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
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <w1@plt>
  mov    $0x0,%edi
  call   <w2@plt>
  mov    $0x0,%edi
  call   <w3@plt>
  mov    $0x0,%edi
  call   <r1@plt>
  mov    $0x0,%edi
  call   <r2@plt>
  mov    $0x0,%edi
  call   <r3@plt>
  mov    $0x0,%edi
  call   <r4@plt>
  mov    $0x0,%edi
  call   <r5@plt>
  mov    $0x0,%edi
  call   <r6@plt>
  mov    $0x0,%edi
  call   <r7@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
__CPROVER_ASYNC_0:
  w1(0);
__CPROVER_ASYNC_1:
  w2(0);
__CPROVER_ASYNC_2:
  w3(0);
__CPROVER_ASYNC_3:
  r1(0);
__CPROVER_ASYNC_4:
  r2(0);
__CPROVER_ASYNC_5:
  r3(0);
__CPROVER_ASYNC_6:
  r4(0);
__CPROVER_ASYNC_7:
  r5(0);
__CPROVER_ASYNC_8:
  r6(0);
__CPROVER_ASYNC_9:
  r7(0);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <w1@plt>
  xor    %edi,%edi
  call   <w2@plt>
  xor    %edi,%edi
  call   <w3@plt>
  xor    %edi,%edi
  call   <r1@plt>
  xor    %edi,%edi
  call   <r2@plt>
  xor    %edi,%edi
  call   <r3@plt>
  xor    %edi,%edi
  call   <r4@plt>
  xor    %edi,%edi
  call   <r5@plt>
  xor    %edi,%edi
  call   <r6@plt>
  xor    %edi,%edi
  call   <r7@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
__CPROVER_ASYNC_0:
  w1(0);
__CPROVER_ASYNC_1:
  w2(0);
__CPROVER_ASYNC_2:
  w3(0);
__CPROVER_ASYNC_3:
  r1(0);
__CPROVER_ASYNC_4:
  r2(0);
__CPROVER_ASYNC_5:
  r3(0);
__CPROVER_ASYNC_6:
  r4(0);
__CPROVER_ASYNC_7:
  r5(0);
__CPROVER_ASYNC_8:
  r6(0);
__CPROVER_ASYNC_9:
  r7(0);
  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
  xor    %edi,%edi
  call   <w1@plt>
  xor    %edi,%edi
  call   <w2@plt>
  xor    %edi,%edi
  call   <w3@plt>
  xor    %edi,%edi
  call   <r1@plt>
  xor    %edi,%edi
  call   <r2@plt>
  xor    %edi,%edi
  call   <r3@plt>
  xor    %edi,%edi
  call   <r4@plt>
  xor    %edi,%edi
  call   <r5@plt>
  xor    %edi,%edi
  call   <r6@plt>
  xor    %edi,%edi
  call   <r7@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
__CPROVER_ASYNC_0:
  w1(0);
__CPROVER_ASYNC_1:
  w2(0);
__CPROVER_ASYNC_2:
  w3(0);
__CPROVER_ASYNC_3:
  r1(0);
__CPROVER_ASYNC_4:
  r2(0);
__CPROVER_ASYNC_5:
  r3(0);
__CPROVER_ASYNC_6:
  r4(0);
__CPROVER_ASYNC_7:
  r5(0);
__CPROVER_ASYNC_8:
  r6(0);
__CPROVER_ASYNC_9:
  r7(0);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x500,%edi
  call   <dac_open@plt>
L6:
  jmp    L6
```

assistant: ```c
int main(void) {
  dac_open(4 * 320);
  while (1)
    ;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x500,%edi
  call   <dac_open@plt>
L2:
  jmp    L2
```

assistant: ```c
int main(void) {
  dac_open(4 * 320);
  while (1)
    ;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x500,%edi
  call   <dac_open@plt>
L1:
  jmp    L1
```

assistant: ```c
int main(void) {
  dac_open(4 * 320);
  while (1)
    ;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x500,%edi
  call   <dac_open@plt>
L7:
  jmp    L7
```

assistant: ```c
int main(void) {
  dac_open(4 * 320);
  while (1)
    ;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  movslq -0x30(%rbp),%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L100:
  cmp    %rdx,%rsp
  je     L68
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L100
L68:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L10
  orq    $0x0,-0x8(%rsp,%rax,1)
L10:
  mov    %rsp,%r12
  lea    -0x30(%rbp),%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  mov    $0x0,%ebx
  lea    D13(%rip),%r13
L42:
  mov    -0x30(%rbp),%eax
  cmp    %ebx,%eax
  jle    L43
  lea    -0x2c(%rbp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    -0x2c(%rbp),%eax
  test   $0x1,%al
  je     L42
  mov    %eax,(%r12,%rbx,4)
  add    $0x1,%rbx
  jmp    L42
L43:
  mov    %eax,%r13d
  shr    $0x1f,%r13d
  add    %eax,%r13d
  sar    %r13d
  cmp    $0x1,%eax
  jle    L11
  mov    $0x0,%ebx
  lea    D15(%rip),%r14
L61:
  mov    %ebx,%eax
  not    %eax
  add    -0x30(%rbp),%eax
  cltq
  mov    (%r12,%rbx,4),%esi
  sub    (%r12,%rax,4),%esi
  mov    %r14,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %ebx,%r13d
  jg     L61
L11:
  mov    -0x30(%rbp),%eax
  test   $0x1,%al
  jne    L106
L8:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L102
  mov    $0x0,%eax
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L106:
  mov    $0x2,%ecx
  cltd
  idiv   %ecx
  cltq
  mov    (%r12,%rax,4),%esi
  lea    D15(%rip),%rdi
  call   <printf@plt>
  jmp    L8
L102:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D13
datatype: string
name: extract_symbolic_data
arguments: tag: D15
datatype: string

tool: D13: %d

tool: D15: %d


assistant: ```c
int main() {
  int n, j = 0, dul, m[n], i, z;
  scanf("%d", &n);
  while (j < n) {
    scanf("%d", &dul);
    if (dul % 2) {
      m[j] = dul;
      j++;
    }
  }
  z = n / 2;
  for (i = 0; i < z; i++) {
    printf("%d\n", m[i] - m[(n - i) - 1]);
  }
  if (n % 2) {
    printf("%d\n", m[n / 2]);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  movslq -0x30(%rbp),%rax
  mov    %rsp,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L30
L115:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L115
L30:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L29
L4:
  lea    -0x30(%rbp),%rsi
  mov    %rsp,%rbx
  lea    -0x2c(%rbp),%r14
  xor    %r12d,%r12d
  lea    D17(%rip),%r13
  mov    %r13,%rdi
  call   <scanf@plt>
  jmp    L109
  nopl   0x0(%rax)
L23:
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    -0x2c(%rbp),%eax
  test   $0x1,%al
  jne    L87
L109:
  mov    -0x30(%rbp),%edx
  mov    %edx,%eax
  cmp    %r12d,%edx
  jg     L23
  mov    %edx,%r12d
  shr    $0x1f,%r12d
  add    %edx,%r12d
  sar    %r12d
  cmp    $0x1,%edx
  jle    L34
  xor    %r14d,%r14d
  lea    D26(%rip),%r13
  jmp    L76
  nopw   0x0(%rax,%rax,1)
L130:
  mov    -0x30(%rbp),%edx
L76:
  sub    %r14d,%edx
  mov    (%rbx,%r14,4),%esi
  mov    %r13,%rdi
  add    $0x1,%r14
  sub    $0x1,%edx
  movslq %edx,%rdx
  sub    (%rbx,%rdx,4),%esi
  call   <printf@plt>
  cmp    %r14d,%r12d
  jg     L130
  mov    -0x30(%rbp),%eax
L34:
  test   $0x1,%al
  jne    L96
L124:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L62
  lea    -0x20(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nopl   0x0(%rax)
L87:
  mov    %eax,(%rbx,%r12,4)
  add    $0x1,%r12
  jmp    L109
L96:
  mov    $0x2,%ecx
  cltd
  lea    D26(%rip),%rdi
  idiv   %ecx
  cltq
  mov    (%rbx,%rax,4),%esi
  call   <printf@plt>
  jmp    L124
L29:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L4
L62:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D17
type_infered: string
name: unpack_data
arguments: tag: D26
type_infered: string

tool: D17: %d

tool: D26: %d


assistant: ```c
int main() {
  int n, j = 0, dul, m[n], i, z;
  scanf("%d", &n);
  while (j < n) {
    scanf("%d", &dul);
    if (dul % 2) {
      m[j] = dul;
      j++;
    }
  }
  z = n / 2;
  for (i = 0; i < z; i++) {
    printf("%d\n", m[i] - m[(n - i) - 1]);
  }
  if (n % 2) {
    printf("%d\n", m[n / 2]);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  movslq -0x30(%rbp),%rax
  mov    %rsp,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L7
L23:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L23
L7:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L19
L96:
  lea    -0x30(%rbp),%rsi
  mov    %rsp,%rbx
  lea    -0x2c(%rbp),%r14
  xor    %r12d,%r12d
  lea    D25(%rip),%r13
  mov    %r13,%rdi
  call   <scanf@plt>
  jmp    L71
  nopl   0x0(%rax)
L36:
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    -0x2c(%rbp),%eax
  test   $0x1,%al
  jne    L56
L71:
  mov    -0x30(%rbp),%edx
  mov    %edx,%eax
  cmp    %r12d,%edx
  jg     L36
  mov    %edx,%r12d
  shr    $0x1f,%r12d
  add    %edx,%r12d
  sar    %r12d
  cmp    $0x1,%edx
  jle    L123
  xor    %r14d,%r14d
  lea    D19(%rip),%r13
  jmp    L129
  nopw   0x0(%rax,%rax,1)
L128:
  mov    -0x30(%rbp),%edx
L129:
  sub    %r14d,%edx
  mov    (%rbx,%r14,4),%esi
  mov    %r13,%rdi
  add    $0x1,%r14
  sub    $0x1,%edx
  movslq %edx,%rdx
  sub    (%rbx,%rdx,4),%esi
  call   <printf@plt>
  cmp    %r14d,%r12d
  jg     L128
  mov    -0x30(%rbp),%eax
L123:
  test   $0x1,%al
  jne    L130
L53:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L49
  lea    -0x20(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nopl   0x0(%rax)
L56:
  mov    %eax,(%rbx,%r12,4)
  add    $0x1,%r12
  jmp    L71
L130:
  mov    $0x2,%ecx
  cltd
  lea    D19(%rip),%rdi
  idiv   %ecx
  cltq
  mov    (%rbx,%rax,4),%esi
  call   <printf@plt>
  jmp    L53
L19:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L96
L49:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D25
type_guessed: string
name: parse_data
arguments: label: D19
type_guessed: string

tool: D25: %d

tool: D19: %d


assistant: ```c
int main() {
  int n, j = 0, dul, m[n], i, z;
  scanf("%d", &n);
  while (j < n) {
    scanf("%d", &dul);
    if (dul % 2) {
      m[j] = dul;
      j++;
    }
  }
  z = n / 2;
  for (i = 0; i < z; i++) {
    printf("%d\n", m[i] - m[(n - i) - 1]);
  }
  if (n % 2) {
    printf("%d\n", m[n / 2]);
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<elapsed_time>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x8,%edi
  call   <xmalloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  movl   $0x0,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jle    L15
  mov    -0x18(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%ecx
  mov    %edx,%eax
  sub    %ecx,%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
  add    %eax,-0x1c(%rbp)
  jmp    L25
L15:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  lea    0xf4240(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%ecx
  mov    %edx,%eax
  sub    %ecx,%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
  add    %eax,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    0x4(%rax),%edx
  sub    $0x1,%edx
  mov    %edx,0x4(%rax)
L25:
  mov    -0x18(%rbp),%rax
  mov    0x4(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    0x4(%rax),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jle    L44
  mov    -0x18(%rbp),%rax
  mov    0x4(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    0x4(%rax),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  imul   $0x3e8,%eax,%eax
  add    %eax,-0x1c(%rbp)
L44:
  cmpl   $0x0,-0x1c(%rbp)
  jg     L49
  movl   $0x1,-0x1c(%rbp)
L49:
  lea    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <_xfree@plt>
  mov    -0x10(%rbp),%rax
  movq   $0x0,(%rax)
  mov    -0x1c(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

assistant: ```c
int elapsed_time(struct timeval *starttime) {
  struct timeval *newtime;
  int elapsed;
  newtime = (struct timeval *)xmalloc(sizeof(struct timeval));
  gettimeofday(newtime, NULL);
  elapsed = 0;
  if ((newtime->tv_usec - starttime->tv_usec) > 0) {
    elapsed += (newtime->tv_usec - starttime->tv_usec) / 1000;
  } else {
    elapsed += (1000000 + newtime->tv_usec - starttime->tv_usec) / 1000;
    newtime->tv_sec--;
  }
  if ((newtime->tv_sec - starttime->tv_sec) > 0) {
    elapsed += 1000 * (newtime->tv_sec - starttime->tv_sec);
  }
  if (elapsed < 1)
    elapsed = 1;
  do {
    char **_p = (char **)&(newtime);
    _xfree(*_p);
    *_p = 0;
  } while (0);
  return (elapsed);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<elapsed_time>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    $0x8,%edi
  call   <xmalloc@plt>
  mov    %rax,%rbp
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    0x0(%rbp),%eax
  mov    (%r12),%ecx
  mov    %eax,%edx
  sub    %ecx,%edx
  test   %edx,%edx
  jle    L8
  movslq %edx,%rbx
  imul   $0x10624dd3,%rbx,%rbx
  sar    $0x26,%rbx
  sar    $0x1f,%edx
  sub    %edx,%ebx
L11:
  mov    0x4(%rbp),%eax
  sub    0x4(%r12),%eax
  test   %eax,%eax
  jle    L22
  imul   $0x3e8,%eax,%eax
  add    %eax,%ebx
L22:
  mov    %rbp,%rdi
  call   <_xfree@plt>
  test   %ebx,%ebx
  mov    $0x1,%eax
  cmovg  %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  lea    0xf4240(%rax),%edx
  sub    %ecx,%edx
  movslq %edx,%rbx
  imul   $0x10624dd3,%rbx,%rbx
  sar    $0x26,%rbx
  sar    $0x1f,%edx
  sub    %edx,%ebx
  subl   $0x1,0x4(%rbp)
  jmp    L11
```

assistant: ```c
int elapsed_time(struct timeval *starttime) {
  struct timeval *newtime;
  int elapsed;
  newtime = (struct timeval *)xmalloc(sizeof(struct timeval));
  gettimeofday(newtime, NULL);
  elapsed = 0;
  if ((newtime->tv_usec - starttime->tv_usec) > 0) {
    elapsed += (newtime->tv_usec - starttime->tv_usec) / 1000;
  } else {
    elapsed += (1000000 + newtime->tv_usec - starttime->tv_usec) / 1000;
    newtime->tv_sec--;
  }
  if ((newtime->tv_sec - starttime->tv_sec) > 0) {
    elapsed += 1000 * (newtime->tv_sec - starttime->tv_sec);
  }
  if (elapsed < 1)
    elapsed = 1;
  do {
    char **_p = (char **)&(newtime);
    _xfree(*_p);
    *_p = 0;
  } while (0);
  return (elapsed);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<elapsed_time>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    $0x8,%edi
  push   %rbp
  push   %rbx
  call   <xmalloc@plt>
  xor    %esi,%esi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    0x0(%rbp),%eax
  mov    (%r12),%ecx
  mov    %eax,%edx
  sub    %ecx,%edx
  test   %edx,%edx
  jle    L16
  movslq %edx,%rbx
  mov    0x4(%rbp),%eax
  sar    $0x1f,%edx
  imul   $0x10624dd3,%rbx,%rbx
  sar    $0x26,%rbx
  sub    %edx,%ebx
L1:
  sub    0x4(%r12),%eax
  test   %eax,%eax
  jle    L25
  imul   $0x3e8,%eax,%eax
  add    %eax,%ebx
L25:
  mov    %rbp,%rdi
  call   <_xfree@plt>
  test   %ebx,%ebx
  mov    $0x1,%eax
  cmovg  %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nop
L16:
  add    $0xf4240,%eax
  sub    %ecx,%eax
  movslq %eax,%rbx
  sar    $0x1f,%eax
  imul   $0x10624dd3,%rbx,%rbx
  sar    $0x26,%rbx
  sub    %eax,%ebx
  mov    0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,0x4(%rbp)
  jmp    L1
```

assistant: ```c
int elapsed_time(struct timeval *starttime) {
  struct timeval *newtime;
  int elapsed;
  newtime = (struct timeval *)xmalloc(sizeof(struct timeval));
  gettimeofday(newtime, NULL);
  elapsed = 0;
  if ((newtime->tv_usec - starttime->tv_usec) > 0) {
    elapsed += (newtime->tv_usec - starttime->tv_usec) / 1000;
  } else {
    elapsed += (1000000 + newtime->tv_usec - starttime->tv_usec) / 1000;
    newtime->tv_sec--;
  }
  if ((newtime->tv_sec - starttime->tv_sec) > 0) {
    elapsed += 1000 * (newtime->tv_sec - starttime->tv_sec);
  }
  if (elapsed < 1)
    elapsed = 1;
  do {
    char **_p = (char **)&(newtime);
    _xfree(*_p);
    *_p = 0;
  } while (0);
  return (elapsed);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<elapsed_time>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    $0x8,%edi
  push   %rbp
  push   %rbx
  call   <xmalloc@plt>
  xor    %esi,%esi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    0x0(%rbp),%eax
  mov    (%r12),%ecx
  mov    %eax,%edx
  sub    %ecx,%edx
  test   %edx,%edx
  jle    L9
  movslq %edx,%rbx
  mov    0x4(%rbp),%eax
  sar    $0x1f,%edx
  imul   $0x10624dd3,%rbx,%rbx
  sar    $0x26,%rbx
  sub    %edx,%ebx
L18:
  sub    0x4(%r12),%eax
  test   %eax,%eax
  jle    L4
  imul   $0x3e8,%eax,%eax
  add    %eax,%ebx
L4:
  mov    %rbp,%rdi
  call   <_xfree@plt>
  test   %ebx,%ebx
  mov    $0x1,%eax
  cmovg  %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nop
L9:
  add    $0xf4240,%eax
  sub    %ecx,%eax
  movslq %eax,%rbx
  sar    $0x1f,%eax
  imul   $0x10624dd3,%rbx,%rbx
  sar    $0x26,%rbx
  sub    %eax,%ebx
  mov    0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,0x4(%rbp)
  jmp    L18
```

assistant: ```c
int elapsed_time(struct timeval *starttime) {
  struct timeval *newtime;
  int elapsed;
  newtime = (struct timeval *)xmalloc(sizeof(struct timeval));
  gettimeofday(newtime, NULL);
  elapsed = 0;
  if ((newtime->tv_usec - starttime->tv_usec) > 0) {
    elapsed += (newtime->tv_usec - starttime->tv_usec) / 1000;
  } else {
    elapsed += (1000000 + newtime->tv_usec - starttime->tv_usec) / 1000;
    newtime->tv_sec--;
  }
  if ((newtime->tv_sec - starttime->tv_sec) > 0) {
    elapsed += 1000 * (newtime->tv_sec - starttime->tv_sec);
  }
  if (elapsed < 1)
    elapsed = 1;
  do {
    char **_p = (char **)&(newtime);
    _xfree(*_p);
    *_p = 0;
  } while (0);
  return (elapsed);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<hashInt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movw   $0x0,-0x2(%rbp)
  mov    -0x14(%rbp),%eax
  shr    $0x18,%eax
  add    %ax,-0x2(%rbp)
  movzwl -0x2(%rbp),%edx
  mov    %edx,%eax
  shl    $0x3,%eax
  add    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  mov    %ax,-0x2(%rbp)
  mov    -0x14(%rbp),%eax
  shr    $0x10,%eax
  movzbl %al,%eax
  add    %ax,-0x2(%rbp)
  movzwl -0x2(%rbp),%edx
  mov    %edx,%eax
  shl    $0x3,%eax
  add    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  mov    %ax,-0x2(%rbp)
  mov    -0x14(%rbp),%eax
  shr    $0x8,%eax
  movzbl %al,%eax
  add    %ax,-0x2(%rbp)
  movzwl -0x2(%rbp),%edx
  mov    %edx,%eax
  shl    $0x3,%eax
  add    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  mov    %ax,-0x2(%rbp)
  mov    -0x14(%rbp),%eax
  movzbl %al,%eax
  add    %ax,-0x2(%rbp)
  movzwl -0x2(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned short hashInt(unsigned int in) {
  unsigned short hash = 0;
  hash += (in >> 24) & 0xFF;
  hash *= 37;
  hash += (in >> 16) & 0xFF;
  hash *= 37;
  hash += (in >> 8) & 0xFF;
  hash *= 37;
  hash += in & 0xFF;
  return hash;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hashInt>:
  endbr64
  mov    %edi,%eax
  shr    $0x18,%eax
  lea    (%rax,%rax,8),%edx
  lea    (%rax,%rdx,4),%edx
  mov    %edi,%eax
  shr    $0x10,%eax
  movzbl %al,%eax
  add    %edx,%eax
  lea    (%rax,%rax,8),%edx
  lea    (%rax,%rdx,4),%edx
  mov    %edi,%eax
  shr    $0x8,%ax
  add    %edx,%eax
  lea    (%rax,%rax,8),%edx
  lea    (%rax,%rdx,4),%eax
  movzbl %dil,%edi
  add    %edi,%eax
  ret
```

assistant: ```c
unsigned short hashInt(unsigned int in) {
  unsigned short hash = 0;
  hash += (in >> 24) & 0xFF;
  hash *= 37;
  hash += (in >> 16) & 0xFF;
  hash *= 37;
  hash += (in >> 8) & 0xFF;
  hash *= 37;
  hash += in & 0xFF;
  return hash;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<hashInt>:
  endbr64
  mov    %edi,%eax
  shr    $0x18,%eax
  lea    (%rax,%rax,8),%edx
  lea    (%rax,%rdx,4),%edx
  mov    %edi,%eax
  shr    $0x10,%eax
  movzbl %al,%eax
  add    %edx,%eax
  lea    (%rax,%rax,8),%edx
  lea    (%rax,%rdx,4),%edx
  mov    %edi,%eax
  movzbl %dil,%edi
  shr    $0x8,%ax
  add    %edx,%eax
  lea    (%rax,%rax,8),%edx
  lea    (%rax,%rdx,4),%eax
  add    %edi,%eax
  ret
```

assistant: ```c
unsigned short hashInt(unsigned int in) {
  unsigned short hash = 0;
  hash += (in >> 24) & 0xFF;
  hash *= 37;
  hash += (in >> 16) & 0xFF;
  hash *= 37;
  hash += (in >> 8) & 0xFF;
  hash *= 37;
  hash += in & 0xFF;
  return hash;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<hashInt>:
  endbr64
  mov    %edi,%eax
  shr    $0x18,%eax
  lea    (%rax,%rax,8),%edx
  lea    (%rax,%rdx,4),%edx
  mov    %edi,%eax
  shr    $0x10,%eax
  movzbl %al,%eax
  add    %edx,%eax
  lea    (%rax,%rax,8),%edx
  lea    (%rax,%rdx,4),%edx
  mov    %edi,%eax
  movzbl %dil,%edi
  shr    $0x8,%ax
  add    %edx,%eax
  lea    (%rax,%rax,8),%edx
  lea    (%rax,%rdx,4),%eax
  add    %edi,%eax
  ret
```

assistant: ```c
unsigned short hashInt(unsigned int in) {
  unsigned short hash = 0;
  hash += (in >> 24) & 0xFF;
  hash *= 37;
  hash += (in >> 16) & 0xFF;
  hash *= 37;
  hash += (in >> 8) & 0xFF;
  hash *= 37;
  hash += in & 0xFF;
  return hash;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strip_trailing_slashes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L5
L8:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L5:
  cmpl   $0x0,-0x4(%rbp)
  jle    L29
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L8
L29:
  nop
  leave
  ret
```

assistant: ```c
void strip_trailing_slashes(char *path) {
  int last;
  last = strlen(path) - 1;
  while (last > 0 && path[last] == '/')
    path[last--] = '\0';
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strip_trailing_slashes>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  test   %edx,%edx
  jle    L18
  mov    %eax,%ecx
  movslq %edx,%rdx
  add    %rbx,%rdx
  cltq
  lea    -0x2(%rbx,%rax,1),%rax
  lea    -0x2(%rcx),%ecx
  sub    %rcx,%rax
L20:
  cmpb   $0x2f,(%rdx)
  jne    L18
  movb   $0x0,(%rdx)
  sub    $0x1,%rdx
  cmp    %rax,%rdx
  jne    L20
L18:
  pop    %rbx
  ret
```

assistant: ```c
void strip_trailing_slashes(char *path) {
  int last;
  last = strlen(path) - 1;
  while (last > 0 && path[last] == '/')
    path[last--] = '\0';
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strip_trailing_slashes>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  test   %edx,%edx
  jle    L23
  mov    %eax,%ecx
  cltq
  movslq %edx,%rdx
  lea    -0x2(%rbx,%rax,1),%rax
  sub    $0x2,%ecx
  add    %rbx,%rdx
  sub    %rcx,%rax
  jmp    L37
  nopl   0x0(%rax,%rax,1)
L25:
  movb   $0x0,(%rdx)
  sub    $0x1,%rdx
  cmp    %rax,%rdx
  je     L23
L37:
  cmpb   $0x2f,(%rdx)
  je     L25
L23:
  pop    %rbx
  ret
```

assistant: ```c
void strip_trailing_slashes(char *path) {
  int last;
  last = strlen(path) - 1;
  while (last > 0 && path[last] == '/')
    path[last--] = '\0';
}
```