ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rdx
  xor    %eax,%eax
  sub    $0x17447d68,%rdx
  je     L0
  mov    %rdx,%rax
  imul   %rdx,%rax
  shl    $0x3,%rax
  and    $0x78,%eax
  or     %rdx,%rax
L0:
  imul   $0x49c8729f,%rax,%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  unsigned short copy11;
  {
    state[0UL] = (input[0UL] - 51238316UL) - 339126204UL;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] > local1) {
        if (state[0UL] <= local1) {
          state[0UL] |= (state[0UL] & 15UL) << 3UL;
        } else {
          state[local1] |= (state[local1] * state[0UL] & 15UL) << 3UL;
        }
      } else if (state[0UL] != local1) {
        state[local1] |= (state[0UL] & 31UL) << 4UL;
      } else {
        copy11 = *((unsigned short *)(&state[local1]) + 1);
        *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 2);
        *((unsigned short *)(&state[local1]) + 2) = copy11;
      }
      local1++;
    }
    output[0UL] = state[0UL] * 1237873311UL;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x3,-0x30(%rbp)
  movl   $0x5,-0x2c(%rbp)
  movl   $0x4,-0x28(%rbp)
  movl   $0x3,-0x24(%rbp)
  movl   $0x3,-0x20(%rbp)
  movl   $0x2,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x8,-0x34(%rbp)
  mov    -0x34(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <print_array@plt>
  mov    -0x34(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <insertion_sort_int@plt>
  mov    -0x34(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <print_array@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
int main() {
  int a[] = {3, 5, 4, 3, 3, 2, 1, 0};
  int count = sizeof(a, sizeof(a[0]));
  print_array(a, count);
  insertion_sort_int(a, count);
  print_array(a, count);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x3,(%rsp)
  movl   $0x5,0x4(%rsp)
  movl   $0x4,0x8(%rsp)
  movl   $0x3,0xc(%rsp)
  movl   $0x3,0x10(%rsp)
  movl   $0x2,0x14(%rsp)
  movl   $0x1,0x18(%rsp)
  movl   $0x0,0x1c(%rsp)
  mov    %rsp,%rbx
  mov    $0x8,%esi
  mov    %rbx,%rdi
  call   <print_array@plt>
  mov    $0x8,%esi
  mov    %rbx,%rdi
  call   <insertion_sort_int@plt>
  mov    $0x8,%esi
  mov    %rbx,%rdi
  call   <print_array@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int a[] = {3, 5, 4, 3, 3, 2, 1, 0};
  int count = sizeof(a, sizeof(a[0]));
  print_array(a, count);
  insertion_sort_int(a, count);
  print_array(a, count);
  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
  push   %rbp
  mov    $0x8,%esi
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x500000003,%rax
  mov    %rsp,%rbp
  movq   $0x1,0x18(%rsp)
  mov    %rbp,%rdi
  mov    %rax,(%rsp)
  movabs $0x300000004,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x200000003,%rax
  mov    %rax,0x10(%rsp)
  call   <print_array@plt>
  mov    $0x8,%esi
  mov    %rbp,%rdi
  call   <insertion_sort_int@plt>
  mov    $0x8,%esi
  mov    %rbp,%rdi
  call   <print_array@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int a[] = {3, 5, 4, 3, 3, 2, 1, 0};
  int count = sizeof(a, sizeof(a[0]));
  print_array(a, count);
  insertion_sort_int(a, count);
  print_array(a, count);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x8,%esi
  sub    $0x30,%rsp
  movdqa D1(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movaps %xmm0,(%rsp)
  movdqa D16(%rip),%xmm0
  mov    %rbp,%rdi
  movaps %xmm0,0x10(%rsp)
  call   <print_array@plt>
  mov    $0x8,%esi
  mov    %rbp,%rdi
  call   <insertion_sort_int@plt>
  mov    $0x8,%esi
  mov    %rbp,%rdi
  call   <print_array@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D1
type: byte[16]
name: read_data
arguments: tag: D16
type: byte[16]

tool: {"D1": [0x03,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D16": [0x03,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int a[] = {3, 5, 4, 3, 3, 2, 1, 0};
  int count = sizeof(a, sizeof(a[0]));
  print_array(a, count);
  insertion_sort_int(a, count);
  print_array(a, count);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<next_generation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L28
L37:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L86
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L32
L86:
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L32:
  addl   $0x1,-0x4(%rbp)
L28:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L37
  movl   $0x1,-0x4(%rbp)
  jmp    L10
L95:
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jne    L46
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  jmp    L52
L46:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
L52:
  addl   $0x1,-0x4(%rbp)
L10:
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jl     L95
  movl   $0x0,-0x4(%rbp)
  jmp    L4
L65:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x4(%rbp)
L4:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L65
  nop
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D12
datatype: string
name: read_data
arguments: tag: D9
datatype: string
name: read_data
arguments: tag: D1
datatype: string

tool: D12: 


tool: D9: .

tool: D1: *

assistant: ```c
void next_generation(int *current, int *next, int length) {
  int i;
  printf("\n");
  for (i = 0; i < length; i++) {
    if (current[i] == 0)
      printf(".");
    else
      printf("*");
  }
  for (i = 1; i < length - 1; i++) {
    if (current[i - 1] == current[i + 1])
      next[i] = 0;
    else
      next[i] = 1;
  }
  for (i = 0; i < length; i++)
    current[i] = next[i];
}
```
user: Translate the below assembly code into C programming language:

```asm
<next_generation>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %edx,%r13d
  lea    D16(%rip),%rdi
  call   <printf@plt>
  test   %r13d,%r13d
  jle    L51
  mov    %rbp,%rbx
  lea    -0x1(%r13),%eax
  lea    0x4(%rbp,%rax,4),%r14
  lea    D3(%rip),%r15
  jmp    L27
L57:
  mov    %r15,%rdi
  call   <printf@plt>
L38:
  add    $0x4,%rbx
  cmp    %r14,%rbx
  je     L3
L27:
  cmpl   $0x0,(%rbx)
  jne    L57
  lea    D20(%rip),%rdi
  call   <printf@plt>
  jmp    L38
L3:
  lea    -0x1(%r13),%ecx
  mov    $0x1,%eax
  cmp    $0x1,%ecx
  jle    L84
L74:
  mov    0x4(%rbp,%rax,4),%esi
  cmp    %esi,-0x4(%rbp,%rax,4)
  setne  %dl
  movzbl %dl,%edx
  mov    %edx,(%r12,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%ecx
  jg     L74
  test   %r13d,%r13d
  jle    L51
L84:
  mov    $0x0,%eax
L41:
  mov    (%r12,%rax,4),%edx
  mov    %edx,0x0(%rbp,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r13d
  jg     L41
L51:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D16
datatype: string
name: extract_symbolic_data
arguments: label: D3
datatype: string
name: extract_symbolic_data
arguments: label: D20
datatype: string

tool: D16: 


tool: D3: *

tool: D20: .

assistant: ```c
void next_generation(int *current, int *next, int length) {
  int i;
  printf("\n");
  for (i = 0; i < length; i++) {
    if (current[i] == 0)
      printf(".");
    else
      printf("*");
  }
  for (i = 1; i < length - 1; i++) {
    if (current[i - 1] == current[i + 1])
      next[i] = 0;
    else
      next[i] = 1;
  }
  for (i = 0; i < length; i++)
    current[i] = next[i];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<next_generation>:
  endbr64
  push   %r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdi,%r13
  lea    D14(%rip),%rdi
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  lea    -0x1(%rbp),%r12d
  sub    $0x18,%rsp
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L89
  mov    %r12d,%eax
  mov    %r13,%rbx
  lea    D15(%rip),%r15
  lea    0x4(%r13,%rax,4),%rax
  mov    %rax,0x8(%rsp)
  jmp    L68
  nopl   0x0(%rax,%rax,1)
L47:
  lea    D8(%rip),%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,0x8(%rsp)
  je     L80
L68:
  mov    (%rbx),%eax
  test   %eax,%eax
  je     L47
  mov    %r15,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,0x8(%rsp)
  jne    L68
L80:
  cmp    $0x1,%r12d
  jle    L28
L57:
  mov    $0x1,%eax
  nopl   0x0(%rax,%rax,1)
L61:
  mov    0x4(%r13,%rax,4),%ecx
  xor    %edx,%edx
  cmp    %ecx,-0x4(%r13,%rax,4)
  setne  %dl
  mov    %edx,(%r14,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r12d
  jg     L61
  test   %ebp,%ebp
  jle    L9
L28:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L93:
  mov    (%r14,%rax,4),%edx
  mov    %edx,0x0(%r13,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%ebp
  jg     L93
L9:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L89:
  cmp    $0x1,%r12d
  jg     L57
  jmp    L9
```

tool_calls: name: unpack_data
arguments: data_tag: D14
type_guessed: string
name: unpack_data
arguments: data_tag: D15
type_guessed: string
name: unpack_data
arguments: data_tag: D8
type_guessed: string

tool: D14: 


tool: D15: *

tool: D8: .

assistant: ```c
void next_generation(int *current, int *next, int length) {
  int i;
  printf("\n");
  for (i = 0; i < length; i++) {
    if (current[i] == 0)
      printf(".");
    else
      printf("*");
  }
  for (i = 1; i < length - 1; i++) {
    if (current[i - 1] == current[i + 1])
      next[i] = 0;
    else
      next[i] = 1;
  }
  for (i = 0; i < length; i++)
    current[i] = next[i];
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<dary_print>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L4
L8:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movq   %rax,%xmm0
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addq   $0x1,-0x8(%rbp)
L4:
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jb     L8
  nop
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D7
type_infered: string

tool: D7: % f


assistant: ```c
void dary_print(double *a, size_t n) {
  size_t i;
  for (i = 0; i < n; i++)
    printf("% f\n", a[i]);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<dary_print>:
  endbr64
  test   %rsi,%rsi
  je     L8
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    $0x0,%ebx
  lea    D9(%rip),%r13
L15:
  movsd  (%r12,%rbx,8),%xmm0
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L15
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D9
type_infered: string

tool: D9: % f


assistant: ```c
void dary_print(double *a, size_t n) {
  size_t i;
  for (i = 0; i < n; i++)
    printf("% f\n", a[i]);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dary_print>:
  endbr64
  test   %rsi,%rsi
  je     L15
  push   %r13
  lea    D6(%rip),%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L9:
  movsd  (%r12,%rbx,8),%xmm0
  mov    %r13,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L9
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  ret
```

tool_calls: name: read_data
arguments: label: D6
data_type: string

tool: D6: % f


assistant: ```c
void dary_print(double *a, size_t n) {
  size_t i;
  for (i = 0; i < n; i++)
    printf("% f\n", a[i]);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<dary_print>:
  endbr64
  test   %rsi,%rsi
  je     L11
  push   %r13
  lea    D5(%rip),%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L4:
  movsd  (%r12,%rbx,8),%xmm0
  mov    %r13,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L4
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  ret
```

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

tool: D5: % f


assistant: ```c
void dary_print(double *a, size_t n) {
  size_t i;
  for (i = 0; i < n; i++)
    printf("% f\n", a[i]);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<no_malloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L8
  mov    $0xffffffff,%edi
  call   <exit@plt>
L8:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *no_malloc(size_t x) {
  void *p = malloc(x);
  if (p)
    exit(-1);
  return p;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<no_malloc>:
  endbr64
  push   %rbx
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L7
  mov    $0xffffffff,%edi
  call   <exit@plt>
L7:
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
void *no_malloc(size_t x) {
  void *p = malloc(x);
  if (p)
    exit(-1);
  return p;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<no_malloc>:
  endbr64
  push   %r12
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L4
  mov    $0xffffffff,%edi
  call   <exit@plt>
L4:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
void *no_malloc(size_t x) {
  void *p = malloc(x);
  if (p)
    exit(-1);
  return p;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<no_malloc>:
  endbr64
  push   %r12
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L1
  mov    $0xffffffff,%edi
  call   <exit@plt>
L1:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
void *no_malloc(size_t x) {
  void *p = malloc(x);
  if (p)
    exit(-1);
  return p;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x120,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x114(%rbp)
  jmp    L73
L3:
  mov    -0x114(%rbp),%eax
  mov    %eax,%edx
  mov    -0x114(%rbp),%eax
  cltq
  mov    %dl,-0x90(%rbp,%rax,1)
  addl   $0x1,-0x114(%rbp)
L73:
  cmpl   $0x80,-0x114(%rbp)
  jle    L3
  movl   $0x1,-0x114(%rbp)
  jmp    L55
L21:
  mov    -0x114(%rbp),%eax
  lea    -0x1(%rax),%ecx
  mov    -0x114(%rbp),%eax
  cltq
  movzbl -0x90(%rbp,%rax,1),%edx
  movslq %ecx,%rax
  mov    %dl,-0x110(%rbp,%rax,1)
  addl   $0x1,-0x114(%rbp)
L55:
  cmpl   $0x80,-0x114(%rbp)
  jle    L21
  movl   $0x1,-0x114(%rbp)
  jmp    L30
L16:
  mov    -0x114(%rbp),%eax
  sub    $0x1,%eax
  cltq
  movzbl -0x110(%rbp,%rax,1),%edx
  mov    -0x114(%rbp),%eax
  cltq
  movzbl -0x90(%rbp,%rax,1),%eax
  cmp    %al,%dl
  je     L34
  mov    $0x0,%eax
  call   <abort@plt>
L34:
  addl   $0x1,-0x114(%rbp)
L30:
  cmpl   $0x80,-0x114(%rbp)
  jle    L16
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L37
  call   <__stack_chk_fail@plt>
L37:
  leave
  ret
```

assistant: ```c
int main1() {
  int i;
  char ia[128];
  char ib[128 + 1];
  for (i = 0; i < 128 + 1; i++) {
    ib[i] = i;
  }
  for (i = 1; i < 128 + 1; i++) {
    ia[i - 1] = ib[i];
  }
  for (i = 1; i <= 128; i++) {
    if (ia[i - 1] != ib[i])
      abort();
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main1>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
L62:
  mov    %al,0x80(%rsp,%rax,1)
  add    $0x1,%rax
  cmp    $0x81,%rax
  jne    L62
  mov    $0x1,%eax
  lea    -0x1(%rsp),%rcx
L52:
  movzbl 0x80(%rsp,%rax,1),%edx
  mov    %dl,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    $0x81,%rax
  jne    L52
  mov    $0x0,%ebx
  lea    0x81(%rsp),%rbp
  jmp    L38
L68:
  add    $0x1,%rbx
  cmp    $0x80,%rbx
  je     L61
L38:
  movzbl 0x0(%rbp,%rbx,1),%eax
  cmp    %al,(%rsp,%rbx,1)
  je     L68
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L68
L61:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  mov    $0x0,%eax
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  ret
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main1() {
  int i;
  char ia[128];
  char ib[128 + 1];
  for (i = 0; i < 128 + 1; i++) {
    ib[i] = i;
  }
  for (i = 1; i < 128 + 1; i++) {
    ia[i - 1] = ib[i];
  }
  for (i = 1; i <= 128; i++) {
    if (ia[i - 1] != ib[i])
      abort();
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main1>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  lea    0x80(%rsp),%rdx
  nopw   0x0(%rax,%rax,1)
L17:
  mov    %al,(%rdx,%rax,1)
  add    $0x1,%rax
  cmp    $0x81,%rax
  jne    L17
  movdqu 0x81(%rsp),%xmm0
  xor    %ebx,%ebx
  mov    %rsp,%r12
  movdqu 0x91(%rsp),%xmm1
  movdqu 0xa1(%rsp),%xmm2
  movdqu 0xb1(%rsp),%xmm3
  lea    0x81(%rsp),%rbp
  movdqu 0xc1(%rsp),%xmm4
  movaps %xmm0,(%rsp)
  movdqu 0xd1(%rsp),%xmm5
  movdqu 0xe1(%rsp),%xmm6
  movaps %xmm1,0x10(%rsp)
  movdqu 0xf1(%rsp),%xmm7
  movaps %xmm2,0x20(%rsp)
  movaps %xmm3,0x30(%rsp)
  movaps %xmm4,0x40(%rsp)
  movaps %xmm5,0x50(%rsp)
  movaps %xmm6,0x60(%rsp)
  movaps %xmm7,0x70(%rsp)
  nopl   0x0(%rax,%rax,1)
L30:
  movzbl 0x0(%rbp,%rbx,1),%eax
  cmp    %al,(%r12,%rbx,1)
  je     L28
  xor    %eax,%eax
  call   <abort@plt>
L28:
  add    $0x1,%rbx
  cmp    $0x80,%rbx
  jne    L30
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x110,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main1() {
  int i;
  char ia[128];
  char ib[128 + 1];
  for (i = 0; i < 128 + 1; i++) {
    ib[i] = i;
  }
  for (i = 1; i < 128 + 1; i++) {
    ia[i - 1] = ib[i];
  }
  for (i = 1; i <= 128; i++) {
    if (ia[i - 1] != ib[i])
      abort();
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main1>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x110,%rsp
  movdqa D63(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  movb   $0x80,0x100(%rsp)
  mov    %rsp,%r12
  lea    0x81(%rsp),%rbp
  movaps %xmm0,0x80(%rsp)
  movdqa D41(%rip),%xmm0
  movaps %xmm0,0x90(%rsp)
  movdqa D59(%rip),%xmm0
  movdqu 0x81(%rsp),%xmm1
  movaps %xmm0,0xa0(%rsp)
  movdqa D40(%rip),%xmm0
  movdqu 0x91(%rsp),%xmm2
  movaps %xmm1,(%rsp)
  movaps %xmm0,0xb0(%rsp)
  movdqa D1(%rip),%xmm0
  movdqu 0xa1(%rsp),%xmm3
  movaps %xmm2,0x10(%rsp)
  movaps %xmm0,0xc0(%rsp)
  movdqa D16(%rip),%xmm0
  movdqu 0xb1(%rsp),%xmm4
  movaps %xmm3,0x20(%rsp)
  movaps %xmm0,0xd0(%rsp)
  movdqa D66(%rip),%xmm0
  movdqu 0xc1(%rsp),%xmm5
  movaps %xmm4,0x30(%rsp)
  movaps %xmm0,0xe0(%rsp)
  movdqa D33(%rip),%xmm0
  movaps %xmm0,0xf0(%rsp)
  movaps %xmm5,0x40(%rsp)
  movdqu 0xd1(%rsp),%xmm6
  movdqu 0xe1(%rsp),%xmm7
  movdqu 0xf1(%rsp),%xmm1
  movaps %xmm6,0x50(%rsp)
  movaps %xmm7,0x60(%rsp)
  movaps %xmm1,0x70(%rsp)
  cs nopw 0x0(%rax,%rax,1)
L8:
  movzbl 0x0(%rbp,%rbx,1),%eax
  cmp    %al,(%r12,%rbx,1)
  je     L5
  xor    %eax,%eax
  call   <abort@plt>
L5:
  add    $0x1,%rbx
  cmp    $0x80,%rbx
  jne    L8
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x110,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D63
type: byte[16]
name: retrieve_data
arguments: data_tag: D41
type: byte[16]
name: retrieve_data
arguments: data_tag: D59
type: byte[16]
name: retrieve_data
arguments: data_tag: D40
type: byte[16]
name: retrieve_data
arguments: data_tag: D1
type: byte[16]
name: retrieve_data
arguments: data_tag: D16
type: byte[16]
name: retrieve_data
arguments: data_tag: D66
type: byte[16]
name: retrieve_data
arguments: data_tag: D33
type: byte[16]

tool: {"D63": [0x00,0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x08,0x09,0x0a,0x0b,0x0c,0x0d,0x0e,0x0f]}

tool: {"D41": [0x10,0x11,0x12,0x13,0x14,0x15,0x16,0x17,0x18,0x19,0x1a,0x1b,0x1c,0x1d,0x1e,0x1f]}

tool: {"D59": [0x20,0x21,0x22,0x23,0x24,0x25,0x26,0x27,0x28,0x29,0x2a,0x2b,0x2c,0x2d,0x2e,0x2f]}

tool: {"D40": [0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x3a,0x3b,0x3c,0x3d,0x3e,0x3f]}

tool: {"D1": [0x40,0x41,0x42,0x43,0x44,0x45,0x46,0x47,0x48,0x49,0x4a,0x4b,0x4c,0x4d,0x4e,0x4f]}

tool: {"D16": [0x50,0x51,0x52,0x53,0x54,0x55,0x56,0x57,0x58,0x59,0x5a,0x5b,0x5c,0x5d,0x5e,0x5f]}

tool: {"D66": [0x60,0x61,0x62,0x63,0x64,0x65,0x66,0x67,0x68,0x69,0x6a,0x6b,0x6c,0x6d,0x6e,0x6f]}

tool: {"D33": [0x70,0x71,0x72,0x73,0x74,0x75,0x76,0x77,0x78,0x79,0x7a,0x7b,0x7c,0x7d,0x7e,0x7f]}

assistant: ```c
int main1() {
  int i;
  char ia[128];
  char ib[128 + 1];
  for (i = 0; i < 128 + 1; i++) {
    ib[i] = i;
  }
  for (i = 1; i < 128 + 1; i++) {
    ia[i - 1] = ib[i];
  }
  for (i = 1; i <= 128; i++) {
    if (ia[i - 1] != ib[i])
      abort();
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<iface1_constructor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D3
data_type: string

tool: D3: A - version 0 constructor context %p


assistant: ```c
int iface1_constructor(void *context) {
  printf("A - version 0 constructor context %p\n", context);
  return (0);
}
```