ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<getpwnam>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <endpwent@plt>
  jmp    L29
  nopw   0x0(%rax,%rax,1)
L25:
  mov    (%rax),%edi
  mov    %rbx,%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L10
L29:
  xor    %eax,%eax
  call   <getpwent@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L25
L10:
  xor    %eax,%eax
  call   <endpwent@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
struct passwd *getpwnam(const char *name) {
  struct passwd *pw;
  endpwent();
  while ((pw = getpwent()) != NULL && strcmp(pw->pw_name, name) != 0) {
  }
  endpwent();
  return pw;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<buf32_to_uint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x18,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x10,%eax
  or     %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  or     %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int buf32_to_uint(const void *buffer) {
  const unsigned char *p = buffer;
  return (((unsigned int)p[0] << 24) | (p[1] << 16) | (p[2] << 8) | p[3]);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<buf32_to_uint>:
  endbr64
  movzbl (%rdi),%eax
  shl    $0x18,%eax
  movzbl 0x3(%rdi),%edx
  or     %edx,%eax
  movzbl 0x1(%rdi),%edx
  shl    $0x10,%edx
  or     %edx,%eax
  movzbl 0x2(%rdi),%edx
  shl    $0x8,%edx
  or     %edx,%eax
  ret
```

assistant: ```c
unsigned int buf32_to_uint(const void *buffer) {
  const unsigned char *p = buffer;
  return (((unsigned int)p[0] << 24) | (p[1] << 16) | (p[2] << 8) | p[3]);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<buf32_to_uint>:
  endbr64
  mov    (%rdi),%eax
  bswap  %eax
  ret
```

assistant: ```c
unsigned int buf32_to_uint(const void *buffer) {
  const unsigned char *p = buffer;
  return (((unsigned int)p[0] << 24) | (p[1] << 16) | (p[2] << 8) | p[3]);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<buf32_to_uint>:
  endbr64
  mov    (%rdi),%eax
  bswap  %eax
  ret
```

assistant: ```c
unsigned int buf32_to_uint(const void *buffer) {
  const unsigned char *p = buffer;
  return (((unsigned int)p[0] << 24) | (p[1] << 16) | (p[2] << 8) | p[3]);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <sum_all@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D27
data_type: string
name: parse_data
arguments: data_label: D19
data_type: string
name: parse_data
arguments: data_label: D8
data_type: string

tool: D27: please input number:

tool: D19: %d

tool: D8: The sum is %d


assistant: ```c
int main(int argc, char *args[]) {
  int number;
  int sum;
  printf("please input number:");
  scanf("%d", &number);
  sum = sum_all(number);
  printf("The sum is %d\n", sum);
  return sum;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D23(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <sum_all@plt>
  mov    %eax,%ebx
  mov    %eax,%esi
  lea    D13(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    %ebx,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D24
data_type: string
name: disassemble_data
arguments: label: D23
data_type: string
name: disassemble_data
arguments: label: D13
data_type: string

tool: D24: please input number:

tool: D23: %d

tool: D13: The sum is %d


assistant: ```c
int main(int argc, char *args[]) {
  int number;
  int sum;
  printf("please input number:");
  scanf("%d", &number);
  sum = sum_all(number);
  printf("The sum is %d\n", sum);
  return sum;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D23(%rip),%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D21(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <sum_all@plt>
  lea    D2(%rip),%rdi
  mov    %eax,%r12d
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D23
data_type: string
name: retrieve_data
arguments: label: D21
data_type: string
name: retrieve_data
arguments: label: D2
data_type: string

tool: D23: please input number:

tool: D21: %d

tool: D2: The sum is %d


assistant: ```c
int main(int argc, char *args[]) {
  int number;
  int sum;
  printf("please input number:");
  scanf("%d", &number);
  sum = sum_all(number);
  printf("The sum is %d\n", sum);
  return sum;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D7(%rip),%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D12(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <sum_all@plt>
  lea    D20(%rip),%rdi
  mov    %eax,%r12d
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
type: string
name: extract_symbolic_data
arguments: tag: D12
type: string
name: extract_symbolic_data
arguments: tag: D20
type: string

tool: D7: please input number:

tool: D12: %d

tool: D20: The sum is %d


assistant: ```c
int main(int argc, char *args[]) {
  int number;
  int sum;
  printf("please input number:");
  scanf("%d", &number);
  sum = sum_all(number);
  printf("The sum is %d\n", sum);
  return sum;
}
```
user: What is the c source code of the assembly code below?

```asm
<wcsncmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x38(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <wcsnlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x38(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <wcsnlen@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  cmp    %rax,%rdx
  cmova  %rax,%rdx
  mov    -0x30(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <wmemcmp@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L31
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  je     L18
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jae    L43
  mov    $0xffffffff,%eax
  jmp    L3
L43:
  mov    $0x1,%eax
  jmp    L3
L18:
  mov    $0x0,%eax
  jmp    L3
L31:
  mov    -0x14(%rbp),%eax
L3:
  leave
  ret
```

assistant: ```c
int wcsncmp(const wchar_t *a, const wchar_t *b, size_t length) {
  size_t n = wcsnlen(a, length);
  size_t m = wcsnlen(b, length);
  int r = wmemcmp(a, b, (n < m ? n : m));
  return r ? r : n == m ? 0 : n < m ? -1 : +1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<wcsncmp>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    %rdx,%rbx
  mov    %rdx,%rsi
  call   <wcsnlen@plt>
  mov    %rax,%rbp
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <wcsnlen@plt>
  mov    %rax,%rbx
  cmp    %rax,%rbp
  mov    %rax,%rdx
  cmovbe %rbp,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <wmemcmp@plt>
  test   %eax,%eax
  jne    L13
  cmp    %rbx,%rbp
  je     L13
  sbb    %eax,%eax
  or     $0x1,%eax
L13:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int wcsncmp(const wchar_t *a, const wchar_t *b, size_t length) {
  size_t n = wcsnlen(a, length);
  size_t m = wcsnlen(b, length);
  int r = wmemcmp(a, b, (n < m ? n : m));
  return r ? r : n == m ? 0 : n < m ? -1 : +1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<wcsncmp>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %rsi,%r13
  mov    %rdx,%rsi
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  call   <wcsnlen@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  mov    %rax,%rbp
  call   <wcsnlen@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  cmp    %rax,%rbp
  mov    %rax,%rdx
  mov    %rax,%rbx
  cmovbe %rbp,%rdx
  call   <wmemcmp@plt>
  test   %eax,%eax
  jne    L9
  cmp    %rbx,%rbp
  je     L9
  sbb    %eax,%eax
  or     $0x1,%eax
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int wcsncmp(const wchar_t *a, const wchar_t *b, size_t length) {
  size_t n = wcsnlen(a, length);
  size_t m = wcsnlen(b, length);
  int r = wmemcmp(a, b, (n < m ? n : m));
  return r ? r : n == m ? 0 : n < m ? -1 : +1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<wcsncmp>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %rsi,%r13
  mov    %rdx,%rsi
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  call   <wcsnlen@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  mov    %rax,%rbp
  call   <wcsnlen@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  cmp    %rax,%rbp
  mov    %rax,%rdx
  mov    %rax,%rbx
  cmovbe %rbp,%rdx
  call   <wmemcmp@plt>
  test   %eax,%eax
  jne    L10
  cmp    %rbx,%rbp
  je     L10
  sbb    %eax,%eax
  or     $0x1,%eax
L10:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int wcsncmp(const wchar_t *a, const wchar_t *b, size_t length) {
  size_t n = wcsnlen(a, length);
  size_t m = wcsnlen(b, length);
  int r = wmemcmp(a, b, (n < m ? n : m));
  return r ? r : n == m ? 0 : n < m ? -1 : +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
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x2,-0x30(%rbp)
  movl   $0xa,-0x2c(%rbp)
  movl   $0x3,-0x28(%rbp)
  movl   $0xe,-0x24(%rbp)
  movl   $0xf,-0x20(%rbp)
  movl   $0x6,-0x1c(%rbp)
  movl   $0x3,-0x18(%rbp)
  movl   $0x9,-0x14(%rbp)
  movl   $0x8,-0x34(%rbp)
  mov    -0x34(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <quick_sort@plt>
  mov    -0x34(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <print@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
int main() {
  int nums[] = {2, 10, 3, 14, 15, 6, 3, 9};
  int n = sizeof(nums) / sizeof(nums[0]);
  quick_sort(nums, n);
  print(nums, n);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x2,(%rsp)
  movl   $0xa,0x4(%rsp)
  movl   $0x3,0x8(%rsp)
  movl   $0xe,0xc(%rsp)
  movl   $0xf,0x10(%rsp)
  movl   $0x6,0x14(%rsp)
  movl   $0x3,0x18(%rsp)
  movl   $0x9,0x1c(%rsp)
  mov    %rsp,%rbx
  mov    $0x8,%esi
  mov    %rbx,%rdi
  call   <quick_sort@plt>
  mov    $0x8,%esi
  mov    %rbx,%rdi
  call   <print@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int nums[] = {2, 10, 3, 14, 15, 6, 3, 9};
  int n = sizeof(nums) / sizeof(nums[0]);
  quick_sort(nums, n);
  print(nums, n);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x8,%esi
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0xa00000002,%rax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  mov    %rax,(%rsp)
  movabs $0xe00000003,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x60000000f,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x900000003,%rax
  mov    %rax,0x18(%rsp)
  call   <quick_sort@plt>
  mov    $0x8,%esi
  mov    %rbp,%rdi
  call   <print@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int nums[] = {2, 10, 3, 14, 15, 6, 3, 9};
  int n = sizeof(nums) / sizeof(nums[0]);
  quick_sort(nums, n);
  print(nums, n);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x8,%esi
  sub    $0x30,%rsp
  movdqa D4(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movaps %xmm0,(%rsp)
  movdqa D7(%rip),%xmm0
  mov    %rbp,%rdi
  movaps %xmm0,0x10(%rsp)
  call   <quick_sort@plt>
  mov    $0x8,%esi
  mov    %rbp,%rdi
  call   <print@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D4
datatype: byte[16]
name: parse_data
arguments: label: D7
datatype: byte[16]

tool: {"D4": [0x02,0x00,0x00,0x00,0x0a,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x0e,0x00,0x00,0x00]}

tool: {"D7": [0x0f,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x09,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int nums[] = {2, 10, 3, 14, 15, 6, 3, 9};
  int n = sizeof(nums) / sizeof(nums[0]);
  quick_sort(nums, n);
  print(nums, n);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<perm>:
  endbr64
  mov    $0x0,%eax
  cmp    $0x1,%esi
  jle    L77
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  movslq %esi,%rax
  shl    $0x2,%rax
  lea    -0x4(%rdi,%rax,1),%rcx
  mov    (%rcx),%edx
  lea    -0x8(%rdi,%rax,1),%rax
  mov    (%rax),%edi
  cmp    %edi,%edx
  jg     L57
  mov    %esi,%ebx
  sub    $0x3,%ebx
  js     L50
  movslq %ebx,%rax
  lea    0x0(%rbp,%rax,4),%rax
L40:
  mov    %rax,%r13
  mov    (%rax),%edi
  lea    0x4(%rax),%r12
  cmp    0x4(%rax),%edi
  jl     L35
  sub    $0x1,%ebx
  sub    $0x4,%rax
  cmp    $0xffffffff,%ebx
  jne    L40
  mov    $0x0,%eax
  jmp    L39
L57:
  mov    %edi,(%rcx)
  mov    %edx,(%rax)
  mov    $0x1,%eax
L39:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L35:
  sub    %ebx,%esi
  lea    -0x1(%rsi),%r14d
  mov    %r14d,%edx
  mov    %r12,%rsi
  call   <find_successor@plt>
  mov    0x0(%r13),%edx
  lea    0x1(%rbx,%rax,1),%eax
  cltq
  lea    0x0(%rbp,%rax,4),%rax
  mov    (%rax),%ecx
  mov    %ecx,0x0(%r13)
  mov    %edx,(%rax)
  mov    %r14d,%esi
  mov    %r12,%rdi
  call   <reverse@plt>
  mov    $0x1,%eax
  jmp    L39
L50:
  mov    $0x0,%eax
  jmp    L39
L77:
  ret
```

assistant: ```c
int perm(int *a, int len) {
  if (len <= 1)
    return 0;
  if (a[len - 1] > a[len - 2]) {
    int temp = a[len - 1];
    a[len - 1] = a[len - 2];
    a[len - 2] = temp;
    return 1;
  }
  for (int ii = len - 3; ii >= 0; ii--) {
    if (a[ii] < a[ii + 1]) {
      const int small_len = len - ii - 1;
      int *small_a = a + ii + 1;
      const int relative_succ_index = find_successor(a[ii], small_a, small_len);
      const int succ_index = relative_succ_index + ii + 1;
      int temp = a[ii];
      a[ii] = a[succ_index];
      a[succ_index] = temp;
      reverse(small_a, small_len);
      return 1;
    }
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<perm>:
  endbr64
  xor    %r8d,%r8d
  cmp    $0x1,%esi
  jle    L19
  movslq %esi,%rax
  push   %r14
  shl    $0x2,%rax
  push   %r13
  mov    %rdi,%r13
  lea    -0x4(%rdi,%rax,1),%rcx
  lea    -0x8(%rdi,%rax,1),%rax
  push   %r12
  push   %rbp
  push   %rbx
  mov    (%rcx),%edx
  mov    (%rax),%edi
  cmp    %edi,%edx
  jg     L46
  lea    -0x3(%rsi),%ebp
  cmp    $0x2,%esi
  je     L42
  movslq %ebp,%rax
  lea    0x0(%r13,%rax,4),%rbx
  jmp    L15
  cs nopw 0x0(%rax,%rax,1)
L56:
  sub    $0x1,%ebp
  sub    $0x4,%rbx
  cmp    $0xffffffff,%ebp
  je     L17
L15:
  mov    (%rbx),%edi
  lea    0x4(%rbx),%r14
  cmp    0x4(%rbx),%edi
  jge    L56
  sub    %ebp,%esi
  lea    -0x1(%rsi),%r12d
  mov    %r14,%rsi
  mov    %r12d,%edx
  call   <find_successor@plt>
  mov    (%rbx),%edx
  mov    %r12d,%esi
  mov    %r14,%rdi
  lea    0x1(%rbp,%rax,1),%eax
  cltq
  lea    0x0(%r13,%rax,4),%rax
  mov    (%rax),%ecx
  mov    %ecx,(%rbx)
  mov    %edx,(%rax)
  call   <reverse@plt>
  mov    $0x1,%r8d
  pop    %rbx
  pop    %rbp
  mov    %r8d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L46:
  mov    %edi,(%rcx)
  mov    $0x1,%r8d
  mov    %edx,(%rax)
L42:
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  cs nopw 0x0(%rax,%rax,1)
L19:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L17:
  xor    %r8d,%r8d
  pop    %rbx
  pop    %rbp
  mov    %r8d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int perm(int *a, int len) {
  if (len <= 1)
    return 0;
  if (a[len - 1] > a[len - 2]) {
    int temp = a[len - 1];
    a[len - 1] = a[len - 2];
    a[len - 2] = temp;
    return 1;
  }
  for (int ii = len - 3; ii >= 0; ii--) {
    if (a[ii] < a[ii + 1]) {
      const int small_len = len - ii - 1;
      int *small_a = a + ii + 1;
      const int relative_succ_index = find_successor(a[ii], small_a, small_len);
      const int succ_index = relative_succ_index + ii + 1;
      int temp = a[ii];
      a[ii] = a[succ_index];
      a[succ_index] = temp;
      reverse(small_a, small_len);
      return 1;
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<perm>:
  endbr64
  xor    %r8d,%r8d
  cmp    $0x1,%esi
  jle    L51
  movslq %esi,%rax
  push   %r14
  shl    $0x2,%rax
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  movq   -0x8(%rdi,%rax,1),%xmm0
  lea    -0x8(%rdi,%rax,1),%rdi
  pshufd $0xe5,%xmm0,%xmm2
  movd   %xmm0,%edx
  movd   %xmm2,%ecx
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %edx,%ecx
  jg     L34
  lea    -0x3(%rsi),%ebp
  cmp    $0x2,%esi
  je     L58
  mov    -0x8(%r13,%rax,1),%edi
  lea    -0xc(%r13,%rax,1),%rbx
  jmp    L19
  nopw   0x0(%rax,%rax,1)
L8:
  sub    $0x1,%ebp
  sub    $0x4,%rbx
  cmp    $0xffffffff,%ebp
  je     L49
L19:
  mov    %edi,%eax
  mov    (%rbx),%edi
  lea    0x4(%rbx),%r14
  cmp    %eax,%edi
  jge    L8
  sub    %ebp,%esi
  lea    -0x1(%rsi),%r12d
  mov    %r14,%rsi
  mov    %r12d,%edx
  call   <find_successor@plt>
  mov    (%rbx),%edx
  mov    %r12d,%esi
  mov    %r14,%rdi
  lea    0x1(%rbp,%rax,1),%eax
  cltq
  lea    0x0(%r13,%rax,4),%rax
  mov    (%rax),%ecx
  mov    %ecx,(%rbx)
  mov    %edx,(%rax)
  call   <reverse@plt>
  mov    $0x1,%r8d
  pop    %rbx
  pop    %rbp
  mov    %r8d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L34:
  movq   %xmm1,(%rdi)
  mov    $0x1,%r8d
L58:
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  cs nopw 0x0(%rax,%rax,1)
L51:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L49:
  xor    %r8d,%r8d
  pop    %rbx
  pop    %rbp
  mov    %r8d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int perm(int *a, int len) {
  if (len <= 1)
    return 0;
  if (a[len - 1] > a[len - 2]) {
    int temp = a[len - 1];
    a[len - 1] = a[len - 2];
    a[len - 2] = temp;
    return 1;
  }
  for (int ii = len - 3; ii >= 0; ii--) {
    if (a[ii] < a[ii + 1]) {
      const int small_len = len - ii - 1;
      int *small_a = a + ii + 1;
      const int relative_succ_index = find_successor(a[ii], small_a, small_len);
      const int succ_index = relative_succ_index + ii + 1;
      int temp = a[ii];
      a[ii] = a[succ_index];
      a[succ_index] = temp;
      reverse(small_a, small_len);
      return 1;
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<GPMCECCSizeValSet>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jne    L21
  mov    -0x4(%rbp),%eax
  add    $0x1fc,%eax
  mov    %eax,%eax
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  add    $0x1fc,%edx
  mov    %edx,%edx
  and    $0xfff00fff,%eax
  mov    %eax,(%rdx)
  mov    -0x4(%rbp),%eax
  add    $0x1fc,%eax
  mov    %eax,%eax
  mov    (%rax),%ecx
  mov    -0xc(%rbp),%eax
  shl    $0xc,%eax
  and    $0xff000,%eax
  mov    %eax,%esi
  mov    -0x4(%rbp),%eax
  add    $0x1fc,%eax
  mov    %eax,%eax
  mov    %rax,%rdx
  mov    %ecx,%eax
  or     %esi,%eax
  mov    %eax,(%rdx)
  jmp    L23
L21:
  cmpl   $0x1,-0x8(%rbp)
  jne    L23
  mov    -0x4(%rbp),%eax
  add    $0x1fc,%eax
  mov    %eax,%eax
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  add    $0x1fc,%edx
  mov    %edx,%edx
  and    $0xc03fffff,%eax
  mov    %eax,(%rdx)
  mov    -0x4(%rbp),%eax
  add    $0x1fc,%eax
  mov    %eax,%eax
  mov    (%rax),%ecx
  mov    -0xc(%rbp),%eax
  shl    $0x16,%eax
  and    $0x3fc00000,%eax
  mov    %eax,%esi
  mov    -0x4(%rbp),%eax
  add    $0x1fc,%eax
  mov    %eax,%eax
  mov    %rax,%rdx
  mov    %ecx,%eax
  or     %esi,%eax
  mov    %eax,(%rdx)
L23:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void GPMCECCSizeValSet(unsigned int baseAddr, unsigned int eccSize, unsigned int eccSizeVal) {
  if (eccSize == (0)) {
    (*((volatile unsigned int *)(baseAddr + (0x1FC)))) &= ~(0x000FF000u);
    (*((volatile unsigned int *)(baseAddr + (0x1FC)))) |= ((eccSizeVal << (0x0000000Cu)) & (0x000FF000u));
  } else if (eccSize == (1)) {
    (*((volatile unsigned int *)(baseAddr + (0x1FC)))) &= ~(0x3FC00000u);
    (*((volatile unsigned int *)(baseAddr + (0x1FC)))) |= ((eccSizeVal << (0x00000016u)) & (0x3FC00000u));
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<GPMCECCSizeValSet>:
  endbr64
  test   %esi,%esi
  jne    L16
  mov    0x1fc(%edi),%ecx
  and    $0xfff00fff,%ecx
  mov    %ecx,0x1fc(%edi)
  mov    0x1fc(%edi),%ecx
  shl    $0xc,%edx
  and    $0xff000,%edx
  or     %ecx,%edx
  mov    %edx,0x1fc(%edi)
  ret
L16:
  cmp    $0x1,%esi
  je     L12
L19:
  ret
L12:
  mov    0x1fc(%edi),%ecx
  and    $0xc03fffff,%ecx
  mov    %ecx,0x1fc(%edi)
  mov    0x1fc(%edi),%ecx
  shl    $0x16,%edx
  and    $0x3fc00000,%edx
  or     %ecx,%edx
  mov    %edx,0x1fc(%edi)
  jmp    L19
```

assistant: ```c
void GPMCECCSizeValSet(unsigned int baseAddr, unsigned int eccSize, unsigned int eccSizeVal) {
  if (eccSize == (0)) {
    (*((volatile unsigned int *)(baseAddr + (0x1FC)))) &= ~(0x000FF000u);
    (*((volatile unsigned int *)(baseAddr + (0x1FC)))) |= ((eccSizeVal << (0x0000000Cu)) & (0x000FF000u));
  } else if (eccSize == (1)) {
    (*((volatile unsigned int *)(baseAddr + (0x1FC)))) &= ~(0x3FC00000u);
    (*((volatile unsigned int *)(baseAddr + (0x1FC)))) |= ((eccSizeVal << (0x00000016u)) & (0x3FC00000u));
  }
}
```