ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的：

```asm
<invert_bit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  xor    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int invert_bit(int n, int number_bit) {
  n ^= 1 << number_bit;
  return (n);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<invert_bit>:
  endbr64
  mov    %edi,%eax
  btc    %esi,%eax
  ret
```

assistant: ```c
int invert_bit(int n, int number_bit) {
  n ^= 1 << number_bit;
  return (n);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<invert_bit>:
  endbr64
  mov    %edi,%eax
  btc    %esi,%eax
  ret
```

assistant: ```c
int invert_bit(int n, int number_bit) {
  n ^= 1 << number_bit;
  return (n);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<invert_bit>:
  endbr64
  mov    %edi,%eax
  btc    %esi,%eax
  ret
```

assistant: ```c
int invert_bit(int n, int number_bit) {
  n ^= 1 << number_bit;
  return (n);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<pptp_gre_bind>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D3(%rip),%rax
  mov    (%rax),%ecx
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  mov    $0x2f,%edx
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <socket@plt>
  mov    %eax,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jns    L37
  mov    $0xffffffff,%eax
  jmp    L17
L37:
  lea    -0x20(%rbp),%rax
  mov    $0x10,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  movq   $0x0,-0x20(%rbp)
  lea    -0x20(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    $0x10,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <connect@plt>
  test   %rax,%rax
  jns    L38
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  mov    $0xffffffff,%eax
  jmp    L17
L38:
  mov    -0x24(%rbp),%eax
L17:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L31
  call   <__stack_chk_fail@plt>
L31:
  leave
  ret
```

assistant: ```c
int pptp_gre_bind(unsigned int ip) {
  struct sockaddr_in addr;
  int s;
  s = socket(AF_INET, SOCK_RAW, 47);
  if (s < 0)
    return -1;
  memset(&addr, 0, sizeof addr);
  addr.sin_family = AF_INET;
  addr.sin_addr.s_addr = ip;
  addr.sin_port = 0;
  if (connect(s, (struct sockaddr *)&addr, sizeof addr) < 0) {
    close(s);
    return -1;
  }
  return s;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<pptp_gre_bind>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %edi,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    $0x2f,%edx
  mov    D15(%rip),%rax
  mov    (%rax),%esi
  mov    D29(%rip),%rax
  mov    (%rax),%edi
  call   <socket@plt>
  test   %eax,%eax
  js     L44
  mov    %eax,%ebx
  mov    %rsp,%rbp
  mov    $0x10,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    D29(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0xc(%rsp)
  mov    %r12d,0x8(%rsp)
  movq   $0x0,(%rsp)
  mov    $0x10,%edx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <connect@plt>
  test   %rax,%rax
  js     L11
L23:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  mov    %ebx,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  mov    %ebx,%edi
  call   <close@plt>
  mov    $0xffffffff,%ebx
  jmp    L23
L44:
  mov    $0xffffffff,%ebx
  jmp    L23
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int pptp_gre_bind(unsigned int ip) {
  struct sockaddr_in addr;
  int s;
  s = socket(AF_INET, SOCK_RAW, 47);
  if (s < 0)
    return -1;
  memset(&addr, 0, sizeof addr);
  addr.sin_family = AF_INET;
  addr.sin_addr.s_addr = ip;
  addr.sin_port = 0;
  if (connect(s, (struct sockaddr *)&addr, sizeof addr) < 0) {
    close(s);
    return -1;
  }
  return s;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<pptp_gre_bind>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  mov    $0x2f,%edx
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    D4(%rip),%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D9(%rip),%rax
  mov    (%rbx),%edi
  mov    (%rax),%esi
  call   <socket@plt>
  test   %eax,%eax
  js     L36
  mov    %rsp,%rbp
  xor    %esi,%esi
  mov    $0x10,%edx
  mov    %eax,%r12d
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    (%rbx),%eax
  mov    $0x10,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  mov    %r13d,0x8(%rsp)
  mov    %eax,0xc(%rsp)
  movq   $0x0,(%rsp)
  call   <connect@plt>
  test   %rax,%rax
  js     L10
L29:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x28,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L36:
  mov    $0xffffffff,%r12d
  jmp    L29
L10:
  mov    %r12d,%edi
  mov    $0xffffffff,%r12d
  call   <close@plt>
  jmp    L29
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int pptp_gre_bind(unsigned int ip) {
  struct sockaddr_in addr;
  int s;
  s = socket(AF_INET, SOCK_RAW, 47);
  if (s < 0)
    return -1;
  memset(&addr, 0, sizeof addr);
  addr.sin_family = AF_INET;
  addr.sin_addr.s_addr = ip;
  addr.sin_port = 0;
  if (connect(s, (struct sockaddr *)&addr, sizeof addr) < 0) {
    close(s);
    return -1;
  }
  return s;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<pptp_gre_bind>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  mov    $0x2f,%edx
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    D10(%rip),%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D5(%rip),%rax
  mov    (%rbx),%edi
  mov    (%rax),%esi
  call   <socket@plt>
  test   %eax,%eax
  js     L40
  mov    %rsp,%rbp
  xor    %esi,%esi
  mov    $0x10,%edx
  mov    %eax,%r12d
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    (%rbx),%eax
  mov    $0x10,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  mov    %r13d,0x8(%rsp)
  mov    %eax,0xc(%rsp)
  movq   $0x0,(%rsp)
  call   <connect@plt>
  test   %rax,%rax
  js     L31
L49:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x28,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L40:
  mov    $0xffffffff,%r12d
  jmp    L49
L31:
  mov    %r12d,%edi
  mov    $0xffffffff,%r12d
  call   <close@plt>
  jmp    L49
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int pptp_gre_bind(unsigned int ip) {
  struct sockaddr_in addr;
  int s;
  s = socket(AF_INET, SOCK_RAW, 47);
  if (s < 0)
    return -1;
  memset(&addr, 0, sizeof addr);
  addr.sin_family = AF_INET;
  addr.sin_addr.s_addr = ip;
  addr.sin_port = 0;
  if (connect(s, (struct sockaddr *)&addr, sizeof addr) < 0) {
    close(s);
    return -1;
  }
  return s;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<make_permutation_table_varlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpl   $0x10000,-0x1c(%rbp)
  setle  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L32
L31:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L32:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L31
  mov    -0x18(%rbp),%rdx
  mov    -0x1c(%rbp),%esi
  mov    -0x8(%rbp),%rax
  mov    $0x40000000,%ecx
  mov    %rax,%rdi
  call   <_internal_simple_quick_sort@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L18
L19:
  mov    -0x1c(%rbp),%eax
  sub    -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  add    %edx,%eax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rdx,%rcx
  cltd
  idivl  -0x1c(%rbp)
  mov    %edx,%eax
  mov    %eax,(%rcx)
  addl   $0x1,-0xc(%rbp)
L18:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L19
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
uint16_t *make_permutation_table_varlen(char **blocks, int nblocks) {
  assert(nblocks <= (1 << 16));
  int i;
  uint16_t *perm = malloc(nblocks * sizeof(uint16_t));
  for (i = 0; i < nblocks; i++)
    perm[i] = i;
  _internal_simple_quick_sort(perm, nblocks, blocks, 1 << 30);
  for (i = 0; i < nblocks; i++)
    perm[i] = (nblocks - i + perm[i]) % nblocks;
  return perm;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<make_permutation_table_varlen>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %esi,%ebp
  cmp    $0x10000,%esi
  setle  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  lea    0x0(,%rbp,4),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %ebp,%ebp
  jle    L12
  mov    %ebp,%ebx
  mov    %ebp,%edx
  mov    $0x0,%eax
L13:
  mov    %eax,(%r12,%rax,4)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L13
  mov    $0x40000000,%ecx
  mov    %r13,%rdx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <_internal_simple_quick_sort@plt>
  mov    %r12,%rcx
L35:
  mov    %ebx,%eax
  add    (%rcx),%eax
  cltd
  idiv   %ebp
  mov    %edx,(%rcx)
  add    $0x4,%rcx
  sub    $0x1,%ebx
  jne    L35
L16:
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L12:
  mov    $0x40000000,%ecx
  mov    %r13,%rdx
  mov    %ebp,%esi
  mov    %rax,%rdi
  call   <_internal_simple_quick_sort@plt>
  jmp    L16
```

assistant: ```c
uint16_t *make_permutation_table_varlen(char **blocks, int nblocks) {
  assert(nblocks <= (1 << 16));
  int i;
  uint16_t *perm = malloc(nblocks * sizeof(uint16_t));
  for (i = 0; i < nblocks; i++)
    perm[i] = i;
  _internal_simple_quick_sort(perm, nblocks, blocks, 1 << 30);
  for (i = 0; i < nblocks; i++)
    perm[i] = (nblocks - i + perm[i]) % nblocks;
  return perm;
}
```
user: Translate the below assembly code into C programming language:

```asm
<make_permutation_table_varlen>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  xor    %edi,%edi
  push   %r13
  movslq %esi,%r13
  push   %r12
  cmp    $0x10000,%r13d
  push   %rbp
  setle  %dil
  mov    %r13,%rbp
  push   %rbx
  call   <assert@plt>
  lea    0x0(,%r13,4),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %r13d,%r13d
  jle    L29
  mov    %r13d,%ebx
  xor    %eax,%eax
  nopl   0x0(%rax)
L22:
  mov    %eax,(%r12,%rax,4)
  add    $0x1,%rax
  cmp    %r13,%rax
  jne    L22
  mov    $0x40000000,%ecx
  mov    %r14,%rdx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <_internal_simple_quick_sort@plt>
  mov    %r12,%rcx
  nopw   0x0(%rax,%rax,1)
L17:
  mov    (%rcx),%eax
  add    $0x4,%rcx
  add    %ebx,%eax
  cltd
  idiv   %ebp
  mov    %edx,-0x4(%rcx)
  sub    $0x1,%ebx
  jne    L17
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L29:
  mov    %r14,%rdx
  mov    %r13d,%esi
  mov    $0x40000000,%ecx
  mov    %rax,%rdi
  call   <_internal_simple_quick_sort@plt>
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
uint16_t *make_permutation_table_varlen(char **blocks, int nblocks) {
  assert(nblocks <= (1 << 16));
  int i;
  uint16_t *perm = malloc(nblocks * sizeof(uint16_t));
  for (i = 0; i < nblocks; i++)
    perm[i] = i;
  _internal_simple_quick_sort(perm, nblocks, blocks, 1 << 30);
  for (i = 0; i < nblocks; i++)
    perm[i] = (nblocks - i + perm[i]) % nblocks;
  return perm;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<make_permutation_table_varlen>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  xor    %edi,%edi
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x10000,%esi
  setle  %dil
  call   <assert@plt>
  lea    0x0(,%rbp,4),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %ebp,%ebp
  jle    L49
  lea    -0x1(%rbp),%eax
  mov    %ebp,%ebx
  cmp    $0x2,%eax
  jbe    L58
  mov    %ebp,%edx
  movdqa D14(%rip),%xmm0
  movdqa D4(%rip),%xmm2
  mov    %r12,%rax
  shr    $0x2,%edx
  shl    $0x4,%rdx
  add    %r12,%rdx
  nopl   0x0(%rax,%rax,1)
L27:
  movdqa %xmm0,%xmm1
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  movups %xmm1,-0x10(%rax)
  cmp    %rdx,%rax
  jne    L27
  mov    %ebp,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%bpl
  je     L26
L0:
  movslq %eax,%rdx
  mov    %eax,(%r12,%rdx,4)
  lea    0x0(,%rdx,4),%rcx
  lea    0x1(%rax),%edx
  cmp    %edx,%ebp
  jle    L26
  add    $0x2,%eax
  mov    %edx,0x4(%r12,%rcx,1)
  cmp    %eax,%ebp
  jle    L26
  mov    %eax,0x8(%r12,%rcx,1)
L26:
  mov    $0x40000000,%ecx
  mov    %r13,%rdx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <_internal_simple_quick_sort@plt>
  mov    %r12,%rcx
  nopl   0x0(%rax)
L76:
  mov    (%rcx),%eax
  add    $0x4,%rcx
  add    %ebx,%eax
  cltd
  idiv   %ebp
  mov    %edx,-0x4(%rcx)
  sub    $0x1,%ebx
  jne    L76
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L49:
  mov    %r13,%rdx
  mov    %ebp,%esi
  mov    $0x40000000,%ecx
  mov    %rax,%rdi
  call   <_internal_simple_quick_sort@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L58:
  xor    %eax,%eax
  jmp    L0
```

tool_calls: name: extract_symbolic_data
arguments: label: D14
type_guessed: byte[16]
name: extract_symbolic_data
arguments: label: D4
type_guessed: byte[16]

tool: {"D14": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D4": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

assistant: ```c
uint16_t *make_permutation_table_varlen(char **blocks, int nblocks) {
  assert(nblocks <= (1 << 16));
  int i;
  uint16_t *perm = malloc(nblocks * sizeof(uint16_t));
  for (i = 0; i < nblocks; i++)
    perm[i] = i;
  _internal_simple_quick_sort(perm, nblocks, blocks, 1 << 30);
  for (i = 0; i < nblocks; i++)
    perm[i] = (nblocks - i + perm[i]) % nblocks;
  return perm;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test_kernels>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x2,-0xa4(%rbp)
  movl   $0x0,-0x94(%rbp)
  movl   $0x0,-0xa0(%rbp)
  jmp    L82
L7:
  mov    -0xa0(%rbp),%eax
  cltq
  movl   $0xffffffff,-0x90(%rbp,%rax,4)
  addl   $0x1,-0xa0(%rbp)
L82:
  cmpl   $0x1f,-0xa0(%rbp)
  jle    L7
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <acc_on_device@plt>
  mov    %eax,-0x94(%rbp)
  movl   $0x0,-0x9c(%rbp)
  jmp    L20
L77:
  mov    -0x9c(%rbp),%eax
  mov    -0xa4(%rbp),%edx
  mov    %edx,-0x90(%rbp,%rax,4)
  addl   $0x1,-0xa4(%rbp)
  addl   $0x1,-0x9c(%rbp)
L20:
  cmpl   $0x1f,-0x9c(%rbp)
  jbe    L77
  cmpl   $0x0,-0x94(%rbp)
  je     L74
  cmpl   $0x22,-0xa4(%rbp)
  je     L10
  mov    $0x1,%eax
  jmp    L37
L10:
  movl   $0x0,-0x98(%rbp)
  jmp    L4
L58:
  mov    -0x98(%rbp),%eax
  cltq
  mov    -0x90(%rbp,%rax,4),%eax
  mov    -0x98(%rbp),%edx
  add    $0x2,%edx
  cmp    %edx,%eax
  je     L116
  mov    $0x1,%eax
  jmp    L37
L116:
  addl   $0x1,-0x98(%rbp)
L4:
  cmpl   $0x1f,-0x98(%rbp)
  jle    L58
L74:
  mov    $0x0,%eax
L37:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
int test_kernels() {
  int val = 2;
  int ary[32];
  int ondev = 0;
  for (int i = 0; i < 32; i++)
    ary[i] = ~0;
#pragma acc kernels copy(ondev)
  {
    ondev = acc_on_device(acc_device_not_host);
#pragma acc loop
    for (unsigned i = 0; i < 32; i++) {
      ary[i] = val;
      val++;
    }
  }
  if (ondev) {
    if (val != 2 + 32)
      return 1;
    for (int i = 0; i < 32; i++)
      if (ary[i] != 2 + i)
        return 1;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<test_kernels>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x80(%rsp),%rbx
  mov    %rbp,%rax
L58:
  movl   $0xffffffff,(%rax)
  add    $0x4,%rax
  cmp    %rbx,%rax
  jne    L58
  mov    D3(%rip),%rax
  mov    (%rax),%edi
  call   <acc_on_device@plt>
  mov    %rbp,%rcx
  mov    $0x2,%edx
L12:
  mov    %edx,(%rcx)
  add    $0x1,%edx
  add    $0x4,%rcx
  cmp    $0x22,%edx
  jne    L12
  test   %eax,%eax
  je     L44
  mov    $0x2,%eax
L35:
  cmp    %eax,0x0(%rbp)
  jne    L37
  add    $0x4,%rbp
  add    $0x1,%eax
  cmp    %rbx,%rbp
  jne    L35
  mov    $0x0,%eax
  jmp    L44
L37:
  mov    $0x1,%eax
L44:
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L66
  add    $0x98,%rsp
  pop    %rbx
  pop    %rbp
  ret
L66:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test_kernels() {
  int val = 2;
  int ary[32];
  int ondev = 0;
  for (int i = 0; i < 32; i++)
    ary[i] = ~0;
#pragma acc kernels copy(ondev)
  {
    ondev = acc_on_device(acc_device_not_host);
#pragma acc loop
    for (unsigned i = 0; i < 32; i++) {
      ary[i] = val;
      val++;
    }
  }
  if (ondev) {
    if (val != 2 + 32)
      return 1;
    for (int i = 0; i < 32; i++)
      if (ary[i] != 2 + i)
        return 1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test_kernels>:
  endbr64
  push   %rbx
  pcmpeqd %xmm0,%xmm0
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    D2(%rip),%rax
  mov    %rsp,%rbx
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  mov    (%rax),%edi
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  movaps %xmm0,0x60(%rsp)
  movaps %xmm0,0x70(%rsp)
  call   <acc_on_device@plt>
  mov    %rbx,%rsi
  mov    %rbx,%rcx
  mov    $0x2,%edx
  cs nopw 0x0(%rax,%rax,1)
L60:
  mov    %edx,(%rcx)
  add    $0x1,%edx
  add    $0x4,%rcx
  cmp    $0x22,%edx
  jne    L60
  test   %eax,%eax
  je     L20
  sub    $0xffffffffffffff80,%rbx
  mov    $0x2,%eax
  cmp    %eax,(%rsi)
  jne    L43
  nop
L46:
  add    $0x4,%rsi
  add    $0x1,%eax
  cmp    %rbx,%rsi
  je     L61
  cmp    %eax,(%rsi)
  je     L46
L43:
  mov    $0x1,%eax
L20:
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L38
  add    $0x90,%rsp
  pop    %rbx
  ret
L61:
  xor    %eax,%eax
  jmp    L20
L38:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test_kernels() {
  int val = 2;
  int ary[32];
  int ondev = 0;
  for (int i = 0; i < 32; i++)
    ary[i] = ~0;
#pragma acc kernels copy(ondev)
  {
    ondev = acc_on_device(acc_device_not_host);
#pragma acc loop
    for (unsigned i = 0; i < 32; i++) {
      ary[i] = val;
      val++;
    }
  }
  if (ondev) {
    if (val != 2 + 32)
      return 1;
    for (int i = 0; i < 32; i++)
      if (ary[i] != 2 + i)
        return 1;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L25
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x1,%eax
  jmp    L19
L25:
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
L19:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D34
data_type: string
name: disassemble_data
arguments: data_tag: D9
data_type: string
name: disassemble_data
arguments: data_tag: D35
data_type: string
name: disassemble_data
arguments: data_tag: D27
data_type: string

tool: D34: Hello. Please include your name: 

tool: D9: %s

tool: D35: Error!

tool: D27: Nice to meet you %s


assistant: ```c
int main(void) {
  char name[100];
  printf("Hello. Please include your name: ");
  if (scanf("%s", name) != 1) {
    printf("Error!\n");
    return 1;
  } else {
    printf("Nice to meet you %s\n", name);
    return 0;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    D31(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D12(%rip),%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L18
  lea    D5(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
L6:
  mov    0x68(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x78,%rsp
  ret
L18:
  mov    %rsp,%rsi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  jmp    L6
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D31
type_guessed: string
name: retrieve_data
arguments: label: D12
type_guessed: string
name: retrieve_data
arguments: label: D5
type_guessed: string
name: retrieve_data
arguments: label: D9
type_guessed: string

tool: D31: Hello. Please include your name: 

tool: D12: %s

tool: D5: Error!

tool: D9: Nice to meet you %s


assistant: ```c
int main(void) {
  char name[100];
  printf("Hello. Please include your name: ");
  if (scanf("%s", name) != 1) {
    printf("Error!\n");
    return 1;
  } else {
    printf("Nice to meet you %s\n", name);
    return 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D34(%rip),%rdi
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D20(%rip),%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L17
  lea    D12(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
L26:
  mov    0x68(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L27
  add    $0x70,%rsp
  pop    %rbp
  ret
L17:
  mov    %rbp,%rsi
  lea    D39(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  jmp    L26
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D34
data_type: string
name: unpack_data
arguments: tag: D20
data_type: string
name: unpack_data
arguments: tag: D12
data_type: string
name: unpack_data
arguments: tag: D39
data_type: string

tool: D34: Hello. Please include your name: 

tool: D20: %s

tool: D12: Error!

tool: D39: Nice to meet you %s


assistant: ```c
int main(void) {
  char name[100];
  printf("Hello. Please include your name: ");
  if (scanf("%s", name) != 1) {
    printf("Error!\n");
    return 1;
  } else {
    printf("Nice to meet you %s\n", name);
    return 0;
  }
}
```