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
<main>:
  endbr64
  push   %r14
  mov    $0x18,%edi
  push   %r13
  lea    D2(%rip),%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  call   <malloc@plt>
  mov    $0x4,%edi
  mov    %rax,%r14
  call   <malloc@plt>
  mov    $0x4,%edi
  mov    %r14,%rbp
  mov    %rax,(%r14)
  call   <malloc@plt>
  mov    $0x4,%edi
  mov    %rax,0x8(%r14)
  call   <malloc@plt>
  mov    %rax,0x10(%r14)
L9:
  xor    %ebx,%ebx
L6:
  mov    0x0(%rbp),%rax
  lea    (%r12,%rbx,1),%edx
  mov    %r13,%rdi
  mov    %dl,(%rax,%rbx,1)
  mov    0x0(%rbp),%rax
  movsbl (%rax,%rbx,1),%esi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x4,%rbx
  jne    L6
  mov    $0xa,%edi
  add    $0x1,%r12d
  add    $0x8,%rbp
  call   <putchar@plt>
  cmp    $0x3,%r12b
  jne    L9
  mov    (%r14),%rdi
  call   <free@plt>
  mov    0x8(%r14),%rdi
  call   <free@plt>
  mov    0x10(%r14),%rdi
  call   <free@plt>
  mov    %r14,%rdi
  call   <free@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: label: D2
type_infered: string

tool: D2: %2d

assistant: ```c
int main(void) {
  char **a;
  char i;
  char j;
  int m = 3;
  int n = 4;
  a = (char **)malloc(m * sizeof(char *));
  for (i = 0; i < m; ++i) {
    a[i] = (char *)malloc(n * sizeof(char));
  }
  for (i = 0; i < m; ++i) {
    for (j = 0; j < n; ++j) {
      a[i][j] = i + j;
      printf("%2d", a[i][j]);
    }
    printf("\n");
  }
  for (i = 0; i < m; ++i) {
    free((void *)a[i]);
  }
  free((void *)a);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<factor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  pxor   %xmm1,%xmm1
  cvtsi2sdq -0x18(%rbp),%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  mov    %eax,%edi
  call   <ceil@plt>
  mov    %rax,-0x8(%rbp)
  movq   $0x2,-0x10(%rbp)
  jmp    L7
L4:
  mov    -0x18(%rbp),%rax
  cqto
  idivq  -0x10(%rbp)
  mov    %rdx,%rax
  test   %rax,%rax
  jne    L22
  mov    -0x10(%rbp),%rax
  jmp    L25
L22:
  addq   $0x1,-0x10(%rbp)
L7:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jle    L4
  mov    -0x18(%rbp),%rax
L25:
  leave
  ret
```

assistant: ```c
long long factor(const long long i) {
  long long max = (long long)ceil(sqrt((double)i));
  long long j = 2;
  while (j <= max) {
    if (i % j == 0)
      return j;
    else
      j++;
  }
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<factor>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  pxor   %xmm0,%xmm0
  cvtsi2sd %rdi,%xmm0
  call   <sqrt@plt>
  mov    %eax,%edi
  call   <ceil@plt>
  cmp    $0x1,%rax
  jle    L18
  mov    %rax,%rsi
  test   $0x1,%bl
  je     L49
  mov    $0x2,%ecx
L41:
  add    $0x1,%rcx
  cmp    %rcx,%rsi
  jl     L18
  mov    %rbx,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  jne    L41
  mov    %rcx,%rbx
L18:
  mov    %rbx,%rax
  pop    %rbx
  ret
L49:
  mov    $0x2,%ebx
  jmp    L18
```

assistant: ```c
long long factor(const long long i) {
  long long max = (long long)ceil(sqrt((double)i));
  long long j = 2;
  while (j <= max) {
    if (i % j == 0)
      return j;
    else
      j++;
  }
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<factor>:
  endbr64
  pxor   %xmm0,%xmm0
  push   %r12
  mov    %rdi,%r12
  cvtsi2sd %rdi,%xmm0
  call   <sqrt@plt>
  mov    %eax,%edi
  call   <ceil@plt>
  cmp    $0x1,%rax
  jle    L44
  test   $0x1,%r12b
  je     L30
  mov    %rax,%rsi
  mov    $0x2,%ecx
  jmp    L32
  nopl   0x0(%rax)
L37:
  mov    %r12,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  je     L33
L32:
  add    $0x1,%rcx
  cmp    %rcx,%rsi
  jge    L37
L44:
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L33:
  mov    %rcx,%r12
  mov    %r12,%rax
  pop    %r12
  ret
L30:
  mov    $0x2,%r12d
  jmp    L44
```

assistant: ```c
long long factor(const long long i) {
  long long max = (long long)ceil(sqrt((double)i));
  long long j = 2;
  while (j <= max) {
    if (i % j == 0)
      return j;
    else
      j++;
  }
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<factor>:
  endbr64
  pxor   %xmm0,%xmm0
  push   %r12
  mov    %rdi,%r12
  cvtsi2sd %rdi,%xmm0
  call   <sqrt@plt>
  mov    %eax,%edi
  call   <ceil@plt>
  cmp    $0x1,%rax
  jle    L33
  test   $0x1,%r12b
  je     L32
  mov    %rax,%rsi
  mov    $0x2,%ecx
  jmp    L54
  nopl   0x0(%rax)
L52:
  mov    %r12,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  je     L12
L54:
  add    $0x1,%rcx
  cmp    %rcx,%rsi
  jge    L52
L33:
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L12:
  mov    %rcx,%r12
  mov    %r12,%rax
  pop    %r12
  ret
L32:
  mov    $0x2,%r12d
  jmp    L33
```

assistant: ```c
long long factor(const long long i) {
  long long max = (long long)ceil(sqrt((double)i));
  long long j = 2;
  while (j <= max) {
    if (i % j == 0)
      return j;
    else
      j++;
  }
  return i;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sum16>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L15
  mov    $0x0,%eax
  jmp    L20
L15:
  movl   $0x0,-0x4(%rbp)
  jmp    L6
L16:
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L6:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jb     L16
  mov    -0x8(%rbp),%eax
L20:
  pop    %rbp
  ret
```

assistant: ```c
unsigned int sum16(unsigned char *crcptr, unsigned int crc_len) {
  unsigned int sum_temp = 0;
  unsigned int i;
  if (crcptr == 0x00)
    return 0;
  for (i = 0; i < crc_len; i++) {
    sum_temp = +*(crcptr + i);
  }
  return sum_temp;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<sum16>:
  endbr64
  test   %rdi,%rdi
  je     L23
  test   %esi,%esi
  je     L29
  mov    $0x0,%edx
L3:
  mov    %edx,%eax
  add    $0x1,%edx
  cmp    %edx,%esi
  jne    L3
  mov    %eax,%eax
  movzbl (%rdi,%rax,1),%eax
  ret
L23:
  mov    $0x0,%eax
  ret
L29:
  mov    %esi,%eax
  ret
```

assistant: ```c
unsigned int sum16(unsigned char *crcptr, unsigned int crc_len) {
  unsigned int sum_temp = 0;
  unsigned int i;
  if (crcptr == 0x00)
    return 0;
  for (i = 0; i < crc_len; i++) {
    sum_temp = +*(crcptr + i);
  }
  return sum_temp;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sum16>:
  endbr64
  test   %rdi,%rdi
  je     L13
  test   %esi,%esi
  je     L13
  lea    -0x1(%rsi),%eax
  movzbl (%rdi,%rax,1),%eax
  ret
  nopl   (%rax)
L13:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned int sum16(unsigned char *crcptr, unsigned int crc_len) {
  unsigned int sum_temp = 0;
  unsigned int i;
  if (crcptr == 0x00)
    return 0;
  for (i = 0; i < crc_len; i++) {
    sum_temp = +*(crcptr + i);
  }
  return sum_temp;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sum16>:
  endbr64
  test   %rdi,%rdi
  je     L9
  test   %esi,%esi
  je     L9
  lea    -0x1(%rsi),%eax
  movzbl (%rdi,%rax,1),%eax
  ret
  nopl   (%rax)
L9:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned int sum16(unsigned char *crcptr, unsigned int crc_len) {
  unsigned int sum_temp = 0;
  unsigned int i;
  if (crcptr == 0x00)
    return 0;
  for (i = 0; i < crc_len; i++) {
    sum_temp = +*(crcptr + i);
  }
  return sum_temp;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D39(%rip),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D5(%rip),%rax
  mov    %rax,%rsi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x14,%esi
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffff9c,%edi
  call   <exit@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D49
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D39
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D29
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D5
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D36
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D58
type_guessed: string

tool: D49: Parameters:

tool: D39: mnt1_1

tool: D29: [ -mount <mount> ]         The mount point (default %s)


tool: D5: this_is_the_default_crc32_test_file_name

tool: D36: [ -file <name> ]           file to do the test on (default %s)


tool: D58: [ -process <nb> ]          The test will be done by <nb> process simultaneously (default %d)


assistant: ```c
void usage() {
  printf("Parameters:\n");
  printf("[ -mount <mount> ]         The mount point (default %s)\n", "mnt1_1");
  printf("[ -file <name> ]           file to do the test on (default %s)\n", "this_is_the_default_crc32_test_file_name");
  printf("[ -process <nb> ]          The test will be done by <nb> process simultaneously (default %d)\n", 20);
  exit(-100);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D54(%rip),%rdi
  call   <puts@plt>
  lea    D32(%rip),%rsi
  lea    D36(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D53(%rip),%rsi
  lea    D29(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x14,%esi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffff9c,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D54
datatype: string
name: extract_symbolic_data
arguments: data_tag: D32
datatype: string
name: extract_symbolic_data
arguments: data_tag: D36
datatype: string
name: extract_symbolic_data
arguments: data_tag: D53
datatype: string
name: extract_symbolic_data
arguments: data_tag: D29
datatype: string
name: extract_symbolic_data
arguments: data_tag: D3
datatype: string

tool: D54: Parameters:

tool: D32: mnt1_1

tool: D36: [ -mount <mount> ]         The mount point (default %s)


tool: D53: this_is_the_default_crc32_test_file_name

tool: D29: [ -file <name> ]           file to do the test on (default %s)


tool: D3: [ -process <nb> ]          The test will be done by <nb> process simultaneously (default %d)


assistant: ```c
void usage() {
  printf("Parameters:\n");
  printf("[ -mount <mount> ]         The mount point (default %s)\n", "mnt1_1");
  printf("[ -file <name> ]           file to do the test on (default %s)\n", "this_is_the_default_crc32_test_file_name");
  printf("[ -process <nb> ]          The test will be done by <nb> process simultaneously (default %d)\n", 20);
  exit(-100);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D22(%rip),%rdi
  call   <puts@plt>
  lea    D28(%rip),%rsi
  lea    D30(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D31(%rip),%rsi
  lea    D56(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D34(%rip),%rdi
  mov    $0x14,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xffffff9c,%edi
  add    $0x8,%rsp
  jmp    1090 <exit@plt>
```

tool_calls: name: read_data
arguments: data_tag: D22
type_guessed: string
name: read_data
arguments: data_tag: D28
type_guessed: string
name: read_data
arguments: data_tag: D30
type_guessed: string
name: read_data
arguments: data_tag: D31
type_guessed: string
name: read_data
arguments: data_tag: D56
type_guessed: string
name: read_data
arguments: data_tag: D34
type_guessed: string

tool: D22: Parameters:

tool: D28: mnt1_1

tool: D30: [ -mount <mount> ]         The mount point (default %s)


tool: D31: this_is_the_default_crc32_test_file_name

tool: D56: [ -file <name> ]           file to do the test on (default %s)


tool: D34: [ -process <nb> ]          The test will be done by <nb> process simultaneously (default %d)


assistant: ```c
void usage() {
  printf("Parameters:\n");
  printf("[ -mount <mount> ]         The mount point (default %s)\n", "mnt1_1");
  printf("[ -file <name> ]           file to do the test on (default %s)\n", "this_is_the_default_crc32_test_file_name");
  printf("[ -process <nb> ]          The test will be done by <nb> process simultaneously (default %d)\n", 20);
  exit(-100);
}
```
user: Translate the below assembly code into C programming language:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D53(%rip),%rdi
  call   <puts@plt>
  lea    D26(%rip),%rsi
  lea    D17(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D44(%rip),%rsi
  lea    D49(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D54(%rip),%rdi
  mov    $0x14,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xffffff9c,%edi
  add    $0x8,%rsp
  jmp    1090 <exit@plt>
```

tool_calls: name: unpack_data
arguments: label: D53
type: string
name: unpack_data
arguments: label: D26
type: string
name: unpack_data
arguments: label: D17
type: string
name: unpack_data
arguments: label: D44
type: string
name: unpack_data
arguments: label: D49
type: string
name: unpack_data
arguments: label: D54
type: string

tool: D53: Parameters:

tool: D26: mnt1_1

tool: D17: [ -mount <mount> ]         The mount point (default %s)


tool: D44: this_is_the_default_crc32_test_file_name

tool: D49: [ -file <name> ]           file to do the test on (default %s)


tool: D54: [ -process <nb> ]          The test will be done by <nb> process simultaneously (default %d)


assistant: ```c
void usage() {
  printf("Parameters:\n");
  printf("[ -mount <mount> ]         The mount point (default %s)\n", "mnt1_1");
  printf("[ -file <name> ]           file to do the test on (default %s)\n", "this_is_the_default_crc32_test_file_name");
  printf("[ -process <nb> ]          The test will be done by <nb> process simultaneously (default %d)\n", 20);
  exit(-100);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<f_b64_ntop_init>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  mov    $0x3,%esi
  mov    %rax,%rdi
  call   <bzero@plt>
  mov    -0x10(%rbp),%rax
  movl   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  movl   $0x0,(%rax)
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int f_b64_ntop_init(unsigned char *tailbuf, int *taillen, int *outlen) {
  bzero(tailbuf, 3);
  *taillen = 0;
  *outlen = 0;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<f_b64_ntop_init>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    $0x3,%esi
  call   <bzero@plt>
  movl   $0x0,0x0(%rbp)
  movl   $0x0,(%rbx)
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int f_b64_ntop_init(unsigned char *tailbuf, int *taillen, int *outlen) {
  bzero(tailbuf, 3);
  *taillen = 0;
  *outlen = 0;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<f_b64_ntop_init>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  mov    $0x3,%esi
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <bzero@plt>
  movl   $0x0,0x0(%rbp)
  xor    %eax,%eax
  movl   $0x0,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int f_b64_ntop_init(unsigned char *tailbuf, int *taillen, int *outlen) {
  bzero(tailbuf, 3);
  *taillen = 0;
  *outlen = 0;
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<f_b64_ntop_init>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  mov    $0x3,%esi
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <bzero@plt>
  movl   $0x0,0x0(%rbp)
  xor    %eax,%eax
  movl   $0x0,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int f_b64_ntop_init(unsigned char *tailbuf, int *taillen, int *outlen) {
  bzero(tailbuf, 3);
  *taillen = 0;
  *outlen = 0;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x74206f6e656c6966,%rax
  movabs $0x6972747320747365,%rdx
  mov    %rax,-0x70(%rbp)
  mov    %rdx,-0x68(%rbp)
  movq   $0xa676e,-0x60(%rbp)
  movq   $0x0,-0x58(%rbp)
  movq   $0x0,-0x50(%rbp)
  movq   $0x0,-0x48(%rbp)
  movq   $0x0,-0x40(%rbp)
  movq   $0x0,-0x38(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  lea    D18(%rip),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x78(%rbp)
  mov    -0x78(%rbp),%rax
  mov    %rax,%rdi
  call   <fileno@plt>
  mov    %eax,-0x7c(%rbp)
  cmpq   $0x0,-0x78(%rbp)
  jne    L22
  mov    $0xffffffff,%eax
  jmp    L2
L22:
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  lea    -0x70(%rbp),%rcx
  mov    -0x7c(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    -0x78(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
L2:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D18
data_type: string
name: unpack_data
arguments: data_tag: D15
data_type: string

tool: D18: w

tool: D15: fileno.txt

assistant: ```c
int main() {
  FILE *fp;
  int fd;
  char str[100] = "fileno test string\n";
  fp = fopen("fileno.txt", "w");
  fd = fileno(fp);
  if (fp == NULL)
    return -1;
  write(fd, str, strlen(str));
  fclose(fp);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  movabs $0x74206f6e656c6966,%rax
  movabs $0x6972747320747365,%rdx
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  movq   $0xa676e,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x38(%rsp)
  movq   $0x0,0x40(%rsp)
  movq   $0x0,0x48(%rsp)
  movq   $0x0,0x50(%rsp)
  movq   $0x0,0x58(%rsp)
  movl   $0x0,0x60(%rsp)
  lea    D6(%rip),%rsi
  lea    D5(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <fileno@plt>
  test   %rbx,%rbx
  je     L16
  mov    %eax,%ebp
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <write@plt>
  mov    %rbx,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
L24:
  mov    0x68(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L20
  add    $0x70,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  mov    $0xffffffff,%eax
  jmp    L24
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
type_guessed: string
name: retrieve_data
arguments: data_tag: D5
type_guessed: string

tool: D6: w

tool: D5: fileno.txt

assistant: ```c
int main() {
  FILE *fp;
  int fd;
  char str[100] = "fileno test string\n";
  fp = fopen("fileno.txt", "w");
  fd = fileno(fp);
  if (fp == NULL)
    return -1;
  write(fd, str, strlen(str));
  fclose(fp);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D20(%rip),%rdi
  lea    D12(%rip),%rsi
  push   %r12
  push   %rbp
  sub    $0x70,%rsp
  movdqa D26(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x60(%rsp)
  movaps %xmm0,(%rsp)
  pxor   %xmm0,%xmm0
  movq   $0xa676e,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  call   <fopen@plt>
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <fileno@plt>
  test   %rbp,%rbp
  je     L20
  mov    %rsp,%r13
  mov    %eax,%r12d
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %r13,%rsi
  mov    %r12d,%edi
  mov    %eax,%edx
  call   <write@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
L6:
  mov    0x68(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x70,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L20:
  or     $0xffffffff,%eax
  jmp    L6
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D20
data_type: string
name: read_data
arguments: data_label: D12
data_type: string
name: read_data
arguments: data_label: D26
data_type: byte[16]

tool: D20: fileno.txt

tool: D12: w

tool: {"D26": [0x66,0x69,0x6c,0x65,0x6e,0x6f,0x20,0x74,0x65,0x73,0x74,0x20,0x73,0x74,0x72,0x69]}

assistant: ```c
int main() {
  FILE *fp;
  int fd;
  char str[100] = "fileno test string\n";
  fp = fopen("fileno.txt", "w");
  fd = fileno(fp);
  if (fp == NULL)
    return -1;
  write(fd, str, strlen(str));
  fclose(fp);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r13
  lea    D9(%rip),%rdi
  lea    D29(%rip),%rsi
  push   %r12
  push   %rbp
  sub    $0x70,%rsp
  movdqa D26(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x60(%rsp)
  movaps %xmm0,(%rsp)
  pxor   %xmm0,%xmm0
  movq   $0xa676e,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  call   <fopen@plt>
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <fileno@plt>
  test   %rbp,%rbp
  je     L2
  mov    %rsp,%r13
  mov    %eax,%r12d
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %r13,%rsi
  mov    %r12d,%edi
  mov    %eax,%edx
  call   <write@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
L6:
  mov    0x68(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L23
  add    $0x70,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  or     $0xffffffff,%eax
  jmp    L6
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D9
datatype: string
name: parse_data
arguments: label: D29
datatype: string
name: parse_data
arguments: label: D26
datatype: byte[16]

tool: D9: fileno.txt

tool: D29: w

tool: {"D26": [0x66,0x69,0x6c,0x65,0x6e,0x6f,0x20,0x74,0x65,0x73,0x74,0x20,0x73,0x74,0x72,0x69]}

assistant: ```c
int main() {
  FILE *fp;
  int fd;
  char str[100] = "fileno test string\n";
  fp = fopen("fileno.txt", "w");
  fd = fileno(fp);
  if (fp == NULL)
    return -1;
  write(fd, str, strlen(str));
  fclose(fp);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<hex_to_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    %ecx,-0x38(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L15
L8:
  mov    -0x14(%rbp),%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <hex_to_dec@plt>
  shl    $0x4,%eax
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  add    %eax,%eax
  add    $0x1,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <hex_to_dec@plt>
  lea    (%rbx,%rax,1),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x14(%rbp)
L15:
  mov    -0x34(%rbp),%eax
  shr    %eax
  cmp    %eax,-0x14(%rbp)
  jbe    L8
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void hex_to_str(char *input, char *output, unsigned input_len, unsigned output_len) {
  for (unsigned int i = 0; i <= input_len / 2; ++i) {
    output[i] = (hex_to_dec(input[2 * i]) * 16 + hex_to_dec(input[2 * i + 1]));
  }
}
```