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

```asm
<qround>:
  endbr64
  mov    %edx,%edx
  mov    %esi,%esi
  push   %r14
  mov    %rdi,%r14
  push   %r13
  lea    (%rdi,%rsi,4),%r13
  mov    %r8d,%r8d
  mov    $0x10,%esi
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  lea    (%r14,%r8,4),%rbp
  lea    (%r14,%r12,4),%r12
  push   %rbx
  lea    (%rdi,%rdx,4),%rbx
  mov    (%rbx),%edi
  add    0x0(%r13),%edi
  mov    %edi,0x0(%r13)
  xor    0x0(%rbp),%edi
  call   <l32@plt>
  mov    $0xc,%esi
  mov    %eax,0x0(%rbp)
  mov    %eax,%edi
  add    (%r12),%edi
  mov    %edi,(%r12)
  xor    (%rbx),%edi
  call   <l32@plt>
  mov    $0x8,%esi
  mov    %eax,(%rbx)
  mov    %eax,%edi
  add    0x0(%r13),%edi
  mov    %edi,0x0(%r13)
  xor    0x0(%rbp),%edi
  call   <l32@plt>
  mov    $0x7,%esi
  mov    %eax,0x0(%rbp)
  mov    %eax,%edi
  add    (%r12),%edi
  mov    %edi,(%r12)
  xor    (%rbx),%edi
  call   <l32@plt>
  mov    %eax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void qround(unsigned int *x, unsigned int a, unsigned int b, unsigned int c, unsigned int d) {
  x[a] = x[a] + x[b];
  x[d] = l32(x[d] ^ x[a], 16);
  x[c] = x[c] + x[d];
  x[b] = l32(x[b] ^ x[c], 12);
  x[a] = x[a] + x[b];
  x[d] = l32(x[d] ^ x[a], 8);
  x[c] = x[c] + x[d];
  x[b] = l32(x[b] ^ x[c], 7);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strcat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <cgc_strlen@plt>
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *strcat(char *destination, const char *source) {
  strcpy(destination + cgc_strlen(destination), source);
  return destination;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<strcat>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <cgc_strlen@plt>
  movslq %eax,%rdi
  add    %rbx,%rdi
  mov    %rbp,%rsi
  call   <strcpy@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *strcat(char *destination, const char *source) {
  strcpy(destination + cgc_strlen(destination), source);
  return destination;
}
```
user: What is the c source code of the assembly code below:

```asm
<strcat>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x8,%rsp
  call   <cgc_strlen@plt>
  mov    %rbp,%rsi
  movslq %eax,%rdi
  add    %r12,%rdi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strcat(char *destination, const char *source) {
  strcpy(destination + cgc_strlen(destination), source);
  return destination;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<strcat>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x8,%rsp
  call   <cgc_strlen@plt>
  mov    %rbp,%rsi
  movslq %eax,%rdi
  add    %r12,%rdi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strcat(char *destination, const char *source) {
  strcpy(destination + cgc_strlen(destination), source);
  return destination;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rsi
  lea    D60(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <bi_input@plt>
  mov    %eax,-0x44(%rbp)
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x48(%rbp),%rax
  mov    %rax,%rsi
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x4c(%rbp),%rax
  mov    %rax,%rsi
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4c(%rbp),%edx
  mov    -0x48(%rbp),%ecx
  mov    -0x44(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <invert@plt>
  mov    %eax,%edx
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %edx,%edi
  call   <to2@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D0
type_guessed: string
name: parse_data
arguments: tag: D60
type_guessed: string
name: parse_data
arguments: tag: D39
type_guessed: string
name: parse_data
arguments: tag: D52
type_guessed: string
name: parse_data
arguments: tag: D50
type_guessed: string
name: parse_data
arguments: tag: D16
type_guessed: string

tool: D0: input x: 

tool: D60: %s

tool: D39: input p: 

tool: D52: %d

tool: D50: input n: 

tool: D16: invert result is :

assistant: ```c
main() {
  int n, p;
  int x;
  char c[20];
  char s[20];
  char d[20];
  unsigned result;
  printf("input x: ");
  scanf("%s", c);
  x = bi_input(c);
  printf("input p: ");
  scanf("%d", &p);
  printf("input n: ");
  scanf("%d", &n);
  printf("invert result is :");
  to2(invert(x, p, n), s);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    D46(%rip),%rdi
  call   <printf@plt>
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rsi
  lea    D13(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <bi_input@plt>
  mov    %eax,%ebx
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D34(%rip),%rbp
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D29(%rip),%rdi
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%edx
  mov    0xc(%rsp),%esi
  mov    %ebx,%edi
  call   <invert@plt>
  mov    %eax,%edi
  lea    0x30(%rsp),%rsi
  call   <to2@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D46
datatype: string
name: parse_data
arguments: data_tag: D13
datatype: string
name: parse_data
arguments: data_tag: D0
datatype: string
name: parse_data
arguments: data_tag: D34
datatype: string
name: parse_data
arguments: data_tag: D29
datatype: string
name: parse_data
arguments: data_tag: D19
datatype: string

tool: D46: input x: 

tool: D13: %s

tool: D0: input p: 

tool: D34: %d

tool: D29: input n: 

tool: D19: invert result is :

assistant: ```c
main() {
  int n, p;
  int x;
  char c[20];
  char s[20];
  char d[20];
  unsigned result;
  printf("input x: ");
  scanf("%s", c);
  x = bi_input(c);
  printf("input p: ");
  scanf("%d", &p);
  printf("input n: ");
  scanf("%d", &n);
  printf("invert result is :");
  to2(invert(x, p, n), s);
}
```
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   %r12
  lea    D27(%rip),%rdi
  lea    D17(%rip),%r12
  push   %rbp
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D37(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <bi_input@plt>
  lea    D1(%rip),%rdi
  mov    %eax,%ebp
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D38(%rip),%rdi
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  mov    0x8(%rsp),%edx
  mov    %ebp,%edi
  call   <invert@plt>
  lea    0x30(%rsp),%rsi
  mov    %eax,%edi
  call   <to2@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x58,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D27
datatype: string
name: retrieve_data
arguments: tag: D17
datatype: string
name: retrieve_data
arguments: tag: D37
datatype: string
name: retrieve_data
arguments: tag: D1
datatype: string
name: retrieve_data
arguments: tag: D38
datatype: string
name: retrieve_data
arguments: tag: D16
datatype: string

tool: D27: input x: 

tool: D17: %d

tool: D37: %s

tool: D1: input p: 

tool: D38: input n: 

tool: D16: invert result is :

assistant: ```c
main() {
  int n, p;
  int x;
  char c[20];
  char s[20];
  char d[20];
  unsigned result;
  printf("input x: ");
  scanf("%s", c);
  x = bi_input(c);
  printf("input p: ");
  scanf("%d", &p);
  printf("input n: ");
  scanf("%d", &n);
  printf("invert result is :");
  to2(invert(x, p, n), s);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  lea    D4(%rip),%rdi
  lea    D17(%rip),%r12
  push   %rbp
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <bi_input@plt>
  lea    D33(%rip),%rdi
  mov    %eax,%ebp
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D28(%rip),%rdi
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D32(%rip),%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  mov    0x8(%rsp),%edx
  mov    %ebp,%edi
  call   <invert@plt>
  lea    0x30(%rsp),%rsi
  mov    %eax,%edi
  call   <to2@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x58,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D4
type_guessed: string
name: retrieve_data
arguments: label: D17
type_guessed: string
name: retrieve_data
arguments: label: D38
type_guessed: string
name: retrieve_data
arguments: label: D33
type_guessed: string
name: retrieve_data
arguments: label: D28
type_guessed: string
name: retrieve_data
arguments: label: D32
type_guessed: string

tool: D4: input x: 

tool: D17: %d

tool: D38: %s

tool: D33: input p: 

tool: D28: input n: 

tool: D32: invert result is :

assistant: ```c
main() {
  int n, p;
  int x;
  char c[20];
  char s[20];
  char d[20];
  unsigned result;
  printf("input x: ");
  scanf("%s", c);
  x = bi_input(c);
  printf("input p: ");
  scanf("%d", &p);
  printf("input n: ");
  scanf("%d", &n);
  printf("invert result is :");
  to2(invert(x, p, n), s);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<RandomFunc>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsi,%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    (%rdi),%eax
  add    $0x136ff696,%eax
  mov    %eax,0x4(%rsp)
  mov    0x4(%rdi),%esi
  test   %esi,%esi
  je     L3
  mov    $0x0,%edx
  lea    0x4(%rsp),%r8
L20:
  mov    0x4(%rsp),%eax
  cmp    %edx,%eax
  jae    L48
  mov    %edx,%edi
  add    %eax,%eax
  mov    %eax,%r10d
  shr    $0x18,%r10d
  mov    %eax,0x4(%rsp,%rdi,4)
  lea    (%r8,%rdi,4),%rdi
  mov    %r10b,0x1(%rdi)
  movzbl %ah,%eax
  mov    %al,0x3(%rdi)
L23:
  add    $0x2,%edx
  cmp    %esi,%edx
  jb     L20
L3:
  mov    0x4(%rsp),%eax
  add    $0x5ca83cf8,%eax
  mov    %eax,(%rcx)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x18,%rsp
  ret
L48:
  mov    %edx,%edi
  shl    $0x2,%eax
  mov    %eax,0x4(%rsp,%rdi,4)
  jmp    L23
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy11;
  char copy12;
  {
    state[0UL] = (input[0UL] + 51238316UL) + 274866410U;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      if (state[0UL] < local1) {
        if (state[0UL] == local1) {
          state[0UL] = state[local1] + state[0UL];
          copy11 = *((unsigned short *)(&state[0UL]) + 0);
          *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = copy11;
        } else {
          state[local1] = state[0UL] + state[local1];
          copy12 = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 3);
          *((char *)(&state[local1]) + 3) = copy12;
        }
      } else {
        state[local1] = state[0UL] + state[local1];
        state[local1] += state[0UL];
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] + 468445703UL) + 1086083825U;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  sub    $0x18,%rsp
  mov    0x4(%rdi),%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    (%rdi),%eax
  add    $0x136ff696,%eax
  mov    %eax,0x4(%rsp)
  test   %ecx,%ecx
  je     L44
  xor    %edx,%edx
  lea    0x4(%rsp),%r8
L31:
  cmp    %edx,%eax
  jae    L8
  add    %eax,%eax
  mov    %edx,%edi
  mov    %eax,%r10d
  mov    %eax,0x4(%rsp,%rdi,4)
  lea    (%r8,%rdi,4),%rdi
  shr    $0x18,%r10d
  mov    %ah,0x3(%rdi)
  mov    %r10b,0x1(%rdi)
  mov    0x4(%rsp),%eax
L46:
  add    $0x2,%edx
  cmp    %ecx,%edx
  jb     L31
L44:
  add    $0x5ca83cf8,%eax
  mov    %eax,(%rsi)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  add    $0x18,%rsp
  ret
  nopl   0x0(%rax)
L8:
  shl    $0x2,%eax
  mov    %edx,%edi
  mov    %eax,0x4(%rsp,%rdi,4)
  jmp    L46
L48:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy11;
  char copy12;
  {
    state[0UL] = (input[0UL] + 51238316UL) + 274866410U;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      if (state[0UL] < local1) {
        if (state[0UL] == local1) {
          state[0UL] = state[local1] + state[0UL];
          copy11 = *((unsigned short *)(&state[0UL]) + 0);
          *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = copy11;
        } else {
          state[local1] = state[0UL] + state[local1];
          copy12 = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 3);
          *((char *)(&state[local1]) + 3) = copy12;
        }
      } else {
        state[local1] = state[0UL] + state[local1];
        state[local1] += state[0UL];
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] + 468445703UL) + 1086083825U;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<RandomFunc>:
  endbr64
  sub    $0x18,%rsp
  mov    0x4(%rdi),%ecx
  mov    %rsi,%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    (%rdi),%eax
  add    $0x136ff696,%eax
  test   %ecx,%ecx
  je     L30
  lea    0x0(,%rax,4),%esi
  mov    %esi,0x4(%rsp)
  cmp    $0x2,%ecx
  jbe    L32
  cmp    $0x1,%esi
  jbe    L98
  shl    $0x4,%eax
  mov    %eax,0xc(%rsp)
L65:
  cmp    $0x4,%ecx
  jbe    L30
  mov    $0x4,%esi
  lea    0x4(%rsp),%r8
L73:
  cmp    %esi,%eax
  jae    L3
  add    %eax,%eax
  mov    %esi,%edi
  mov    %eax,%r10d
  mov    %eax,0x4(%rsp,%rdi,4)
  lea    (%r8,%rdi,4),%rdi
  shr    $0x18,%r10d
  mov    %ah,0x3(%rdi)
  mov    %r10b,0x1(%rdi)
  mov    0x4(%rsp),%eax
L7:
  add    $0x2,%esi
  cmp    %ecx,%esi
  jb     L73
  nopl   0x0(%rax,%rax,1)
L30:
  add    $0x5ca83cf8,%eax
  mov    %eax,(%rdx)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L95
  add    $0x18,%rsp
  ret
  nopl   0x0(%rax)
L32:
  mov    %esi,%eax
  jmp    L30
  nopl   0x0(%rax)
L98:
  shl    $0x3,%eax
  mov    %eax,%edi
  mov    %eax,0xc(%rsp)
  sar    $0x18,%edi
  mov    %ah,0xf(%rsp)
  mov    %esi,%eax
  mov    %dil,0xd(%rsp)
  jmp    L65
  nopl   0x0(%rax)
L3:
  shl    $0x2,%eax
  mov    %esi,%edi
  mov    %eax,0x4(%rsp,%rdi,4)
  jmp    L7
L95:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy11;
  char copy12;
  {
    state[0UL] = (input[0UL] + 51238316UL) + 274866410U;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      if (state[0UL] < local1) {
        if (state[0UL] == local1) {
          state[0UL] = state[local1] + state[0UL];
          copy11 = *((unsigned short *)(&state[0UL]) + 0);
          *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
          *((unsigned short *)(&state[0UL]) + 1) = copy11;
        } else {
          state[local1] = state[0UL] + state[local1];
          copy12 = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 3);
          *((char *)(&state[local1]) + 3) = copy12;
        }
      } else {
        state[local1] = state[0UL] + state[local1];
        state[local1] += state[0UL];
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] + 468445703UL) + 1086083825U;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<choose>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <imin@plt>
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jns    L33
  movl   $0x0,-0xc(%rbp)
  jmp    L20
L33:
  cmpl   $0x0,-0x8(%rbp)
  jne    L18
  movl   $0x1,-0xc(%rbp)
  jmp    L20
L18:
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <imax@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x2,-0x10(%rbp)
  jmp    L6
L58:
  mov    -0x4(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  imul   -0xc(%rbp),%eax
  cltd
  idivl  -0x10(%rbp)
  mov    %eax,-0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
L6:
  mov    -0x10(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L58
L20:
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
int choose(unsigned int n, unsigned int k) {
  int i;
  int mn;
  int mx;
  int value;
  mn = imin(k, n - k);
  if (mn < 0) {
    value = 0;
  } else if (mn == 0) {
    value = 1;
  } else {
    mx = imax(k, n - k);
    value = mx + 1;
    for (i = 2; i <= mn; i++) {
      value = (value * (mx + i)) / i;
    }
  }
  return value;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<choose>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebp
  sub    %esi,%edi
  mov    %edi,%r12d
  mov    %edi,%esi
  mov    %ebp,%edi
  call   <imin@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  test   %ebx,%ebx
  js     L16
  mov    $0x1,%eax
  jne    L2
L16:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <imax@plt>
  mov    %eax,%esi
  lea    0x1(%rax),%eax
  cmp    $0x1,%ebx
  jle    L16
  add    $0x1,%ebx
  mov    $0x2,%ecx
L18:
  lea    (%rsi,%rcx,1),%edx
  imul   %edx,%eax
  cltd
  idiv   %ecx
  add    $0x1,%ecx
  cmp    %ebx,%ecx
  jne    L18
  jmp    L16
```

assistant: ```c
int choose(unsigned int n, unsigned int k) {
  int i;
  int mn;
  int mx;
  int value;
  mn = imin(k, n - k);
  if (mn < 0) {
    value = 0;
  } else if (mn == 0) {
    value = 1;
  } else {
    mx = imax(k, n - k);
    value = mx + 1;
    for (i = 2; i <= mn; i++) {
      value = (value * (mx + i)) / i;
    }
  }
  return value;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<choose>:
  endbr64
  push   %r12
  sub    %esi,%edi
  push   %rbp
  mov    %esi,%ebp
  mov    %edi,%r12d
  mov    %edi,%esi
  push   %rbx
  mov    %ebp,%edi
  call   <imin@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  test   %ebx,%ebx
  js     L10
  mov    $0x1,%eax
  jne    L27
L10:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L27:
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <imax@plt>
  mov    %eax,%esi
  lea    0x1(%rax),%eax
  cmp    $0x1,%ebx
  je     L10
  add    $0x1,%ebx
  mov    $0x2,%ecx
  nopl   0x0(%rax)
L2:
  lea    (%rcx,%rsi,1),%edx
  imul   %edx,%eax
  cltd
  idiv   %ecx
  add    $0x1,%ecx
  cmp    %ebx,%ecx
  jne    L2
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int choose(unsigned int n, unsigned int k) {
  int i;
  int mn;
  int mx;
  int value;
  mn = imin(k, n - k);
  if (mn < 0) {
    value = 0;
  } else if (mn == 0) {
    value = 1;
  } else {
    mx = imax(k, n - k);
    value = mx + 1;
    for (i = 2; i <= mn; i++) {
      value = (value * (mx + i)) / i;
    }
  }
  return value;
}
```
user: What is the c source code of the assembly code below:

```asm
<choose>:
  endbr64
  push   %r12
  sub    %esi,%edi
  push   %rbp
  mov    %esi,%ebp
  mov    %edi,%r12d
  mov    %edi,%esi
  push   %rbx
  mov    %ebp,%edi
  call   <imin@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  test   %ebx,%ebx
  js     L8
  mov    $0x1,%eax
  jne    L35
L8:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L35:
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <imax@plt>
  mov    %eax,%esi
  lea    0x1(%rax),%eax
  cmp    $0x1,%ebx
  je     L8
  add    $0x1,%ebx
  mov    $0x2,%ecx
  nopl   0x0(%rax)
L22:
  lea    (%rcx,%rsi,1),%edx
  imul   %edx,%eax
  cltd
  idiv   %ecx
  add    $0x1,%ecx
  cmp    %ecx,%ebx
  jne    L22
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int choose(unsigned int n, unsigned int k) {
  int i;
  int mn;
  int mx;
  int value;
  mn = imin(k, n - k);
  if (mn < 0) {
    value = 0;
  } else if (mn == 0) {
    value = 1;
  } else {
    mx = imax(k, n - k);
    value = mx + 1;
    for (i = 2; i <= mn; i++) {
      value = (value * (mx + i)) / i;
    }
  }
  return value;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<set_nonce_in_tweak>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  and    $0xf0,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  sar    $0x4,%eax
  xor    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  shl    $0x4,%eax
  movzbl %al,%edx
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  sar    $0x4,%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  xor    %ecx,%edx
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  shl    $0x4,%eax
  movzbl %al,%edx
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%eax
  sar    $0x4,%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  xor    %ecx,%edx
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%eax
  shl    $0x4,%eax
  movzbl %al,%edx
  mov    -0x10(%rbp),%rax
  add    $0xc,%rax
  mov    (%rax),%eax
  sar    $0x4,%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%rax
  add    $0xc,%rax
  xor    %ecx,%edx
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  add    $0xc,%rax
  mov    (%rax),%eax
  shl    $0x4,%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%rax
  lea    0x10(%rax),%rdx
  movzbl %cl,%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void set_nonce_in_tweak(uint8_t *tweak, const uint8_t *nonce) {
  tweak[0] = (tweak[0] & 0xf0) ^ (nonce[0] >> 4);
  tweak[1] = (nonce[0] & 0xf) << 4 ^ (nonce[1] >> 4);
  tweak[2] = (nonce[1] & 0xf) << 4 ^ (nonce[2] >> 4);
  tweak[3] = (nonce[2] & 0xf) << 4 ^ (nonce[3] >> 4);
  tweak[4] = (nonce[3] & 0xf) << 4;
}
```