ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<qcount_update>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <qcount_read@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  add    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <qcount_save@plt>
  cmp    $0x1,%eax
  jne    L9
  mov    -0x8(%rbp),%rax
  jmp    L2
L9:
  mov    $0x0,%eax
L2:
  leave
  ret
```

assistant: ```c
int64_t qcount_update(const char *filepath, int64_t number) {
  int64_t counter = qcount_read(filepath);
  counter += number;
  if (qcount_save(filepath, counter) == true) {
    return counter;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<qcount_update>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  call   <qcount_read@plt>
  mov    %rax,%rbx
  lea    (%rax,%rbp,1),%rax
  mov    %rax,%rsi
  mov    %r12,%rdi
  call   <qcount_save@plt>
  add    %rbp,%rbx
  cmp    $0x1,%eax
  mov    $0x0,%eax
  cmove  %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int64_t qcount_update(const char *filepath, int64_t number) {
  int64_t counter = qcount_read(filepath);
  counter += number;
  if (qcount_save(filepath, counter) == true) {
    return counter;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<qcount_update>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <qcount_read@plt>
  mov    %rbp,%rdi
  add    %rax,%r12
  mov    %r12,%rsi
  call   <qcount_save@plt>
  cmp    $0x1,%eax
  mov    $0x0,%eax
  cmovne %rax,%r12
  add    $0x8,%rsp
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
int64_t qcount_update(const char *filepath, int64_t number) {
  int64_t counter = qcount_read(filepath);
  counter += number;
  if (qcount_save(filepath, counter) == true) {
    return counter;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<qcount_update>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <qcount_read@plt>
  mov    %rbp,%rdi
  add    %rax,%r12
  mov    %r12,%rsi
  call   <qcount_save@plt>
  cmp    $0x1,%eax
  mov    $0x0,%eax
  cmovne %rax,%r12
  add    $0x8,%rsp
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
int64_t qcount_update(const char *filepath, int64_t number) {
  int64_t counter = qcount_read(filepath);
  counter += number;
  if (qcount_save(filepath, counter) == true) {
    return counter;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xmalloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L4
  mov    $0x7,%edi
  call   <inccount@plt>
L4:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *xmalloc(size_t size) {
  void *p = malloc(size);
  if (p) {
    inccount(7);
  }
  return (p);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<xmalloc>:
  endbr64
  push   %rbx
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L2
  mov    $0x7,%edi
  call   <inccount@plt>
L2:
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
void *xmalloc(size_t size) {
  void *p = malloc(size);
  if (p) {
    inccount(7);
  }
  return (p);
}
```
user: Translate the below assembly code into C programming language:

```asm
<xmalloc>:
  endbr64
  push   %r12
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L3
  mov    $0x7,%edi
  call   <inccount@plt>
L3:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
void *xmalloc(size_t size) {
  void *p = malloc(size);
  if (p) {
    inccount(7);
  }
  return (p);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<xmalloc>:
  endbr64
  push   %r12
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L4
  mov    $0x7,%edi
  call   <inccount@plt>
L4:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
void *xmalloc(size_t size) {
  void *p = malloc(size);
  if (p) {
    inccount(7);
  }
  return (p);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<Fatal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <Error@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

assistant: ```c
void Fatal(char *msg) {
  Error(msg);
  exit(1);
}
```
user: What is the c source code of the assembly code below:

```asm
<Fatal>:
  endbr64
  sub    $0x8,%rsp
  call   <Error@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void Fatal(char *msg) {
  Error(msg);
  exit(1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<Fatal>:
  endbr64
  sub    $0x8,%rsp
  call   <Error@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void Fatal(char *msg) {
  Error(msg);
  exit(1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Fatal>:
  endbr64
  sub    $0x8,%rsp
  call   <Error@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void Fatal(char *msg) {
  Error(msg);
  exit(1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  shr    $0x3,%eax
  or     $0x61bab5,%eax
  mov    %eax,%edx
  movsbl %ah,%ecx
  shl    $0x8,%eax
  sar    $0x18,%eax
  mov    %al,%dh
  movzbl %dl,%eax
  shl    $0x10,%eax
  and    $0xff00ffff,%edx
  or     %eax,%edx
  mov    %cl,%dl
  and    $0x369937ec,%edx
  mov    %edx,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local2;
  unsigned int local1;
  char copy11;
  char copy12;
  char copy13;
  char copy14;
  {
    state[0UL] = (input[0UL] | 51238316UL) >> 3U;
    local1 = 0UL;
    while (local1 < 1UL) {
      local2 = 0UL;
      while (local2 < 1UL) {
        if (state[0UL] > (local2 & local1)) {
          copy11 = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = copy11;
          copy11 = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy11;
          copy12 = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 3);
          *((char *)(&state[0UL]) + 3) = copy12;
          copy12 = *((char *)(&state[0UL]) + 3);
          *((char *)(&state[0UL]) + 3) = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = copy12;
        } else {
          copy13 = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = copy13;
          copy14 = *((char *)(&state[0UL]) + 2);
          *((char *)(&state[0UL]) + 2) = *((char *)(&state[0UL]) + 3);
          *((char *)(&state[0UL]) + 3) = copy14;
        }
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] & 916010988UL;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  shr    $0x3,%eax
  or     $0x61bab5,%eax
  mov    %eax,%ecx
  mov    %eax,%edx
  shl    $0x8,%ecx
  sar    $0x18,%ecx
  mov    %cl,%dh
  movzbl %al,%ecx
  movsbl %ah,%eax
  shl    $0x10,%ecx
  and    $0xff00ffff,%edx
  or     %ecx,%edx
  mov    %al,%dl
  and    $0x369937ec,%edx
  mov    %edx,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local2;
  unsigned int local1;
  char copy11;
  char copy12;
  char copy13;
  char copy14;
  {
    state[0UL] = (input[0UL] | 51238316UL) >> 3U;
    local1 = 0UL;
    while (local1 < 1UL) {
      local2 = 0UL;
      while (local2 < 1UL) {
        if (state[0UL] > (local2 & local1)) {
          copy11 = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = copy11;
          copy11 = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy11;
          copy12 = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 3);
          *((char *)(&state[0UL]) + 3) = copy12;
          copy12 = *((char *)(&state[0UL]) + 3);
          *((char *)(&state[0UL]) + 3) = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = copy12;
        } else {
          copy13 = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = copy13;
          copy14 = *((char *)(&state[0UL]) + 2);
          *((char *)(&state[0UL]) + 2) = *((char *)(&state[0UL]) + 3);
          *((char *)(&state[0UL]) + 3) = copy14;
        }
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] & 916010988UL;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  shr    $0x3,%eax
  or     $0x61bab5,%eax
  movzbl %al,%ecx
  mov    %eax,%edx
  movsbl %ah,%edi
  shl    $0x8,%eax
  shl    $0x10,%ecx
  and    $0xff00ffff,%edx
  sar    $0x18,%eax
  or     %ecx,%edx
  xor    %ecx,%ecx
  mov    %dil,%cl
  mov    %al,%ch
  mov    %cx,%dx
  and    $0x369937ec,%edx
  mov    %edx,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local2;
  unsigned int local1;
  char copy11;
  char copy12;
  char copy13;
  char copy14;
  {
    state[0UL] = (input[0UL] | 51238316UL) >> 3U;
    local1 = 0UL;
    while (local1 < 1UL) {
      local2 = 0UL;
      while (local2 < 1UL) {
        if (state[0UL] > (local2 & local1)) {
          copy11 = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = copy11;
          copy11 = *((char *)(&state[local1]) + 2);
          *((char *)(&state[local1]) + 2) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy11;
          copy12 = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 3);
          *((char *)(&state[0UL]) + 3) = copy12;
          copy12 = *((char *)(&state[0UL]) + 3);
          *((char *)(&state[0UL]) + 3) = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = copy12;
        } else {
          copy13 = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = copy13;
          copy14 = *((char *)(&state[0UL]) + 2);
          *((char *)(&state[0UL]) + 2) = *((char *)(&state[0UL]) + 3);
          *((char *)(&state[0UL]) + 3) = copy14;
        }
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] & 916010988UL;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L8
L1:
  lea    D6(%rip),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <benchmark@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <print_results@plt>
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L8:
  cmpl   $0x2,-0xc(%rbp)
  jle    L1
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D6
type_guessed: string
name: disassemble_data
arguments: data_label: D8
type_guessed: string
name: disassemble_data
arguments: data_label: D1
type_guessed: string

tool: D6: ramp

tool: D8: # c::native::%s


tool: D1: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::native::%s\n", "ramp");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  mov    $0x0,%ebx
  lea    D24(%rip),%r13
  lea    D28(%rip),%r12
  lea    D15(%rip),%rbp
L9:
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <benchmark@plt>
  call   <print_results@plt>
  add    $0x1,%ebx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L9
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D24
data_type: string
name: retrieve_data
arguments: data_tag: D28
data_type: string
name: retrieve_data
arguments: data_tag: D15
data_type: string

tool: D24: ramp

tool: D28: # c::native::%s


tool: D15: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::native::%s\n", "ramp");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  xor    %edi,%edi
  lea    D17(%rip),%r13
  push   %r12
  lea    D26(%rip),%r12
  push   %rbp
  lea    D24(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
L8:
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  xor    %eax,%eax
  call   <benchmark@plt>
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L8
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D17
data_type: string
name: extract_symbolic_data
arguments: tag: D26
data_type: string
name: extract_symbolic_data
arguments: tag: D24
data_type: string

tool: D17: ramp

tool: D26: # c::native::%s


tool: D24: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::native::%s\n", "ramp");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r13
  xor    %edi,%edi
  lea    D11(%rip),%r13
  push   %r12
  lea    D18(%rip),%r12
  push   %rbp
  lea    D27(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
L4:
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  xor    %eax,%eax
  call   <benchmark@plt>
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L4
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D11
type_infered: string
name: retrieve_data
arguments: data_tag: D18
type_infered: string
name: retrieve_data
arguments: data_tag: D27
type_infered: string

tool: D11: ramp

tool: D18: # c::native::%s


tool: D27: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::native::%s\n", "ramp");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<move>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    %ecx,-0x34(%rbp)
  mov    -0x2c(%rbp),%eax
  cltd
  idivl  -0x30(%rbp)
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x1,-0x10(%rbp)
  jmp    L1
L11:
  mov    -0x10(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jne    L8
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x18(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x18(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
L8:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  cltd
  idivl  -0x30(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jge    L1
  mov    -0x34(%rbp),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x2c(%rbp),%esi
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <next_index@plt>
  mov    %eax,-0x14(%rbp)
L1:
  mov    -0x18(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L11
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *move(int *cards, int card_num, int player_num, int skip) {
  int cards_per_player = card_num / player_num;
  int *cand = malloc(sizeof(int) * cards_per_player);
  int i = 0, index = 0;
  int count = 1;
  while (i < cards_per_player) {
    if (count == player_num) {
      cand[i++] = cards[index];
    }
    cards[index] = 0;
    count = (count++) % player_num;
    if (i < cards_per_player)
      index = next_index(cards, card_num, index, skip);
  }
  return cand;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<move>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r13
  mov    %esi,%eax
  mov    %esi,0x8(%rsp)
  mov    %edx,%ebx
  mov    %ecx,0xc(%rsp)
  cltd
  idiv   %ebx
  mov    %eax,%r14d
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%r15
  test   %r14d,%r14d
  jle    L32
  mov    $0x1,%r12d
  mov    $0x0,%esi
  mov    $0x0,%ebp
  jmp    L13
L27:
  movslq %esi,%rax
  mov    0x0(%r13,%rax,4),%edx
  movslq %ebp,%rax
  mov    %edx,(%r15,%rax,4)
  lea    0x1(%rbp),%ebp
  jmp    L22
L6:
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  mov    %edx,%r12d
  mov    0xc(%rsp),%ecx
  mov    %esi,%edx
  mov    0x8(%rsp),%esi
  mov    %r13,%rdi
  call   <next_index@plt>
  mov    %eax,%esi
L13:
  cmp    %r12d,%ebx
  je     L27
L22:
  movslq %esi,%rax
  movl   $0x0,0x0(%r13,%rax,4)
  cmp    %r14d,%ebp
  jl     L6
L32:
  mov    %r15,%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int *move(int *cards, int card_num, int player_num, int skip) {
  int cards_per_player = card_num / player_num;
  int *cand = malloc(sizeof(int) * cards_per_player);
  int i = 0, index = 0;
  int count = 1;
  while (i < cards_per_player) {
    if (count == player_num) {
      cand[i++] = cards[index];
    }
    cards[index] = 0;
    count = (count++) % player_num;
    if (i < cards_per_player)
      index = next_index(cards, card_num, index, skip);
  }
  return cand;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<move>:
  endbr64
  mov    %esi,%eax
  push   %r15
  mov    $0x1,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  mov    %edx,%ebx
  cltd
  idiv   %ebx
  sub    $0x18,%rsp
  mov    %ecx,0x4(%rsp)
  lea    0x0(,%rax,4),%edi
  mov    %eax,%ebp
  call   <malloc@plt>
  xor    %r8d,%r8d
  mov    %rax,0x8(%rsp)
  test   %ebp,%ebp
  jg     L53
  jmp    L15
  nopl   0x0(%rax,%rax,1)
L31:
  movl   $0x0,(%rax)
  mov    %r15d,%eax
  mov    0x4(%rsp),%ecx
  mov    %r12d,%esi
  cltd
  mov    %r13,%rdi
  idiv   %ebx
  mov    %edx,%r15d
  mov    %r8d,%edx
  call   <next_index@plt>
  mov    %eax,%r8d
  cmp    %r14d,%ebp
  jle    L15
L53:
  movslq %r8d,%rax
  lea    0x0(%r13,%rax,4),%rax
  cmp    %r15d,%ebx
  jne    L31
  mov    (%rax),%edx
  mov    0x8(%rsp),%rcx
  movslq %r14d,%r9
  lea    0x1(%r14),%r15d
  mov    %edx,(%rcx,%r9,4)
  movl   $0x0,(%rax)
  cmp    %r15d,%ebp
  jg     L54
L15:
  mov    0x8(%rsp),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L54:
  mov    0x4(%rsp),%ecx
  mov    %r8d,%edx
  mov    %r12d,%esi
  mov    %r13,%rdi
  mov    %r15d,%r14d
  xor    %r15d,%r15d
  call   <next_index@plt>
  mov    %eax,%r8d
  jmp    L53
```

assistant: ```c
int *move(int *cards, int card_num, int player_num, int skip) {
  int cards_per_player = card_num / player_num;
  int *cand = malloc(sizeof(int) * cards_per_player);
  int i = 0, index = 0;
  int count = 1;
  while (i < cards_per_player) {
    if (count == player_num) {
      cand[i++] = cards[index];
    }
    cards[index] = 0;
    count = (count++) % player_num;
    if (i < cards_per_player)
      index = next_index(cards, card_num, index, skip);
  }
  return cand;
}
```