ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<strupper>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L45
  jmp    L40
  nopw   0x0(%rax,%rax,1)
L49:
  add    %rax,%rbx
  movsbl (%rbx),%edi
  test   %dil,%dil
  je     L40
L45:
  call   <skip_multibyte_char@plt>
  test   %rax,%rax
  jne    L49
  movsbl (%rbx),%edi
  call   <islower@plt>
  test   %rax,%rax
  jne    L54
L61:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L45
L40:
  pop    %rbx
  ret
  nopl   0x0(%rax)
L54:
  movsbl (%rbx),%edi
  call   <toupper@plt>
  mov    %al,(%rbx)
  jmp    L61
```

assistant: ```c
void strupper(char *s) {
  while (*s) {
    {
      size_t skip = skip_multibyte_char(*s);
      if (skip != 0)
        s += skip;
      else {
        if (islower((int)*s))
          *s = toupper((int)*s);
        s++;
      }
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pathconf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <_pathconf@plt>
  leave
  ret
```

assistant: ```c
long pathconf(const char *p, int name) {
  p;
  return _pathconf(name);
}
```
user: Translate the below assembly code into C programming language:

```asm
<pathconf>:
  endbr64
  sub    $0x8,%rsp
  mov    %esi,%edi
  call   <_pathconf@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
long pathconf(const char *p, int name) {
  p;
  return _pathconf(name);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<pathconf>:
  endbr64
  mov    %esi,%edi
  jmp    1050 <_pathconf@plt>
```

assistant: ```c
long pathconf(const char *p, int name) {
  p;
  return _pathconf(name);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pathconf>:
  endbr64
  mov    %esi,%edi
  jmp    1050 <_pathconf@plt>
```

assistant: ```c
long pathconf(const char *p, int name) {
  p;
  return _pathconf(name);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<Realloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L1
  cmpl   $0x0,-0x1c(%rbp)
  jle    L1
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <CodingError@plt>
L1:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D15
type_guessed: string
name: disassemble_data
arguments: tag: D1
type_guessed: string

tool: D15: Realloc

tool: D1: Realloc failed

assistant: ```c
void *Realloc(void *OldMemory, int Size) {
  void *Memory;
  if ((Memory = realloc(OldMemory, Size)) == NULL && Size > 0) {
    perror("Realloc");
    CodingError("Realloc failed");
  }
  return (Memory);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<Realloc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%ebx
  call   <realloc@plt>
  mov    %rax,%rbp
  test   %ebx,%ebx
  jle    L12
  test   %rax,%rax
  je     L0
L12:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  lea    D18(%rip),%rdi
  call   <perror@plt>
  lea    D0(%rip),%rdi
  call   <CodingError@plt>
  jmp    L12
```

tool_calls: name: extract_symbolic_data
arguments: tag: D18
type_infered: string
name: extract_symbolic_data
arguments: tag: D0
type_infered: string

tool: D18: Realloc

tool: D0: Realloc failed

assistant: ```c
void *Realloc(void *OldMemory, int Size) {
  void *Memory;
  if ((Memory = realloc(OldMemory, Size)) == NULL && Size > 0) {
    perror("Realloc");
    CodingError("Realloc failed");
  }
  return (Memory);
}
```
user: Translate the below assembly code into C programming language:

```asm
<Realloc>:
  endbr64
  push   %r12
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <realloc@plt>
  mov    %rax,%r12
  test   %ebx,%ebx
  jle    L3
  test   %rax,%rax
  je     L12
L3:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax)
L12:
  lea    D17(%rip),%rdi
  call   <perror@plt>
  lea    D4(%rip),%rdi
  call   <CodingError@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: label: D17
type_infered: string
name: disassemble_data
arguments: label: D4
type_infered: string

tool: D17: Realloc

tool: D4: Realloc failed

assistant: ```c
void *Realloc(void *OldMemory, int Size) {
  void *Memory;
  if ((Memory = realloc(OldMemory, Size)) == NULL && Size > 0) {
    perror("Realloc");
    CodingError("Realloc failed");
  }
  return (Memory);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Realloc>:
  endbr64
  push   %r12
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <realloc@plt>
  mov    %rax,%r12
  test   %ebx,%ebx
  jle    L10
  test   %rax,%rax
  je     L7
L10:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax)
L7:
  lea    D3(%rip),%rdi
  call   <perror@plt>
  lea    D1(%rip),%rdi
  call   <CodingError@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: tag: D3
type: string
name: parse_data
arguments: tag: D1
type: string

tool: D3: Realloc

tool: D1: Realloc failed

assistant: ```c
void *Realloc(void *OldMemory, int Size) {
  void *Memory;
  if ((Memory = realloc(OldMemory, Size)) == NULL && Size > 0) {
    perror("Realloc");
    CodingError("Realloc failed");
  }
  return (Memory);
}
```
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
  lea    -0xc3000(%rsp),%r11
L32:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L32
  sub    $0x540,%rsp
  mov    %edi,-0xc3534(%rbp)
  mov    %rsi,-0xc3540(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0xc3540(%rbp),%rdx
  mov    -0xc3534(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <set_problem_size@plt>
  mov    %eax,-0xc3528(%rbp)
  lea    -0xc3510(%rbp),%rdx
  mov    -0xc3528(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <get_sub_problem_num@plt>
  mov    %eax,-0xc3524(%rbp)
  mov    $0x0,%eax
  call   <get_time@plt>
  mov    %rax,-0xc3518(%rbp)
  movq   $0x0,-0xc3520(%rbp)
  movl   $0x1,-0xc352c(%rbp)
  jmp    L4
L21:
  lea    -0xc3510(%rbp),%rcx
  mov    -0xc3524(%rbp),%edx
  mov    -0xc352c(%rbp),%esi
  mov    -0xc3528(%rbp),%eax
  mov    %eax,%edi
  call   <solver@plt>
  mov    -0xc352c(%rbp),%edx
  movslq %edx,%rdx
  mov    %eax,-0x61a90(%rbp,%rdx,4)
  addl   $0x1,-0xc352c(%rbp)
L4:
  mov    -0xc352c(%rbp),%eax
  cmp    -0xc3524(%rbp),%eax
  jle    L21
  movl   $0x1,-0xc352c(%rbp)
  jmp    L2
L48:
  mov    -0xc352c(%rbp),%eax
  cltq
  mov    -0x61a90(%rbp,%rax,4),%eax
  cltq
  add    %rax,-0xc3520(%rbp)
  addl   $0x1,-0xc352c(%rbp)
L2:
  mov    -0xc352c(%rbp),%eax
  cmp    -0xc3524(%rbp),%eax
  jle    L48
  mov    $0x0,%eax
  call   <get_time@plt>
  sub    -0xc3518(%rbp),%rax
  mov    %rax,%rcx
  mov    -0xc3520(%rbp),%rdx
  mov    -0xc3528(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <print_result@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int prob[100000];
  int rets[100000];
  int i;
  int n = set_problem_size(argc, argv);
  int jobs = get_sub_problem_num(n, prob);
  long long usec = get_time();
  long long answers = 0;
#pragma omp parallel for schedule(dynamic)
  for (i = 1; i <= jobs; i++) {
    rets[i] = solver(n, i, jobs, prob);
  }
  for (i = 1; i <= jobs; i++)
    answers += rets[i];
  print_result(n, get_time() - usec, answers);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xc3000(%rsp),%r11
L40:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L40
  sub    $0x510,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc3508(%rsp)
  xor    %eax,%eax
  call   <set_problem_size@plt>
  mov    %eax,%r12d
  mov    %rsp,%rsi
  mov    %eax,%edi
  call   <get_sub_problem_num@plt>
  mov    %eax,%ebp
  mov    $0x0,%eax
  call   <get_time@plt>
  mov    %rax,%r13
  test   %ebp,%ebp
  jle    L21
  mov    $0x1,%ebx
  mov    %rsp,%r14
L8:
  mov    %r14,%rcx
  mov    %ebp,%edx
  mov    %ebx,%esi
  mov    %r12d,%edi
  call   <solver@plt>
  mov    %eax,0x61a80(%rsp,%rbx,4)
  add    $0x1,%rbx
  cmp    %ebx,%ebp
  jge    L8
  lea    0x61a84(%rsp),%rax
  lea    -0x1(%rbp),%edx
  lea    0x61a88(%rsp,%rdx,4),%rcx
  mov    $0x0,%ebx
L7:
  movslq (%rax),%rdx
  add    %rdx,%rbx
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L7
L30:
  mov    $0x0,%eax
  call   <get_time@plt>
  sub    %r13,%rax
  mov    %rax,%rsi
  mov    %rbx,%rdx
  mov    %r12d,%edi
  call   <print_result@plt>
  mov    0xc3508(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  mov    $0x0,%eax
  add    $0xc3510,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L21:
  mov    $0x0,%ebx
  jmp    L30
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int prob[100000];
  int rets[100000];
  int i;
  int n = set_problem_size(argc, argv);
  int jobs = get_sub_problem_num(n, prob);
  long long usec = get_time();
  long long answers = 0;
#pragma omp parallel for schedule(dynamic)
  for (i = 1; i <= jobs; i++) {
    rets[i] = solver(n, i, jobs, prob);
  }
  for (i = 1; i <= jobs; i++)
    answers += rets[i];
  print_result(n, get_time() - usec, answers);
  return 0;
}
```
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   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xc3000(%rsp),%r11
L54:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L54
  sub    $0x518,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc3508(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  call   <set_problem_size@plt>
  mov    %r12,%rsi
  mov    %eax,%edi
  mov    %eax,%r15d
  call   <get_sub_problem_num@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <get_time@plt>
  mov    %rax,%r13
  test   %ebp,%ebp
  jle    L23
  mov    $0x1,%ebx
  lea    0x61a80(%rsp),%r14
  nopw   0x0(%rax,%rax,1)
L42:
  mov    %ebx,%esi
  mov    %r12,%rcx
  mov    %ebp,%edx
  mov    %r15d,%edi
  call   <solver@plt>
  mov    %eax,(%r14,%rbx,4)
  add    $0x1,%rbx
  cmp    %ebx,%ebp
  jge    L42
  lea    -0x1(%rbp),%edx
  lea    0x61a84(%rsp),%rax
  xor    %r12d,%r12d
  lea    0x61a88(%rsp,%rdx,4),%rcx
  nopl   0x0(%rax)
L53:
  movslq (%rax),%rdx
  add    $0x4,%rax
  add    %rdx,%r12
  cmp    %rcx,%rax
  jne    L53
L10:
  xor    %eax,%eax
  call   <get_time@plt>
  mov    %r12,%rdx
  mov    %r15d,%edi
  sub    %r13,%rax
  mov    %rax,%rsi
  call   <print_result@plt>
  mov    0xc3508(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0xc3518,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L23:
  xor    %r12d,%r12d
  jmp    L10
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int prob[100000];
  int rets[100000];
  int i;
  int n = set_problem_size(argc, argv);
  int jobs = get_sub_problem_num(n, prob);
  long long usec = get_time();
  long long answers = 0;
#pragma omp parallel for schedule(dynamic)
  for (i = 1; i <= jobs; i++) {
    rets[i] = solver(n, i, jobs, prob);
  }
  for (i = 1; i <= jobs; i++)
    answers += rets[i];
  print_result(n, get_time() - usec, answers);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xc3000(%rsp),%r11
L15:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L15
  sub    $0x518,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc3508(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  call   <set_problem_size@plt>
  mov    %r12,%rsi
  mov    %eax,%edi
  mov    %eax,%r15d
  call   <get_sub_problem_num@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <get_time@plt>
  mov    %rax,%r13
  test   %ebp,%ebp
  jle    L43
  mov    $0x1,%ebx
  lea    0x61a80(%rsp),%r14
  nopw   0x0(%rax,%rax,1)
L36:
  mov    %ebx,%esi
  mov    %r12,%rcx
  mov    %ebp,%edx
  mov    %r15d,%edi
  call   <solver@plt>
  mov    %eax,(%r14,%rbx,4)
  add    $0x1,%rbx
  cmp    %ebx,%ebp
  jge    L36
  lea    -0x1(%rbp),%eax
  cmp    $0x2,%eax
  jbe    L41
  mov    %ebp,%edx
  lea    0x61a84(%rsp),%rax
  pxor   %xmm1,%xmm1
  shr    $0x2,%edx
  pxor   %xmm4,%xmm4
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x61a94(%rsp,%rdx,1),%rdx
  nopl   0x0(%rax,%rax,1)
L107:
  movdqu (%rax),%xmm0
  movdqa %xmm4,%xmm2
  add    $0x10,%rax
  pcmpgtd %xmm0,%xmm2
  movdqa %xmm0,%xmm3
  punpckldq %xmm2,%xmm3
  punpckhdq %xmm2,%xmm0
  paddq  %xmm3,%xmm1
  paddq  %xmm0,%xmm1
  cmp    %rdx,%rax
  jne    L107
  movdqa %xmm1,%xmm0
  mov    %ebp,%edx
  psrldq $0x8,%xmm0
  and    $0xfffffffc,%edx
  paddq  %xmm0,%xmm1
  lea    0x1(%rdx),%eax
  movq   %xmm1,%r12
  cmp    %edx,%ebp
  je     L8
L51:
  movslq %eax,%rdx
  movslq 0x61a80(%rsp,%rdx,4),%rdx
  add    %rdx,%r12
  lea    0x1(%rax),%edx
  cmp    %edx,%ebp
  jl     L8
  movslq %edx,%rdx
  add    $0x2,%eax
  movslq 0x61a80(%rsp,%rdx,4),%rdx
  add    %rdx,%r12
  cmp    %eax,%ebp
  jl     L8
  cltq
  movslq 0x61a80(%rsp,%rax,4),%rax
  add    %rax,%r12
L8:
  xor    %eax,%eax
  call   <get_time@plt>
  mov    %r12,%rdx
  mov    %r15d,%edi
  sub    %r13,%rax
  mov    %rax,%rsi
  call   <print_result@plt>
  mov    0xc3508(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L93
  add    $0xc3518,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L43:
  xor    %r12d,%r12d
  jmp    L8
L41:
  xor    %r12d,%r12d
  mov    $0x1,%eax
  jmp    L51
L93:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int prob[100000];
  int rets[100000];
  int i;
  int n = set_problem_size(argc, argv);
  int jobs = get_sub_problem_num(n, prob);
  long long usec = get_time();
  long long answers = 0;
#pragma omp parallel for schedule(dynamic)
  for (i = 1; i <= jobs; i++) {
    rets[i] = solver(n, i, jobs, prob);
  }
  for (i = 1; i <= jobs; i++)
    answers += rets[i];
  print_result(n, get_time() - usec, answers);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <make_strings@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D9
type_guessed: string

tool: D9: string 1 [%s] string 2 [%s] 


assistant: ```c
int main() {
  char *s1, *s2;
  make_strings(s1, s2);
  printf("string 1 [%s] string 2 [%s] \n", s1, s2);
  return 0;
}
```
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
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%ebp
  mov    %rbp,%rsi
  mov    $0x0,%ebx
  mov    %rbx,%rdi
  call   <make_strings@plt>
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D2
data_type: string

tool: D2: string 1 [%s] string 2 [%s] 


assistant: ```c
int main() {
  char *s1, *s2;
  make_strings(s1, s2);
  printf("string 1 [%s] string 2 [%s] \n", s1, s2);
  return 0;
}
```
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
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  sub    $0x8,%rsp
  call   <make_strings@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D5
type: string

tool: D5: string 1 [%s] string 2 [%s] 


assistant: ```c
int main() {
  char *s1, *s2;
  make_strings(s1, s2);
  printf("string 1 [%s] string 2 [%s] \n", s1, s2);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  sub    $0x8,%rsp
  call   <make_strings@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: label: D4
datatype: string

tool: D4: string 1 [%s] string 2 [%s] 


assistant: ```c
int main() {
  char *s1, *s2;
  make_strings(s1, s2);
  printf("string 1 [%s] string 2 [%s] \n", s1, s2);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<getFirstHand>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L39
L60:
  movl   $0x0,-0x4(%rbp)
L29:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%ecx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  movslq %ecx,%rax
  imul   $0x4ec4ec4f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,(%rsi)
  mov    $0x0,%eax
  call   <rand@plt>
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  shr    $0x1e,%ecx
  add    %ecx,%eax
  and    $0x3,%eax
  sub    %ecx,%eax
  mov    %eax,(%rdx)
  movl   $0x0,-0x8(%rbp)
  jmp    L33
L64:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jne    L47
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jne    L47
  movl   $0x1,-0x4(%rbp)
L47:
  addl   $0x1,-0x8(%rbp)
L33:
  mov    -0x8(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L64
  cmpl   $0x1,-0x4(%rbp)
  je     L29
  addl   $0x1,-0xc(%rbp)
L39:
  cmpl   $0x4,-0xc(%rbp)
  jle    L60
  nop
  nop
  leave
  ret
```

assistant: ```c
void getFirstHand(int cardRank[], int cardSuit[]) {
  int i, j;
  int cardDup;
  for (i = 0; i < 5; i += 1) {
    cardDup = 0;
    do {
      cardRank[i] = (rand() % 13);
      cardSuit[i] = (rand() % 4);
      for (j = 0; j < i; j += 1) {
        if ((cardRank[i] == cardRank[j]) && (cardSuit[i] == cardSuit[j])) {
          cardDup = 1;
        }
      }
    } while (cardDup == 1);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getFirstHand>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r14
  mov    $0x0,%ebp
  mov    $0x1,%r15d
  jmp    L29
L25:
  add    $0x1,%rdx
  cmp    %edx,%r12d
  jle    L66
L87:
  cmp    (%rbx,%rdx,4),%ecx
  jne    L25
  cmp    (%r14,%rdx,4),%eax
  cmove  %r15d,%r13d
  jmp    L25
L66:
  cmp    $0x1,%r13d
  jne    L74
L44:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,2),%ecx
  lea    (%rdx,%rcx,4),%edx
  sub    %edx,%eax
  mov    %eax,(%rbx,%rbp,4)
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  shr    $0x1e,%edx
  add    %edx,%eax
  and    $0x3,%eax
  sub    %edx,%eax
  mov    %eax,(%r14,%rbp,4)
  test   %ebp,%ebp
  jle    L66
  mov    (%rbx,%rbp,4),%ecx
  mov    $0x0,%edx
  jmp    L87
L74:
  add    $0x1,%rbp
  cmp    $0x5,%rbp
  je     L26
L29:
  mov    $0x0,%r13d
  mov    %ebp,%r12d
  jmp    L44
L26:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void getFirstHand(int cardRank[], int cardSuit[]) {
  int i, j;
  int cardDup;
  for (i = 0; i < 5; i += 1) {
    cardDup = 0;
    do {
      cardRank[i] = (rand() % 13);
      cardSuit[i] = (rand() % 4);
      for (j = 0; j < i; j += 1) {
        if ((cardRank[i] == cardRank[j]) && (cardSuit[i] == cardSuit[j])) {
          cardDup = 1;
        }
      }
    } while (cardDup == 1);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getFirstHand>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
L16:
  xor    %r14d,%r14d
  mov    %r12d,%r13d
  cs nopw 0x0(%rax,%rax,1)
L40:
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,2),%ecx
  lea    (%rdx,%rcx,4),%edx
  sub    %edx,%eax
  mov    %eax,0x0(%rbp,%r12,4)
  xor    %eax,%eax
  call   <rand@plt>
  cltd
  shr    $0x1e,%edx
  add    %edx,%eax
  and    $0x3,%eax
  sub    %edx,%eax
  mov    %eax,(%r15,%r12,4)
  test   %r12,%r12
  je     L42
  mov    0x0(%rbp,%r12,4),%ecx
  xor    %edx,%edx
L8:
  cmp    0x0(%rbp,%rdx,4),%ecx
  jne    L15
  cmp    (%r15,%rdx,4),%eax
  cmove  %ebx,%r14d
L15:
  add    $0x1,%rdx
  cmp    %edx,%r13d
  jg     L8
L42:
  cmp    $0x1,%r14d
  je     L40
  add    $0x1,%r12
  cmp    $0x5,%r12
  jne    L16
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void getFirstHand(int cardRank[], int cardSuit[]) {
  int i, j;
  int cardDup;
  for (i = 0; i < 5; i += 1) {
    cardDup = 0;
    do {
      cardRank[i] = (rand() % 13);
      cardSuit[i] = (rand() % 4);
      for (j = 0; j < i; j += 1) {
        if ((cardRank[i] == cardRank[j]) && (cardSuit[i] == cardSuit[j])) {
          cardDup = 1;
        }
      }
    } while (cardDup == 1);
  }
}
```