ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<b6b_offload>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    -0x10(%rbp),%rdx
  mov    %rax,%rdi
  call   *%rdx
  mov    $0x1,%eax
  leave
  ret
```

assistant: ```c
int b6b_offload(struct b6b_interp *interp, void (*fn)(void *), void *arg) {
  fn(arg);
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<b6b_offload>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdx,%rdi
  call   *%rsi
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int b6b_offload(struct b6b_interp *interp, void (*fn)(void *), void *arg) {
  fn(arg);
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<b6b_offload>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdx,%rdi
  call   *%rsi
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int b6b_offload(struct b6b_interp *interp, void (*fn)(void *), void *arg) {
  fn(arg);
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<b6b_offload>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdx,%rdi
  call   *%rsi
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int b6b_offload(struct b6b_interp *interp, void (*fn)(void *), void *arg) {
  fn(arg);
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<fill_one>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L39:
  mov    -0x4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L10
  mov    -0x20(%rbp),%eax
  cltq
  shl    $0x5,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  movl   $0x1,(%rdx,%rax,4)
  jmp    L35
L10:
  mov    -0x20(%rbp),%eax
  cltq
  shl    $0x5,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  movl   $0x0,(%rdx,%rax,4)
L35:
  addl   $0x1,-0x4(%rbp)
L11:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L39
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void fill_one(int A[][8], int columns, int number_row) {
  int i;
  for (i = 0; i < columns; i++)
    if (i % 2 == 0)
      A[number_row][i] = 1;
    else
      A[number_row][i] = 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fill_one>:
  endbr64
  test   %esi,%esi
  jle    L11
  movslq %edx,%rdx
  shl    $0x5,%rdx
  add    %rdx,%rdi
  mov    $0x0,%eax
L10:
  test   $0x1,%al
  sete   %dl
  movzbl %dl,%edx
  mov    %edx,(%rdi,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%esi
  jg     L10
L11:
  ret
```

assistant: ```c
void fill_one(int A[][8], int columns, int number_row) {
  int i;
  for (i = 0; i < columns; i++)
    if (i % 2 == 0)
      A[number_row][i] = 1;
    else
      A[number_row][i] = 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fill_one>:
  endbr64
  test   %esi,%esi
  jle    L2
  movslq %edx,%rdx
  xor    %eax,%eax
  mov    $0x1,%r8d
  shl    $0x5,%rdx
  add    %rdx,%rdi
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %eax,%ecx
  mov    %r8d,%edx
  and    $0x1,%ecx
  sub    %ecx,%edx
  mov    %edx,(%rdi,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%esi
  jg     L1
L2:
  ret
```

assistant: ```c
void fill_one(int A[][8], int columns, int number_row) {
  int i;
  for (i = 0; i < columns; i++)
    if (i % 2 == 0)
      A[number_row][i] = 1;
    else
      A[number_row][i] = 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fill_one>:
  endbr64
  mov    %esi,%eax
  test   %esi,%esi
  jle    L56
  movslq %edx,%rdx
  shl    $0x5,%rdx
  add    %rdx,%rdi
  lea    -0x1(%rsi),%edx
  cmp    $0x2,%edx
  jbe    L9
  movdqa D30(%rip),%xmm0
  mov    %esi,%edx
  shr    $0x2,%edx
  movups %xmm0,(%rdi)
  cmp    $0x1,%edx
  je     L101
  movups %xmm0,0x10(%rdi)
  cmp    $0x2,%edx
  je     L101
  movdqa D7(%rip),%xmm1
  movdqa D25(%rip),%xmm4
  xor    %ecx,%ecx
  pxor   %xmm3,%xmm3
  movdqa D21(%rip),%xmm2
L73:
  movdqa %xmm1,%xmm0
  mov    %rcx,%rsi
  paddd  %xmm4,%xmm1
  add    $0x1,%rcx
  pand   %xmm2,%xmm0
  shl    $0x4,%rsi
  pcmpeqd %xmm3,%xmm0
  pand   %xmm2,%xmm0
  movups %xmm0,0x20(%rdi,%rsi,1)
  lea    0x2(%rcx),%esi
  cmp    %esi,%edx
  ja     L73
L101:
  mov    %eax,%edx
  and    $0xfffffffc,%edx
  test   $0x3,%al
  je     L56
  movslq %edx,%rcx
  add    $0x1,%edx
  lea    (%rdi,%rcx,4),%rdi
  movl   $0x1,(%rdi)
  cmp    %edx,%eax
  jle    L56
  mov    $0x1,%ecx
  test   $0x1,%dl
  jne    L22
L71:
  add    $0x1,%edx
  mov    %ecx,0x4(%rdi)
  cmp    %edx,%eax
  jle    L56
  not    %edx
  and    $0x1,%edx
  mov    %edx,0x8(%rdi)
  ret
L9:
  movl   $0x1,(%rdi)
  mov    $0x1,%edx
  cmp    $0x1,%esi
  jg     L22
L56:
  ret
  cs nopw 0x0(%rax,%rax,1)
L22:
  xor    %ecx,%ecx
  jmp    L71
```

tool_calls: name: retrieve_data
arguments: tag: D30
type_guessed: byte[16]
name: retrieve_data
arguments: tag: D7
type_guessed: byte[16]
name: retrieve_data
arguments: tag: D25
type_guessed: byte[16]
name: retrieve_data
arguments: tag: D21
type_guessed: byte[16]

tool: {"D30": [0x01,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: {"D7": [0x08,0x00,0x00,0x00,0x09,0x00,0x00,0x00,0x0a,0x00,0x00,0x00,0x0b,0x00,0x00,0x00]}

tool: {"D25": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D21": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
void fill_one(int A[][8], int columns, int number_row) {
  int i;
  for (i = 0; i < columns; i++)
    if (i % 2 == 0)
      A[number_row][i] = 1;
    else
      A[number_row][i] = 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<_cc3_camera_reset>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x3fffc01c,%eax
  movq   $0x2000,(%rax)
  mov    $0x1,%edi
  call   <cc3_timer_wait_ms@plt>
  mov    $0x3fffc018,%eax
  movq   $0x2000,(%rax)
  mov    $0x1,%edi
  call   <cc3_timer_wait_ms@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void _cc3_camera_reset() {
  (*(volatile unsigned long *)((0x3FFFC01C))) = 0x2000;
  cc3_timer_wait_ms(1);
  (*(volatile unsigned long *)((0x3FFFC018))) = 0x2000;
  cc3_timer_wait_ms(1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_cc3_camera_reset>:
  endbr64
  sub    $0x8,%rsp
  movq   $0x2000,0x3fffc01c
  mov    $0x1,%edi
  call   <cc3_timer_wait_ms@plt>
  movq   $0x2000,0x3fffc018
  mov    $0x1,%edi
  call   <cc3_timer_wait_ms@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void _cc3_camera_reset() {
  (*(volatile unsigned long *)((0x3FFFC01C))) = 0x2000;
  cc3_timer_wait_ms(1);
  (*(volatile unsigned long *)((0x3FFFC018))) = 0x2000;
  cc3_timer_wait_ms(1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<_cc3_camera_reset>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  movq   $0x2000,0x3fffc01c
  call   <cc3_timer_wait_ms@plt>
  mov    $0x1,%edi
  movq   $0x2000,0x3fffc018
  add    $0x8,%rsp
  jmp    1050 <cc3_timer_wait_ms@plt>
```

assistant: ```c
void _cc3_camera_reset() {
  (*(volatile unsigned long *)((0x3FFFC01C))) = 0x2000;
  cc3_timer_wait_ms(1);
  (*(volatile unsigned long *)((0x3FFFC018))) = 0x2000;
  cc3_timer_wait_ms(1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_cc3_camera_reset>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  movq   $0x2000,0x3fffc01c
  call   <cc3_timer_wait_ms@plt>
  mov    $0x1,%edi
  movq   $0x2000,0x3fffc018
  add    $0x8,%rsp
  jmp    1050 <cc3_timer_wait_ms@plt>
```

assistant: ```c
void _cc3_camera_reset() {
  (*(volatile unsigned long *)((0x3FFFC01C))) = 0x2000;
  cc3_timer_wait_ms(1);
  (*(volatile unsigned long *)((0x3FFFC018))) = 0x2000;
  cc3_timer_wait_ms(1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<numberdivisors>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  movq   $0x1,-0x8(%rbp)
  movq   $0x1,-0x8(%rbp)
  jmp    L15
L12:
  mov    -0x28(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x8(%rbp)
  mov    %rdx,%rax
  test   %rax,%rax
  jne    L13
  mov    -0x28(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x8(%rbp)
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  je     L20
  addl   $0x1,-0x14(%rbp)
L20:
  addl   $0x1,-0x14(%rbp)
L13:
  addq   $0x1,-0x8(%rbp)
L15:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jb     L12
  mov    -0x14(%rbp),%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
type_infered: string

tool: D8: limit = %llu | i = %llu


assistant: ```c
int numberdivisors(unsigned long long x) {
  unsigned long long limit = x;
  int numberOfDivisors = 0;
  unsigned long long i = 1;
  for (i = 1; i < limit; ++i) {
    if (x % i == 0) {
      limit = x / i;
      printf("limit = %llu | i = %llu\n", limit, i);
      if (limit != i) {
        numberOfDivisors++;
      }
      numberOfDivisors++;
    }
  }
  return numberOfDivisors;
}
```
user: What is the c source code of the assembly code below?

```asm
<numberdivisors>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%rdi
  jbe    L35
  mov    %rdi,%rbp
  mov    %rdi,%r12
  mov    $0x1,%ebx
  mov    $0x0,%r13d
  lea    D6(%rip),%r14
  jmp    L27
L49:
  add    $0x1,%rbx
  cmp    %rbx,%r12
  jbe    L29
L27:
  mov    %rbp,%rax
  mov    $0x0,%edx
  div    %rbx
  test   %rdx,%rdx
  jne    L49
  mov    %rbp,%rax
  mov    $0x0,%edx
  div    %rbx
  mov    %rax,%r15
  mov    %rax,%r12
  mov    %rbx,%rdx
  mov    %rax,%rsi
  mov    %r14,%rdi
  call   <printf@plt>
  cmp    %rbx,%r15
  setne  %al
  movzbl %al,%eax
  lea    0x1(%r13,%rax,1),%r13d
  jmp    L49
L35:
  mov    $0x0,%r13d
L29:
  mov    %r13d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: unpack_data
arguments: label: D6
type: string

tool: D6: limit = %llu | i = %llu


assistant: ```c
int numberdivisors(unsigned long long x) {
  unsigned long long limit = x;
  int numberOfDivisors = 0;
  unsigned long long i = 1;
  for (i = 1; i < limit; ++i) {
    if (x % i == 0) {
      limit = x / i;
      printf("limit = %llu | i = %llu\n", limit, i);
      if (limit != i) {
        numberOfDivisors++;
      }
      numberOfDivisors++;
    }
  }
  return numberOfDivisors;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<numberdivisors>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x1,%rdi
  jbe    L39
  mov    %rdi,%rbp
  mov    %rdi,%r12
  mov    $0x1,%ebx
  xor    %r14d,%r14d
  lea    D8(%rip),%r13
  jmp    L15
  nopl   0x0(%rax)
L26:
  add    $0x1,%rbx
  cmp    %rbx,%r12
  jbe    L46
L15:
  mov    %rbp,%rax
  xor    %edx,%edx
  div    %rbx
  test   %rdx,%rdx
  jne    L26
  mov    %rax,%r12
  mov    %rbx,%rdx
  mov    %rax,%rsi
  mov    %r13,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  cmp    %rbx,%r12
  setne  %al
  add    $0x1,%rbx
  lea    0x1(%r14,%rax,1),%r14d
  cmp    %rbx,%r12
  ja     L15
L46:
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L39:
  xor    %r14d,%r14d
  pop    %rbx
  pop    %rbp
  mov    %r14d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: data_tag: D8
data_type: string

tool: D8: limit = %llu | i = %llu


assistant: ```c
int numberdivisors(unsigned long long x) {
  unsigned long long limit = x;
  int numberOfDivisors = 0;
  unsigned long long i = 1;
  for (i = 1; i < limit; ++i) {
    if (x % i == 0) {
      limit = x / i;
      printf("limit = %llu | i = %llu\n", limit, i);
      if (limit != i) {
        numberOfDivisors++;
      }
      numberOfDivisors++;
    }
  }
  return numberOfDivisors;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<numberdivisors>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x1,%rdi
  jbe    L49
  mov    %rdi,%rbp
  mov    %rdi,%r12
  mov    $0x1,%ebx
  xor    %r14d,%r14d
  lea    D8(%rip),%r13
  jmp    L6
  nopl   0x0(%rax)
L23:
  add    $0x1,%rbx
  cmp    %rbx,%r12
  jbe    L13
L6:
  mov    %rbp,%rax
  xor    %edx,%edx
  div    %rbx
  test   %rdx,%rdx
  jne    L23
  mov    %rax,%r12
  mov    %rbx,%rdx
  mov    %rax,%rsi
  mov    %r13,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  cmp    %rbx,%r12
  setne  %al
  add    $0x1,%rbx
  lea    0x1(%r14,%rax,1),%r14d
  cmp    %rbx,%r12
  ja     L6
L13:
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L49:
  xor    %r14d,%r14d
  pop    %rbx
  pop    %rbp
  mov    %r14d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: data_label: D8
data_type: string

tool: D8: limit = %llu | i = %llu


assistant: ```c
int numberdivisors(unsigned long long x) {
  unsigned long long limit = x;
  int numberOfDivisors = 0;
  unsigned long long i = 1;
  for (i = 1; i < limit; ++i) {
    if (x % i == 0) {
      limit = x / i;
      printf("limit = %llu | i = %llu\n", limit, i);
      if (limit != i) {
        numberOfDivisors++;
      }
      numberOfDivisors++;
    }
  }
  return numberOfDivisors;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<test5>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x34,-0xd0(%rbp)
  movq   $0x18,-0xc8(%rbp)
  movabs $0x7766554433221100,%rax
  movabs $0xffeeddccbbaa9988,%rdx
  mov    %rax,-0xc0(%rbp)
  mov    %rdx,-0xb8(%rbp)
  mov    %rax,-0xa0(%rbp)
  mov    %rdx,-0x98(%rbp)
  movabs $0xbebaafdeefbeadde,%rax
  mov    %rax,-0x90(%rbp)
  movabs $0x706050403020100,%rax
  movabs $0xf0e0d0c0b0a0908,%rdx
  mov    %rax,-0x80(%rbp)
  mov    %rdx,-0x78(%rbp)
  movabs $0x1716151413121110,%rax
  movabs $0x1f1e1d1c1b1a1918,%rdx
  mov    %rax,-0x70(%rbp)
  mov    %rdx,-0x68(%rbp)
  movabs $0x2726252423222120,%rax
  movabs $0x2f2e2d2c2b2a2928,%rdx
  mov    %rax,-0x60(%rbp)
  mov    %rdx,-0x58(%rbp)
  movl   $0xbebafefe,-0x50(%rbp)
  movabs $0x79798c7dd051da3b,%rax
  movabs $0x3f9a6e3ea46401cd,%rdx
  mov    %rax,-0x40(%rbp)
  mov    %rdx,-0x38(%rbp)
  movabs $0x20714ef230a88b7f,%rax
  movabs $0x62a0310b307bfa52,%rdx
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
  movabs $0xed603456b9832393,%rax
  movabs $0xdde85d6764bba21d,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  movl   $0xaffbb4f5,-0x10(%rbp)
  movabs $0xecf0bbbe4a65b868,%rax
  movabs $0xf09d03c0f502705b,%rdx
  mov    %rax,-0xb0(%rbp)
  mov    %rdx,-0xa8(%rbp)
  lea    -0x40(%rbp),%r9
  mov    -0xd0(%rbp),%r8
  lea    -0x80(%rbp),%rcx
  mov    -0xc8(%rbp),%rdx
  lea    -0xa0(%rbp),%rsi
  lea    -0xc0(%rbp),%rax
  sub    $0x8,%rsp
  lea    -0xb0(%rbp),%rdi
  push   %rdi
  mov    %rax,%rdi
  call   <run_test@plt>
  add    $0x10,%rsp
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
int test5() {
  unsigned long long mlen = 52;
  const unsigned long long hlen = 24;
  const unsigned char k[16] = {0x00, 0x11, 0x22, 0x33, 0x44, 0x55, 0x66, 0x77, 0x88, 0x99, 0xaa, 0xbb, 0xcc, 0xdd, 0xee, 0xff};
  const unsigned char h[24] = {0x00, 0x11, 0x22, 0x33, 0x44, 0x55, 0x66, 0x77, 0x88, 0x99, 0xaa, 0xbb, 0xcc, 0xdd, 0xee, 0xff, 0xde, 0xad, 0xbe, 0xef, 0xde, 0xaf, 0xba, 0xbe};
  const unsigned char m[52] = {0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07, 0x08, 0x09, 0x0a, 0x0b, 0x0c, 0x0d, 0x0e, 0x0f, 0x10, 0x11, 0x12, 0x13, 0x14, 0x15, 0x16, 0x17, 0x18, 0x19, 0x1a, 0x1b, 0x1c, 0x1d, 0x1e, 0x1f, 0x20, 0x21, 0x22, 0x23, 0x24, 0x25, 0x26, 0x27, 0x28, 0x29, 0x2a, 0x2b, 0x2c, 0x2d, 0x2e, 0x2f, 0xfe, 0xfe, 0xba, 0xbe};
  const unsigned char c[52] = {0x3b, 0xda, 0x51, 0xd0, 0x7d, 0x8c, 0x79, 0x79, 0xcd, 0x01, 0x64, 0xa4, 0x3e, 0x6e, 0x9a, 0x3f, 0x7f, 0x8b, 0xa8, 0x30, 0xf2, 0x4e, 0x71, 0x20, 0x52, 0xfa, 0x7b, 0x30, 0x0b, 0x31, 0xa0, 0x62, 0x93, 0x23, 0x83, 0xb9, 0x56, 0x34, 0x60, 0xed, 0x1d, 0xa2, 0xbb, 0x64, 0x67, 0x5d, 0xe8, 0xdd, 0xf5, 0xb4, 0xfb, 0xaf};
  const unsigned char t[16] = {0x68, 0xb8, 0x65, 0x4a, 0xbe, 0xbb, 0xf0, 0xec, 0x5b, 0x70, 0x02, 0xf5, 0xc0, 0x03, 0x9d, 0xf0};
  return run_test(k, h, hlen, m, mlen, c, t);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test5>:
  endbr64
  sub    $0xc8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xb8(%rsp)
  xor    %eax,%eax
  movabs $0x7766554433221100,%rax
  movabs $0xffeeddccbbaa9988,%rdx
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  mov    %rax,0x20(%rsp)
  mov    %rdx,0x28(%rsp)
  movabs $0xbebaafdeefbeadde,%rax
  mov    %rax,0x30(%rsp)
  movabs $0x706050403020100,%rax
  movabs $0xf0e0d0c0b0a0908,%rdx
  mov    %rax,0x40(%rsp)
  mov    %rdx,0x48(%rsp)
  movabs $0x1716151413121110,%rax
  movabs $0x1f1e1d1c1b1a1918,%rdx
  mov    %rax,0x50(%rsp)
  mov    %rdx,0x58(%rsp)
  movabs $0x2726252423222120,%rax
  movabs $0x2f2e2d2c2b2a2928,%rdx
  mov    %rax,0x60(%rsp)
  mov    %rdx,0x68(%rsp)
  movl   $0xbebafefe,0x70(%rsp)
  movabs $0x79798c7dd051da3b,%rax
  movabs $0x3f9a6e3ea46401cd,%rdx
  mov    %rax,0x80(%rsp)
  mov    %rdx,0x88(%rsp)
  movabs $0x20714ef230a88b7f,%rax
  movabs $0x62a0310b307bfa52,%rdx
  mov    %rax,0x90(%rsp)
  mov    %rdx,0x98(%rsp)
  movabs $0xed603456b9832393,%rax
  movabs $0xdde85d6764bba21d,%rdx
  mov    %rax,0xa0(%rsp)
  mov    %rdx,0xa8(%rsp)
  movl   $0xaffbb4f5,0xb0(%rsp)
  movabs $0xecf0bbbe4a65b868,%rax
  movabs $0xf09d03c0f502705b,%rdx
  mov    %rax,0x10(%rsp)
  mov    %rdx,0x18(%rsp)
  lea    0x40(%rsp),%rcx
  lea    0x20(%rsp),%rsi
  mov    %rsp,%rdi
  sub    $0x8,%rsp
  lea    0x18(%rsp),%rax
  push   %rax
  lea    0x90(%rsp),%r9
  mov    $0x34,%r8d
  mov    $0x18,%edx
  call   <run_test@plt>
  add    $0x10,%rsp
  mov    0xb8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0xc8,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test5() {
  unsigned long long mlen = 52;
  const unsigned long long hlen = 24;
  const unsigned char k[16] = {0x00, 0x11, 0x22, 0x33, 0x44, 0x55, 0x66, 0x77, 0x88, 0x99, 0xaa, 0xbb, 0xcc, 0xdd, 0xee, 0xff};
  const unsigned char h[24] = {0x00, 0x11, 0x22, 0x33, 0x44, 0x55, 0x66, 0x77, 0x88, 0x99, 0xaa, 0xbb, 0xcc, 0xdd, 0xee, 0xff, 0xde, 0xad, 0xbe, 0xef, 0xde, 0xaf, 0xba, 0xbe};
  const unsigned char m[52] = {0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07, 0x08, 0x09, 0x0a, 0x0b, 0x0c, 0x0d, 0x0e, 0x0f, 0x10, 0x11, 0x12, 0x13, 0x14, 0x15, 0x16, 0x17, 0x18, 0x19, 0x1a, 0x1b, 0x1c, 0x1d, 0x1e, 0x1f, 0x20, 0x21, 0x22, 0x23, 0x24, 0x25, 0x26, 0x27, 0x28, 0x29, 0x2a, 0x2b, 0x2c, 0x2d, 0x2e, 0x2f, 0xfe, 0xfe, 0xba, 0xbe};
  const unsigned char c[52] = {0x3b, 0xda, 0x51, 0xd0, 0x7d, 0x8c, 0x79, 0x79, 0xcd, 0x01, 0x64, 0xa4, 0x3e, 0x6e, 0x9a, 0x3f, 0x7f, 0x8b, 0xa8, 0x30, 0xf2, 0x4e, 0x71, 0x20, 0x52, 0xfa, 0x7b, 0x30, 0x0b, 0x31, 0xa0, 0x62, 0x93, 0x23, 0x83, 0xb9, 0x56, 0x34, 0x60, 0xed, 0x1d, 0xa2, 0xbb, 0x64, 0x67, 0x5d, 0xe8, 0xdd, 0xf5, 0xb4, 0xfb, 0xaf};
  const unsigned char t[16] = {0x68, 0xb8, 0x65, 0x4a, 0xbe, 0xbb, 0xf0, 0xec, 0x5b, 0x70, 0x02, 0xf5, 0xc0, 0x03, 0x9d, 0xf0};
  return run_test(k, h, hlen, m, mlen, c, t);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<emit_sse_get_sign_mask>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D1(%rip),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L13
L19:
  addq   $0x1,-0x10(%rbp)
L13:
  mov    -0x10(%rbp),%rax
  and    $0xf,%eax
  test   %rax,%rax
  jne    L19
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movabs $0x8000000000000000,%rdx
  mov    %rdx,(%rax)
  mov    -0x10(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned char *emit_sse_get_sign_mask() {
  static unsigned char bufx[32];
  unsigned char *buf = bufx + 1;
  while ((long)buf % 16)
    buf++;
  unsigned long long *bit_mask = (unsigned long long *)buf;
  *bit_mask = (unsigned long long)1 << 63;
  return buf;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<emit_sse_get_sign_mask>:
  endbr64
  lea    D0(%rip),%rax
L4:
  add    $0x1,%rax
  test   $0xf,%al
  jne    L4
  movabs $0x8000000000000000,%rdx
  mov    %rdx,(%rax)
  ret
```

assistant: ```c
unsigned char *emit_sse_get_sign_mask() {
  static unsigned char bufx[32];
  unsigned char *buf = bufx + 1;
  while ((long)buf % 16)
    buf++;
  unsigned long long *bit_mask = (unsigned long long *)buf;
  *bit_mask = (unsigned long long)1 << 63;
  return buf;
}
```
user: What is the c source code of the assembly code below:

```asm
<emit_sse_get_sign_mask>:
  endbr64
  lea    D1(%rip),%rax
  nopl   0x0(%rax,%rax,1)
L0:
  add    $0x1,%rax
  test   $0xf,%al
  jne    L0
  movabs $0x8000000000000000,%rdx
  mov    %rdx,(%rax)
  ret
```

assistant: ```c
unsigned char *emit_sse_get_sign_mask() {
  static unsigned char bufx[32];
  unsigned char *buf = bufx + 1;
  while ((long)buf % 16)
    buf++;
  unsigned long long *bit_mask = (unsigned long long *)buf;
  *bit_mask = (unsigned long long)1 << 63;
  return buf;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<emit_sse_get_sign_mask>:
  endbr64
  lea    D7(%rip),%rax
  nopl   0x0(%rax,%rax,1)
L8:
  add    $0x1,%rax
  test   $0xf,%al
  jne    L8
  movabs $0x8000000000000000,%rdx
  mov    %rdx,(%rax)
  ret
```

assistant: ```c
unsigned char *emit_sse_get_sign_mask() {
  static unsigned char bufx[32];
  unsigned char *buf = bufx + 1;
  while ((long)buf % 16)
    buf++;
  unsigned long long *bit_mask = (unsigned long long *)buf;
  *bit_mask = (unsigned long long)1 << 63;
  return buf;
}
```
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    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D10(%rip),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <replacement@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D10
type_infered: string
name: unpack_data
arguments: label: D13
type_infered: string

tool: D10: rb

tool: D13: db

assistant: ```c
int main(void) {
  FILE *fp = fopen("db", "rb");
  replacement(fp);
  fclose(fp);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbx
  lea    D19(%rip),%rsi
  lea    D1(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <replacement@plt>
  mov    %rbx,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: label: D19
data_type: string
name: parse_data
arguments: label: D1
data_type: string

tool: D19: rb

tool: D1: db

assistant: ```c
int main(void) {
  FILE *fp = fopen("db", "rb");
  replacement(fp);
  fclose(fp);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D0(%rip),%rsi
  lea    D12(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <replacement@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D0
data_type: string
name: disassemble_data
arguments: tag: D12
data_type: string

tool: D0: rb

tool: D12: db

assistant: ```c
int main(void) {
  FILE *fp = fopen("db", "rb");
  replacement(fp);
  fclose(fp);
  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
  lea    D2(%rip),%rsi
  lea    D5(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <replacement@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D2
type_infered: string
name: parse_data
arguments: label: D5
type_infered: string

tool: D2: rb

tool: D5: db

assistant: ```c
int main(void) {
  FILE *fp = fopen("db", "rb");
  replacement(fp);
  fclose(fp);
  return 0;
}
```