ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<hx4_x4djbx33a_128_ref>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %r8,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x20(%rsp)
  movabs $0x150500001505,%rax
  mov    %rax,0x8(%rsp)
  mov    %rax,0x10(%rsp)
  push   %r9
  mov    %r8,%r9
  mov    %rcx,%r8
  mov    %rdx,%rcx
  mov    %rsi,%rdx
  mov    %rdi,%rsi
  mov    $0x10,%edi
  call   <hx4_check_params@plt>
  pop    %rdx
  pop    %rcx
  mov    %eax,%r13d
  test   %eax,%eax
  jne    L4
  lea    (%rbx,%r14,4),%rdi
  cmp    %rdi,%rbx
  jae    L19
  xor    %edx,%edx
  nopl   (%rax)
L38:
  movslq %edx,%rcx
  add    $0x4,%rbx
  add    $0x1,%edx
  mov    (%rsp,%rcx,4),%esi
  and    $0x3,%edx
  mov    %esi,%eax
  shl    $0x5,%eax
  add    %esi,%eax
  add    -0x4(%rbx),%eax
  mov    %eax,(%rsp,%rcx,4)
  cmp    %rbx,%rdi
  ja     L38
L19:
  mov    %rsp,%r14
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <hx4_xor_cookie_128@plt>
  mov    $0x10,%edx
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
L4:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x20,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int hx4_x4djbx33a_128_ref(const void *buffer, size_t buffer_size, const void *cookie, size_t cookie_sz, void *out_hash, size_t out_hash_size) {
  const uint8_t *p;
  const uint8_t *const buffer_end = (uint8_t *)buffer + buffer_size;
  uint32_t state[] = {5381, 5381, 5381, 5381};
  int state_i = 0;
  int rc;
  rc = hx4_check_params(sizeof(state), buffer, buffer_size, cookie, cookie_sz, out_hash, out_hash_size);
  if (rc != (0)) {
    return rc;
  }
  p = buffer;
  while (p < buffer_end) {
    state[state_i] = state[state_i] * 33 + *p;
    p++;
    state_i = (state_i + 1) % 4;
  }
  hx4_xor_cookie_128(state, cookie);
  memcpy(out_hash, state, sizeof(state));
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<hx4_x4djbx33a_128_ref>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %r8,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x28,%rsp
  movdqa D2(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x20(%rsp)
  xor    %eax,%eax
  movaps %xmm0,0x8(%rsp)
  push   %r9
  mov    %r8,%r9
  mov    %rcx,%r8
  mov    %rdx,%rcx
  mov    %rsi,%rdx
  mov    %rdi,%rsi
  mov    $0x10,%edi
  call   <hx4_check_params@plt>
  pop    %rdx
  pop    %rcx
  mov    %eax,%r13d
  test   %eax,%eax
  jne    L35
  lea    (%rbx,%r14,4),%rdi
  cmp    %rdi,%rbx
  jae    L0
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L2:
  movslq %edx,%rcx
  add    $0x4,%rbx
  add    $0x1,%edx
  mov    (%rsp,%rcx,4),%esi
  and    $0x3,%edx
  mov    %esi,%eax
  shl    $0x5,%eax
  add    %esi,%eax
  add    -0x4(%rbx),%eax
  mov    %eax,(%rsp,%rcx,4)
  cmp    %rbx,%rdi
  ja     L2
L0:
  mov    %rsp,%r14
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <hx4_xor_cookie_128@plt>
  mov    $0x10,%edx
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
L35:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x20,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D2
datatype: byte[16]

tool: {"D2": [0x05,0x15,0x00,0x00,0x05,0x15,0x00,0x00,0x05,0x15,0x00,0x00,0x05,0x15,0x00,0x00]}

assistant: ```c
int hx4_x4djbx33a_128_ref(const void *buffer, size_t buffer_size, const void *cookie, size_t cookie_sz, void *out_hash, size_t out_hash_size) {
  const uint8_t *p;
  const uint8_t *const buffer_end = (uint8_t *)buffer + buffer_size;
  uint32_t state[] = {5381, 5381, 5381, 5381};
  int state_i = 0;
  int rc;
  rc = hx4_check_params(sizeof(state), buffer, buffer_size, cookie, cookie_sz, out_hash, out_hash_size);
  if (rc != (0)) {
    return rc;
  }
  p = buffer;
  while (p < buffer_end) {
    state[state_i] = state[state_i] * 33 + *p;
    p++;
    state_i = (state_i + 1) % 4;
  }
  hx4_xor_cookie_128(state, cookie);
  memcpy(out_hash, state, sizeof(state));
  return (0);
}
```
user: What is the c source code of the assembly code below?

```asm
<func>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D0
type: string

tool: D0: 
 Thread one is running...


assistant: ```c
void *func(void *arg) {
  printf("\n Thread one is running...\n");
  exit(1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<func>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D6
type_infered: string

tool: D6: 
 Thread one is running...


assistant: ```c
void *func(void *arg) {
  printf("\n Thread one is running...\n");
  exit(1);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<func>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

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

tool: D6: 
 Thread one is running...


assistant: ```c
void *func(void *arg) {
  printf("\n Thread one is running...\n");
  exit(1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<func>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
datatype: string

tool: D9: 
 Thread one is running...


assistant: ```c
void *func(void *arg) {
  printf("\n Thread one is running...\n");
  exit(1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<matrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %edx,-0x2c(%rbp)
  mov    %ecx,-0x30(%rbp)
  mov    -0x28(%rbp),%eax
  sub    -0x24(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x30(%rbp),%eax
  sub    -0x2c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  cltq
  shl    $0x3,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  imul   -0x14(%rbp),%eax
  cltq
  shl    $0x3,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x2c(%rbp),%eax
  cltq
  shl    $0x3,%rax
  neg    %rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,(%rax)
  movl   $0x1,-0x1c(%rbp)
  jmp    L1
L3:
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rsi
  mov    -0x10(%rbp),%rax
  add    %rsi,%rax
  add    %rcx,%rdx
  mov    %rdx,(%rax)
  addl   $0x1,-0x1c(%rbp)
L1:
  mov    -0x1c(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L3
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x3,%rax
  neg    %rax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  leave
  ret
```

assistant: ```c
double **matrix(int nrl, int nrh, int ncl, int nch) {
  int i;
  int nrow = nrh - nrl + 1;
  int ncol = nch - ncl + 1;
  double **pArray = (double **)malloc((size_t)(nrow * sizeof(double *)));
  double *pMatrix = (double *)malloc((size_t)(nrow * ncol * sizeof(double)));
  pArray[0] = pMatrix - ncl;
  for (i = 1; i < nrow; i++) {
    pArray[i] = pArray[i - 1] + ncol;
  }
  return pArray - nrl;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<matrix>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %edx,%r14d
  sub    %edi,%esi
  mov    %esi,%r13d
  lea    0x1(%rsi),%r15d
  sub    %edx,%ecx
  lea    0x1(%rcx),%r12d
  movslq %r15d,%rdi
  shl    $0x3,%rdi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %r15d,%edi
  imul   %r12d,%edi
  movslq %edi,%rdi
  shl    $0x3,%rdi
  call   <malloc@plt>
  movslq %r14d,%r14
  shl    $0x3,%r14
  sub    %r14,%rax
  mov    %rax,0x0(%rbp)
  cmp    $0x1,%r15d
  jle    L3
  movslq %r12d,%rcx
  shl    $0x3,%rcx
  mov    %rbp,%rax
  lea    -0x1(%r13),%edx
  lea    0x8(%rbp,%rdx,8),%rsi
L18:
  mov    (%rax),%rdx
  add    $0x8,%rax
  add    %rcx,%rdx
  mov    %rdx,(%rax)
  cmp    %rsi,%rax
  jne    L18
L3:
  movslq %ebx,%rbx
  shl    $0x3,%rbx
  mov    %rbp,%rax
  sub    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
double **matrix(int nrl, int nrh, int ncl, int nch) {
  int i;
  int nrow = nrh - nrl + 1;
  int ncol = nch - ncl + 1;
  double **pArray = (double **)malloc((size_t)(nrow * sizeof(double *)));
  double *pMatrix = (double *)malloc((size_t)(nrow * ncol * sizeof(double)));
  pArray[0] = pMatrix - ncl;
  for (i = 1; i < nrow; i++) {
    pArray[i] = pArray[i - 1] + ncol;
  }
  return pArray - nrl;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<matrix>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  movslq %edx,%r12
  push   %rbp
  movslq %edi,%rbp
  sub    %r12d,%ecx
  shl    $0x3,%r12
  sub    %ebp,%esi
  push   %rbx
  lea    0x1(%rcx),%ebx
  lea    0x1(%rsi),%r13d
  mov    %esi,%r15d
  movslq %r13d,%rdi
  sub    $0x8,%rsp
  shl    $0x3,%rdi
  call   <malloc@plt>
  mov    %r13d,%edi
  imul   %ebx,%edi
  mov    %rax,%r14
  movslq %edi,%rdi
  shl    $0x3,%rdi
  call   <malloc@plt>
  sub    %r12,%rax
  mov    %rax,(%r14)
  cmp    $0x1,%r13d
  jle    L9
  movslq %ebx,%rcx
  lea    -0x1(%r15),%esi
  lea    0x8(%r14),%rdx
  shl    $0x3,%rcx
  lea    0x8(%r14,%rsi,8),%rsi
  jmp    L2
  xchg   %ax,%ax
L1:
  add    $0x8,%rdx
L2:
  add    %rcx,%rax
  mov    %rax,(%rdx)
  cmp    %rsi,%rdx
  jne    L1
L9:
  add    $0x8,%rsp
  shl    $0x3,%rbp
  mov    %r14,%rax
  pop    %rbx
  sub    %rbp,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
double **matrix(int nrl, int nrh, int ncl, int nch) {
  int i;
  int nrow = nrh - nrl + 1;
  int ncol = nch - ncl + 1;
  double **pArray = (double **)malloc((size_t)(nrow * sizeof(double *)));
  double *pMatrix = (double *)malloc((size_t)(nrow * ncol * sizeof(double)));
  pArray[0] = pMatrix - ncl;
  for (i = 1; i < nrow; i++) {
    pArray[i] = pArray[i - 1] + ncol;
  }
  return pArray - nrl;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<matrix>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  movslq %edx,%r12
  push   %rbp
  movslq %edi,%rbp
  sub    %r12d,%ecx
  shl    $0x3,%r12
  sub    %ebp,%esi
  push   %rbx
  lea    0x1(%rcx),%ebx
  lea    0x1(%rsi),%r13d
  mov    %esi,%r15d
  movslq %r13d,%rdi
  sub    $0x8,%rsp
  shl    $0x3,%rdi
  call   <malloc@plt>
  mov    %r13d,%edi
  imul   %ebx,%edi
  mov    %rax,%r14
  movslq %edi,%rdi
  shl    $0x3,%rdi
  call   <malloc@plt>
  sub    %r12,%rax
  mov    %rax,(%r14)
  cmp    $0x1,%r13d
  jle    L0
  movslq %ebx,%rcx
  lea    -0x1(%r15),%esi
  lea    0x8(%r14),%rdx
  shl    $0x3,%rcx
  lea    0x10(%r14,%rsi,8),%rsi
  nopl   0x0(%rax)
L18:
  add    %rcx,%rax
  add    $0x8,%rdx
  mov    %rax,-0x8(%rdx)
  cmp    %rsi,%rdx
  jne    L18
L0:
  add    $0x8,%rsp
  shl    $0x3,%rbp
  mov    %r14,%rax
  pop    %rbx
  sub    %rbp,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
double **matrix(int nrl, int nrh, int ncl, int nch) {
  int i;
  int nrow = nrh - nrl + 1;
  int ncol = nch - ncl + 1;
  double **pArray = (double **)malloc((size_t)(nrow * sizeof(double *)));
  double *pMatrix = (double *)malloc((size_t)(nrow * ncol * sizeof(double)));
  pArray[0] = pMatrix - ncl;
  for (i = 1; i < nrow; i++) {
    pArray[i] = pArray[i - 1] + ncol;
  }
  return pArray - nrl;
}
```
user: What is the c source code of the assembly code below:

```asm
<ft_flagconv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x68,%al
  jne    L49
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x68,%al
  jne    L49
  movl   $0x1,-0x4(%rbp)
  jmp    L50
L49:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x68,%al
  jne    L51
  movl   $0x2,-0x4(%rbp)
  jmp    L50
L51:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x6c,%al
  jne    L62
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x6c,%al
  jne    L62
  movl   $0x3,-0x4(%rbp)
  jmp    L50
L62:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x6c,%al
  jne    L79
  movl   $0x4,-0x4(%rbp)
  jmp    L50
L79:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x6a,%al
  jne    L5
  movl   $0x5,-0x4(%rbp)
  jmp    L50
L5:
  movl   $0x6,-0x4(%rbp)
L50:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int ft_flagconv(char *str) {
  int value;
  if (*str == 'h' && *(str + 1) == 'h')
    value = 1;
  else if (*str == 'h')
    value = 2;
  else if (*str == 'l' && *(str + 1) == 'l')
    value = 3;
  else if (*str == 'l')
    value = 4;
  else if (*str == 'j')
    value = 5;
  else
    value = 6;
  return (value);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ft_flagconv>:
  endbr64
  movzbl (%rdi),%eax
  cmp    $0x68,%al
  je     L17
  cmp    $0x6c,%al
  je     L4
  cmp    $0x6a,%al
  setne  %al
  movzbl %al,%eax
  add    $0x5,%eax
  ret
L17:
  cmpb   $0x68,0x1(%rdi)
  setne  %al
  movzbl %al,%eax
  add    $0x1,%eax
  ret
L4:
  cmpb   $0x6c,0x1(%rdi)
  setne  %al
  movzbl %al,%eax
  add    $0x3,%eax
  ret
```

assistant: ```c
int ft_flagconv(char *str) {
  int value;
  if (*str == 'h' && *(str + 1) == 'h')
    value = 1;
  else if (*str == 'h')
    value = 2;
  else if (*str == 'l' && *(str + 1) == 'l')
    value = 3;
  else if (*str == 'l')
    value = 4;
  else if (*str == 'j')
    value = 5;
  else
    value = 6;
  return (value);
}
```
user: What is the c source code of the assembly code below?

```asm
<ft_flagconv>:
  endbr64
  movzbl (%rdi),%eax
  cmp    $0x68,%al
  je     L11
  cmp    $0x6c,%al
  je     L3
  cmp    $0x6a,%al
  setne  %al
  movzbl %al,%eax
  add    $0x5,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  cmpb   $0x6c,0x1(%rdi)
  setne  %al
  add    $0x3,%eax
  ret
  nopl   (%rax)
L11:
  xor    %eax,%eax
  cmpb   $0x68,0x1(%rdi)
  setne  %al
  add    $0x1,%eax
  ret
```

assistant: ```c
int ft_flagconv(char *str) {
  int value;
  if (*str == 'h' && *(str + 1) == 'h')
    value = 1;
  else if (*str == 'h')
    value = 2;
  else if (*str == 'l' && *(str + 1) == 'l')
    value = 3;
  else if (*str == 'l')
    value = 4;
  else if (*str == 'j')
    value = 5;
  else
    value = 6;
  return (value);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_flagconv>:
  endbr64
  movzbl (%rdi),%eax
  cmp    $0x68,%al
  je     L4
  cmp    $0x6c,%al
  je     L16
  cmp    $0x6a,%al
  setne  %al
  movzbl %al,%eax
  add    $0x5,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L16:
  xor    %eax,%eax
  cmpb   $0x6c,0x1(%rdi)
  setne  %al
  add    $0x3,%eax
  ret
  nopl   (%rax)
L4:
  xor    %eax,%eax
  cmpb   $0x68,0x1(%rdi)
  setne  %al
  add    $0x1,%eax
  ret
```

assistant: ```c
int ft_flagconv(char *str) {
  int value;
  if (*str == 'h' && *(str + 1) == 'h')
    value = 1;
  else if (*str == 'h')
    value = 2;
  else if (*str == 'l' && *(str + 1) == 'l')
    value = 3;
  else if (*str == 'l')
    value = 4;
  else if (*str == 'j')
    value = 5;
  else
    value = 6;
  return (value);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<arg_get_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L17
L7:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L6
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L21
  mov    $0xffffffff,%eax
  jmp    L9
L21:
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    -0x30(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    $0x0,%eax
  jmp    L9
L6:
  nop
  addl   $0x1,-0x4(%rbp)
L17:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L7
  mov    $0xffffffff,%eax
L9:
  leave
  ret
```

assistant: ```c
int arg_get_int(int argc, char *argv[], const char *key, int *value) {
  int i = 0;
  for (i = 1; i < argc; i++) {
    if (strcmp(key, argv[i]) != 0) {
      continue;
    }
    if (argv[i + 1] == NULL) {
      return -1;
    }
    *value = atoi(argv[i + 1]);
    return 0;
  }
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<arg_get_int>:
  endbr64
  cmp    $0x1,%edi
  jle    L1
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbp
  mov    %rdx,%r13
  mov    %rcx,%r14
  mov    %edi,%ebx
  mov    $0x1,%r12d
L6:
  mov    0x0(%rbp,%r12,8),%rsi
  mov    %r13,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L58
  mov    0x8(%rbp,%r12,8),%rdi
  test   %rdi,%rdi
  je     L19
  call   <atoi@plt>
  mov    %eax,(%r14)
  mov    $0x0,%eax
L3:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L58:
  add    $0x1,%r12
  cmp    %rbx,%r12
  jne    L6
  mov    $0xffffffff,%eax
  jmp    L3
L1:
  mov    $0xffffffff,%eax
  ret
L19:
  mov    $0xffffffff,%eax
  jmp    L3
```

assistant: ```c
int arg_get_int(int argc, char *argv[], const char *key, int *value) {
  int i = 0;
  for (i = 1; i < argc; i++) {
    if (strcmp(key, argv[i]) != 0) {
      continue;
    }
    if (argv[i + 1] == NULL) {
      return -1;
    }
    *value = atoi(argv[i + 1]);
    return 0;
  }
  return -1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<arg_get_int>:
  endbr64
  cmp    $0x1,%edi
  jle    L2
  push   %r14
  movslq %edi,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  mov    %rsi,%rbx
L20:
  mov    (%rbx,%rbp,8),%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L4
  mov    0x8(%rbx,%rbp,8),%rdi
  test   %rdi,%rdi
  je     L18
  call   <atoi@plt>
  mov    %eax,0x0(%r13)
  xor    %eax,%eax
L47:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L4:
  add    $0x1,%rbp
  cmp    %r14,%rbp
  jne    L20
L18:
  mov    $0xffffffff,%eax
  jmp    L47
  nopl   0x0(%rax,%rax,1)
L2:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int arg_get_int(int argc, char *argv[], const char *key, int *value) {
  int i = 0;
  for (i = 1; i < argc; i++) {
    if (strcmp(key, argv[i]) != 0) {
      continue;
    }
    if (argv[i + 1] == NULL) {
      return -1;
    }
    *value = atoi(argv[i + 1]);
    return 0;
  }
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<arg_get_int>:
  endbr64
  cmp    $0x1,%edi
  jle    L16
  push   %r14
  movslq %edi,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  mov    %rsi,%rbx
L45:
  mov    (%rbx,%rbp,8),%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L30
  mov    0x8(%rbx,%rbp,8),%rdi
  test   %rdi,%rdi
  je     L37
  call   <atoi@plt>
  mov    %eax,0x0(%r13)
  xor    %eax,%eax
L10:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L30:
  add    $0x1,%rbp
  cmp    %r14,%rbp
  jne    L45
L37:
  mov    $0xffffffff,%eax
  jmp    L10
  nopl   0x0(%rax,%rax,1)
L16:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int arg_get_int(int argc, char *argv[], const char *key, int *value) {
  int i = 0;
  for (i = 1; i < argc; i++) {
    if (strcmp(key, argv[i]) != 0) {
      continue;
    }
    if (argv[i + 1] == NULL) {
      return -1;
    }
    *value = atoi(argv[i + 1]);
    return 0;
  }
  return -1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  shr    $0x3,%rax
  or     $0x61bab5,%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x18(%rbp)
  nop
  mov    -0x10(%rbp),%rax
  add    %rax,%rax
  or     $0xa23d3e6,%rax
  mov    %rax,%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  unsigned int copy12;
  unsigned short copy13;
  {
    state[0UL] = (input[0UL] | 51238316UL) >> 3UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      if (state[0UL] > local1) {
        if (state[0UL] > local1) {
          state[local1] = state[0UL] >> ((state[0UL] & 7UL) | 1UL);
        }
      } else if (state[0UL] == local1) {
        copy12 = *((unsigned int *)(&state[local1]) + 0);
        *((unsigned int *)(&state[local1]) + 0) = *((unsigned int *)(&state[local1]) + 1);
        *((unsigned int *)(&state[local1]) + 1) = copy12;
      } else {
        copy13 = *((unsigned short *)(&state[local1]) + 2);
        *((unsigned short *)(&state[local1]) + 2) = *((unsigned short *)(&state[local1]) + 3);
        *((unsigned short *)(&state[local1]) + 3) = copy13;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] | 85060083UL) << 1UL;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  shr    $0x3,%rax
  add    %rax,%rax
  or     $0xae3f7ee,%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 int copy12;
  unsigned short copy13;
  {
    state[0UL] = (input[0UL] | 51238316UL) >> 3UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      if (state[0UL] > local1) {
        if (state[0UL] > local1) {
          state[local1] = state[0UL] >> ((state[0UL] & 7UL) | 1UL);
        }
      } else if (state[0UL] == local1) {
        copy12 = *((unsigned int *)(&state[local1]) + 0);
        *((unsigned int *)(&state[local1]) + 0) = *((unsigned int *)(&state[local1]) + 1);
        *((unsigned int *)(&state[local1]) + 1) = copy12;
      } else {
        copy13 = *((unsigned short *)(&state[local1]) + 2);
        *((unsigned short *)(&state[local1]) + 2) = *((unsigned short *)(&state[local1]) + 3);
        *((unsigned short *)(&state[local1]) + 3) = copy13;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] | 85060083UL) << 1UL;
  }
}
```