ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<get_last_winner_str>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edi,%ebx
  mov    $0x28,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L40
  cmp    $0x2,%ebx
  je     L15
  test   %ebx,%ebx
  jne    L31
  mov    %r12,%rdi
  lea    D22(%rip),%rsi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L40:
  cmp    $0x2,%ebx
  je     L15
  test   %ebx,%ebx
  jne    L38
  mov    %r12,%rdi
  lea    D14(%rip),%rsi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L15:
  mov    %r12,%rdi
  lea    D37(%rip),%rsi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L31:
  mov    %r12,%rdi
  lea    D6(%rip),%rsi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L38:
  mov    %r12,%rdi
  lea    D15(%rip),%rsi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D22
type_guessed: string
name: extract_symbolic_data
arguments: label: D14
type_guessed: string
name: extract_symbolic_data
arguments: label: D37
type_guessed: string
name: extract_symbolic_data
arguments: label: D6
type_guessed: string
name: extract_symbolic_data
arguments: label: D15
type_guessed: string

tool: D22: You Won

tool: D14: They Won

tool: D37: Waiting

tool: D6: They won

tool: D15: You won

assistant: ```c
char *get_last_winner_str(int last, char *name, char *u1) {
  char *last_str = malloc(40);
  if (strcmp(name, u1) == 0) {
    if (last == 2)
      strcpy(last_str, "Waiting");
    else if (last == 0)
      strcpy(last_str, "You Won");
    else
      strcpy(last_str, "They won");
  } else {
    if (last == 2)
      strcpy(last_str, "Waiting");
    else if (last == 0)
      strcpy(last_str, "They Won");
    else
      strcpy(last_str, "You won");
  }
  return last_str;
}
```
user: Translate the below assembly code into C programming language:

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdx
  lea    D3(%rip),%rax
  mov    %rax,%rsi
  mov    $0x2,%edi
  call   <errx@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D3
type_infered: string

tool: D3: usage: %s [-b BITS] [-s SEED] [-j JUMPS] [-r RUNS] [-m MASK_HEX] [-c CPU]

assistant: ```c
void usage(char **argv) {
  errx(2,
       "usage: %s [-b BITS] [-s SEED] [-j JUMPS] [-r RUNS] [-m MASK_HEX]"
       " [-c CPU]",
       argv[0]);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  mov    (%rdi),%rdx
  lea    D2(%rip),%rsi
  mov    $0x2,%edi
  call   <errx@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D2
datatype: string

tool: D2: usage: %s [-b BITS] [-s SEED] [-j JUMPS] [-r RUNS] [-m MASK_HEX] [-c CPU]

assistant: ```c
void usage(char **argv) {
  errx(2,
       "usage: %s [-b BITS] [-s SEED] [-j JUMPS] [-r RUNS] [-m MASK_HEX]"
       " [-c CPU]",
       argv[0]);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<usage>:
  endbr64
  mov    (%rdi),%rdx
  lea    D4(%rip),%rsi
  mov    $0x2,%edi
  jmp    1050 <errx@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D4
type: string

tool: D4: usage: %s [-b BITS] [-s SEED] [-j JUMPS] [-r RUNS] [-m MASK_HEX] [-c CPU]

assistant: ```c
void usage(char **argv) {
  errx(2,
       "usage: %s [-b BITS] [-s SEED] [-j JUMPS] [-r RUNS] [-m MASK_HEX]"
       " [-c CPU]",
       argv[0]);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<usage>:
  endbr64
  mov    (%rdi),%rdx
  lea    D1(%rip),%rsi
  mov    $0x2,%edi
  jmp    1050 <errx@plt>
```

tool_calls: name: read_data
arguments: label: D1
type_guessed: string

tool: D1: usage: %s [-b BITS] [-s SEED] [-j JUMPS] [-r RUNS] [-m MASK_HEX] [-c CPU]

assistant: ```c
void usage(char **argv) {
  errx(2,
       "usage: %s [-b BITS] [-s SEED] [-j JUMPS] [-r RUNS] [-m MASK_HEX]"
       " [-c CPU]",
       argv[0]);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<settimeout>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x3e8,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  imul   $0x3e8,%edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <getmonotime@plt>
  mov    -0x18(%rbp),%rdx
  lea    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <timeradd@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
void settimeout(struct timeval *timeout, int msec) {
  struct timeval s = {
      .tv_sec = (msec / 1000),
      .tv_usec = (msec % 1000) * 1000,
  };
  getmonotime(timeout);
  timeradd(timeout, &s, timeout);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<settimeout>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movslq %esi,%rax
  imul   $0x10624dd3,%rax,%rax
  sar    $0x26,%rax
  mov    %esi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,(%rsp)
  imul   $0x3e8,%eax,%eax
  sub    %eax,%esi
  imul   $0x3e8,%esi,%esi
  mov    %esi,0x4(%rsp)
  call   <getmonotime@plt>
  mov    %rsp,%rsi
  mov    %rbx,%rdx
  mov    %rbx,%rdi
  call   <timeradd@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x10,%rsp
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void settimeout(struct timeval *timeout, int msec) {
  struct timeval s = {
      .tv_sec = (msec / 1000),
      .tv_usec = (msec % 1000) * 1000,
  };
  getmonotime(timeout);
  timeradd(timeout, &s, timeout);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<settimeout>:
  endbr64
  push   %rbp
  mov    %esi,%edx
  mov    %rdi,%rbp
  sar    $0x1f,%edx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movslq %esi,%rax
  imul   $0x10624dd3,%rax,%rax
  sar    $0x26,%rax
  sub    %edx,%eax
  mov    %eax,(%rsp)
  imul   $0x3e8,%eax,%eax
  sub    %eax,%esi
  imul   $0x3e8,%esi,%esi
  mov    %esi,0x4(%rsp)
  call   <getmonotime@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdx
  mov    %rbp,%rdi
  call   <timeradd@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x10,%rsp
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void settimeout(struct timeval *timeout, int msec) {
  struct timeval s = {
      .tv_sec = (msec / 1000),
      .tv_usec = (msec % 1000) * 1000,
  };
  getmonotime(timeout);
  timeradd(timeout, &s, timeout);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<settimeout>:
  endbr64
  push   %rbp
  mov    %esi,%edx
  mov    %rdi,%rbp
  sar    $0x1f,%edx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movslq %esi,%rax
  imul   $0x10624dd3,%rax,%rax
  sar    $0x26,%rax
  sub    %edx,%eax
  imul   $0x3e8,%eax,%edx
  movd   %eax,%xmm0
  sub    %edx,%esi
  imul   $0x3e8,%esi,%esi
  movd   %esi,%xmm1
  punpckldq %xmm1,%xmm0
  movq   %xmm0,(%rsp)
  call   <getmonotime@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdx
  mov    %rbp,%rdi
  call   <timeradd@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x10,%rsp
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void settimeout(struct timeval *timeout, int msec) {
  struct timeval s = {
      .tv_sec = (msec / 1000),
      .tv_usec = (msec % 1000) * 1000,
  };
  getmonotime(timeout);
  timeradd(timeout, &s, timeout);
}
```
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
  mov    $0x5,%esi
  mov    $0x4,%edi
  call   <adder@plt>
  mov    %eax,%ecx
  mov    $0x5,%edx
  mov    $0x4,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x5,%esi
  mov    $0x4,%edi
  call   <subber@plt>
  mov    %eax,%ecx
  mov    $0x5,%edx
  mov    $0x4,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x5,%esi
  mov    $0x4,%edi
  call   <multer@plt>
  mov    %eax,%ecx
  mov    $0x5,%edx
  mov    $0x4,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D22
type_guessed: string
name: read_data
arguments: data_tag: D27
type_guessed: string
name: read_data
arguments: data_tag: D29
type_guessed: string

tool: D22: adder:
	%d - %d = %d


tool: D27: subber:
	%d - %d = %d


tool: D29: multer:
	%d - %d = %d


assistant: ```c
int main(void) {
  printf("adder:\n\t%d - %d = %d\n", 4, 5, adder(4, 5));
  printf("subber:\n\t%d - %d = %d\n", 4, 5, subber(4, 5));
  printf("multer:\n\t%d - %d = %d\n", 4, 5, multer(4, 5));
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%esi
  mov    $0x4,%edi
  call   <adder@plt>
  mov    %eax,%ecx
  mov    $0x5,%edx
  mov    $0x4,%esi
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    $0x5,%esi
  mov    $0x4,%edi
  call   <subber@plt>
  mov    %eax,%ecx
  mov    $0x5,%edx
  mov    $0x4,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x5,%esi
  mov    $0x4,%edi
  call   <multer@plt>
  mov    %eax,%ecx
  mov    $0x5,%edx
  mov    $0x4,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D17
type_infered: string
name: retrieve_data
arguments: data_tag: D9
type_infered: string
name: retrieve_data
arguments: data_tag: D7
type_infered: string

tool: D17: adder:
	%d - %d = %d


tool: D9: subber:
	%d - %d = %d


tool: D7: multer:
	%d - %d = %d


assistant: ```c
int main(void) {
  printf("adder:\n\t%d - %d = %d\n", 4, 5, adder(4, 5));
  printf("subber:\n\t%d - %d = %d\n", 4, 5, subber(4, 5));
  printf("multer:\n\t%d - %d = %d\n", 4, 5, multer(4, 5));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%esi
  mov    $0x4,%edi
  call   <adder@plt>
  mov    $0x5,%edx
  mov    $0x4,%esi
  lea    D20(%rip),%rdi
  mov    %eax,%ecx
  call   <printf@plt>
  mov    $0x5,%esi
  mov    $0x4,%edi
  call   <subber@plt>
  mov    $0x5,%edx
  mov    $0x4,%esi
  lea    D10(%rip),%rdi
  mov    %eax,%ecx
  call   <printf@plt>
  mov    $0x5,%esi
  mov    $0x4,%edi
  call   <multer@plt>
  mov    $0x5,%edx
  mov    $0x4,%esi
  lea    D28(%rip),%rdi
  mov    %eax,%ecx
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D20
data_type: string
name: read_data
arguments: tag: D10
data_type: string
name: read_data
arguments: tag: D28
data_type: string

tool: D20: adder:
	%d - %d = %d


tool: D10: subber:
	%d - %d = %d


tool: D28: multer:
	%d - %d = %d


assistant: ```c
int main(void) {
  printf("adder:\n\t%d - %d = %d\n", 4, 5, adder(4, 5));
  printf("subber:\n\t%d - %d = %d\n", 4, 5, subber(4, 5));
  printf("multer:\n\t%d - %d = %d\n", 4, 5, multer(4, 5));
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%esi
  mov    $0x4,%edi
  call   <adder@plt>
  mov    $0x5,%edx
  mov    $0x4,%esi
  lea    D11(%rip),%rdi
  mov    %eax,%ecx
  call   <printf@plt>
  mov    $0x5,%esi
  mov    $0x4,%edi
  call   <subber@plt>
  mov    $0x5,%edx
  mov    $0x4,%esi
  lea    D21(%rip),%rdi
  mov    %eax,%ecx
  call   <printf@plt>
  mov    $0x5,%esi
  mov    $0x4,%edi
  call   <multer@plt>
  mov    $0x5,%edx
  mov    $0x4,%esi
  lea    D19(%rip),%rdi
  mov    %eax,%ecx
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D11
type_guessed: string
name: disassemble_data
arguments: tag: D21
type_guessed: string
name: disassemble_data
arguments: tag: D19
type_guessed: string

tool: D11: adder:
	%d - %d = %d


tool: D21: subber:
	%d - %d = %d


tool: D19: multer:
	%d - %d = %d


assistant: ```c
int main(void) {
  printf("adder:\n\t%d - %d = %d\n", 4, 5, adder(4, 5));
  printf("subber:\n\t%d - %d = %d\n", 4, 5, subber(4, 5));
  printf("multer:\n\t%d - %d = %d\n", 4, 5, multer(4, 5));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<simplify>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  test   %eax,%eax
  je     L120
  mov    %eax,%esi
  mov    $0x2,%edx
  mov    $0x1,%ebx
  mov    $0x0,%eax
  mov    $0x0,%r8d
  mov    $0x0,%r10d
  jmp    L7
L66:
  mov    %edx,%edi
  add    %rbp,%rdi
  cmpb   $0x2e,(%rdi)
  je     L117
  movb   $0x0,0x1(%rbp,%rax,1)
  cmpb   $0x2f,(%rdi)
  je     L159
  add    $0x1,%r8d
  mov    %eax,%r10d
L151:
  add    $0x1,%rax
  lea    0x1(%rbx),%ecx
  add    $0x1,%edx
  cmp    %ebx,%esi
  je     L1
  mov    %ecx,%ebx
L7:
  mov    %eax,%ecx
  cmpb   $0x2f,0x0(%rbp,%rax,1)
  jne    L151
  cmpb   $0x2e,0x1(%rbp,%rax,1)
  je     L66
  mov    %eax,%r10d
  jmp    L151
L117:
  test   %eax,%eax
  je     L151
  lea    0x1(%r10),%edi
  cmp    %edx,%edi
  ja     L150
  movslq %edi,%rdi
  sub    %r10d,%r8d
L70:
  movb   $0x0,0x0(%rbp,%rdi,1)
  lea    (%r8,%rdi,1),%r9d
  add    $0x1,%rdi
  cmp    %edi,%edx
  jae    L70
L39:
  lea    0x3(%rcx),%edi
  add    %rbp,%rdi
  mov    %r9d,%r8d
  mov    %ecx,%r10d
  cmpb   $0x2f,(%rdi)
  jne    L151
  movb   $0x0,(%rdi)
  lea    0x1(%r9),%r8d
  jmp    L151
L150:
  mov    %r8d,%r9d
  jmp    L39
L159:
  movb   $0x0,(%rdi)
  add    $0x2,%r8d
  mov    %eax,%r10d
  jmp    L151
L115:
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  je     L24
L0:
  movzbl (%rdx),%ecx
  test   %cl,%cl
  je     L115
  mov    %esi,%r8d
  mov    %cl,(%rax,%r8,1)
  lea    0x1(%rsi),%esi
  jmp    L115
L1:
  mov    %ebx,%edi
  sub    %r8d,%edi
  call   <malloc@plt>
  mov    %rbp,%rdx
  lea    -0x1(%rbx),%ecx
  lea    0x1(%rbp,%rcx,1),%rdi
  mov    $0x0,%esi
  jmp    L0
L120:
  mov    $0x0,%edi
  call   <malloc@plt>
L24:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *simplify(char *path) {
  unsigned int len_path = strlen(path);
  unsigned int last_slash = 0;
  unsigned int pointer = 0;
  unsigned int removed = 0;
  while (pointer != len_path) {
    if (path[pointer] == '/') {
      if (path[pointer + 1] == '.' && path[pointer + 2] == '.') {
        if (pointer == 0) {
          pointer++;
          continue;
        }
        for (int i = last_slash + 1; i <= pointer + 2; i++) {
          path[i] = 0;
          removed++;
        }
        if (path[pointer + 3] == '/') {
          path[pointer + 3] = 0;
          removed++;
        }
      } else if (path[pointer + 1] == '.') {
        path[pointer + 1] = 0;
        removed++;
        if (path[pointer + 2] == '/') {
          path[pointer + 2] = 0;
          removed++;
        }
      }
      last_slash = pointer;
    }
    pointer++;
  }
  char *tmp = malloc(len_path - removed);
  unsigned int j = 0;
  for (int i = 0; i < len_path; i++) {
    if (path[i] != 0) {
      tmp[j++] = path[i];
    }
  }
  return tmp;
}
```
user: What is the c source code of the assembly code below:

```asm
<simplify>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  test   %eax,%eax
  je     L59
  mov    %eax,%ecx
  xor    %esi,%esi
  xor    %ebp,%ebp
  xor    %r9d,%r9d
  nopl   0x0(%rax,%rax,1)
L66:
  mov    %ebp,%edx
  lea    0x1(%rbp),%eax
  cmpb   $0x2f,(%rbx,%rdx,1)
  je     L12
L17:
  mov    %eax,%ebp
L50:
  cmp    %ecx,%ebp
  jne    L66
  mov    %ebp,%edi
  sub    %esi,%edi
  call   <malloc@plt>
  lea    -0x1(%rbp),%ecx
  mov    %rbx,%rdx
  xor    %esi,%esi
  lea    0x1(%rbx,%rcx,1),%rdi
  nop
L55:
  movzbl (%rdx),%ecx
  test   %cl,%cl
  je     L103
  mov    %esi,%r8d
  add    $0x1,%esi
  mov    %cl,(%rax,%r8,1)
L103:
  add    $0x1,%rdx
  cmp    %rdx,%rdi
  jne    L55
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L12:
  mov    %eax,%edx
  add    %rbx,%rdx
  cmpb   $0x2e,(%rdx)
  je     L43
  mov    %ebp,%r9d
  jmp    L17
  nopw   0x0(%rax,%rax,1)
L43:
  lea    0x2(%rbp),%edi
  mov    %rdi,%r8
  add    %rbx,%rdi
  cmpb   $0x2e,(%rdi)
  je     L154
  movb   $0x0,(%rdx)
  cmpb   $0x2f,(%rdi)
  je     L129
  add    $0x1,%esi
  mov    %ebp,%r9d
  jmp    L17
  xchg   %ax,%ax
L154:
  test   %ebp,%ebp
  jne    L148
  mov    $0x1,%ebp
  jmp    L50
  xchg   %ax,%ax
L129:
  movb   $0x0,(%rdi)
  add    $0x2,%esi
  mov    %ebp,%r9d
  jmp    L17
  xchg   %ax,%ax
L148:
  lea    0x1(%r9),%edx
  cmp    %edx,%r8d
  jb     L159
  sub    %r9d,%esi
  movslq %edx,%rdx
  mov    %esi,%edi
  nopl   0x0(%rax)
L168:
  movb   $0x0,(%rbx,%rdx,1)
  lea    (%rdi,%rdx,1),%esi
  add    $0x1,%rdx
  cmp    %edx,%r8d
  jae    L168
L159:
  lea    0x3(%rbp),%edx
  mov    %ebp,%r9d
  add    %rbx,%rdx
  cmpb   $0x2f,(%rdx)
  jne    L17
  movb   $0x0,(%rdx)
  add    $0x1,%esi
  jmp    L17
  nopl   (%rax)
L59:
  add    $0x8,%rsp
  xor    %edi,%edi
  pop    %rbx
  pop    %rbp
  jmp    1070 <malloc@plt>
```

assistant: ```c
char *simplify(char *path) {
  unsigned int len_path = strlen(path);
  unsigned int last_slash = 0;
  unsigned int pointer = 0;
  unsigned int removed = 0;
  while (pointer != len_path) {
    if (path[pointer] == '/') {
      if (path[pointer + 1] == '.' && path[pointer + 2] == '.') {
        if (pointer == 0) {
          pointer++;
          continue;
        }
        for (int i = last_slash + 1; i <= pointer + 2; i++) {
          path[i] = 0;
          removed++;
        }
        if (path[pointer + 3] == '/') {
          path[pointer + 3] = 0;
          removed++;
        }
      } else if (path[pointer + 1] == '.') {
        path[pointer + 1] = 0;
        removed++;
        if (path[pointer + 2] == '/') {
          path[pointer + 2] = 0;
          removed++;
        }
      }
      last_slash = pointer;
    }
    pointer++;
  }
  char *tmp = malloc(len_path - removed);
  unsigned int j = 0;
  for (int i = 0; i < len_path; i++) {
    if (path[i] != 0) {
      tmp[j++] = path[i];
    }
  }
  return tmp;
}
```
user: Translate the below assembly code into C programming language:

```asm
<recursiveBinarySearch>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jle    L38
  mov    $0xffffffff,%eax
  jmp    L29
L38:
  mov    -0x20(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x24(%rbp)
  jne    L7
  mov    -0x4(%rbp),%eax
  jmp    L29
L7:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x24(%rbp)
  jge    L42
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x20(%rbp)
  jmp    L56
L42:
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x1c(%rbp)
L56:
  mov    -0x24(%rbp),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <recursiveBinarySearch>
L29:
  leave
  ret
```

assistant: ```c
int recursiveBinarySearch(int list[], int left, int right, int key) {
  int middle;
  if (left > right)
    return -1;
  else {
    middle = left + (right - left) / 2;
    if (key == list[middle])
      return middle;
    else if (key < list[middle])
      right = middle - 1;
    else
      left = middle + 1;
    return recursiveBinarySearch(list, left, right, key);
  }
  return -1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<recursiveBinarySearch>:
  endbr64
  cmp    %edx,%esi
  jg     L16
  mov    %edx,%r8d
  sub    %esi,%r8d
  mov    %r8d,%eax
  shr    $0x1f,%eax
  add    %r8d,%eax
  sar    %eax
  add    %esi,%eax
  movslq %eax,%r8
  mov    (%rdi,%r8,4),%r8d
  cmp    %ecx,%r8d
  je     L3
  jle    L11
  lea    -0x1(%rax),%edx
L24:
  sub    $0x8,%rsp
  call   <recursiveBinarySearch>
  add    $0x8,%rsp
  ret
L11:
  lea    0x1(%rax),%esi
  jmp    L24
L16:
  mov    $0xffffffff,%eax
  ret
L3:
  ret
```

assistant: ```c
int recursiveBinarySearch(int list[], int left, int right, int key) {
  int middle;
  if (left > right)
    return -1;
  else {
    middle = left + (right - left) / 2;
    if (key == list[middle])
      return middle;
    else if (key < list[middle])
      right = middle - 1;
    else
      left = middle + 1;
    return recursiveBinarySearch(list, left, right, key);
  }
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<recursiveBinarySearch>:
  endbr64
  nopl   0x0(%rax)
L31:
  cmp    %edx,%esi
  jg     L34
  mov    %edx,%eax
  sub    %esi,%eax
  sar    %eax
  add    %esi,%eax
  movslq %eax,%r8
  cmp    %ecx,(%rdi,%r8,4)
  je     L8
  jle    L5
  lea    -0x1(%rax),%edx
  jmp    L31
  nopl   0x0(%rax)
L5:
  lea    0x1(%rax),%esi
  jmp    L31
  nopl   (%rax)
L34:
  mov    $0xffffffff,%eax
L8:
  ret
```

assistant: ```c
int recursiveBinarySearch(int list[], int left, int right, int key) {
  int middle;
  if (left > right)
    return -1;
  else {
    middle = left + (right - left) / 2;
    if (key == list[middle])
      return middle;
    else if (key < list[middle])
      right = middle - 1;
    else
      left = middle + 1;
    return recursiveBinarySearch(list, left, right, key);
  }
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<recursiveBinarySearch>:
  endbr64
  nopl   0x0(%rax)
L8:
  cmp    %edx,%esi
  jg     L13
L15:
  mov    %edx,%eax
  sub    %esi,%eax
  sar    %eax
  add    %esi,%eax
  movslq %eax,%r8
  cmp    %ecx,(%rdi,%r8,4)
  je     L7
  jle    L0
  lea    -0x1(%rax),%edx
  cmp    %esi,%edx
  jge    L15
L13:
  mov    $0xffffffff,%eax
L7:
  ret
  nopl   0x0(%rax)
L0:
  lea    0x1(%rax),%esi
  jmp    L8
```

assistant: ```c
int recursiveBinarySearch(int list[], int left, int right, int key) {
  int middle;
  if (left > right)
    return -1;
  else {
    middle = left + (right - left) / 2;
    if (key == list[middle])
      return middle;
    else if (key < list[middle])
      right = middle - 1;
    else
      left = middle + 1;
    return recursiveBinarySearch(list, left, right, key);
  }
  return -1;
}
```