ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x186000(%rsp),%r11
L36:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L36
  sub    $0xa70,%rsp
  lea    D0(%rip),%r14
  mov    %r14,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x186a68(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  cmpq   $0x0,0x8(%rsp)
  jle    L19
  lea    0x10(%rsp),%rbp
  xor    %ebx,%ebx
  lea    0xc3540(%rsp),%r12
  mov    %rbp,%r13
  nopl   0x0(%rax)
L41:
  mov    %r13,%rsi
  mov    %r14,%rdi
  movq   $0x0,(%r12,%rbx,8)
  add    $0x1,%rbx
  call   <scanf@plt>
  mov    0x8(%rsp),%rcx
  add    $0x8,%r13
  cmp    %rbx,%rcx
  jg     L41
  movq   $0x0,0x10(%rsp,%rcx,8)
  test   %rcx,%rcx
  jle    L19
  mov    0x10(%rsp),%rsi
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L8:
  add    $0x1,%rax
  mov    %rsi,%rdx
  mov    0x0(%rbp,%rax,8),%rsi
  add    %rsi,%rdx
  mov    %rdx,-0x8(%r12,%rax,8)
  cmp    %rax,%rcx
  jne    L8
  xor    %ebx,%ebx
  lea    D1(%rip),%rbp
  nopw   0x0(%rax,%rax,1)
L30:
  mov    (%r12,%rbx,8),%rsi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,0x8(%rsp)
  jg     L30
L19:
  mov    0x186a68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L59
  add    $0x186a70,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L59:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D0
type_infered: string
name: unpack_data
arguments: data_label: D1
type_infered: string

tool: D0: %lld

tool: D1: %lld 

assistant: ```c
int main(void) {
  long long n, i, ar[100005], br[100005];
  scanf("%lld", &n);
  for (i = 0; i < n; i++) {
    br[i] = 0;
    scanf("%lld", &ar[i]);
  }
  ar[n] = 0;
  for (i = 0; i < n; i++) {
    br[i] = ar[i] + ar[i + 1];
  }
  for (i = 0; i < n; i++)
    printf("%lld ", br[i]);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x186000(%rsp),%r11
L40:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L40
  sub    $0xa70,%rsp
  lea    D7(%rip),%r14
  mov    %r14,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x186a68(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  cmpq   $0x0,0x8(%rsp)
  jle    L48
  lea    0x10(%rsp),%rbp
  xor    %ebx,%ebx
  lea    0xc3540(%rsp),%r13
  mov    %rbp,%r12
  nopl   0x0(%rax)
L56:
  movq   $0x0,0x0(%r13,%rbx,8)
  mov    %r12,%rsi
  mov    %r14,%rdi
  add    $0x1,%rbx
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  add    $0x8,%r12
  cmp    %rbx,%rax
  jg     L56
  movq   $0x0,0x10(%rsp,%rax,8)
  test   %rax,%rax
  jle    L48
  cmp    $0x1,%rax
  je     L74
  mov    %rax,%rcx
  mov    %r13,%rdx
  shr    %rcx
  shl    $0x4,%rcx
  add    %rbp,%rcx
  nopw   0x0(%rax,%rax,1)
L83:
  movdqu 0x8(%rbp),%xmm0
  paddq  0x0(%rbp),%xmm0
  add    $0x10,%rbp
  add    $0x10,%rdx
  movaps %xmm0,-0x10(%rdx)
  cmp    %rcx,%rbp
  jne    L83
  mov    %rax,%rdx
  and    $0xfffffffffffffffe,%rdx
  test   $0x1,%al
  je     L4
L31:
  mov    0x10(%rsp,%rdx,8),%rax
  add    0x18(%rsp,%rdx,8),%rax
  mov    %rax,0xc3540(%rsp,%rdx,8)
L4:
  xor    %ebx,%ebx
  lea    D2(%rip),%rbp
  nopl   0x0(%rax)
L78:
  mov    0x0(%r13,%rbx,8),%rsi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,0x8(%rsp)
  jg     L78
L48:
  mov    0x186a68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  add    $0x186a70,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L74:
  xor    %edx,%edx
  jmp    L31
L36:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D7
type_infered: string
name: parse_data
arguments: label: D2
type_infered: string

tool: D7: %lld

tool: D2: %lld 

assistant: ```c
int main(void) {
  long long n, i, ar[100005], br[100005];
  scanf("%lld", &n);
  for (i = 0; i < n; i++) {
    br[i] = 0;
    scanf("%lld", &ar[i]);
  }
  ar[n] = 0;
  for (i = 0; i < n; i++) {
    br[i] = ar[i] + ar[i + 1];
  }
  for (i = 0; i < n; i++)
    printf("%lld ", br[i]);
  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
  mov    %rsp,%rbp
  movl   $0x15,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  int twentyone = 21;
  return twentyone << 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  mov    $0x2a,%eax
  ret
```

assistant: ```c
int main(void) {
  int twentyone = 21;
  return twentyone << 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  mov    $0x2a,%eax
  ret
```

assistant: ```c
int main(void) {
  int twentyone = 21;
  return twentyone << 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  mov    $0x2a,%eax
  ret
```

assistant: ```c
int main(void) {
  int twentyone = 21;
  return twentyone << 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<must_strcat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <must_malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x30(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *must_strcat(char *a, char *b) {
  size_t al;
  size_t bl;
  char *answer;
  al = strlen(a);
  bl = strlen(b);
  answer = must_malloc(al + bl + 1);
  strcpy(answer, a);
  strcpy(answer + al, b);
  return answer;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<must_strcat>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r12
  call   <strlen@plt>
  mov    %rax,%rbx
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x1(%rbx,%rax,1),%rdi
  call   <must_malloc@plt>
  mov    %rax,%rbp
  mov    %r13,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  lea    0x0(%rbp,%rbx,1),%rdi
  mov    %r12,%rsi
  call   <strcpy@plt>
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *must_strcat(char *a, char *b) {
  size_t al;
  size_t bl;
  char *answer;
  al = strlen(a);
  bl = strlen(b);
  answer = must_malloc(al + bl + 1);
  strcpy(answer, a);
  strcpy(answer + al, b);
  return answer;
}
```
user: Translate the below assembly code into C programming language:

```asm
<must_strcat>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %rax,%rbx
  call   <strlen@plt>
  lea    0x1(%rbx,%rax,1),%rdi
  call   <must_malloc@plt>
  mov    %r13,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strcpy@plt>
  lea    (%r12,%rbx,1),%rdi
  mov    %rbp,%rsi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *must_strcat(char *a, char *b) {
  size_t al;
  size_t bl;
  char *answer;
  al = strlen(a);
  bl = strlen(b);
  answer = must_malloc(al + bl + 1);
  strcpy(answer, a);
  strcpy(answer + al, b);
  return answer;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  shr    $0x3,%eax
  or     $0x61bab5,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  and    $0x8,%eax
  test   %eax,%eax
  je     L0
  mov    -0xc(%rbp),%eax
  and    $0x10,%eax
  test   %eax,%eax
  je     L19
  mov    -0xc(%rbp),%edx
  mov    -0xc(%rbp),%eax
  shr    %eax
  and    $0x6,%eax
  or     $0x1,%eax
  mov    %eax,%ecx
  shr    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L47
L19:
  lea    -0xc(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  lea    -0xc(%rbp),%rax
  add    $0x3,%rax
  lea    -0xc(%rbp),%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  lea    -0xc(%rbp),%rax
  add    $0x3,%rax
  movzbl -0xd(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0xc(%rbp),%edx
  mov    -0xc(%rbp),%eax
  shr    $0x3,%eax
  and    $0x6,%eax
  or     $0x1,%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L47
L0:
  mov    -0xc(%rbp),%edx
  mov    -0xc(%rbp),%eax
  shr    $0x3,%eax
  and    $0x6,%eax
  or     $0x1,%eax
  mov    %eax,%ecx
  shr    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0xc(%rbp),%eax
  shr    $0x4,%eax
  and    $0x6,%eax
  or     $0x1,%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0xc(%rbp)
L47:
  mov    -0xc(%rbp),%eax
  or     $0x36f5edda,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  char copy12;
  {
    state[0UL] = (input[0UL] | 51238316UL) >> 3U;
    if ((state[0UL] >> 3U) & 1U) {
      if ((state[0UL] >> 4U) & 1U) {
        state[0UL] >>= ((state[0UL] >> 1U) & 7U) | 1UL;
      } else {
        copy12 = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 3);
        *((char *)(&state[0UL]) + 3) = copy12;
        state[0UL] <<= ((state[0UL] >> 3U) & 7U) | 1UL;
      }
    } else {
      state[0UL] >>= ((state[0UL] >> 3U) & 7U) | 1UL;
      state[0UL] <<= ((state[0UL] >> 4U) & 7U) | 1UL;
    }
    output[0UL] = state[0UL] | 922086874UL;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%edx
  mov    %edx,%eax
  shr    $0x3,%eax
  test   $0x40,%dl
  je     L5
  or     $0x61bab5,%eax
  mov    %eax,%ecx
  shr    %ecx
  and    $0x6,%ecx
  or     $0x1,%ecx
  shr    %cl,%eax
  jmp    L14
L5:
  and    $0xffffff80,%eax
  or     $0x61ba80,%eax
L14:
  or     $0x36f5edda,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  char copy12;
  {
    state[0UL] = (input[0UL] | 51238316UL) >> 3U;
    if ((state[0UL] >> 3U) & 1U) {
      if ((state[0UL] >> 4U) & 1U) {
        state[0UL] >>= ((state[0UL] >> 1U) & 7U) | 1UL;
      } else {
        copy12 = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 3);
        *((char *)(&state[0UL]) + 3) = copy12;
        state[0UL] <<= ((state[0UL] >> 3U) & 7U) | 1UL;
      }
    } else {
      state[0UL] >>= ((state[0UL] >> 3U) & 7U) | 1UL;
      state[0UL] <<= ((state[0UL] >> 4U) & 7U) | 1UL;
    }
    output[0UL] = state[0UL] | 922086874UL;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%edx
  mov    %edx,%eax
  shr    $0x3,%eax
  and    $0x40,%edx
  je     L6
  or     $0x61bab5,%eax
  mov    %eax,%ecx
  shr    %ecx
  and    $0x6,%ecx
  or     $0x1,%ecx
  shr    %cl,%eax
  or     $0x36f5edda,%eax
  mov    %eax,(%rsi)
  ret
  nopl   0x0(%rax)
L6:
  and    $0xffffff80,%eax
  or     $0x61ba80,%eax
  or     $0x36f5edda,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  char copy12;
  {
    state[0UL] = (input[0UL] | 51238316UL) >> 3U;
    if ((state[0UL] >> 3U) & 1U) {
      if ((state[0UL] >> 4U) & 1U) {
        state[0UL] >>= ((state[0UL] >> 1U) & 7U) | 1UL;
      } else {
        copy12 = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 3);
        *((char *)(&state[0UL]) + 3) = copy12;
        state[0UL] <<= ((state[0UL] >> 3U) & 7U) | 1UL;
      }
    } else {
      state[0UL] >>= ((state[0UL] >> 3U) & 7U) | 1UL;
      state[0UL] <<= ((state[0UL] >> 4U) & 7U) | 1UL;
    }
    output[0UL] = state[0UL] | 922086874UL;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%edx
  mov    %edx,%eax
  shr    $0x3,%eax
  and    $0x40,%edx
  je     L3
  or     $0x61bab5,%eax
  mov    %eax,%ecx
  shr    %ecx
  and    $0x6,%ecx
  or     $0x1,%ecx
  shr    %cl,%eax
  or     $0x36f5edda,%eax
  mov    %eax,(%rsi)
  ret
  nopl   0x0(%rax)
L3:
  and    $0xffffff80,%eax
  or     $0x61ba80,%eax
  or     $0x36f5edda,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  char copy12;
  {
    state[0UL] = (input[0UL] | 51238316UL) >> 3U;
    if ((state[0UL] >> 3U) & 1U) {
      if ((state[0UL] >> 4U) & 1U) {
        state[0UL] >>= ((state[0UL] >> 1U) & 7U) | 1UL;
      } else {
        copy12 = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 3);
        *((char *)(&state[0UL]) + 3) = copy12;
        state[0UL] <<= ((state[0UL] >> 3U) & 7U) | 1UL;
      }
    } else {
      state[0UL] >>= ((state[0UL] >> 3U) & 7U) | 1UL;
      state[0UL] <<= ((state[0UL] >> 4U) & 7U) | 1UL;
    }
    output[0UL] = state[0UL] | 922086874UL;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<AESDataProcess>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x18(%rbp)
  cmpl   $0x44036000,-0x14(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <ASSERT@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    $0x44036000,%edi
  call   <AESLengthSet@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L5:
  mov    -0x4(%rbp),%eax
  lea    0x3(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x2,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <AESDataWrite@plt>
  mov    -0x4(%rbp),%eax
  lea    0x3(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x2,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <AESDataRead@plt>
  addl   $0x10,-0x4(%rbp)
L2:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L5
  mov    $0x1,%eax
  leave
  ret
```

assistant: ```c
bool AESDataProcess(uint32_t ui32Base, uint32_t *pui32Src, uint32_t *pui32Dest, uint32_t ui32Length) {
  uint32_t ui32Count;
  ASSERT(ui32Base == 0x44036000);
  AESLengthSet(0x44036000, (uint64_t)ui32Length);
  for (ui32Count = 0; ui32Count < ui32Length; ui32Count += 16) {
    AESDataWrite(ui32Base, pui32Src + (ui32Count / 4));
    AESDataRead(ui32Base, pui32Dest + (ui32Count / 4));
  }
  return (true);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<AESDataProcess>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r12d
  mov    %rsi,%r14
  mov    %rdx,%r15
  mov    %ecx,%r13d
  cmp    $0x44036000,%edi
  sete   %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  mov    %r13d,%esi
  mov    $0x44036000,%edi
  call   <AESLengthSet@plt>
  test   %r13d,%r13d
  jle    L5
  sub    $0x1,%r13d
  and    $0xfffffff0,%r13d
  add    $0x10,%r13d
  mov    $0x0,%ebp
L9:
  lea    0x3(%rbp),%ebx
  test   %ebp,%ebp
  cmovns %ebp,%ebx
  sar    $0x2,%ebx
  movslq %ebx,%rbx
  shl    $0x2,%rbx
  lea    (%r14,%rbx,1),%rsi
  mov    %r12d,%edi
  call   <AESDataWrite@plt>
  lea    (%r15,%rbx,1),%rsi
  mov    %r12d,%edi
  call   <AESDataRead@plt>
  add    $0x10,%ebp
  cmp    %r13d,%ebp
  jne    L9
L5:
  mov    $0x1,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
bool AESDataProcess(uint32_t ui32Base, uint32_t *pui32Src, uint32_t *pui32Dest, uint32_t ui32Length) {
  uint32_t ui32Count;
  ASSERT(ui32Base == 0x44036000);
  AESLengthSet(0x44036000, (uint64_t)ui32Length);
  for (ui32Count = 0; ui32Count < ui32Length; ui32Count += 16) {
    AESDataWrite(ui32Base, pui32Src + (ui32Count / 4));
    AESDataRead(ui32Base, pui32Dest + (ui32Count / 4));
  }
  return (true);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<AESDataProcess>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x44036000,%edi
  sete   %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  mov    %r12d,%esi
  mov    $0x44036000,%edi
  call   <AESLengthSet@plt>
  test   %r12d,%r12d
  jle    L8
  sub    $0x1,%r12d
  xor    %r15d,%r15d
  and    $0xfffffff0,%r12d
  add    $0x10,%r12d
  nopl   0x0(%rax)
L13:
  mov    %r15d,%ebx
  mov    %ebp,%edi
  add    $0x10,%r15d
  sar    $0x2,%ebx
  movslq %ebx,%rbx
  shl    $0x2,%rbx
  lea    0x0(%r13,%rbx,1),%rsi
  call   <AESDataWrite@plt>
  lea    (%r14,%rbx,1),%rsi
  mov    %ebp,%edi
  call   <AESDataRead@plt>
  cmp    %r12d,%r15d
  jne    L13
L8:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
bool AESDataProcess(uint32_t ui32Base, uint32_t *pui32Src, uint32_t *pui32Dest, uint32_t ui32Length) {
  uint32_t ui32Count;
  ASSERT(ui32Base == 0x44036000);
  AESLengthSet(0x44036000, (uint64_t)ui32Length);
  for (ui32Count = 0; ui32Count < ui32Length; ui32Count += 16) {
    AESDataWrite(ui32Base, pui32Src + (ui32Count / 4));
    AESDataRead(ui32Base, pui32Dest + (ui32Count / 4));
  }
  return (true);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<AESDataProcess>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x44036000,%edi
  sete   %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  mov    %r12d,%esi
  mov    $0x44036000,%edi
  call   <AESLengthSet@plt>
  test   %r12d,%r12d
  jle    L3
  sub    $0x1,%r12d
  xor    %r15d,%r15d
  and    $0xfffffff0,%r12d
  add    $0x10,%r12d
  nopl   0x0(%rax)
L13:
  mov    %r15d,%ebx
  mov    %ebp,%edi
  add    $0x10,%r15d
  sar    $0x2,%ebx
  movslq %ebx,%rbx
  shl    $0x2,%rbx
  lea    0x0(%r13,%rbx,1),%rsi
  call   <AESDataWrite@plt>
  lea    (%r14,%rbx,1),%rsi
  mov    %ebp,%edi
  call   <AESDataRead@plt>
  cmp    %r12d,%r15d
  jne    L13
L3:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
bool AESDataProcess(uint32_t ui32Base, uint32_t *pui32Src, uint32_t *pui32Dest, uint32_t ui32Length) {
  uint32_t ui32Count;
  ASSERT(ui32Base == 0x44036000);
  AESLengthSet(0x44036000, (uint64_t)ui32Length);
  for (ui32Count = 0; ui32Count < ui32Length; ui32Count += 16) {
    AESDataWrite(ui32Base, pui32Src + (ui32Count / 4));
    AESDataRead(ui32Base, pui32Dest + (ui32Count / 4));
  }
  return (true);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<isInCharArr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L1
L39:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L10
  mov    -0x4(%rbp),%eax
  jmp    L22
L10:
  addl   $0x1,-0x4(%rbp)
L1:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L39
  mov    $0xffffffff,%eax
L22:
  leave
  ret
```

assistant: ```c
int isInCharArr(char **arr, char *str) {
  int i = 0;
  while (arr[i] != NULL) {
    if (strcmp(arr[i], str) == 0) {
      return i;
    }
    i++;
  }
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<isInCharArr>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    (%rdi),%rdi
  test   %rdi,%rdi
  je     L39
  mov    %rsi,%r12
  mov    $0x0,%ebx
L31:
  mov    %ebx,%r13d
  mov    %r12,%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L25
  add    $0x1,%rbx
  mov    0x0(%rbp,%rbx,8),%rdi
  test   %rdi,%rdi
  jne    L31
  mov    $0xffffffff,%r13d
L25:
  mov    %r13d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L39:
  mov    $0xffffffff,%r13d
  jmp    L25
```

assistant: ```c
int isInCharArr(char **arr, char *str) {
  int i = 0;
  while (arr[i] != NULL) {
    if (strcmp(arr[i], str) == 0) {
      return i;
    }
    i++;
  }
  return -1;
}
```
user: What is the c source code of the assembly code below?

```asm
<isInCharArr>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%rdi
  test   %rdi,%rdi
  je     L11
  mov    %rsi,%r12
  xor    %ebx,%ebx
  jmp    L12
L21:
  add    $0x1,%rbx
  mov    0x0(%rbp,%rbx,8),%rdi
  test   %rdi,%rdi
  je     L11
L12:
  mov    %r12,%rsi
  mov    %ebx,%r13d
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L21
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L11:
  add    $0x8,%rsp
  mov    $0xffffffff,%r13d
  pop    %rbx
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int isInCharArr(char **arr, char *str) {
  int i = 0;
  while (arr[i] != NULL) {
    if (strcmp(arr[i], str) == 0) {
      return i;
    }
    i++;
  }
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<isInCharArr>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%rdi
  test   %rdi,%rdi
  je     L0
  mov    %rsi,%r12
  xor    %ebx,%ebx
  jmp    L6
L13:
  add    $0x1,%rbx
  mov    0x0(%rbp,%rbx,8),%rdi
  test   %rdi,%rdi
  je     L0
L6:
  mov    %r12,%rsi
  mov    %ebx,%r13d
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L13
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L0:
  add    $0x8,%rsp
  mov    $0xffffffff,%r13d
  pop    %rbx
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int isInCharArr(char **arr, char *str) {
  int i = 0;
  while (arr[i] != NULL) {
    if (strcmp(arr[i], str) == 0) {
      return i;
    }
    i++;
  }
  return -1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<umax_vector>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L13
L2:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jae    L26
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
L26:
  addl   $0x1,-0x4(%rbp)
L13:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L2
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned umax_vector(const unsigned *array, int n) {
  unsigned max = array[0];
  for (int i = 1; i < n; i++) {
    if (array[i] > max)
      max = array[i];
  }
  return max;
}
```