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
<qfits_get_hdrinfo>:
  endbr64
  test   %rdi,%rdi
  je     L47
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  test   %esi,%esi
  js     L60
  mov    %rdx,%rax
  mov    %rdx,%rbp
  mov    %rcx,%r12
  or     %rcx,%rax
  je     L60
  mov    %rdi,%r13
  test   %rdx,%rdx
  je     L16
  or     $0x20000000,%esi
  call   <qfits_query@plt>
  mov    %eax,0x0(%rbp)
  test   %eax,%eax
  js     L60
L16:
  xor    %eax,%eax
  test   %r12,%r12
  je     L4
  or     $0x8000000,%ebx
  mov    %r13,%rdi
  mov    %ebx,%esi
  call   <qfits_query@plt>
  mov    %eax,(%r12)
  sar    $0x1f,%eax
L4:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L60:
  mov    $0xffffffff,%eax
  jmp    L4
L47:
  or     $0xffffffff,%eax
  ret
```

assistant: ```c
int qfits_get_hdrinfo(const char *filename, int xtnum, int *seg_start, int *seg_size) {
  if (filename == NULL || xtnum < 0 || (seg_start == NULL && seg_size == NULL)) {
    return -1;
  }
  if (seg_start != NULL) {
    *seg_start = qfits_query(filename, (1 << 29) | xtnum);
    if (*seg_start < 0)
      return -1;
  }
  if (seg_size != NULL) {
    *seg_size = qfits_query(filename, (1 << 27) | xtnum);
    if (*seg_size < 0)
      return -1;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<delta_time>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movsd  %xmm0,-0x18(%rbp)
  mov    %edi,-0x1c(%rbp)
  mov    %rsi,-0x28(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x1c(%rbp),%xmm0
  movsd  -0x18(%rbp),%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  cvttsd2si %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <write_var_len@plt>
  leave
  ret
```

assistant: ```c
int delta_time(double beat, int nticks, FILE *fp) {
  double dval = beat * nticks;
  long val = (long)dval;
  return write_var_len(val, fp);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<delta_time>:
  endbr64
  sub    $0x8,%rsp
  movapd %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %edi,%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%rdi
  call   <write_var_len@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int delta_time(double beat, int nticks, FILE *fp) {
  double dval = beat * nticks;
  long val = (long)dval;
  return write_var_len(val, fp);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<delta_time>:
  endbr64
  movapd %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %edi,%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%rdi
  jmp    1050 <write_var_len@plt>
```

assistant: ```c
int delta_time(double beat, int nticks, FILE *fp) {
  double dval = beat * nticks;
  long val = (long)dval;
  return write_var_len(val, fp);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<delta_time>:
  endbr64
  movapd %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %edi,%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%rdi
  jmp    1050 <write_var_len@plt>
```

assistant: ```c
int delta_time(double beat, int nticks, FILE *fp) {
  double dval = beat * nticks;
  long val = (long)dval;
  return write_var_len(val, fp);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<from_digits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L11
L17:
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%ebx
  mov    -0x2c(%rbp),%eax
  sub    $0x1,%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  imul   %ebx,%eax
  add    %eax,-0x14(%rbp)
  addl   $0x1,-0x18(%rbp)
L11:
  mov    -0x18(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L17
  mov    -0x14(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int from_digits(int digit[], int n) {
  int j;
  int num = 0;
  for (j = 0; j < n; j++) {
    num += digit[j] * pow(10, n - 1 - j);
  }
  return num;
}
```
user: Translate the below assembly code into C programming language:

```asm
<from_digits>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L24
  mov    %rdi,%r14
  mov    %esi,%r13d
  lea    0x1(%rsi),%r15d
  mov    $0x1,%ebx
  mov    $0x0,%r12d
L25:
  mov    -0x4(%r14,%rbx,4),%ebp
  mov    %r13d,%esi
  sub    %ebx,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  imul   %eax,%ebp
  add    %ebp,%r12d
  add    $0x1,%rbx
  cmp    %r15,%rbx
  jne    L25
L22:
  mov    %r12d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L24:
  mov    $0x0,%r12d
  jmp    L22
```

assistant: ```c
int from_digits(int digit[], int n) {
  int j;
  int num = 0;
  for (j = 0; j < n; j++) {
    num += digit[j] * pow(10, n - 1 - j);
  }
  return num;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<from_digits>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L5
  mov    %rdi,%r12
  lea    -0x1(%rsi),%ebx
  xor    %r13d,%r13d
  nopl   0x0(%rax,%rax,1)
L13:
  mov    (%r12),%ebp
  mov    %ebx,%esi
  mov    $0xa,%edi
  add    $0x4,%r12
  call   <pow@plt>
  imul   %eax,%ebp
  add    %ebp,%r13d
  sub    $0x1,%ebx
  jae    L13
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L5:
  add    $0x8,%rsp
  xor    %r13d,%r13d
  pop    %rbx
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int from_digits(int digit[], int n) {
  int j;
  int num = 0;
  for (j = 0; j < n; j++) {
    num += digit[j] * pow(10, n - 1 - j);
  }
  return num;
}
```
user: Translate the below assembly code into C programming language:

```asm
<from_digits>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L0
  mov    %rdi,%r12
  lea    -0x1(%rsi),%ebx
  xor    %r13d,%r13d
  nopl   0x0(%rax,%rax,1)
L13:
  mov    (%r12),%ebp
  mov    %ebx,%esi
  mov    $0xa,%edi
  add    $0x4,%r12
  call   <pow@plt>
  imul   %eax,%ebp
  add    %ebp,%r13d
  sub    $0x1,%ebx
  jae    L13
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L0:
  add    $0x8,%rsp
  xor    %r13d,%r13d
  pop    %rbx
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int from_digits(int digit[], int n) {
  int j;
  int num = 0;
  for (j = 0; j < n; j++) {
    num += digit[j] * pow(10, n - 1 - j);
  }
  return num;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<make_wave>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x1,-0x10(%rbp)
  jmp    L58
L3:
  movl   $0x1,-0xc(%rbp)
  jmp    L12
L47:
  mov    -0x10(%rbp),%eax
  lea    0x30(%rax),%ecx
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0xc(%rbp)
L12:
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jle    L47
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0xa,(%rax)
  addl   $0x1,-0x10(%rbp)
L58:
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L3
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L45
L51:
  movl   $0x1,-0xc(%rbp)
  jmp    L7
L59:
  mov    -0x10(%rbp),%eax
  lea    0x30(%rax),%ecx
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0xc(%rbp)
L7:
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jle    L59
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0xa,(%rax)
  subl   $0x1,-0x10(%rbp)
L45:
  cmpl   $0x0,-0x10(%rbp)
  jg     L51
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void make_wave(int amp, int freq, char buf[]) {
  char *ptr;
  int i, j;
  ptr = &buf[0];
  for (i = 1; i <= amp; i++) {
    for (j = 1; j <= i; j++) {
      *ptr++ = i + '0';
    }
    *ptr++ = '\n';
  }
  for (i = amp - 1; i >= 1; i--) {
    for (j = 1; j <= i; j++) {
      *ptr++ = i + '0';
    }
    *ptr++ = '\n';
  }
  *ptr = '\0';
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<make_wave>:
  endbr64
  test   %edi,%edi
  jle    L4
  lea    0x1(%rdi),%r8d
  mov    $0x31,%ecx
  mov    $0x1,%esi
  jmp    L88
L18:
  add    $0x1,%rdx
  mov    %cl,-0x1(%rdx)
  cmp    %rax,%rdx
  jne    L18
L55:
  lea    0x1(%rax),%rdx
  movb   $0xa,(%rax)
  add    $0x1,%rsi
  add    $0x1,%ecx
  cmp    %r8,%rsi
  je     L4
L88:
  lea    (%rdx,%rsi,1),%rax
  test   %esi,%esi
  jg     L18
  mov    %rdx,%rax
  jmp    L55
L4:
  lea    -0x1(%rdi),%ecx
  test   %ecx,%ecx
  jle    L19
  lea    -0x2(%rdi),%esi
  add    $0x30,%ecx
  jmp    L62
L100:
  add    $0x1,%rdx
  mov    %cl,-0x1(%rdx)
  cmp    %rdx,%rax
  jne    L100
L117:
  lea    0x1(%rax),%rdx
  movb   $0xa,(%rax)
  sub    $0x1,%esi
  sub    $0x1,%ecx
  cmp    $0xffffffff,%esi
  je     L19
L62:
  lea    0x1(%rsi),%edi
  mov    %esi,%eax
  lea    0x1(%rdx,%rax,1),%rax
  test   %edi,%edi
  jg     L100
  mov    %rdx,%rax
  jmp    L117
L19:
  movb   $0x0,(%rdx)
  ret
```

assistant: ```c
void make_wave(int amp, int freq, char buf[]) {
  char *ptr;
  int i, j;
  ptr = &buf[0];
  for (i = 1; i <= amp; i++) {
    for (j = 1; j <= i; j++) {
      *ptr++ = i + '0';
    }
    *ptr++ = '\n';
  }
  for (i = amp - 1; i >= 1; i--) {
    for (j = 1; j <= i; j++) {
      *ptr++ = i + '0';
    }
    *ptr++ = '\n';
  }
  *ptr = '\0';
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<make_wave>:
  endbr64
  mov    %rdx,%rcx
  test   %edi,%edi
  jle    L28
  push   %r14
  push   %r13
  push   %r12
  lea    0x1(%rdi),%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L5:
  lea    0x30(%rbx),%esi
  mov    %rbx,%rdx
  mov    %rcx,%rdi
  movsbl %sil,%esi
  call   <memset@plt>
  mov    %rax,%rcx
  lea    (%rax,%rbx,1),%rax
  add    $0x1,%rbx
  movb   $0xa,(%rax)
  add    %rbx,%rcx
  cmp    %r12,%rbx
  jne    L5
  cmp    $0x1,%ebp
  jle    L2
  sub    $0x2,%ebp
  mov    $0x1,%r13d
  xchg   %ax,%ax
L22:
  lea    0x1(%rbp),%r14d
  mov    %ebp,%r12d
  lea    0x31(%rbp),%esi
  mov    %rcx,%rdi
  test   %r14d,%r14d
  lea    0x1(%r12),%rbx
  movsbl %sil,%esi
  cmovle %r13,%rbx
  mov    %rbx,%rdx
  call   <memset@plt>
  add    %rax,%rbx
  mov    %rax,%rcx
  xor    %eax,%eax
  test   %r14d,%r14d
  cmovle %rax,%r12
  movb   $0xa,(%rbx)
  lea    0x2(%rcx,%r12,1),%rcx
  sub    $0x1,%ebp
  jae    L22
L2:
  movb   $0x0,(%rcx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L28:
  movb   $0x0,(%rdx)
  ret
```

assistant: ```c
void make_wave(int amp, int freq, char buf[]) {
  char *ptr;
  int i, j;
  ptr = &buf[0];
  for (i = 1; i <= amp; i++) {
    for (j = 1; j <= i; j++) {
      *ptr++ = i + '0';
    }
    *ptr++ = '\n';
  }
  for (i = amp - 1; i >= 1; i--) {
    for (j = 1; j <= i; j++) {
      *ptr++ = i + '0';
    }
    *ptr++ = '\n';
  }
  *ptr = '\0';
  return;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<make_wave>:
  endbr64
  mov    %rdx,%rcx
  test   %edi,%edi
  jle    L18
  push   %r14
  push   %r13
  push   %r12
  lea    0x1(%rdi),%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L32:
  lea    0x30(%rbx),%esi
  mov    %rbx,%rdx
  mov    %rcx,%rdi
  movsbl %sil,%esi
  call   <memset@plt>
  mov    %rax,%rcx
  lea    (%rax,%rbx,1),%rax
  add    $0x1,%rbx
  movb   $0xa,(%rax)
  add    %rbx,%rcx
  cmp    %r12,%rbx
  jne    L32
  cmp    $0x1,%ebp
  jle    L35
  sub    $0x2,%ebp
  mov    $0x1,%r13d
  xchg   %ax,%ax
L1:
  lea    0x1(%rbp),%r14d
  mov    %ebp,%r12d
  lea    0x31(%rbp),%esi
  mov    %rcx,%rdi
  test   %r14d,%r14d
  lea    0x1(%r12),%rbx
  movsbl %sil,%esi
  cmovle %r13,%rbx
  mov    %rbx,%rdx
  call   <memset@plt>
  add    %rax,%rbx
  mov    %rax,%rcx
  xor    %eax,%eax
  test   %r14d,%r14d
  cmovle %rax,%r12
  movb   $0xa,(%rbx)
  lea    0x2(%rcx,%r12,1),%rcx
  sub    $0x1,%ebp
  jae    L1
L35:
  movb   $0x0,(%rcx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L18:
  movb   $0x0,(%rdx)
  ret
```

assistant: ```c
void make_wave(int amp, int freq, char buf[]) {
  char *ptr;
  int i, j;
  ptr = &buf[0];
  for (i = 1; i <= amp; i++) {
    for (j = 1; j <= i; j++) {
      *ptr++ = i + '0';
    }
    *ptr++ = '\n';
  }
  for (i = amp - 1; i >= 1; i--) {
    for (j = 1; j <= i; j++) {
      *ptr++ = i + '0';
    }
    *ptr++ = '\n';
  }
  *ptr = '\0';
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <testcase@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  mov    %eax,-0x4(%rbp)
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(void) {
  int x;
  x = testcase() ? 1 : 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
  mov    $0x0,%eax
  call   <testcase@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int x;
  x = testcase() ? 1 : 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    %eax,%eax
  call   <testcase@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int x;
  x = testcase() ? 1 : 0;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <testcase@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int x;
  x = testcase() ? 1 : 0;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ft_strlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L12
L20:
  addl   $0x1,-0x4(%rbp)
L12:
  cmpq   $0x0,-0x18(%rbp)
  je     L5
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L20
L5:
  mov    -0x4(%rbp),%eax
  cltq
  pop    %rbp
  ret
```

assistant: ```c
size_t ft_strlen(const char *s) {
  int i;
  i = 0;
  while (s && s[i] != '\0')
    i++;
  return (i);
}
```
user: What is the c source code of the assembly code below?

```asm
<ft_strlen>:
  endbr64
  test   %rdi,%rdi
  je     L12
  mov    $0x0,%edx
L20:
  mov    %edx,%eax
  add    $0x1,%rdx
  cmpb   $0x0,-0x1(%rdi,%rdx,1)
  jne    L20
L0:
  cltq
  ret
L12:
  mov    $0x0,%eax
  jmp    L0
```

assistant: ```c
size_t ft_strlen(const char *s) {
  int i;
  i = 0;
  while (s && s[i] != '\0')
    i++;
  return (i);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ft_strlen>:
  endbr64
  test   %rdi,%rdi
  je     L1
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L24:
  mov    %rax,%r8
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L24
L13:
  mov    %r8,%rax
  ret
L1:
  xor    %r8d,%r8d
  jmp    L13
```

assistant: ```c
size_t ft_strlen(const char *s) {
  int i;
  i = 0;
  while (s && s[i] != '\0')
    i++;
  return (i);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_strlen>:
  endbr64
  test   %rdi,%rdi
  je     L15
  cmpb   $0x0,(%rdi)
  je     L15
  mov    $0x1,%eax
  nopl   0x0(%rax,%rax,1)
L11:
  mov    %rax,%r8
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L11
  mov    %r8,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  xor    %r8d,%r8d
  mov    %r8,%rax
  ret
```

assistant: ```c
size_t ft_strlen(const char *s) {
  int i;
  i = 0;
  while (s && s[i] != '\0')
    i++;
  return (i);
}
```
user: Translate the below assembly code into C programming language:

```asm
<compute>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1c0,%rsp
  mov    %rdi,-0x1b8(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1ac(%rbp)
  jmp    L17
L35:
  mov    -0x1b8(%rbp),%rax
  mov    %rax,%rdi
  call   <pthread_mutex_lock@plt>
  mov    $0x0,%eax
  call   <rand@plt>
  mov    -0x1ac(%rbp),%edx
  movslq %edx,%rdx
  mov    %eax,-0x1a0(%rbp,%rdx,4)
  mov    -0x1b8(%rbp),%rax
  mov    %rax,%rdi
  call   <pthread_mutex_unlock@plt>
  addl   $0x1,-0x1ac(%rbp)
L17:
  cmpl   $0x63,-0x1ac(%rbp)
  jle    L35
  movl   $0x0,-0x1a8(%rbp)
  movl   $0x0,-0x1ac(%rbp)
  jmp    L23
L18:
  mov    -0x1ac(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  add    %eax,-0x1a8(%rbp)
  addl   $0x1,-0x1ac(%rbp)
L23:
  cmpl   $0x63,-0x1ac(%rbp)
  jle    L18
  mov    -0x1a8(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x1a4(%rbp)
  addl   $0x1,-0x1a4(%rbp)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

assistant: ```c
void compute(pthread_mutex_t *l) {
  int a[100];
  int i;
  for (i = 0; i < 100; i++) {
    pthread_mutex_lock(l);
    a[i] = rand();
    pthread_mutex_unlock(l);
  }
  int sum = 0;
  for (i = 0; i < 100; i++) {
    sum += a[i];
  }
  int m = sum / 100;
  m++;
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<compute>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x64,%ebx
L11:
  mov    %rbp,%rdi
  call   <pthread_mutex_lock@plt>
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %rbp,%rdi
  call   <pthread_mutex_unlock@plt>
  sub    $0x1,%ebx
  jne    L11
  mov    $0x64,%eax
L19:
  sub    $0x1,%eax
  jne    L19
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void compute(pthread_mutex_t *l) {
  int a[100];
  int i;
  for (i = 0; i < 100; i++) {
    pthread_mutex_lock(l);
    a[i] = rand();
    pthread_mutex_unlock(l);
  }
  int sum = 0;
  for (i = 0; i < 100; i++) {
    sum += a[i];
  }
  int m = sum / 100;
  m++;
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<compute>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    $0x64,%ebx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %rbp,%rdi
  call   <pthread_mutex_lock@plt>
  xor    %eax,%eax
  call   <rand@plt>
  mov    %rbp,%rdi
  call   <pthread_mutex_unlock@plt>
  sub    $0x1,%ebx
  jne    L5
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void compute(pthread_mutex_t *l) {
  int a[100];
  int i;
  for (i = 0; i < 100; i++) {
    pthread_mutex_lock(l);
    a[i] = rand();
    pthread_mutex_unlock(l);
  }
  int sum = 0;
  for (i = 0; i < 100; i++) {
    sum += a[i];
  }
  int m = sum / 100;
  m++;
  return;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<compute>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    $0x64,%ebx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %rbp,%rdi
  call   <pthread_mutex_lock@plt>
  xor    %eax,%eax
  call   <rand@plt>
  mov    %rbp,%rdi
  call   <pthread_mutex_unlock@plt>
  sub    $0x1,%ebx
  jne    L6
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void compute(pthread_mutex_t *l) {
  int a[100];
  int i;
  for (i = 0; i < 100; i++) {
    pthread_mutex_lock(l);
    a[i] = rand();
    pthread_mutex_unlock(l);
  }
  int sum = 0;
  for (i = 0; i < 100; i++) {
    sum += a[i];
  }
  int m = sum / 100;
  m++;
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<getline>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
L15:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x5(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L11
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x5(%rbp),%eax
  mov    %al,(%rdx)
L11:
  cmpb   $0xa,-0x5(%rbp)
  jne    L15
  mov    -0x4(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void getline(char buffer[], int n) {
  char character;
  int i = 0;
  do {
    character = getchar();
    if (i < n)
      buffer[i++] = character;
  } while (character != '\n');
  buffer[i - 1] = '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<getline>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    $0x0,%ebx
  jmp    L1
L30:
  movslq %ebx,%rdx
  mov    %al,(%r12,%rdx,1)
  lea    0x1(%rbx),%ebx
L13:
  cmp    $0xa,%al
  je     L12
L1:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    %ebp,%ebx
  jge    L13
  jmp    L30
L12:
  movslq %ebx,%rbx
  movb   $0x0,-0x1(%r12,%rbx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void getline(char buffer[], int n) {
  char character;
  int i = 0;
  do {
    character = getchar();
    if (i < n)
      buffer[i++] = character;
  } while (character != '\n');
  buffer[i - 1] = '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<getline>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nop
L3:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %ebp,%ebx
  jge    L7
  movslq %ebx,%rdx
  add    $0x1,%ebx
  mov    %al,(%r12,%rdx,1)
L7:
  cmp    $0xa,%al
  jne    L3
  movslq %ebx,%rbx
  movb   $0x0,-0x1(%r12,%rbx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void getline(char buffer[], int n) {
  char character;
  int i = 0;
  do {
    character = getchar();
    if (i < n)
      buffer[i++] = character;
  } while (character != '\n');
  buffer[i - 1] = '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<getline>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
L11:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %r12d,%ebx
  jge    L9
  lea    0x1(%rbx),%edx
  mov    %al,0x0(%rbp,%rbx,1)
  movslq %edx,%rbx
  cmp    $0xa,%al
  jne    L11
L22:
  movb   $0x0,-0x1(%rbp,%rbx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L9:
  cmp    $0xa,%al
  je     L22
L35:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L35
  movb   $0x0,-0x1(%rbp,%rbx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void getline(char buffer[], int n) {
  char character;
  int i = 0;
  do {
    character = getchar();
    if (i < n)
      buffer[i++] = character;
  } while (character != '\n');
  buffer[i - 1] = '\0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    D4(%rip),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  void *p;
  (void)argc;
  (void)argv;
  p = (void *)(flock);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  void *p;
  (void)argc;
  (void)argv;
  p = (void *)(flock);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  void *p;
  (void)argc;
  (void)argv;
  p = (void *)(flock);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  void *p;
  (void)argc;
  (void)argv;
  p = (void *)(flock);
  return 0;
}
```