0';
  } else {
    strcpy(directorySpec, ".");
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<getDirectory>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    $0x2f,%esi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L13
  mov    $0x2f,%esi
  mov    %rbx,%rdi
  call   <strrchr@plt>
  sub    %rbx,%rax
  mov    %rax,%rbp
  mov    %eax,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <strncpy@plt>
  movslq %ebp,%rbp
  movb   $0x0,(%r12,%rbp,1)
L18:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L13:
  lea    D2(%rip),%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  jmp    L18
```

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

tool: D2: .

assistant: ```c
void getDirectory(const char *filename, char *directorySpec) {
  int lengthOfDirectory;
  if (strrchr(filename, '/') != 0) {
    lengthOfDirectory = strrchr(filename, '/') - filename;
    strncpy(directorySpec, filename, lengthOfDirectory);
    directorySpec[lengthOfDirectory] = '\0';
  } else {
    strcpy(directorySpec, ".");
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<getDirectory>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  mov    $0x2f,%esi
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strrchr@plt>
  test   %rax,%rax
  je     L2
  mov    %rbp,%rdi
  mov    $0x2f,%esi
  call   <strrchr@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  sub    %rbp,%rax
  mov    %rax,%rbx
  mov    %eax,%edx
  call   <strncpy@plt>
  movslq %ebx,%rbx
  movb   $0x0,(%r12,%rbx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  pop    %rbx
  mov    %r12,%rdi
  lea    D2(%rip),%rsi
  pop    %rbp
  pop    %r12
  jmp    1080 <strcpy@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
type: string

tool: D2: .

assistant: ```c
void getDirectory(const char *filename, char *directorySpec) {
  int lengthOfDirectory;
  if (strrchr(filename, '/') != 0) {
    lengthOfDirectory = strrchr(filename, '/') - filename;
    strncpy(directorySpec, filename, lengthOfDirectory);
    directorySpec[lengthOfDirectory] = '\0';
  } else {
    strcpy(directorySpec, ".");
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getDirectory>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  mov    $0x2f,%esi
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strrchr@plt>
  test   %rax,%rax
  je     L8
  mov    %rbp,%rdi
  mov    $0x2f,%esi
  call   <strrchr@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  sub    %rbp,%rax
  mov    %rax,%rbx
  mov    %eax,%edx
  call   <strncpy@plt>
  movslq %ebx,%rbx
  movb   $0x0,(%r12,%rbx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  pop    %rbx
  mov    %r12,%rdi
  lea    D4(%rip),%rsi
  pop    %rbp
  pop    %r12
  jmp    1080 <strcpy@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D4
datatype: string

tool: D4: .

assistant: ```c
void getDirectory(const char *filename, char *directorySpec) {
  int lengthOfDirectory;
  if (strrchr(filename, '/') != 0) {
    lengthOfDirectory = strrchr(filename, '/') - filename;
    strncpy(directorySpec, filename, lengthOfDirectory);
    directorySpec[lengthOfDirectory] = '\0';
  } else {
    strcpy(directorySpec, ".");
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mpeg_h_3d_audio_level>:
  endbr64
  cmp    $0x7,%dil
  ja     L2
  movzbl %dil,%edi
  lea    D83(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D32(%rip),%rax
  ret
  lea    D46(%rip),%rax
  ret
  lea    D36(%rip),%rax
  ret
  lea    D19(%rip),%rax
  ret
  lea    D53(%rip),%rax
  ret
  lea    D62(%rip),%rax
  ret
  lea    D62(%rip),%rax
  ret
L2:
  lea    D16(%rip),%rax
  ret
  lea    D11(%rip),%rax
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D32
data_type: string
name: retrieve_data
arguments: data_label: D46
data_type: string
name: retrieve_data
arguments: data_label: D36
data_type: string
name: retrieve_data
arguments: data_label: D19
data_type: string
name: retrieve_data
arguments: data_label: D53
data_type: string
name: retrieve_data
arguments: data_label: D62
data_type: string
name: retrieve_data
arguments: data_label: D16
data_type: string
name: retrieve_data
arguments: data_label: D11
data_type: string

tool: D32: Level 1

tool: D46: Level 2

tool: D36: Level 3

tool: D19: Level 4

tool: D53: Level 5

tool: D62: Reserved

tool: D16: BROKEN

tool: D11: Unspecified

assistant: ```c
const char *mpeg_h_3d_audio_level(unsigned char x) {
  switch (x) {
  case 0:
    return "Unspecified";
  case 1:
    return "Level 1";
  case 2:
    return "Level 2";
  case 3:
    return "Level 3";
  case 4:
    return "Level 4";
  case 5:
    return "Level 5";
  case 6:
    return "Reserved";
  case 7:
    return "Reserved";
  }
  return "BROKEN";
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mpeg_h_3d_audio_level>:
  endbr64
  cmp    $0x7,%dil
  ja     L8
  lea    D64(%rip),%rdx
  movzbl %dil,%edi
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nop
  lea    D17(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D11(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D6(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D60(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D63(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D76(%rip),%rax
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D56(%rip),%rax
  ret
L8:
  lea    D62(%rip),%rax
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D17
datatype: string
name: unpack_data
arguments: data_tag: D11
datatype: string
name: unpack_data
arguments: data_tag: D6
datatype: string
name: unpack_data
arguments: data_tag: D60
datatype: string
name: unpack_data
arguments: data_tag: D63
datatype: string
name: unpack_data
arguments: data_tag: D76
datatype: string
name: unpack_data
arguments: data_tag: D56
datatype: string
name: unpack_data
arguments: data_tag: D62
datatype: string

tool: D17: Reserved

tool: D11: Level 1

tool: D6: Unspecified

tool: D60: Level 5

tool: D63: Level 2

tool: D76: Level 3

tool: D56: Level 4

tool: D62: BROKEN

assistant: ```c
const char *mpeg_h_3d_audio_level(unsigned char x) {
  switch (x) {
  case 0:
    return "Unspecified";
  case 1:
    return "Level 1";
  case 2:
    return "Level 2";
  case 3:
    return "Level 3";
  case 4:
    return "Level 4";
  case 5:
    return "Level 5";
  case 6:
    return "Reserved";
  case 7:
    return "Reserved";
  }
  return "BROKEN";
}
```