ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbx
  cmp    $0x3,%edi
  jne    L4
L17:
  mov    0x10(%rbx),%rdi
  call   <atoi@plt>
  mov    %eax,%ebp
  mov    0x8(%rbx),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
  mov    %ebp,%esi
  call   <my_dup2@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  je     L15
L14:
  mov    %ebx,%esi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L4:
  lea    D19(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L17
L15:
  lea    D26(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L14
```

tool_calls: name: parse_data
arguments: tag: D16
type_guessed: string
name: parse_data
arguments: tag: D19
type_guessed: string
name: parse_data
arguments: tag: D26
type_guessed: string

tool: D16: 
[main]: new fd is %d


tool: D19: usage: dup2 src_fd new_fd

tool: D26: dup to new_fd error!

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 3) {
    printf("usage: dup2 src_fd new_fd\n");
    exit(1);
  }
  int new_fd = new_fd = my_dup2(atoi(argv[1]), atoi(argv[2]));
  if (new_fd == -1) {
    printf("dup to new_fd error!\n");
    exit(1);
  }
  printf("\n[main]: new fd is %d\n", new_fd);
  return new_fd;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  sub    $0x8,%rsp
  cmp    $0x3,%edi
  je     L20
  lea    D27(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L20:
  mov    0x10(%r12),%rdi
  call   <atoi@plt>
  mov    0x8(%r12),%rdi
  mov    %eax,%ebp
  call   <atoi@plt>
  mov    %ebp,%esi
  mov    %eax,%edi
  call   <my_dup2@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L15
L21:
  mov    %r12d,%esi
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
L15:
  lea    D28(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L21
```

tool_calls: name: parse_data
arguments: data_label: D27
type_guessed: string
name: parse_data
arguments: data_label: D11
type_guessed: string
name: parse_data
arguments: data_label: D28
type_guessed: string

tool: D27: usage: dup2 src_fd new_fd

tool: D11: 
[main]: new fd is %d


tool: D28: dup to new_fd error!

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 3) {
    printf("usage: dup2 src_fd new_fd\n");
    exit(1);
  }
  int new_fd = new_fd = my_dup2(atoi(argv[1]), atoi(argv[2]));
  if (new_fd == -1) {
    printf("dup to new_fd error!\n");
    exit(1);
  }
  printf("\n[main]: new fd is %d\n", new_fd);
  return new_fd;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  sub    $0x8,%rsp
  cmp    $0x3,%edi
  je     L7
  lea    D29(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L7:
  mov    0x10(%r12),%rdi
  call   <atoi@plt>
  mov    0x8(%r12),%rdi
  mov    %eax,%ebp
  call   <atoi@plt>
  mov    %ebp,%esi
  mov    %eax,%edi
  call   <my_dup2@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L4
L20:
  mov    %r12d,%esi
  lea    D17(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
L4:
  lea    D3(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L20
```

tool_calls: name: extract_symbolic_data
arguments: tag: D29
type_infered: string
name: extract_symbolic_data
arguments: tag: D17
type_infered: string
name: extract_symbolic_data
arguments: tag: D3
type_infered: string

tool: D29: usage: dup2 src_fd new_fd

tool: D17: 
[main]: new fd is %d


tool: D3: dup to new_fd error!

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 3) {
    printf("usage: dup2 src_fd new_fd\n");
    exit(1);
  }
  int new_fd = new_fd = my_dup2(atoi(argv[1]), atoi(argv[2]));
  if (new_fd == -1) {
    printf("dup to new_fd error!\n");
    exit(1);
  }
  printf("\n[main]: new fd is %d\n", new_fd);
  return new_fd;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<insl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L13
L12:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  lea    0x8(%rax),%rcx
  mov    %rcx,-0x10(%rbp)
  mov    %rdx,(%rax)
L13:
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  test   %rax,%rax
  jne    L12
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void insl(unsigned long port, void *dst, unsigned long count) {
  unsigned long *p = dst;
  while (count--)
    *p++ = ({
      unsigned long __v = (*(volatile unsigned long *)(port));
      __v;
    });
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<insl>:
  endbr64
  test   %rdx,%rdx
  je     L11
  mov    $0x0,%eax
L17:
  mov    (%rdi),%rcx
  mov    %rcx,(%rsi,%rax,8)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L17
L11:
  ret
```

assistant: ```c
void insl(unsigned long port, void *dst, unsigned long count) {
  unsigned long *p = dst;
  while (count--)
    *p++ = ({
      unsigned long __v = (*(volatile unsigned long *)(port));
      __v;
    });
}
```
user: What is the c source code of the assembly code below:

```asm
<insl>:
  endbr64
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L15
  nopl   0x0(%rax,%rax,1)
L10:
  mov    (%rdi),%rcx
  mov    %rcx,(%rsi,%rax,8)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L10
L15:
  ret
```

assistant: ```c
void insl(unsigned long port, void *dst, unsigned long count) {
  unsigned long *p = dst;
  while (count--)
    *p++ = ({
      unsigned long __v = (*(volatile unsigned long *)(port));
      __v;
    });
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<insl>:
  endbr64
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L19
  nopl   0x0(%rax,%rax,1)
L15:
  mov    (%rdi),%rcx
  mov    %rcx,(%rsi,%rax,8)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L15
L19:
  ret
```

assistant: ```c
void insl(unsigned long port, void *dst, unsigned long count) {
  unsigned long *p = dst;
  while (count--)
    *p++ = ({
      unsigned long __v = (*(volatile unsigned long *)(port));
      __v;
    });
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sort3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <sort2@plt>
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <sort2@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <sort2@plt>
  nop
  leave
  ret
```

assistant: ```c
void sort3(int *a, int *b, int *c) {
  sort2(a, b);
  sort2(b, c);
  sort2(a, b);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sort3>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r12
  call   <sort2@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <sort2@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <sort2@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void sort3(int *a, int *b, int *c) {
  sort2(a, b);
  sort2(b, c);
  sort2(a, b);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sort3>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  call   <sort2@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <sort2@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1050 <sort2@plt>
```

assistant: ```c
void sort3(int *a, int *b, int *c) {
  sort2(a, b);
  sort2(b, c);
  sort2(a, b);
}
```
user: What is the c source code of the assembly code below?

```asm
<sort3>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  call   <sort2@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <sort2@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1050 <sort2@plt>
```

assistant: ```c
void sort3(int *a, int *b, int *c) {
  sort2(a, b);
  sort2(b, c);
  sort2(a, b);
}
```
user: What is the c source code of the assembly code below:

```asm
<menu_load>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D18
type_infered: string
name: disassemble_data
arguments: data_label: D35
type_infered: string
name: disassemble_data
arguments: data_label: D8
type_infered: string
name: disassemble_data
arguments: data_label: D32
type_infered: string
name: disassemble_data
arguments: data_label: D40
type_infered: string

tool: D18: 
	************ Type of Load ************


tool: D35: 	1. Centrifugal Pump, Fan, Light machine tool, Conveyors.


tool: D8: 	2. Heavy duty fan, Blower, Compressor, Reciprocating Pump, Line shaft, Heavy duty Machine.


tool: D32: 	3. Vacuum Pump, Rolling Mills, Hammers, Grinders 


tool: D40: 	4. Others


assistant: ```c
void menu_load() {
  printf("\n\t************ Type of Load ************\n");
  printf("\t1. Centrifugal Pump, Fan, Light machine tool, Conveyors.\n");
  printf("\t2. Heavy duty fan, Blower, Compressor, Reciprocating Pump, Line shaft, Heavy duty Machine.\n");
  printf("\t3. Vacuum Pump, Rolling Mills, Hammers, Grinders \n");
  printf("\t4. Others\n");
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<menu_load>:
  endbr64
  sub    $0x8,%rsp
  lea    D21(%rip),%rdi
  call   <printf@plt>
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    D15(%rip),%rdi
  call   <printf@plt>
  lea    D48(%rip),%rdi
  call   <printf@plt>
  lea    D13(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D21
type_infered: string
name: parse_data
arguments: data_label: D4
type_infered: string
name: parse_data
arguments: data_label: D15
type_infered: string
name: parse_data
arguments: data_label: D48
type_infered: string
name: parse_data
arguments: data_label: D13
type_infered: string

tool: D21: 
	************ Type of Load ************


tool: D4: 	1. Centrifugal Pump, Fan, Light machine tool, Conveyors.


tool: D15: 	2. Heavy duty fan, Blower, Compressor, Reciprocating Pump, Line shaft, Heavy duty Machine.


tool: D48: 	3. Vacuum Pump, Rolling Mills, Hammers, Grinders 


tool: D13: 	4. Others


assistant: ```c
void menu_load() {
  printf("\n\t************ Type of Load ************\n");
  printf("\t1. Centrifugal Pump, Fan, Light machine tool, Conveyors.\n");
  printf("\t2. Heavy duty fan, Blower, Compressor, Reciprocating Pump, Line shaft, Heavy duty Machine.\n");
  printf("\t3. Vacuum Pump, Rolling Mills, Hammers, Grinders \n");
  printf("\t4. Others\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<menu_load>:
  endbr64
  sub    $0x8,%rsp
  lea    D17(%rip),%rdi
  call   <printf@plt>
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    D39(%rip),%rdi
  call   <printf@plt>
  lea    D3(%rip),%rdi
  call   <printf@plt>
  lea    D46(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D17
type_guessed: string
name: retrieve_data
arguments: tag: D4
type_guessed: string
name: retrieve_data
arguments: tag: D39
type_guessed: string
name: retrieve_data
arguments: tag: D3
type_guessed: string
name: retrieve_data
arguments: tag: D46
type_guessed: string

tool: D17: 
	************ Type of Load ************


tool: D4: 	1. Centrifugal Pump, Fan, Light machine tool, Conveyors.


tool: D39: 	2. Heavy duty fan, Blower, Compressor, Reciprocating Pump, Line shaft, Heavy duty Machine.


tool: D3: 	3. Vacuum Pump, Rolling Mills, Hammers, Grinders 


tool: D46: 	4. Others


assistant: ```c
void menu_load() {
  printf("\n\t************ Type of Load ************\n");
  printf("\t1. Centrifugal Pump, Fan, Light machine tool, Conveyors.\n");
  printf("\t2. Heavy duty fan, Blower, Compressor, Reciprocating Pump, Line shaft, Heavy duty Machine.\n");
  printf("\t3. Vacuum Pump, Rolling Mills, Hammers, Grinders \n");
  printf("\t4. Others\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<menu_load>:
  endbr64
  sub    $0x8,%rsp
  lea    D33(%rip),%rdi
  call   <printf@plt>
  lea    D2(%rip),%rdi
  call   <printf@plt>
  lea    D41(%rip),%rdi
  call   <printf@plt>
  lea    D15(%rip),%rdi
  call   <printf@plt>
  lea    D29(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D33
datatype: string
name: retrieve_data
arguments: tag: D2
datatype: string
name: retrieve_data
arguments: tag: D41
datatype: string
name: retrieve_data
arguments: tag: D15
datatype: string
name: retrieve_data
arguments: tag: D29
datatype: string

tool: D33: 
	************ Type of Load ************


tool: D2: 	1. Centrifugal Pump, Fan, Light machine tool, Conveyors.


tool: D41: 	2. Heavy duty fan, Blower, Compressor, Reciprocating Pump, Line shaft, Heavy duty Machine.


tool: D15: 	3. Vacuum Pump, Rolling Mills, Hammers, Grinders 


tool: D29: 	4. Others


assistant: ```c
void menu_load() {
  printf("\n\t************ Type of Load ************\n");
  printf("\t1. Centrifugal Pump, Fan, Light machine tool, Conveyors.\n");
  printf("\t2. Heavy duty fan, Blower, Compressor, Reciprocating Pump, Line shaft, Heavy duty Machine.\n");
  printf("\t3. Vacuum Pump, Rolling Mills, Hammers, Grinders \n");
  printf("\t4. Others\n");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<FindPathInPaths>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x13000(%rsp),%r11
L109:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L109
  sub    $0x8c0,%rsp
  mov    %rdi,-0x138a8(%rbp)
  mov    %rsi,-0x138b0(%rbp)
  mov    %edx,-0x138b4(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x13890(%rbp),%rax
  mov    $0x13880,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0xffffffff,-0x13890(%rbp)
  movl   $0x0,-0x13894(%rbp)
  movl   $0x0,-0x1389c(%rbp)
  jmp    L71
L25:
  mov    -0x1389c(%rbp),%eax
  cltq
  movl   $0xffffffff,-0x13890(%rbp,%rax,4)
  addl   $0x1,-0x1389c(%rbp)
L71:
  cmpl   $0x4e1f,-0x1389c(%rbp)
  jle    L25
  movl   $0x0,-0x1389c(%rbp)
  jmp    L116
L76:
  movl   $0x0,-0x13898(%rbp)
  jmp    L20
L91:
  mov    -0x1389c(%rbp),%eax
  cltq
  imul   $0x4e20,%rax,%rdx
  mov    -0x138a8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x13898(%rbp),%eax
  cltq
  movzwl (%rdx,%rax,2),%eax
  cwtl
  cmp    %eax,-0x138b4(%rbp)
  jne    L90
  mov    -0x13894(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x13894(%rbp)
  cltq
  mov    -0x1389c(%rbp),%edx
  mov    %edx,-0x13890(%rbp,%rax,4)
L90:
  addl   $0x1,-0x13898(%rbp)
L20:
  cmpl   $0x270f,-0x13898(%rbp)
  jle    L91
  addl   $0x1,-0x1389c(%rbp)
L116:
  cmpl   $0x4e1f,-0x1389c(%rbp)
  jle    L76
  movl   $0x0,-0x1389c(%rbp)
  jmp    L31
L17:
  mov    -0x1389c(%rbp),%eax
  cltq
  mov    -0x13890(%rbp,%rax,4),%eax
  cmp    $0xffffffff,%eax
  je     L3
  mov    -0x1389c(%rbp),%eax
  cltq
  mov    -0x13890(%rbp,%rax,4),%esi
  mov    -0x138b4(%rbp),%ecx
  mov    -0x138b0(%rbp),%rdx
  mov    -0x138a8(%rbp),%rax
  mov    %rax,%rdi
  call   <PrintSinglePath@plt>
L3:
  addl   $0x1,-0x1389c(%rbp)
L31:
  cmpl   $0x4e1f,-0x1389c(%rbp)
  jle    L17
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L43
  call   <__stack_chk_fail@plt>
L43:
  leave
  ret
```

assistant: ```c
void FindPathInPaths(short int paths[20000][10000], char words[2400][5], int endVertice) {
  int i, j;
  int finalPaths[20000] = {-1};
  int finalPathCount = 0;
  for (i = 0; i < 20000; i++)
    finalPaths[i] = -1;
  for (i = 0; i < 20000; i++) {
    for (j = 0; j < 10000; j++) {
      if (paths[i][j] == endVertice)
        finalPaths[finalPathCount++] = i;
    }
  }
  for (i = 0; i < 20000; i++) {
    if (finalPaths[i] != -1) {
      PrintSinglePath(paths, finalPaths[i], words, endVertice);
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<FindPathInPaths>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x13000(%rsp),%r11
L4:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L4
  sub    $0x898,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r14
  mov    %edx,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x13888(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r15
  mov    $0x13880,%edx
  mov    $0x0,%esi
  mov    %r15,%rdi
  call   <memset@plt>
  movl   $0xffffffff,(%rsp)
  mov    %r15,%rbp
  lea    0x13880(%rsp),%r12
L28:
  movl   $0xffffffff,(%rax)
  add    $0x4,%rax
  cmp    %r12,%rax
  jne    L28
  lea    0x4e20(%r13),%rcx
  mov    $0x0,%esi
  mov    $0x0,%edi
  jmp    L21
L67:
  add    $0x2,%rax
  cmp    %rcx,%rax
  je     L5
L73:
  movswl (%rax),%edx
  cmp    %ebx,%edx
  jne    L67
  movslq %esi,%rdx
  mov    %edi,(%rsp,%rdx,4)
  lea    0x1(%rsi),%esi
  jmp    L67
L5:
  add    $0x1,%edi
  add    $0x4e20,%rcx
  cmp    $0x4e20,%edi
  je     L49
L21:
  lea    -0x4e20(%rcx),%rax
  jmp    L73
L46:
  add    $0x4,%rbp
  cmp    %r12,%rbp
  je     L29
L49:
  mov    0x0(%rbp),%esi
  cmp    $0xffffffff,%esi
  je     L46
  mov    %ebx,%ecx
  mov    %r14,%rdx
  mov    %r13,%rdi
  call   <PrintSinglePath@plt>
  jmp    L46
L29:
  mov    0x13888(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x13898,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void FindPathInPaths(short int paths[20000][10000], char words[2400][5], int endVertice) {
  int i, j;
  int finalPaths[20000] = {-1};
  int finalPathCount = 0;
  for (i = 0; i < 20000; i++)
    finalPaths[i] = -1;
  for (i = 0; i < 20000; i++) {
    for (j = 0; j < 10000; j++) {
      if (paths[i][j] == endVertice)
        finalPaths[finalPathCount++] = i;
    }
  }
  for (i = 0; i < 20000; i++) {
    if (finalPaths[i] != -1) {
      PrintSinglePath(paths, finalPaths[i], words, endVertice);
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<FindPathInPaths>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x13000(%rsp),%r11
L42:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L42
  sub    $0x890,%rsp
  mov    %rsp,%r13
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %edx,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x13888(%rsp)
  xor    %eax,%eax
  mov    $0xff,%esi
  mov    %r13,%rdi
  mov    $0x13880,%edx
  call   <memset@plt>
  lea    0x4e20(%rbx),%rcx
  xor    %esi,%esi
  xor    %edi,%edi
L16:
  lea    -0x4e20(%rcx),%rax
  nop
L31:
  movswl (%rax),%edx
  cmp    %r12d,%edx
  jne    L67
  movslq %esi,%rdx
  add    $0x1,%esi
  mov    %edi,(%rsp,%rdx,4)
L67:
  add    $0x2,%rax
  cmp    %rax,%rcx
  jne    L31
  add    $0x1,%edi
  add    $0x4e20,%rcx
  cmp    $0x4e20,%edi
  jne    L16
  lea    0x13880(%rsp),%r14
L2:
  mov    0x0(%r13),%esi
  cmp    $0xffffffff,%esi
  je     L8
  mov    %r12d,%ecx
  mov    %rbp,%rdx
  mov    %rbx,%rdi
  call   <PrintSinglePath@plt>
L8:
  add    $0x4,%r13
  cmp    %r13,%r14
  jne    L2
  mov    0x13888(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L57
  add    $0x13890,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L57:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void FindPathInPaths(short int paths[20000][10000], char words[2400][5], int endVertice) {
  int i, j;
  int finalPaths[20000] = {-1};
  int finalPathCount = 0;
  for (i = 0; i < 20000; i++)
    finalPaths[i] = -1;
  for (i = 0; i < 20000; i++) {
    for (j = 0; j < 10000; j++) {
      if (paths[i][j] == endVertice)
        finalPaths[finalPathCount++] = i;
    }
  }
  for (i = 0; i < 20000; i++) {
    if (finalPaths[i] != -1) {
      PrintSinglePath(paths, finalPaths[i], words, endVertice);
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<FindPathInPaths>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x13000(%rsp),%r11
L54:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L54
  sub    $0x890,%rsp
  mov    %rsp,%r13
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %edx,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x13888(%rsp)
  xor    %eax,%eax
  mov    $0xff,%esi
  mov    %r13,%rdi
  mov    $0x13880,%edx
  call   <memset@plt>
  lea    0x4e20(%rbx),%rcx
  xor    %esi,%esi
  xor    %edi,%edi
L8:
  lea    -0x4e20(%rcx),%rax
  nop
L61:
  movswl (%rax),%edx
  cmp    %r12d,%edx
  jne    L57
  movslq %esi,%rdx
  add    $0x1,%esi
  mov    %edi,(%rsp,%rdx,4)
L57:
  add    $0x2,%rax
  cmp    %rax,%rcx
  jne    L61
  add    $0x1,%edi
  add    $0x4e20,%rcx
  cmp    $0x4e20,%edi
  jne    L8
  lea    0x13880(%rsp),%r14
L5:
  mov    0x0(%r13),%esi
  cmp    $0xffffffff,%esi
  je     L0
  mov    %r12d,%ecx
  mov    %rbp,%rdx
  mov    %rbx,%rdi
  call   <PrintSinglePath@plt>
L0:
  add    $0x4,%r13
  cmp    %r13,%r14
  jne    L5
  mov    0x13888(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  add    $0x13890,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L37:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void FindPathInPaths(short int paths[20000][10000], char words[2400][5], int endVertice) {
  int i, j;
  int finalPaths[20000] = {-1};
  int finalPathCount = 0;
  for (i = 0; i < 20000; i++)
    finalPaths[i] = -1;
  for (i = 0; i < 20000; i++) {
    for (j = 0; j < 10000; j++) {
      if (paths[i][j] == endVertice)
        finalPaths[finalPathCount++] = i;
    }
  }
  for (i = 0; i < 20000; i++) {
    if (finalPaths[i] != -1) {
      PrintSinglePath(paths, finalPaths[i], words, endVertice);
    }
  }
}
```