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
<wheel_product>:
  endbr64
  cmp    $0x1,%edi
  jle    L6
  add    $0x1,%edi
  mov    $0x2,%eax
  mov    $0x1,%r8d
  nopw   0x0(%rax,%rax,1)
L11:
  imul   %eax,%r8d
  add    $0x1,%eax
  cmp    %edi,%eax
  jne    L11
  mov    %r8d,%eax
  ret
  nop
L6:
  mov    $0x1,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int64_t wheel_product(int upto) {
  int64_t product = 1;
  int i;
  for (i = 2; i <= upto; i++) {
    product *= i;
  }
  return product;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L31
L10:
  addl   $0x1,-0x4(%rbp)
L31:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L10
  subl   $0x1,-0x4(%rbp)
  jmp    L5
L14:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x9(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x8(%rbp)
  subl   $0x1,-0x4(%rbp)
L5:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L14
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void reverse(char s[]) {
  int begin, end;
  char tmp;
  begin = 0;
  end = 0;
  while (s[end] != '\0') {
    end++;
  }
  end--;
  while (begin < end) {
    tmp = s[begin];
    s[begin] = s[end];
    s[end] = tmp;
    begin++;
    end--;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<reverse>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L38
  mov    $0x0,%edx
L29:
  mov    %edx,%eax
  add    $0x1,%rdx
  cmpb   $0x0,(%rdi,%rdx,1)
  jne    L29
  test   %eax,%eax
  jle    L38
  cltq
  mov    $0x0,%edx
L31:
  movzbl (%rdi,%rdx,1),%ecx
  movzbl (%rdi,%rax,1),%esi
  mov    %sil,(%rdi,%rdx,1)
  mov    %cl,(%rdi,%rax,1)
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %edx,%eax
  jg     L31
L38:
  ret
```

assistant: ```c
void reverse(char s[]) {
  int begin, end;
  char tmp;
  begin = 0;
  end = 0;
  while (s[end] != '\0') {
    end++;
  }
  end--;
  while (begin < end) {
    tmp = s[begin];
    s[begin] = s[end];
    s[end] = tmp;
    begin++;
    end--;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<reverse>:
  endbr64
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L16
  xor    %edx,%edx
  nopl   (%rax)
L26:
  movslq %edx,%rax
  add    $0x1,%rdx
  cmpb   $0x0,(%rdi,%rdx,1)
  jne    L26
  test   %eax,%eax
  je     L16
  xor    %edx,%edx
  jmp    L22
  nopl   (%rax)
L8:
  movzbl (%rdi,%rdx,1),%ecx
L22:
  movzbl (%rdi,%rax,1),%esi
  mov    %sil,(%rdi,%rdx,1)
  add    $0x1,%rdx
  mov    %cl,(%rdi,%rax,1)
  sub    $0x1,%rax
  cmp    %edx,%eax
  jg     L8
L16:
  ret
```

assistant: ```c
void reverse(char s[]) {
  int begin, end;
  char tmp;
  begin = 0;
  end = 0;
  while (s[end] != '\0') {
    end++;
  }
  end--;
  while (begin < end) {
    tmp = s[begin];
    s[begin] = s[end];
    s[end] = tmp;
    begin++;
    end--;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<reverse>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L35
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L33:
  movslq %edx,%rax
  add    $0x1,%rdx
  cmpb   $0x0,(%rdi,%rdx,1)
  jne    L33
  test   %eax,%eax
  je     L35
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L5:
  movzbl (%rdi,%rdx,1),%ecx
  movzbl (%rdi,%rax,1),%esi
  mov    %sil,(%rdi,%rdx,1)
  add    $0x1,%rdx
  mov    %cl,(%rdi,%rax,1)
  sub    $0x1,%rax
  cmp    %edx,%eax
  jg     L5
L35:
  ret
```

assistant: ```c
void reverse(char s[]) {
  int begin, end;
  char tmp;
  begin = 0;
  end = 0;
  while (s[end] != '\0') {
    end++;
  }
  end--;
  while (begin < end) {
    tmp = s[begin];
    s[begin] = s[end];
    s[end] = tmp;
    begin++;
    end--;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hanoi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    %ecx,-0x10(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L0
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edi
  mov    -0xc(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  call   <hanoi>
  mov    -0xc(%rbp),%ecx
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edi
  mov    -0x8(%rbp),%ecx
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  call   <hanoi>
  jmp    L14
L0:
  nop
L14:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D0
data_type: string

tool: D0: %d.%d->%d


assistant: ```c
void hanoi(int n, int start, int focus, int temp) {
  if (0 == n) {
    return;
  } else {
    hanoi(n - 1, start, temp, focus);
    printf("%d.%d->%d\n", n, start, focus);
    hanoi(n - 1, temp, focus, start);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<hanoi>:
  endbr64
  test   %edi,%edi
  jne    L7
  ret
L7:
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%r13d
  mov    %edx,%ebp
  mov    %ecx,%r12d
  lea    -0x1(%rdi),%r14d
  mov    %edx,%ecx
  mov    %r12d,%edx
  mov    %r14d,%edi
  call   <hanoi>
  mov    %ebp,%ecx
  mov    %r13d,%edx
  mov    %ebx,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    %r13d,%ecx
  mov    %ebp,%edx
  mov    %r12d,%esi
  mov    %r14d,%edi
  call   <hanoi>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D0
datatype: string

tool: D0: %d.%d->%d


assistant: ```c
void hanoi(int n, int start, int focus, int temp) {
  if (0 == n) {
    return;
  } else {
    hanoi(n - 1, start, temp, focus);
    printf("%d.%d->%d\n", n, start, focus);
    hanoi(n - 1, temp, focus, start);
  }
}
```
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
  lea    -0xa000(%rsp),%r11
L6:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L6
  sub    $0xdc0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xadc0(%rbp),%rax
  mov    %rax,%rsi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0xadb8(%rbp)
  jmp    L21
L47:
  lea    -0xadbc(%rbp),%rax
  mov    %rax,%rsi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xadbc(%rbp),%edx
  mov    -0xadb8(%rbp),%eax
  cltq
  mov    %edx,-0xadb0(%rbp,%rax,4)
  addl   $0x1,-0xadb8(%rbp)
L21:
  mov    -0xadc0(%rbp),%eax
  cmp    %eax,-0xadb8(%rbp)
  jl     L47
  movl   $0x0,-0xadb8(%rbp)
  jmp    L76
L63:
  mov    -0xadc0(%rbp),%eax
  sub    $0x1,%eax
  sub    -0xadb8(%rbp),%eax
  cltq
  mov    -0xadb0(%rbp,%rax,4),%edx
  mov    -0xadb8(%rbp),%eax
  cltq
  mov    %edx,-0x56e0(%rbp,%rax,4)
  addl   $0x1,-0xadb8(%rbp)
L76:
  mov    -0xadc0(%rbp),%eax
  cmp    %eax,-0xadb8(%rbp)
  jl     L63
  movl   $0x0,-0xadb8(%rbp)
  jmp    L97
L91:
  movl   $0x0,-0xadb4(%rbp)
  jmp    L53
L51:
  mov    -0xadb4(%rbp),%eax
  cltq
  mov    -0x56e0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xadb4(%rbp)
L53:
  mov    -0xadb4(%rbp),%eax
  cmp    -0xadb8(%rbp),%eax
  jle    L51
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0xadb8(%rbp)
L97:
  mov    -0xadc0(%rbp),%eax
  cmp    %eax,-0xadb8(%rbp)
  jl     L91
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L85
  call   <__stack_chk_fail@plt>
L85:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D27
type_guessed: string
name: retrieve_data
arguments: tag: D0
type_guessed: string

tool: D27: %d

tool: D0:  %d

assistant: ```c
int main() {
  int i, N, n, j, x;
  int a[5555], b[5555];
  scanf("%d", &N);
  for (i = 0; i <= N - 1; i++) {
    scanf("%d", &n);
    a[i] = n;
  }
  for (i = 0; i < N; i++) {
    b[i] = a[N - 1 - i];
  }
  for (i = 0; i < N; i++) {
    for (j = 0; j <= i; j++) {
      printf(" %d", b[j]);
    }
    printf("\n");
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xa000(%rsp),%r11
L102:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L102
  sub    $0xdc0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xadb8(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x8(%rsp)
  jle    L107
  mov    $0x0,%ebx
  lea    D25(%rip),%rbp
L24:
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  mov    %eax,0x10(%rsp,%rbx,4)
  mov    0x8(%rsp),%ecx
  add    $0x1,%rbx
  cmp    %ebx,%ecx
  jg     L24
  test   %ecx,%ecx
  jle    L107
  movslq %ecx,%rax
  lea    0xc(%rsp,%rax,4),%rdx
  lea    0x56e0(%rsp),%rax
  lea    -0x1(%rcx),%ecx
  lea    0x56e4(%rsp,%rcx,4),%rsi
L70:
  mov    (%rdx),%ecx
  mov    %ecx,(%rax)
  sub    $0x4,%rdx
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L70
  mov    $0x0,%ebp
  lea    D8(%rip),%r12
  jmp    L63
L67:
  mov    0x56e0(%rsp,%rbx,4),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %ebx,%ebp
  jge    L67
L84:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%ebp
  cmp    %ebp,0x8(%rsp)
  jle    L107
L63:
  mov    $0x0,%ebx
  test   %ebp,%ebp
  jns    L67
  jmp    L84
L107:
  mov    0xadb8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    $0x0,%eax
  add    $0xadc0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D25
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D8
type_guessed: string

tool: D25: %d

tool: D8:  %d

assistant: ```c
int main() {
  int i, N, n, j, x;
  int a[5555], b[5555];
  scanf("%d", &N);
  for (i = 0; i <= N - 1; i++) {
    scanf("%d", &n);
    a[i] = n;
  }
  for (i = 0; i < N; i++) {
    b[i] = a[N - 1 - i];
  }
  for (i = 0; i < N; i++) {
    for (j = 0; j <= i; j++) {
      printf(" %d", b[j]);
    }
    printf("\n");
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xa000(%rsp),%r11
L10:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L10
  sub    $0xdc0,%rsp
  lea    D14(%rip),%rbp
  mov    %rbp,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0xadb8(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  test   %eax,%eax
  jle    L57
  xor    %ebx,%ebx
  lea    0xc(%rsp),%r13
  lea    0x10(%rsp),%r12
  nopl   0x0(%rax,%rax,1)
L13:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  mov    0x8(%rsp),%ecx
  mov    %eax,(%r12,%rbx,4)
  add    $0x1,%rbx
  cmp    %ebx,%ecx
  jg     L13
  test   %ecx,%ecx
  jle    L57
  movslq %ecx,%rax
  lea    0x56e0(%rsp),%r13
  sub    $0x1,%ecx
  lea    0x56e4(%rsp),%rbp
  lea    -0x4(%r12,%rax,4),%rdx
  mov    %r13,%rax
  lea    0x0(%rbp,%rcx,4),%rsi
  nopw   0x0(%rax,%rax,1)
L12:
  mov    (%rdx),%ecx
  add    $0x4,%rax
  sub    $0x4,%rdx
  mov    %ecx,-0x4(%rax)
  cmp    %rsi,%rax
  jne    L12
  xor    %r14d,%r14d
  lea    D15(%rip),%r12
  nopl   0x0(%rax)
L61:
  mov    %r13,%rbx
  nopl   0x0(%rax,%rax,1)
L69:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L69
  mov    $0xa,%edi
  add    $0x1,%r14d
  add    $0x4,%rbp
  call   <putchar@plt>
  cmp    %r14d,0x8(%rsp)
  jg     L61
L57:
  mov    0xadb8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  add    $0xadc0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D14
data_type: string
name: retrieve_data
arguments: data_tag: D15
data_type: string

tool: D14: %d

tool: D15:  %d

assistant: ```c
int main() {
  int i, N, n, j, x;
  int a[5555], b[5555];
  scanf("%d", &N);
  for (i = 0; i <= N - 1; i++) {
    scanf("%d", &n);
    a[i] = n;
  }
  for (i = 0; i < N; i++) {
    b[i] = a[N - 1 - i];
  }
  for (i = 0; i < N; i++) {
    for (j = 0; j <= i; j++) {
      printf(" %d", b[j]);
    }
    printf("\n");
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xa000(%rsp),%r11
L58:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L58
  sub    $0xdc0,%rsp
  lea    D11(%rip),%rbp
  mov    %rbp,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0xadb8(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  test   %eax,%eax
  jle    L37
  xor    %ebx,%ebx
  lea    0xc(%rsp),%r13
  lea    0x10(%rsp),%r12
  nopl   0x0(%rax,%rax,1)
L15:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  mov    %eax,(%r12,%rbx,4)
  mov    0x8(%rsp),%eax
  add    $0x1,%rbx
  cmp    %ebx,%eax
  jg     L15
  test   %eax,%eax
  jle    L37
  lea    -0x1(%rax),%edi
  cmp    $0x2,%edi
  jbe    L51
  movslq %eax,%rdx
  mov    %eax,%r8d
  lea    0x56e0(%rsp),%r14
  shr    $0x2,%r8d
  lea    -0x10(%r12,%rdx,4),%rsi
  mov    %r14,%rcx
  lea    -0x1(%r8),%r9d
  lea    -0x10(%rsi),%rdx
  shl    $0x4,%r9
  mov    %rdx,%r8
  sub    %r9,%r8
  jmp    L124
  nopl   0x0(%rax)
L90:
  sub    $0x10,%rdx
L124:
  movdqu (%rsi),%xmm0
  add    $0x10,%rcx
  mov    %rdx,%rsi
  pshufd $0x1b,%xmm0,%xmm0
  movaps %xmm0,-0x10(%rcx)
  cmp    %r8,%rdx
  jne    L90
  mov    %eax,%edx
  and    $0xfffffffc,%edx
  test   $0x3,%al
  je     L60
L136:
  mov    %edi,%ecx
  movslq %edx,%rsi
  sub    %edx,%ecx
  movslq %ecx,%rcx
  mov    0x10(%rsp,%rcx,4),%ecx
  mov    %ecx,0x56e0(%rsp,%rsi,4)
  lea    0x1(%rdx),%ecx
  cmp    %ecx,%eax
  jle    L60
  mov    %edi,%esi
  movslq %ecx,%r8
  add    $0x2,%edx
  sub    %ecx,%esi
  movslq %esi,%rcx
  mov    0x10(%rsp,%rcx,4),%ecx
  mov    %ecx,0x56e0(%rsp,%r8,4)
  cmp    %edx,%eax
  jle    L60
  sub    %edx,%edi
  movslq %edx,%rcx
  movslq %edi,%rax
  mov    0x10(%rsp,%rax,4),%eax
  mov    %eax,0x56e0(%rsp,%rcx,4)
L60:
  lea    0x56e4(%rsp),%rbp
  xor    %r13d,%r13d
  lea    D5(%rip),%r12
  nopl   0x0(%rax)
L43:
  mov    %r14,%rbx
  nopl   0x0(%rax,%rax,1)
L111:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L111
  mov    $0xa,%edi
  add    $0x1,%r13d
  add    $0x4,%rbp
  call   <putchar@plt>
  cmp    %r13d,0x8(%rsp)
  jg     L43
L37:
  mov    0xadb8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L113
  add    $0xadc0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L51:
  xor    %edx,%edx
  lea    0x56e0(%rsp),%r14
  jmp    L136
L113:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D11
type_guessed: string
name: parse_data
arguments: label: D5
type_guessed: string

tool: D11: %d

tool: D5:  %d

assistant: ```c
int main() {
  int i, N, n, j, x;
  int a[5555], b[5555];
  scanf("%d", &N);
  for (i = 0; i <= N - 1; i++) {
    scanf("%d", &n);
    a[i] = n;
  }
  for (i = 0; i < N; i++) {
    b[i] = a[N - 1 - i];
  }
  for (i = 0; i < N; i++) {
    for (j = 0; j <= i; j++) {
      printf(" %d", b[j]);
    }
    printf("\n");
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<wBasFGets>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <wPopNumber@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <wFileLineInput@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <wPushString@plt>
  nop
  leave
  ret
```

assistant: ```c
void wBasFGets() {
  int handle;
  char *buffer;
  handle = wPopNumber();
  buffer = wFileLineInput(handle);
  wPushString(buffer);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<wBasFGets>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <wPopNumber@plt>
  mov    %eax,%edi
  call   <wFileLineInput@plt>
  mov    %rax,%rdi
  call   <wPushString@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void wBasFGets() {
  int handle;
  char *buffer;
  handle = wPopNumber();
  buffer = wFileLineInput(handle);
  wPushString(buffer);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<wBasFGets>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <wPopNumber@plt>
  mov    %eax,%edi
  call   <wFileLineInput@plt>
  add    $0x8,%rsp
  mov    %rax,%rdi
  jmp    1080 <wPushString@plt>
```

assistant: ```c
void wBasFGets() {
  int handle;
  char *buffer;
  handle = wPopNumber();
  buffer = wFileLineInput(handle);
  wPushString(buffer);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ro2omLoopf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xb8(%rbp)
  mov    %rsi,-0xc0(%rbp)
  mov    %rdx,-0xc8(%rbp)
  mov    %ecx,-0xcc(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x4,-0x90(%rbp)
  movq   $0x9,-0x88(%rbp)
  movq   $0x0,-0x98(%rbp)
  jmp    L31
L66:
  mov    -0x98(%rbp),%rax
  imul   -0x90(%rbp),%rax
  mov    %rax,-0x80(%rbp)
  mov    -0x98(%rbp),%rax
  imul   -0x88(%rbp),%rax
  mov    %rax,-0x78(%rbp)
  movq   $0x0,-0xa0(%rbp)
  jmp    L50
L26:
  mov    -0x80(%rbp),%rdx
  mov    -0xa0(%rbp),%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0xb8(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  cvtss2sd %xmm0,%xmm0
  mov    -0xa0(%rbp),%rax
  movsd  %xmm0,-0x70(%rbp,%rax,8)
  addq   $0x1,-0xa0(%rbp)
L50:
  mov    -0xa0(%rbp),%rax
  cmp    -0x90(%rbp),%rax
  jb     L26
  mov    -0xcc(%rbp),%edx
  lea    -0x50(%rbp),%rcx
  lea    -0x70(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <ro2om@plt>
  mov    %eax,-0xa4(%rbp)
  movq   $0x0,-0xa0(%rbp)
  jmp    L56
L10:
  mov    -0xa0(%rbp),%rax
  movsd  -0x50(%rbp,%rax,8),%xmm0
  mov    -0x78(%rbp),%rdx
  mov    -0xa0(%rbp),%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0xc0(%rbp),%rax
  add    %rdx,%rax
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,(%rax)
  addq   $0x1,-0xa0(%rbp)
L56:
  mov    -0xa0(%rbp),%rax
  cmp    -0x88(%rbp),%rax
  jb     L10
  addq   $0x1,-0x98(%rbp)
L31:
  mov    -0x98(%rbp),%rax
  cmp    -0xc8(%rbp),%rax
  jb     L66
  mov    $0x1,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

assistant: ```c
int ro2omLoopf(float *ro, float *om, unsigned long long n, int p) {
  int trash;
  unsigned long long j, i, iIn, iOut, rostep = 4, omstep = 9;
  double roFlt[4], omFlt[9];
#pragma omp parallel for default(shared) private(j, i, iIn, iOut, trash, roFlt, omFlt) firstprivate(rostep, omstep) schedule(static)
  for (i = 0; i < n; i++) {
    iIn = i * rostep;
    iOut = i * omstep;
    for (j = 0; j < rostep; j++) {
      roFlt[j] = (double)ro[iIn + j];
    }
    trash = ro2om(roFlt, omFlt, p);
    for (j = 0; j < omstep; j++) {
      om[iOut + j] = (float)omFlt[j];
    }
  }
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ro2omLoopf>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L30
  mov    %rdx,%r14
  mov    %ecx,%r15d
  mov    %rsi,%rbx
  mov    %rdi,%r12
  mov    $0x0,%r13d
  lea    0x20(%rsp),%rbp
  jmp    L10
L36:
  mov    %rsp,%rdi
  mov    %r15d,%edx
  mov    %rbp,%rsi
  call   <ro2om@plt>
  mov    $0x0,%eax
L23:
  pxor   %xmm0,%xmm0
  cvtsd2ss 0x0(%rbp,%rax,8),%xmm0
  movss  %xmm0,(%rbx,%rax,4)
  add    $0x1,%rax
  cmp    $0x9,%rax
  jne    L23
  add    $0x1,%r13
  add    $0x24,%rbx
  add    $0x10,%r12
  cmp    %r13,%r14
  je     L30
L10:
  mov    $0x0,%eax
L44:
  pxor   %xmm0,%xmm0
  cvtss2sd (%r12,%rax,4),%xmm0
  movsd  %xmm0,(%rsp,%rax,8)
  add    $0x1,%rax
  cmp    $0x4,%rax
  jne    L44
  jmp    L36
L30:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x1,%eax
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ro2omLoopf(float *ro, float *om, unsigned long long n, int p) {
  int trash;
  unsigned long long j, i, iIn, iOut, rostep = 4, omstep = 9;
  double roFlt[4], omFlt[9];
#pragma omp parallel for default(shared) private(j, i, iIn, iOut, trash, roFlt, omFlt) firstprivate(rostep, omstep) schedule(static)
  for (i = 0; i < n; i++) {
    iIn = i * rostep;
    iOut = i * omstep;
    for (j = 0; j < rostep; j++) {
      roFlt[j] = (double)ro[iIn + j];
    }
    trash = ro2om(roFlt, omFlt, p);
    for (j = 0; j < omstep; j++) {
      om[iOut + j] = (float)omFlt[j];
    }
  }
  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
<ro2omLoopf>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %rdx,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L30
  mov    %ecx,%r12d
  mov    %rsi,%rbx
  mov    %rdi,%r14
  xor    %ebp,%ebp
  lea    0x10(%rsp),%r15
  lea    0x30(%rsp),%r13
  nopl   0x0(%rax)
L15:
  xor    %eax,%eax
L26:
  pxor   %xmm0,%xmm0
  cvtss2sd (%r14,%rax,4),%xmm0
  movsd  %xmm0,(%r15,%rax,8)
  add    $0x1,%rax
  cmp    $0x4,%rax
  jne    L26
  mov    %r12d,%edx
  mov    %r13,%rsi
  mov    %r15,%rdi
  call   <ro2om@plt>
  xor    %eax,%eax
  nopl   0x0(%rax)
L31:
  pxor   %xmm0,%xmm0
  cvtsd2ss 0x0(%r13,%rax,8),%xmm0
  movss  %xmm0,(%rbx,%rax,4)
  add    $0x1,%rax
  cmp    $0x9,%rax
  jne    L31
  add    $0x1,%rbp
  add    $0x24,%rbx
  add    $0x10,%r14
  cmp    %rbp,0x8(%rsp)
  jne    L15
L30:
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x88,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ro2omLoopf(float *ro, float *om, unsigned long long n, int p) {
  int trash;
  unsigned long long j, i, iIn, iOut, rostep = 4, omstep = 9;
  double roFlt[4], omFlt[9];
#pragma omp parallel for default(shared) private(j, i, iIn, iOut, trash, roFlt, omFlt) firstprivate(rostep, omstep) schedule(static)
  for (i = 0; i < n; i++) {
    iIn = i * rostep;
    iOut = i * omstep;
    for (j = 0; j < rostep; j++) {
      roFlt[j] = (double)ro[iIn + j];
    }
    trash = ro2om(roFlt, omFlt, p);
    for (j = 0; j < omstep; j++) {
      om[iOut + j] = (float)omFlt[j];
    }
  }
  return 1;
}
```