ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<create_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L39
  movl   $0x0,-0xc(%rbp)
  jmp    L9
L5:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L9:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L5
  mov    -0x8(%rbp),%rax
  jmp    L18
L39:
  mov    $0x0,%eax
L18:
  leave
  ret
```

assistant: ```c
int *create_array(int n, int initial_value) {
  int *a = malloc(n * sizeof(int));
  if (a != NULL) {
    int i;
    for (i = 0; i < n; i++)
      a[i] = initial_value;
    return a;
  } else
    return NULL;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<create_array>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
  lea    0x0(,%rdi,4),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L1
  test   %ebp,%ebp
  jle    L1
  mov    %rax,%rdx
  mov    %ebp,%ebp
  lea    (%rax,%rbp,4),%rcx
L10:
  mov    %ebx,(%rdx)
  add    $0x4,%rdx
  cmp    %rcx,%rdx
  jne    L10
L1:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int *create_array(int n, int initial_value) {
  int *a = malloc(n * sizeof(int));
  if (a != NULL) {
    int i;
    for (i = 0; i < n; i++)
      a[i] = initial_value;
    return a;
  } else
    return NULL;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<create_array>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  lea    0x0(,%rbp,4),%edi
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <malloc@plt>
  test   %rax,%rax
  je     L2
  test   %ebp,%ebp
  jle    L2
  mov    %rax,%rdx
  lea    (%rax,%rbp,4),%rcx
  nopw   0x0(%rax,%rax,1)
L13:
  mov    %ebx,(%rdx)
  add    $0x4,%rdx
  cmp    %rcx,%rdx
  jne    L13
L2:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int *create_array(int n, int initial_value) {
  int *a = malloc(n * sizeof(int));
  if (a != NULL) {
    int i;
    for (i = 0; i < n; i++)
      a[i] = initial_value;
    return a;
  } else
    return NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<create_array>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  shl    $0x2,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  test   %rax,%rax
  je     L44
  test   %ebx,%ebx
  jle    L44
  lea    -0x1(%rbx),%edx
  cmp    $0x2,%edx
  jbe    L1
  mov    %ebx,%ecx
  movd   %ebp,%xmm1
  mov    %rax,%rdx
  shr    $0x2,%ecx
  pshufd $0x0,%xmm1,%xmm0
  shl    $0x4,%rcx
  add    %rax,%rcx
  nop
L13:
  movups %xmm0,(%rdx)
  add    $0x10,%rdx
  cmp    %rcx,%rdx
  jne    L13
  mov    %ebx,%edx
  and    $0xfffffffc,%edx
  test   $0x3,%bl
  je     L44
L18:
  movslq %edx,%rcx
  mov    %ebp,(%rax,%rcx,4)
  lea    0x0(,%rcx,4),%rsi
  lea    0x1(%rdx),%ecx
  cmp    %ebx,%ecx
  jge    L44
  add    $0x2,%edx
  mov    %ebp,0x4(%rax,%rsi,1)
  cmp    %edx,%ebx
  jle    L44
  mov    %ebp,0x8(%rax,%rsi,1)
L44:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  xor    %edx,%edx
  jmp    L18
```

assistant: ```c
int *create_array(int n, int initial_value) {
  int *a = malloc(n * sizeof(int));
  if (a != NULL) {
    int i;
    for (i = 0; i < n; i++)
      a[i] = initial_value;
    return a;
  } else
    return NULL;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<compute_match_rating_codex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  movb   $0x0,-0x1a(%rbp)
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L108
L38:
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <toupper@plt>
  mov    %al,-0x19(%rbp)
  cmpb   $0x20,-0x19(%rbp)
  je     L39
  cmpq   $0x0,-0x18(%rbp)
  je     L40
  cmpb   $0x41,-0x19(%rbp)
  je     L39
  cmpb   $0x45,-0x19(%rbp)
  je     L39
  cmpb   $0x49,-0x19(%rbp)
  je     L39
  cmpb   $0x4f,-0x19(%rbp)
  je     L39
  cmpb   $0x55,-0x19(%rbp)
  je     L39
L40:
  movzbl -0x19(%rbp),%eax
  cmp    -0x1a(%rbp),%al
  je     L117
  cmpq   $0x6,-0x10(%rbp)
  jne    L132
  mov    -0x30(%rbp),%rax
  lea    0x3(%rax),%rdx
  mov    -0x30(%rbp),%rax
  movzbl 0x4(%rax),%eax
  mov    %al,(%rdx)
  mov    -0x30(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    -0x30(%rbp),%rax
  movzbl 0x5(%rax),%eax
  mov    %al,(%rdx)
  movq   $0x5,-0x10(%rbp)
L132:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x30(%rbp),%rdx
  add    %rax,%rdx
  movzbl -0x19(%rbp),%eax
  mov    %al,(%rdx)
  jmp    L20
L39:
  nop
  jmp    L20
L117:
  nop
L20:
  addq   $0x1,-0x18(%rbp)
L108:
  mov    -0x18(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jae    L71
  cmpq   $0x6,-0x10(%rbp)
  jbe    L38
L71:
  mov    -0x30(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
  leave
  ret
```

assistant: ```c
size_t compute_match_rating_codex(const char *str, char codex[7]) {
  size_t len = strlen(str);
  size_t i, j;
  char c, prev;
  prev = '\0';
  for (i = 0, j = 0; i < len && j < 7; i++) {
    c = toupper(str[i]);
    if (c == ' ' || (i != 0 && (c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U'))) {
      continue;
    }
    if (c == prev) {
      continue;
    }
    if (j == 6) {
      codex[3] = codex[4];
      codex[4] = codex[5];
      j = 5;
    }
    codex[j++] = c;
  }
  codex[j] = '\0';
  return j;
}
```
user: What is the c source code of the assembly code below?

```asm
<compute_match_rating_codex>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r14
  mov    %rsi,%r12
  call   <strlen@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L45
  mov    $0x0,%ebp
  mov    $0x0,%ebx
  mov    $0x5,%r15d
  jmp    L34
L102:
  test   %al,%al
  je     L44
L29:
  cmp    $0x6,%rbp
  je     L92
L97:
  mov    %al,(%r12,%rbp,1)
  lea    0x1(%rbp),%rbp
L44:
  add    $0x1,%rbx
  cmp    %rbx,%r13
  jbe    L38
  cmp    $0x6,%rbp
  ja     L38
L34:
  movsbl (%r14,%rbx,1),%edi
  call   <toupper@plt>
  cmp    $0x20,%al
  je     L44
  test   %rbx,%rbx
  je     L102
  lea    -0x41(%rax),%edx
  cmp    $0x14,%dl
  ja     L102
  mov    $0x104111,%ecx
  bt     %rdx,%rcx
  jae    L29
  jmp    L44
L92:
  movzbl 0x4(%r12),%edx
  mov    %dl,0x3(%r12)
  movzbl 0x5(%r12),%edx
  mov    %dl,0x4(%r12)
  mov    %r15,%rbp
  jmp    L97
L45:
  mov    %rax,%rbp
L38:
  movb   $0x0,(%r12,%rbp,1)
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
size_t compute_match_rating_codex(const char *str, char codex[7]) {
  size_t len = strlen(str);
  size_t i, j;
  char c, prev;
  prev = '\0';
  for (i = 0, j = 0; i < len && j < 7; i++) {
    c = toupper(str[i]);
    if (c == ' ' || (i != 0 && (c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U'))) {
      continue;
    }
    if (c == prev) {
      continue;
    }
    if (j == 6) {
      codex[3] = codex[4];
      codex[4] = codex[5];
      j = 5;
    }
    codex[j++] = c;
  }
  codex[j] = '\0';
  return j;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<compute_match_rating_codex>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <strlen@plt>
  test   %rax,%rax
  je     L36
  mov    %rax,%r13
  xor    %r12d,%r12d
  xor    %r14d,%r14d
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L31:
  lea    -0x41(%rax),%edx
  cmp    $0x14,%dl
  ja     L89
  mov    $0x104111,%ecx
  bt     %rdx,%rcx
  jb     L18
L7:
  cmp    $0x6,%r12
  je     L32
  mov    %r12,%rdx
  add    $0x1,%r12
L53:
  mov    %al,(%rbx,%rdx,1)
L18:
  add    $0x1,%r14
  cmp    %r14,%r13
  je     L64
L33:
  movsbl 0x0(%rbp,%r14,1),%edi
  call   <toupper@plt>
  cmp    $0x20,%al
  je     L18
  test   %r14,%r14
  jne    L31
L89:
  test   %al,%al
  jne    L7
  add    $0x1,%r14
  cmp    %r14,%r13
  jne    L33
L64:
  add    %r12,%rbx
  mov    %r12,%rax
  movb   $0x0,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L32:
  movzbl 0x4(%rbx),%edx
  mov    %dl,0x3(%rbx)
  movzbl 0x5(%rbx),%edx
  mov    %dl,0x4(%rbx)
  mov    $0x5,%edx
  jmp    L53
  nopl   (%rax)
L36:
  xor    %r12d,%r12d
  movb   $0x0,(%rbx)
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
size_t compute_match_rating_codex(const char *str, char codex[7]) {
  size_t len = strlen(str);
  size_t i, j;
  char c, prev;
  prev = '\0';
  for (i = 0, j = 0; i < len && j < 7; i++) {
    c = toupper(str[i]);
    if (c == ' ' || (i != 0 && (c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U'))) {
      continue;
    }
    if (c == prev) {
      continue;
    }
    if (j == 6) {
      codex[3] = codex[4];
      codex[4] = codex[5];
      j = 5;
    }
    codex[j++] = c;
  }
  codex[j] = '\0';
  return j;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<compute_match_rating_codex>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <strlen@plt>
  test   %rax,%rax
  je     L81
  mov    %rax,%r13
  xor    %r12d,%r12d
  xor    %r14d,%r14d
  jmp    L100
  nopl   0x0(%rax,%rax,1)
L36:
  lea    -0x41(%rax),%edx
  cmp    $0x14,%dl
  ja     L44
  mov    $0x104111,%ecx
  bt     %rdx,%rcx
  jb     L18
L22:
  cmp    $0x6,%r12
  je     L89
  mov    %r12,%rdx
  add    $0x1,%r12
L9:
  mov    %al,(%rbx,%rdx,1)
L18:
  add    $0x1,%r14
  cmp    %r14,%r13
  je     L55
L100:
  movsbl 0x0(%rbp,%r14,1),%edi
  call   <toupper@plt>
  cmp    $0x20,%al
  je     L18
  test   %r14,%r14
  jne    L36
L44:
  test   %al,%al
  jne    L22
  add    $0x1,%r14
  cmp    %r14,%r13
  jne    L100
L55:
  add    %r12,%rbx
  mov    %r12,%rax
  movb   $0x0,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L89:
  movzbl 0x4(%rbx),%edx
  mov    %dl,0x3(%rbx)
  movzbl 0x5(%rbx),%edx
  mov    %dl,0x4(%rbx)
  mov    $0x5,%edx
  jmp    L9
  nopl   (%rax)
L81:
  xor    %r12d,%r12d
  movb   $0x0,(%rbx)
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
size_t compute_match_rating_codex(const char *str, char codex[7]) {
  size_t len = strlen(str);
  size_t i, j;
  char c, prev;
  prev = '\0';
  for (i = 0, j = 0; i < len && j < 7; i++) {
    c = toupper(str[i]);
    if (c == ' ' || (i != 0 && (c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U'))) {
      continue;
    }
    if (c == prev) {
      continue;
    }
    if (j == 6) {
      codex[3] = codex[4];
      codex[4] = codex[5];
      j = 5;
    }
    codex[j++] = c;
  }
  codex[j] = '\0';
  return j;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L28
L7:
  mov    -0x4(%rbp),%eax
  and    $0x7,%eax
  test   %eax,%eax
  jne    L34
  lea    D60(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L34:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <q0@plt>
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L28:
  cmpl   $0xff,-0x4(%rbp)
  jbe    L7
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D79(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L4
L11:
  mov    -0x4(%rbp),%eax
  and    $0x7,%eax
  test   %eax,%eax
  jne    L13
  lea    D60(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L13:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <q1@plt>
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L4:
  cmpl   $0xff,-0x4(%rbp)
  jbe    L11
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D31
type: string
name: unpack_data
arguments: label: D60
type: string
name: unpack_data
arguments: label: D24
type: string
name: unpack_data
arguments: label: D3
type: string
name: unpack_data
arguments: label: D79
type: string
name: unpack_data
arguments: label: D12
type: string

tool: D31: static const uint8_t q0[256] = {

tool: D60: 
  

tool: D24: 0x%02X,

tool: D3: 
};


tool: D79: static const uint8_t q1[256] = {

tool: D12: 
};

assistant: ```c
int main(void) {
  unsigned i;
  printf("static const uint8_t q0[256] = {");
  for (i = 0; i < 256; i++) {
    if ((i % 8) == 0)
      printf("\n  ");
    printf("0x%02X,", q0(i));
  }
  printf("\n};\n\n");
  printf("static const uint8_t q1[256] = {");
  for (i = 0; i < 256; i++) {
    if ((i % 8) == 0)
      printf("\n  ");
    printf("0x%02X,", q1(i));
  }
  printf("\n};\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    D48(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%ebx
  lea    D1(%rip),%r12
  lea    D73(%rip),%rbp
  jmp    L55
L52:
  mov    %ebx,%edi
  call   <q0@plt>
  mov    %eax,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0x100,%ebx
  je     L78
L55:
  test   $0x7,%bl
  jne    L52
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L52
L78:
  lea    D66(%rip),%rdi
  call   <puts@plt>
  lea    D41(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%ebx
  lea    D1(%rip),%r12
  lea    D73(%rip),%rbp
  jmp    L12
L40:
  mov    %ebx,%edi
  call   <q1@plt>
  mov    %eax,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0x100,%ebx
  je     L24
L12:
  test   $0x7,%bl
  jne    L40
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L40
L24:
  lea    D14(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: tag: D48
datatype: string
name: parse_data
arguments: tag: D1
datatype: string
name: parse_data
arguments: tag: D73
datatype: string
name: parse_data
arguments: tag: D66
datatype: string
name: parse_data
arguments: tag: D41
datatype: string
name: parse_data
arguments: tag: D14
datatype: string

tool: D48: static const uint8_t q0[256] = {

tool: D1: 
  

tool: D73: 0x%02X,

tool: D66: 
};


tool: D41: static const uint8_t q1[256] = {

tool: D14: 
};

assistant: ```c
int main(void) {
  unsigned i;
  printf("static const uint8_t q0[256] = {");
  for (i = 0; i < 256; i++) {
    if ((i % 8) == 0)
      printf("\n  ");
    printf("0x%02X,", q0(i));
  }
  printf("\n};\n\n");
  printf("static const uint8_t q1[256] = {");
  for (i = 0; i < 256; i++) {
    if ((i % 8) == 0)
      printf("\n  ");
    printf("0x%02X,", q1(i));
  }
  printf("\n};\n");
  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   %r12
  lea    D24(%rip),%rdi
  xor    %eax,%eax
  lea    D13(%rip),%r12
  push   %rbp
  lea    D1(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  call   <printf@plt>
  jmp    L77
  nopl   0x0(%rax,%rax,1)
L1:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <q0@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  cmp    $0x100,%ebx
  je     L20
L77:
  test   $0x7,%bl
  jne    L1
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L1
L20:
  lea    D25(%rip),%rdi
  xor    %ebx,%ebx
  lea    D13(%rip),%r12
  call   <puts@plt>
  lea    D63(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L58
  nopl   0x0(%rax)
L19:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <q1@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  cmp    $0x100,%ebx
  je     L41
L58:
  test   $0x7,%bl
  jne    L19
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L19
L41:
  lea    D34(%rip),%rdi
  call   <puts@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D24
datatype: string
name: extract_symbolic_data
arguments: tag: D13
datatype: string
name: extract_symbolic_data
arguments: tag: D1
datatype: string
name: extract_symbolic_data
arguments: tag: D25
datatype: string
name: extract_symbolic_data
arguments: tag: D63
datatype: string
name: extract_symbolic_data
arguments: tag: D34
datatype: string

tool: D24: static const uint8_t q0[256] = {

tool: D13: 
  

tool: D1: 0x%02X,

tool: D25: 
};


tool: D63: static const uint8_t q1[256] = {

tool: D34: 
};

assistant: ```c
int main(void) {
  unsigned i;
  printf("static const uint8_t q0[256] = {");
  for (i = 0; i < 256; i++) {
    if ((i % 8) == 0)
      printf("\n  ");
    printf("0x%02X,", q0(i));
  }
  printf("\n};\n\n");
  printf("static const uint8_t q1[256] = {");
  for (i = 0; i < 256; i++) {
    if ((i % 8) == 0)
      printf("\n  ");
    printf("0x%02X,", q1(i));
  }
  printf("\n};\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  lea    D17(%rip),%rdi
  xor    %eax,%eax
  lea    D26(%rip),%r12
  push   %rbp
  lea    D58(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  call   <printf@plt>
  jmp    L77
  nopl   0x0(%rax,%rax,1)
L10:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <q0@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  cmp    $0x100,%ebx
  je     L14
L77:
  test   $0x7,%bl
  jne    L10
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L10
L14:
  lea    D52(%rip),%rdi
  xor    %ebx,%ebx
  lea    D26(%rip),%r12
  call   <puts@plt>
  lea    D2(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L55
  nopl   0x0(%rax)
L52:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <q1@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  cmp    $0x100,%ebx
  je     L66
L55:
  test   $0x7,%bl
  jne    L52
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L52
L66:
  lea    D33(%rip),%rdi
  call   <puts@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D17
type_guessed: string
name: unpack_data
arguments: data_label: D26
type_guessed: string
name: unpack_data
arguments: data_label: D58
type_guessed: string
name: unpack_data
arguments: data_label: D52
type_guessed: string
name: unpack_data
arguments: data_label: D2
type_guessed: string
name: unpack_data
arguments: data_label: D33
type_guessed: string

tool: D17: static const uint8_t q0[256] = {

tool: D26: 
  

tool: D58: 0x%02X,

tool: D52: 
};


tool: D2: static const uint8_t q1[256] = {

tool: D33: 
};

assistant: ```c
int main(void) {
  unsigned i;
  printf("static const uint8_t q0[256] = {");
  for (i = 0; i < 256; i++) {
    if ((i % 8) == 0)
      printf("\n  ");
    printf("0x%02X,", q0(i));
  }
  printf("\n};\n\n");
  printf("static const uint8_t q1[256] = {");
  for (i = 0; i < 256; i++) {
    if ((i % 8) == 0)
      printf("\n  ");
    printf("0x%02X,", q1(i));
  }
  printf("\n};\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<old_arr3alloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    -0x18(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L30
L17:
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rax,(%rdx)
  movl   $0x0,-0xc(%rbp)
  jmp    L37
L20:
  mov    -0x20(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%rcx
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  shl    $0x3,%rdx
  add    %rcx,%rdx
  mov    %rax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L37:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L20
  addl   $0x1,-0x10(%rbp)
L30:
  mov    -0x10(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L17
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void ***old_arr3alloc(const int size, const int n1, const int n2, const int n3) {
  int i, j;
  int ***aa;
  aa = (int ***)malloc(n1 * sizeof(int **));
  for (i = 0; i < n1; i++) {
    aa[i] = (int **)malloc(n2 * sizeof(int *));
    for (j = 0; j < n2; j++)
      aa[i][j] = (int *)malloc(n3 * sizeof(size));
  }
  return (void ***)aa;
}
```
user: What is the c source code of the assembly code below:

```asm
<old_arr3alloc>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %esi,%ebx
  mov    %edx,%r15d
  mov    %ecx,%r12d
  lea    0x0(,%rsi,8),%edi
  call   <malloc@plt>
  mov    %rax,0x18(%rsp)
  test   %ebx,%ebx
  jle    L15
  lea    0x0(,%r15,8),%ecx
  mov    %ecx,0xc(%rsp)
  shl    $0x2,%r12d
  mov    %rax,%r14
  mov    %ebx,%ebx
  lea    (%rax,%rbx,8),%rax
  mov    %rax,0x10(%rsp)
  lea    -0x1(%r15),%eax
  lea    0x8(,%rax,8),%r13
L4:
  mov    0xc(%rsp),%edi
  call   <malloc@plt>
  mov    %r14,%rbp
  mov    %rax,(%r14)
  test   %r15d,%r15d
  jle    L26
  mov    $0x0,%ebx
L29:
  mov    %r12d,%edi
  call   <malloc@plt>
  mov    0x0(%rbp),%rdx
  mov    %rax,(%rdx,%rbx,1)
  add    $0x8,%rbx
  cmp    %r13,%rbx
  jne    L29
L26:
  add    $0x8,%r14
  cmp    0x10(%rsp),%r14
  jne    L4
L15:
  mov    0x18(%rsp),%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void ***old_arr3alloc(const int size, const int n1, const int n2, const int n3) {
  int i, j;
  int ***aa;
  aa = (int ***)malloc(n1 * sizeof(int **));
  for (i = 0; i < n1; i++) {
    aa[i] = (int **)malloc(n2 * sizeof(int *));
    for (j = 0; j < n2; j++)
      aa[i][j] = (int *)malloc(n3 * sizeof(size));
  }
  return (void ***)aa;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<old_arr3alloc>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  lea    0x0(,%rbp,8),%edi
  mov    %ecx,%ebx
  sub    $0x18,%rsp
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  test   %ebp,%ebp
  jle    L18
  mov    %rax,%r15
  lea    (%rax,%rbp,8),%rax
  lea    0x0(,%r12,8),%r14d
  shl    $0x2,%ebx
  mov    %rax,(%rsp)
  lea    -0x1(%r12),%eax
  lea    0x8(,%rax,8),%rbp
  nopl   0x0(%rax)
L19:
  mov    %r14d,%edi
  call   <malloc@plt>
  mov    %rax,(%r15)
  test   %r12d,%r12d
  jle    L29
  xor    %r13d,%r13d
  nopl   0x0(%rax,%rax,1)
L21:
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    (%r15),%rdx
  mov    %rax,(%rdx,%r13,1)
  add    $0x8,%r13
  cmp    %r13,%rbp
  jne    L21
L29:
  add    $0x8,%r15
  cmp    (%rsp),%r15
  jne    L19
L18:
  mov    0x8(%rsp),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void ***old_arr3alloc(const int size, const int n1, const int n2, const int n3) {
  int i, j;
  int ***aa;
  aa = (int ***)malloc(n1 * sizeof(int **));
  for (i = 0; i < n1; i++) {
    aa[i] = (int **)malloc(n2 * sizeof(int *));
    for (j = 0; j < n2; j++)
      aa[i][j] = (int *)malloc(n3 * sizeof(size));
  }
  return (void ***)aa;
}
```