ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D6(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  js     L6
  xor    %ebx,%ebx
  lea    D11(%rip),%rbp
  nopw   0x0(%rax,%rax,1)
L0:
  mov    (%rsp),%esi
  mov    %rbp,%rdi
  imul   %ebx,%esi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jge    L0
L6:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D6
type: string
name: retrieve_data
arguments: tag: D11
type: string

tool: D6: %d

tool: D11: %d


assistant: ```c
void main() {
  int num, num1, i, mul;
  scanf("%d", &num);
  scanf("%d", &num1);
  for (i = 0; i <= num1; i++) {
    mul = num * i;
    printf("%d\n", mul);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D9(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  js     L7
  xor    %ebx,%ebx
  lea    D8(%rip),%rbp
  nopw   0x0(%rax,%rax,1)
L23:
  mov    (%rsp),%esi
  mov    %rbp,%rdi
  imul   %ebx,%esi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jge    L23
L7:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D9
datatype: string
name: disassemble_data
arguments: label: D8
datatype: string

tool: D9: %d

tool: D8: %d


assistant: ```c
void main() {
  int num, num1, i, mul;
  scanf("%d", &num);
  scanf("%d", &num1);
  for (i = 0; i <= num1; i++) {
    mul = num * i;
    printf("%d\n", mul);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L29
L33:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0xf4240,%eax
  cltd
  idiv   %ecx
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L18
L36:
  addl   $0x1,-0x1c(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  lea    D3(%rip),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <benchmark@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%edi
  call   <print_results@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L18:
  cmpl   $0x2,-0x14(%rbp)
  jle    L36
  addl   $0x1,-0x18(%rbp)
L29:
  cmpl   $0x6,-0x18(%rbp)
  jle    L33
  mov    -0x1c(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D3
type_infered: string
name: parse_data
arguments: data_label: D19
type_infered: string
name: parse_data
arguments: data_label: D14
type_infered: string

tool: D3: scuminabs

tool: D19: # c::%s:len=%d


tool: D14: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "scuminabs", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  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    $0x3,%r12d
  mov    $0x1,%r14d
  lea    D9(%rip),%r15
  jmp    L35
L19:
  add    $0x1,%r14d
  add    $0x3,%r12d
  cmp    $0x15,%r12d
  je     L13
L35:
  lea    -0x3(%r12),%ebx
  mov    %r14d,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ebp
  lea    -0x1(%r14),%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0xf4240,%eax
  cltd
  idiv   %ecx
  mov    %eax,%r13d
L26:
  add    $0x1,%ebx
  mov    %ebp,%edx
  mov    %r15,%rsi
  lea    D14(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  mov    %r13d,%edi
  call   <benchmark@plt>
  mov    %r13d,%edi
  call   <print_results@plt>
  mov    %ebx,%esi
  lea    D26(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %r12d,%ebx
  jne    L26
  jmp    L19
L13:
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
data_type: string
name: extract_symbolic_data
arguments: label: D14
data_type: string
name: extract_symbolic_data
arguments: label: D26
data_type: string

tool: D9: scuminabs

tool: D14: # c::%s:len=%d


tool: D26: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "scuminabs", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D26(%rip),%r15
  push   %r14
  lea    D17(%rip),%r14
  push   %r13
  lea    D7(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L6:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0xf4240,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L18:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L18
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L6
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: data_label: D26
datatype: string
name: parse_data
arguments: data_label: D17
datatype: string
name: parse_data
arguments: data_label: D7
datatype: string

tool: D26: scuminabs

tool: D17: # c::%s:len=%d


tool: D7: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "scuminabs", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D11(%rip),%r15
  push   %r14
  lea    D5(%rip),%r14
  push   %r13
  lea    D21(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L15:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0xf4240,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L3:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L3
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L15
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: label: D11
type_guessed: string
name: disassemble_data
arguments: label: D5
type_guessed: string
name: disassemble_data
arguments: label: D21
type_guessed: string

tool: D11: scuminabs

tool: D5: # c::%s:len=%d


tool: D21: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "scuminabs", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<GP_add_vertex_s>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D8
type_infered: string

tool: D8: v %d %s


assistant: ```c
void GP_add_vertex_s(void *arg, int v, char *label) {
  printf("v %d %s\n", v, label);
  free(label);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<GP_add_vertex_s>:
  endbr64
  push   %rbx
  mov    %rdx,%rbx
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D2
type: string

tool: D2: v %d %s


assistant: ```c
void GP_add_vertex_s(void *arg, int v, char *label) {
  printf("v %d %s\n", v, label);
  free(label);
}
```
user: Translate the below assembly code into C programming language:

```asm
<GP_add_vertex_s>:
  endbr64
  push   %rbp
  lea    D9(%rip),%rdi
  mov    %rdx,%rbp
  call   <printf@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <free@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
datatype: string

tool: D9: v %d %s


assistant: ```c
void GP_add_vertex_s(void *arg, int v, char *label) {
  printf("v %d %s\n", v, label);
  free(label);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<GP_add_vertex_s>:
  endbr64
  push   %rbp
  lea    D3(%rip),%rdi
  mov    %rdx,%rbp
  call   <printf@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <free@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D3
data_type: string

tool: D3: v %d %s


assistant: ```c
void GP_add_vertex_s(void *arg, int v, char *label) {
  printf("v %d %s\n", v, label);
  free(label);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<print_usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D1
datatype: string

tool: D1: usage: sudoku [OPTIONS] PUZZLE_FILE

Options:
  -n    print only the number of solutions found
  -h    show this message and exit


assistant: ```c
void print_usage(void) {
  printf("usage: sudoku [OPTIONS] PUZZLE_FILE\n"
         "\n"
         "Options:\n"
         "  -n    print only the number of solutions found\n"
         "  -h    show this message and exit\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<print_usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D8
data_type: string

tool: D8: usage: sudoku [OPTIONS] PUZZLE_FILE

Options:
  -n    print only the number of solutions found
  -h    show this message and exit


assistant: ```c
void print_usage(void) {
  printf("usage: sudoku [OPTIONS] PUZZLE_FILE\n"
         "\n"
         "Options:\n"
         "  -n    print only the number of solutions found\n"
         "  -h    show this message and exit\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<print_usage>:
  endbr64
  lea    D1(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: unpack_data
arguments: label: D1
type: string

tool: D1: usage: sudoku [OPTIONS] PUZZLE_FILE

Options:
  -n    print only the number of solutions found
  -h    show this message and exit


assistant: ```c
void print_usage(void) {
  printf("usage: sudoku [OPTIONS] PUZZLE_FILE\n"
         "\n"
         "Options:\n"
         "  -n    print only the number of solutions found\n"
         "  -h    show this message and exit\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<print_usage>:
  endbr64
  lea    D1(%rip),%rdi
  jmp    1050 <printf@plt>
```

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

tool: D1: usage: sudoku [OPTIONS] PUZZLE_FILE

Options:
  -n    print only the number of solutions found
  -h    show this message and exit


assistant: ```c
void print_usage(void) {
  printf("usage: sudoku [OPTIONS] PUZZLE_FILE\n"
         "\n"
         "Options:\n"
         "  -n    print only the number of solutions found\n"
         "  -h    show this message and exit\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<produce1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x190,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L19
L10:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L19:
  cmpl   $0x63,-0xc(%rbp)
  jle    L10
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *produce1() {
  int i;
  int *result;
  result = (int *)malloc(sizeof(int) * 100);
  for (i = 0; i < 100; i++) {
    result[i] = i;
  }
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<produce1>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x190,%edi
  call   <malloc@plt>
  mov    %rax,%rcx
  mov    $0x0,%edx
L0:
  mov    %edx,(%rcx)
  add    $0x1,%edx
  add    $0x4,%rcx
  cmp    $0x64,%edx
  jne    L0
  add    $0x8,%rsp
  ret
```

assistant: ```c
int *produce1() {
  int i;
  int *result;
  result = (int *)malloc(sizeof(int) * 100);
  for (i = 0; i < 100; i++) {
    result[i] = i;
  }
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<produce1>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x190,%edi
  call   <malloc@plt>
  xor    %edx,%edx
  mov    %rax,%rcx
  nopw   0x0(%rax,%rax,1)
L4:
  mov    %edx,(%rcx)
  add    $0x1,%edx
  add    $0x4,%rcx
  cmp    $0x64,%edx
  jne    L4
  add    $0x8,%rsp
  ret
```

assistant: ```c
int *produce1() {
  int i;
  int *result;
  result = (int *)malloc(sizeof(int) * 100);
  for (i = 0; i < 100; i++) {
    result[i] = i;
  }
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<produce1>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x190,%edi
  call   <malloc@plt>
  movdqa D15(%rip),%xmm0
  movdqa D5(%rip),%xmm2
  mov    %rax,%rdx
  lea    0x190(%rax),%rcx
  nopl   0x0(%rax)
L2:
  movdqa %xmm0,%xmm1
  add    $0x10,%rdx
  paddd  %xmm2,%xmm0
  movups %xmm1,-0x10(%rdx)
  cmp    %rdx,%rcx
  jne    L2
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D15
type: byte[16]
name: retrieve_data
arguments: data_label: D5
type: byte[16]

tool: {"D15": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D5": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

assistant: ```c
int *produce1() {
  int i;
  int *result;
  result = (int *)malloc(sizeof(int) * 100);
  for (i = 0; i < 100; i++) {
    result[i] = i;
  }
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<COM_SplitPath>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x440,%rsp
  mov    %rdi,-0x428(%rbp)
  mov    %rsi,-0x430(%rbp)
  mov    %rdx,-0x438(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xffffffff,-0x414(%rbp)
  mov    -0x430(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x438(%rbp),%rax
  movb   $0x0,(%rax)
  movl   $0x0,-0x418(%rbp)
  jmp    L55
L59:
  mov    -0x418(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x428(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x419(%rbp)
  mov    -0x418(%rbp),%eax
  cltq
  movzbl -0x419(%rbp),%edx
  mov    %dl,-0x410(%rbp,%rax,1)
  mov    -0x418(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x430(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x419(%rbp),%eax
  mov    %al,(%rdx)
  cmpb   $0x0,-0x419(%rbp)
  je     L11
  cmpb   $0x2f,-0x419(%rbp)
  jne    L37
  mov    -0x418(%rbp),%eax
  mov    %eax,-0x414(%rbp)
L37:
  addl   $0x1,-0x418(%rbp)
L55:
  cmpl   $0x3ff,-0x418(%rbp)
  jle    L59
  jmp    L33
L11:
  nop
L33:
  mov    -0x414(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x430(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x414(%rbp),%eax
  add    $0x1,%eax
  lea    -0x410(%rbp),%rdx
  cltq
  add    %rax,%rdx
  mov    -0x438(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  leave
  ret
```

assistant: ```c
void COM_SplitPath(const char *path, char *dir, char *fileName) {
  int i;
  int lastSlash = -1;
  char buf[1024];
  *dir = '\0';
  *fileName = '\0';
  for (i = 0; i < 1024; i++) {
    char c = path[i];
    buf[i] = c;
    dir[i] = c;
    if (!c)
      break;
    if (c == '/') {
      lastSlash = i;
    } else if (c == '.') {
    }
  }
  dir[lastSlash + 1] = '\0';
  strcpy(fileName, &buf[lastSlash + 1]);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<COM_SplitPath>:
  endbr64
  sub    $0x418,%rsp
  mov    %rdi,%r10
  mov    %rdx,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  movb   $0x0,(%rsi)
  movb   $0x0,(%rdx)
  mov    $0xffffffff,%r8d
L29:
  movzbl (%r10,%rax,1),%ecx
  mov    %cl,(%rsp,%rax,1)
  mov    %cl,(%rsi,%rax,1)
  test   %cl,%cl
  je     L13
  cmp    $0x2f,%cl
  cmove  %eax,%r8d
  add    $0x1,%rax
  cmp    $0x400,%rax
  jne    L29
L13:
  movslq %r8d,%rax
  movb   $0x0,0x1(%rsi,%rax,1)
  add    $0x1,%r8d
  movslq %r8d,%r8
  lea    (%rsp,%r8,1),%rsi
  call   <strcpy@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x418,%rsp
  ret
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void COM_SplitPath(const char *path, char *dir, char *fileName) {
  int i;
  int lastSlash = -1;
  char buf[1024];
  *dir = '\0';
  *fileName = '\0';
  for (i = 0; i < 1024; i++) {
    char c = path[i];
    buf[i] = c;
    dir[i] = c;
    if (!c)
      break;
    if (c == '/') {
      lastSlash = i;
    } else if (c == '.') {
    }
  }
  dir[lastSlash + 1] = '\0';
  strcpy(fileName, &buf[lastSlash + 1]);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<COM_SplitPath>:
  endbr64
  sub    $0x418,%rsp
  mov    %rdi,%r10
  mov    $0xffffffff,%r8d
  mov    %rdx,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r9
  movb   $0x0,(%rsi)
  movb   $0x0,(%rdx)
  jmp    L3
  nopl   (%rax)
L30:
  cmp    $0x2f,%cl
  cmove  %eax,%r8d
  add    $0x1,%rax
  cmp    $0x400,%rax
  je     L18
L3:
  movzbl (%r10,%rax,1),%ecx
  mov    %cl,(%r9,%rax,1)
  mov    %cl,(%rsi,%rax,1)
  test   %cl,%cl
  jne    L30
L18:
  add    $0x1,%r8d
  movslq %r8d,%r8
  movb   $0x0,(%rsi,%r8,1)
  lea    (%r9,%r8,1),%rsi
  call   <strcpy@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  add    $0x418,%rsp
  ret
L39:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void COM_SplitPath(const char *path, char *dir, char *fileName) {
  int i;
  int lastSlash = -1;
  char buf[1024];
  *dir = '\0';
  *fileName = '\0';
  for (i = 0; i < 1024; i++) {
    char c = path[i];
    buf[i] = c;
    dir[i] = c;
    if (!c)
      break;
    if (c == '/') {
      lastSlash = i;
    } else if (c == '.') {
    }
  }
  dir[lastSlash + 1] = '\0';
  strcpy(fileName, &buf[lastSlash + 1]);
}
```
user: What is the c source code of the assembly code below?

```asm
<COM_SplitPath>:
  endbr64
  sub    $0x418,%rsp
  mov    %rdi,%r10
  mov    $0xffffffff,%r8d
  mov    %rdx,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r9
  movb   $0x0,(%rsi)
  movb   $0x0,(%rdx)
  jmp    L24
  nopl   (%rax)
L35:
  cmp    $0x2f,%cl
  cmove  %eax,%r8d
  add    $0x1,%rax
  cmp    $0x400,%rax
  je     L16
L24:
  movzbl (%r10,%rax,1),%ecx
  mov    %cl,(%r9,%rax,1)
  mov    %cl,(%rsi,%rax,1)
  test   %cl,%cl
  jne    L35
L16:
  add    $0x1,%r8d
  movslq %r8d,%r8
  movb   $0x0,(%rsi,%r8,1)
  lea    (%r9,%r8,1),%rsi
  call   <strcpy@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x418,%rsp
  ret
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void COM_SplitPath(const char *path, char *dir, char *fileName) {
  int i;
  int lastSlash = -1;
  char buf[1024];
  *dir = '\0';
  *fileName = '\0';
  for (i = 0; i < 1024; i++) {
    char c = path[i];
    buf[i] = c;
    dir[i] = c;
    if (!c)
      break;
    if (c == '/') {
      lastSlash = i;
    } else if (c == '.') {
    }
  }
  dir[lastSlash + 1] = '\0';
  strcpy(fileName, &buf[lastSlash + 1]);
}
```