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

```asm
<startsWith_nocase>:
  endbr64
  test   %rdi,%rdi
  je     L35
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  test   %rsi,%rsi
  je     L38
  mov    %rdi,%rbx
  cmp    %rsi,%rdi
  je     L9
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  xor    %r8d,%r8d
  cmp    %rax,%r12
  jb     L22
  test   %rax,%rax
  je     L9
  xor    %esi,%esi
  jmp    L76
  nopw   0x0(%rax,%rax,1)
L11:
  add    $0x1,%rsi
  cmp    %rsi,%rax
  je     L9
L76:
  movzbl (%rbx,%rsi,1),%ecx
  lea    -0x41(%rcx),%edi
  lea    0x20(%rcx),%edx
  cmp    $0x1a,%dil
  cmovb  %edx,%ecx
  movzbl 0x0(%rbp,%rsi,1),%edx
  lea    -0x41(%rdx),%r8d
  lea    0x20(%rdx),%edi
  cmp    $0x1a,%r8b
  cmovb  %edi,%edx
  cmp    %dl,%cl
  je     L11
L38:
  xor    %r8d,%r8d
L22:
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L9:
  mov    $0x1,%r8d
  pop    %rbx
  pop    %rbp
  mov    %r8d,%eax
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L35:
  xor    %eax,%eax
  ret
```

assistant: ```c
int startsWith_nocase(const char *str, const char *with) {
  if (str == NULL || with == NULL)
    return 0;
  if (str == with)
    return 1;
  size_t l1 = strlen(str);
  size_t l2 = strlen(with);
  if (l1 < l2)
    return 0;
  for (int i = 0; i < l2; i++) {
    char s1 = str[i];
    if (s1 >= 'A' && s1 <= 'Z')
      s1 += ' ';
    char s2 = with[i];
    if (s2 >= 'A' && s2 <= 'Z')
      s2 += ' ';
    if (s1 != s2) {
      return 0;
    }
  }
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<startsWith_nocase>:
  endbr64
  test   %rdi,%rdi
  je     L78
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  test   %rsi,%rsi
  je     L45
  mov    %rdi,%rbx
  cmp    %rsi,%rdi
  je     L22
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  xor    %r8d,%r8d
  cmp    %rax,%r12
  jb     L39
  test   %rax,%rax
  je     L22
  xor    %esi,%esi
  jmp    L73
  nopw   0x0(%rax,%rax,1)
L53:
  add    $0x1,%rsi
  cmp    %rsi,%rax
  je     L22
L73:
  movzbl (%rbx,%rsi,1),%ecx
  lea    -0x41(%rcx),%edi
  lea    0x20(%rcx),%edx
  cmp    $0x1a,%dil
  cmovb  %edx,%ecx
  movzbl 0x0(%rbp,%rsi,1),%edx
  lea    -0x41(%rdx),%r8d
  lea    0x20(%rdx),%edi
  cmp    $0x1a,%r8b
  cmovb  %edi,%edx
  cmp    %dl,%cl
  je     L53
L45:
  xor    %r8d,%r8d
L39:
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L22:
  mov    $0x1,%r8d
  pop    %rbx
  pop    %rbp
  mov    %r8d,%eax
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L78:
  xor    %eax,%eax
  ret
```

assistant: ```c
int startsWith_nocase(const char *str, const char *with) {
  if (str == NULL || with == NULL)
    return 0;
  if (str == with)
    return 1;
  size_t l1 = strlen(str);
  size_t l2 = strlen(with);
  if (l1 < l2)
    return 0;
  for (int i = 0; i < l2; i++) {
    char s1 = str[i];
    if (s1 >= 'A' && s1 <= 'Z')
      s1 += ' ';
    char s2 = with[i];
    if (s2 >= 'A' && s2 <= 'Z')
      s2 += ' ';
    if (s1 != s2) {
      return 0;
    }
  }
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  movl   $0x7b,-0x24(%rbp)
  movl   $0x0,-0x28(%rbp)
  jmp    L0
L3:
  mov    $0x0,%eax
  call   <omp_get_num_teams@plt>
  mov    %eax,%r12d
  mov    $0x0,%eax
  call   <omp_get_num_threads@plt>
  mov    %eax,%r13d
  mov    $0x0,%eax
  call   <omp_get_thread_num@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <omp_get_team_num@plt>
  mov    %eax,%esi
  mov    -0x24(%rbp),%edx
  mov    -0x28(%rbp),%eax
  sub    $0x8,%rsp
  push   %r12
  mov    %r13d,%r9d
  mov    %edx,%r8d
  mov    %eax,%ecx
  mov    %ebx,%edx
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x10,%rsp
  addl   $0x1,-0x28(%rbp)
L0:
  cmpl   $0x23,-0x28(%rbp)
  jle    L3
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  lea    -0x18(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D2
type_infered: string
name: disassemble_data
arguments: tag: D19
type_infered: string

tool: D2: team %d thread %d doing iteration %d and x is %d num_threads %d num_teams is %d


tool: D19: end of function foo

assistant: ```c
void foo() {
  int x = 123;
  int i;
#pragma omp target teams distribute num_teams(7)
  for (i = 0; i < 36; i++) {
    printf("team %d thread %d doing iteration %d and x is %d num_threads %d num_teams is %d\n", (int)omp_get_team_num(), (int)omp_get_thread_num(), i, x, (int)omp_get_num_threads(), (int)omp_get_num_teams());
  }
  printf("end of function foo\n");
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<foo>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%ebp
  lea    D6(%rip),%r13
L3:
  mov    $0x0,%eax
  call   <omp_get_num_teams@plt>
  mov    %rax,%r14
  mov    $0x0,%eax
  call   <omp_get_num_threads@plt>
  mov    %rax,%r12
  mov    $0x0,%eax
  call   <omp_get_thread_num@plt>
  mov    %rax,%rbx
  mov    $0x0,%eax
  call   <omp_get_team_num@plt>
  sub    $0x8,%rsp
  push   %r14
  mov    %r12d,%r9d
  mov    $0x7b,%r8d
  mov    %ebp,%ecx
  mov    %ebx,%edx
  mov    %eax,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebp
  add    $0x10,%rsp
  cmp    $0x24,%ebp
  jne    L3
  lea    D18(%rip),%rdi
  call   <puts@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D6
type: string
name: unpack_data
arguments: data_label: D18
type: string

tool: D6: team %d thread %d doing iteration %d and x is %d num_threads %d num_teams is %d


tool: D18: end of function foo

assistant: ```c
void foo() {
  int x = 123;
  int i;
#pragma omp target teams distribute num_teams(7)
  for (i = 0; i < 36; i++) {
    printf("team %d thread %d doing iteration %d and x is %d num_threads %d num_teams is %d\n", (int)omp_get_team_num(), (int)omp_get_thread_num(), i, x, (int)omp_get_num_threads(), (int)omp_get_num_teams());
  }
  printf("end of function foo\n");
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<foo>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  lea    D9(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  nopl   (%rax)
L9:
  xor    %eax,%eax
  call   <omp_get_num_teams@plt>
  mov    %rax,%r14
  xor    %eax,%eax
  call   <omp_get_num_threads@plt>
  mov    %rax,%r13
  xor    %eax,%eax
  call   <omp_get_thread_num@plt>
  mov    %rax,%rbx
  xor    %eax,%eax
  call   <omp_get_team_num@plt>
  sub    $0x8,%rsp
  mov    %ebp,%ecx
  mov    %ebx,%edx
  push   %r14
  mov    %eax,%esi
  mov    %r13d,%r9d
  mov    $0x7b,%r8d
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <printf@plt>
  pop    %rax
  pop    %rdx
  cmp    $0x24,%ebp
  jne    L9
  pop    %rbx
  lea    D5(%rip),%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    10a0 <puts@plt>
```

tool_calls: name: read_data
arguments: tag: D9
type: string
name: read_data
arguments: tag: D5
type: string

tool: D9: team %d thread %d doing iteration %d and x is %d num_threads %d num_teams is %d


tool: D5: end of function foo

assistant: ```c
void foo() {
  int x = 123;
  int i;
#pragma omp target teams distribute num_teams(7)
  for (i = 0; i < 36; i++) {
    printf("team %d thread %d doing iteration %d and x is %d num_threads %d num_teams is %d\n", (int)omp_get_team_num(), (int)omp_get_thread_num(), i, x, (int)omp_get_num_threads(), (int)omp_get_num_teams());
  }
  printf("end of function foo\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<foo>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  lea    D18(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  nopl   (%rax)
L2:
  xor    %eax,%eax
  call   <omp_get_num_teams@plt>
  mov    %rax,%r14
  xor    %eax,%eax
  call   <omp_get_num_threads@plt>
  mov    %rax,%r13
  xor    %eax,%eax
  call   <omp_get_thread_num@plt>
  mov    %rax,%rbx
  xor    %eax,%eax
  call   <omp_get_team_num@plt>
  sub    $0x8,%rsp
  mov    %ebp,%ecx
  mov    %ebx,%edx
  push   %r14
  mov    %eax,%esi
  mov    %r13d,%r9d
  mov    $0x7b,%r8d
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <printf@plt>
  pop    %rax
  pop    %rdx
  cmp    $0x24,%ebp
  jne    L2
  pop    %rbx
  lea    D6(%rip),%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    10a0 <puts@plt>
```

tool_calls: name: disassemble_data
arguments: label: D18
type_guessed: string
name: disassemble_data
arguments: label: D6
type_guessed: string

tool: D18: team %d thread %d doing iteration %d and x is %d num_threads %d num_teams is %d


tool: D6: end of function foo

assistant: ```c
void foo() {
  int x = 123;
  int i;
#pragma omp target teams distribute num_teams(7)
  for (i = 0; i < 36; i++) {
    printf("team %d thread %d doing iteration %d and x is %d num_threads %d num_teams is %d\n", (int)omp_get_team_num(), (int)omp_get_thread_num(), i, x, (int)omp_get_num_threads(), (int)omp_get_num_teams());
  }
  printf("end of function foo\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<crl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %esi,-0x38(%rbp)
  mov    %edx,-0x3c(%rbp)
  mov    %ecx,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x17,-0x14(%rbp)
  movl   $0x1c,-0x10(%rbp)
  movl   $0x21,-0xc(%rbp)
  movl   $0x5304,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffffb21642c9,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x4,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x4,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x3e0f83e1,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x3,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x18(%rbp)
  lea    -0x2c(%rbp),%rcx
  lea    -0x30(%rbp),%rdx
  mov    -0x20(%rbp),%eax
  mov    $0x17,%esi
  mov    %eax,%edi
  call   <egcd@plt>
  imul   -0x20(%rbp),%eax
  cltd
  idivl  -0x24(%rbp)
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  imul   -0x34(%rbp),%eax
  add    %eax,-0x28(%rbp)
  lea    -0x2c(%rbp),%rcx
  lea    -0x30(%rbp),%rdx
  mov    -0x1c(%rbp),%eax
  mov    $0x1c,%esi
  mov    %eax,%edi
  call   <egcd@plt>
  imul   -0x1c(%rbp),%eax
  cltd
  idivl  -0x24(%rbp)
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  imul   -0x38(%rbp),%eax
  add    %eax,-0x28(%rbp)
  lea    -0x2c(%rbp),%rcx
  lea    -0x30(%rbp),%rdx
  mov    -0x18(%rbp),%eax
  mov    $0x21,%esi
  mov    %eax,%edi
  call   <egcd@plt>
  imul   -0x18(%rbp),%eax
  cltd
  idivl  -0x24(%rbp)
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  imul   -0x3c(%rbp),%eax
  add    %eax,-0x28(%rbp)
  mov    -0x28(%rbp),%eax
  sub    -0x40(%rbp),%eax
  mov    %eax,%edx
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  cltd
  idivl  -0x24(%rbp)
  mov    %edx,-0x28(%rbp)
  cmpl   $0x0,-0x28(%rbp)
  jne    L2
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x28(%rbp)
L2:
  mov    -0x28(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
int crl(int p, int e, int i, int d) {
  int next, x, y;
  int m[] = {23, 28, 33};
  int M = 23 * 28 * 33;
  int M1 = M / 23;
  int M2 = M / 28;
  int M3 = M / 33;
  next += (M1 * egcd(M1, 23, &x, &y) % M + M) * p;
  next += (M2 * egcd(M2, 28, &x, &y) % M + M) * e;
  next += (M3 * egcd(M3, 33, &x, &y) % M + M) * i;
  next = (next - d + M) % M;
  if (next == 0)
    next = M;
  return next;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<crl>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%r13d
  mov    %esi,%ebx
  mov    %edx,%r12d
  mov    %ecx,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r15
  mov    %rsp,%r14
  mov    %r15,%rcx
  mov    %r14,%rdx
  mov    $0x17,%esi
  mov    $0x39c,%edi
  call   <egcd@plt>
  imul   $0x39c,%eax,%eax
  movslq %eax,%rdx
  imul   $0x31571165,%rdx,%rdx
  sar    $0x2c,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x5304,%edx,%edx
  sub    %edx,%eax
  add    $0x5304,%eax
  imul   %eax,%r13d
  mov    %r15,%rcx
  mov    %r14,%rdx
  mov    $0x1c,%esi
  mov    $0x2f7,%edi
  call   <egcd@plt>
  imul   $0x2f7,%eax,%eax
  movslq %eax,%rdx
  imul   $0x31571165,%rdx,%rdx
  sar    $0x2c,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x5304,%edx,%edx
  sub    %edx,%eax
  add    $0x5304,%eax
  imul   %eax,%ebx
  add    %r13d,%ebx
  mov    %r15,%rcx
  mov    %r14,%rdx
  mov    $0x21,%esi
  mov    $0x284,%edi
  call   <egcd@plt>
  imul   $0x284,%eax,%eax
  movslq %eax,%rdx
  imul   $0x31571165,%rdx,%rdx
  sar    $0x2c,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x5304,%edx,%edx
  sub    %edx,%eax
  add    $0x5304,%eax
  imul   %r12d,%eax
  sub    %ebp,%ebx
  lea    0x5304(%rbx,%rax,1),%edx
  movslq %edx,%rax
  imul   $0x31571165,%rax,%rax
  sar    $0x2c,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x5304,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    $0x5304,%edx
  cmove  %edx,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int crl(int p, int e, int i, int d) {
  int next, x, y;
  int m[] = {23, 28, 33};
  int M = 23 * 28 * 33;
  int M1 = M / 23;
  int M2 = M / 28;
  int M3 = M / 33;
  next += (M1 * egcd(M1, 23, &x, &y) % M + M) * p;
  next += (M2 * egcd(M2, 28, &x, &y) % M + M) * e;
  next += (M3 * egcd(M3, 33, &x, &y) % M + M) * i;
  next = (next - d + M) % M;
  if (next == 0)
    next = M;
  return next;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<crl>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %edi,%ebx
  mov    $0x39c,%edi
  sub    $0x28,%rsp
  mov    %esi,0xc(%rsp)
  lea    0x14(%rsp),%r14
  lea    0x10(%rsp),%r15
  mov    $0x17,%esi
  mov    %r14,%rcx
  mov    %r15,%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <egcd@plt>
  mov    %r14,%rcx
  mov    $0x1c,%esi
  mov    $0x2f7,%edi
  imul   $0x39c,%eax,%r12d
  movslq %r12d,%rax
  mov    %r12d,%edx
  imul   $0x31571165,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x2c,%rax
  sub    %edx,%eax
  mov    %r15,%rdx
  imul   $0x5304,%eax,%eax
  sub    %eax,%r12d
  call   <egcd@plt>
  add    $0x5304,%r12d
  mov    %r14,%rcx
  mov    $0x21,%esi
  imul   $0x2f7,%eax,%eax
  imul   %ebx,%r12d
  movslq %eax,%rdx
  mov    %eax,%edi
  imul   $0x31571165,%rdx,%rdx
  sar    $0x1f,%edi
  sar    $0x2c,%rdx
  sub    %edi,%edx
  mov    $0x284,%edi
  imul   $0x5304,%edx,%edx
  sub    %edx,%eax
  mov    %r15,%rdx
  lea    0x5304(%rax),%ebx
  imul   0xc(%rsp),%ebx
  call   <egcd@plt>
  imul   $0x284,%eax,%eax
  add    %r12d,%ebx
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x31571165,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x2c,%rdx
  sub    %ecx,%edx
  imul   $0x5304,%edx,%edx
  sub    %edx,%eax
  add    $0x5304,%eax
  imul   %ebp,%eax
  sub    %r13d,%ebx
  lea    0x5304(%rbx,%rax,1),%edx
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x31571165,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x2c,%rax
  sub    %ecx,%eax
  imul   $0x5304,%eax,%ecx
  mov    %edx,%eax
  mov    $0x5304,%edx
  sub    %ecx,%eax
  cmove  %edx,%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int crl(int p, int e, int i, int d) {
  int next, x, y;
  int m[] = {23, 28, 33};
  int M = 23 * 28 * 33;
  int M1 = M / 23;
  int M2 = M / 28;
  int M3 = M / 33;
  next += (M1 * egcd(M1, 23, &x, &y) % M + M) * p;
  next += (M2 * egcd(M2, 28, &x, &y) % M + M) * e;
  next += (M3 * egcd(M3, 33, &x, &y) % M + M) * i;
  next = (next - d + M) % M;
  if (next == 0)
    next = M;
  return next;
}
```
user: Translate the below assembly code into C programming language:

```asm
<crl>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %edi,%ebx
  mov    $0x39c,%edi
  sub    $0x28,%rsp
  mov    %esi,0xc(%rsp)
  lea    0x14(%rsp),%r14
  lea    0x10(%rsp),%r15
  mov    $0x17,%esi
  mov    %r14,%rcx
  mov    %r15,%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <egcd@plt>
  mov    %r14,%rcx
  mov    $0x1c,%esi
  mov    $0x2f7,%edi
  imul   $0x39c,%eax,%r12d
  movslq %r12d,%rax
  mov    %r12d,%edx
  imul   $0x31571165,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x2c,%rax
  sub    %edx,%eax
  mov    %r15,%rdx
  imul   $0x5304,%eax,%eax
  sub    %eax,%r12d
  call   <egcd@plt>
  add    $0x5304,%r12d
  mov    %r14,%rcx
  mov    $0x21,%esi
  imul   $0x2f7,%eax,%eax
  imul   %ebx,%r12d
  movslq %eax,%rdx
  mov    %eax,%edi
  imul   $0x31571165,%rdx,%rdx
  sar    $0x1f,%edi
  sar    $0x2c,%rdx
  sub    %edi,%edx
  mov    $0x284,%edi
  imul   $0x5304,%edx,%edx
  sub    %edx,%eax
  mov    %r15,%rdx
  lea    0x5304(%rax),%ebx
  imul   0xc(%rsp),%ebx
  call   <egcd@plt>
  imul   $0x284,%eax,%eax
  add    %r12d,%ebx
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x31571165,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x2c,%rdx
  sub    %ecx,%edx
  imul   $0x5304,%edx,%edx
  sub    %edx,%eax
  add    $0x5304,%eax
  imul   %ebp,%eax
  sub    %r13d,%ebx
  lea    0x5304(%rbx,%rax,1),%edx
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x31571165,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x2c,%rax
  sub    %ecx,%eax
  imul   $0x5304,%eax,%ecx
  mov    %edx,%eax
  mov    $0x5304,%edx
  sub    %ecx,%eax
  cmove  %edx,%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int crl(int p, int e, int i, int d) {
  int next, x, y;
  int m[] = {23, 28, 33};
  int M = 23 * 28 * 33;
  int M1 = M / 23;
  int M2 = M / 28;
  int M3 = M / 33;
  next += (M1 * egcd(M1, 23, &x, &y) % M + M) * p;
  next += (M2 * egcd(M2, 28, &x, &y) % M + M) * e;
  next += (M3 * egcd(M3, 33, &x, &y) % M + M) * i;
  next = (next - d + M) % M;
  if (next == 0)
    next = M;
  return next;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L8
L12:
  mov    -0x14(%rbp),%eax
  mov    -0x10(%rbp),%edx
  sub    %edx,%eax
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0xc(%rbp)
L8:
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  jg     L12
  mov    -0x14(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D25
type_guessed: string
name: disassemble_data
arguments: data_label: D5
type_guessed: string
name: disassemble_data
arguments: data_label: D15
type_guessed: string

tool: D25: Enter N and K:

tool: D5: %d %d

tool: D15: Result of the division is %d and the residue is %d

assistant: ```c
int main() {
  int n, k, i;
  printf("Enter N and K:\n");
  scanf("%d %d", &n, &k);
  i = 0;
  while (n > k) {
    n -= k;
    i++;
  }
  printf("Result of the division is %d and the residue is %d", i, n);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D14(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%edx
  mov    0x4(%rsp),%eax
  cmp    %eax,%edx
  jle    L18
  mov    $0x0,%esi
L35:
  sub    %eax,%edx
  add    $0x1,%esi
  cmp    %eax,%edx
  jg     L35
  mov    %edx,(%rsp)
L6:
  lea    D23(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L18:
  mov    $0x0,%esi
  jmp    L6
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D14
datatype: string
name: read_data
arguments: tag: D13
datatype: string
name: read_data
arguments: tag: D23
datatype: string

tool: D14: Enter N and K:

tool: D13: %d %d

tool: D23: Result of the division is %d and the residue is %d

assistant: ```c
int main() {
  int n, k, i;
  printf("Enter N and K:\n");
  scanf("%d %d", &n, &k);
  i = 0;
  while (n > k) {
    n -= k;
    i++;
  }
  printf("Result of the division is %d and the residue is %d", i, n);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D2(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D12(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%edx
  mov    0x4(%rsp),%eax
  xor    %esi,%esi
  cmp    %eax,%edx
  jle    L0
  nopl   (%rax)
L23:
  sub    %eax,%edx
  add    $0x1,%esi
  cmp    %eax,%edx
  jg     L23
  mov    %edx,(%rsp)
L0:
  xor    %eax,%eax
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D2
datatype: string
name: read_data
arguments: data_tag: D12
datatype: string
name: read_data
arguments: data_tag: D16
datatype: string

tool: D2: Enter N and K:

tool: D12: %d %d

tool: D16: Result of the division is %d and the residue is %d

assistant: ```c
int main() {
  int n, k, i;
  printf("Enter N and K:\n");
  scanf("%d %d", &n, &k);
  i = 0;
  while (n > k) {
    n -= k;
    i++;
  }
  printf("Result of the division is %d and the residue is %d", i, n);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D9(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D21(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%edx
  mov    0x4(%rsp),%eax
  xor    %esi,%esi
  cmp    %eax,%edx
  jle    L28
  nopl   (%rax)
L29:
  sub    %eax,%edx
  add    $0x1,%esi
  cmp    %eax,%edx
  jg     L29
  mov    %edx,(%rsp)
L28:
  xor    %eax,%eax
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D9
datatype: string
name: unpack_data
arguments: data_label: D21
datatype: string
name: unpack_data
arguments: data_label: D5
datatype: string

tool: D9: Enter N and K:

tool: D21: %d %d

tool: D5: Result of the division is %d and the residue is %d

assistant: ```c
int main() {
  int n, k, i;
  printf("Enter N and K:\n");
  scanf("%d %d", &n, &k);
  i = 0;
  while (n > k) {
    n -= k;
    i++;
  }
  printf("Result of the division is %d and the residue is %d", i, n);
  return 0;
}
```