ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rbx
  jmp    L2
L22:
  movzbl 0x6(%rsp),%edi
  call   <font_reverse_c@plt>
  mov    %al,0x7(%rsp)
  lea    0x7(%rsp),%rsi
  mov    $0x1,%edx
  mov    $0x1,%edi
  call   <write@plt>
L2:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <read@plt>
  test   %eax,%eax
  jg     L22
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  while (1) {
    unsigned char ic;
    if (read(0, &ic, 1) < 1)
      break;
    unsigned char oc = font_reverse_c(ic);
    write(1, &oc, 1);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rbx
  lea    0x7(%rsp),%rbp
  jmp    L5
  cs nopw 0x0(%rax,%rax,1)
L8:
  movzbl 0x6(%rsp),%edi
  call   <font_reverse_c@plt>
  mov    $0x1,%edx
  mov    %rbp,%rsi
  mov    $0x1,%edi
  mov    %al,0x7(%rsp)
  call   <write@plt>
L5:
  xor    %edi,%edi
  mov    $0x1,%edx
  mov    %rbx,%rsi
  call   <read@plt>
  test   %eax,%eax
  jg     L8
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L21:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  while (1) {
    unsigned char ic;
    if (read(0, &ic, 1) < 1)
      break;
    unsigned char oc = font_reverse_c(ic);
    write(1, &oc, 1);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rbx
  lea    0x7(%rsp),%rbp
  jmp    L22
  cs nopw 0x0(%rax,%rax,1)
L9:
  movzbl 0x6(%rsp),%edi
  call   <font_reverse_c@plt>
  mov    $0x1,%edx
  mov    %rbp,%rsi
  mov    $0x1,%edi
  mov    %al,0x7(%rsp)
  call   <write@plt>
L22:
  xor    %edi,%edi
  mov    $0x1,%edx
  mov    %rbx,%rsi
  call   <read@plt>
  test   %eax,%eax
  jg     L9
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  while (1) {
    unsigned char ic;
    if (read(0, &ic, 1) < 1)
      break;
    unsigned char oc = font_reverse_c(ic);
    write(1, &oc, 1);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<strcasecmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  jmp    L34
L14:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L21
  mov    $0x0,%eax
  jmp    L38
L21:
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x20(%rbp)
L34:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <toupper@plt>
  mov    %eax,%ebx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <toupper@plt>
  cmp    %eax,%ebx
  je     L14
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <toupper@plt>
  mov    %eax,%ebx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <toupper@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
L38:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int strcasecmp(const char *a, const char *b) {
  while (toupper(*a) == toupper(*b)) {
    if (!*a)
      return 0;
    a++;
    b++;
  }
  return toupper(*a) - toupper(*b);
}
```
user: Translate the below assembly code into C programming language:

```asm
<strcasecmp>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  jmp    L16
L36:
  add    $0x1,%rbx
  add    $0x1,%r12
L16:
  movsbl (%rbx),%edi
  call   <toupper@plt>
  mov    %eax,%ebp
  movsbl (%r12),%edi
  call   <toupper@plt>
  cmp    %eax,%ebp
  jne    L35
  cmpb   $0x0,(%rbx)
  jne    L36
  mov    $0x0,%eax
  jmp    L19
L35:
  movsbl (%rbx),%edi
  call   <toupper@plt>
  mov    %eax,%ebx
  movsbl (%r12),%edi
  call   <toupper@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
L19:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int strcasecmp(const char *a, const char *b) {
  while (toupper(*a) == toupper(*b)) {
    if (!*a)
      return 0;
    a++;
    b++;
  }
  return toupper(*a) - toupper(*b);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<strcasecmp>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  jmp    L7
L14:
  cmpb   $0x0,(%rbx)
  je     L17
  add    $0x1,%rbx
  add    $0x1,%r12
L7:
  movsbl (%rbx),%edi
  call   <toupper@plt>
  movsbl (%r12),%edi
  mov    %eax,%ebp
  call   <toupper@plt>
  cmp    %eax,%ebp
  je     L14
  movsbl (%rbx),%edi
  call   <toupper@plt>
  movsbl (%r12),%edi
  mov    %eax,%ebx
  call   <toupper@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  sub    %r8d,%eax
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L17:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int strcasecmp(const char *a, const char *b) {
  while (toupper(*a) == toupper(*b)) {
    if (!*a)
      return 0;
    a++;
    b++;
  }
  return toupper(*a) - toupper(*b);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strcasecmp>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  jmp    L23
L7:
  cmpb   $0x0,(%rbx)
  je     L19
  add    $0x1,%rbx
  add    $0x1,%r12
L23:
  movsbl (%rbx),%edi
  call   <toupper@plt>
  movsbl (%r12),%edi
  mov    %eax,%ebp
  call   <toupper@plt>
  cmp    %eax,%ebp
  je     L7
  movsbl (%rbx),%edi
  call   <toupper@plt>
  movsbl (%r12),%edi
  mov    %eax,%ebx
  call   <toupper@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  sub    %r8d,%eax
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L19:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int strcasecmp(const char *a, const char *b) {
  while (toupper(*a) == toupper(*b)) {
    if (!*a)
      return 0;
    a++;
    b++;
  }
  return toupper(*a) - toupper(*b);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<strlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %rdi,-0x10(%rbp)
  mov    $0x0,%ebx
  jmp    L6
L4:
  add    $0x1,%ebx
L6:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L4
  movslq %ebx,%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  register int i = 0;
  while (*str++)
    i++;
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strlen>:
  endbr64
  lea    0x1(%rdi),%rax
  cmpb   $0x0,(%rdi)
  je     L7
  mov    %rax,%rdx
  mov    $0x1,%ecx
  sub    %eax,%ecx
L16:
  lea    (%rcx,%rdx,1),%eax
  add    $0x1,%rdx
  cmpb   $0x0,-0x1(%rdx)
  jne    L16
L6:
  cltq
  ret
L7:
  mov    $0x0,%eax
  jmp    L6
```

assistant: ```c
size_t strlen(const char *str) {
  register int i = 0;
  while (*str++)
    i++;
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  lea    0x1(%rdi),%rdx
  je     L13
  mov    $0x1,%ecx
  sub    %edx,%ecx
  nopl   0x0(%rax)
L8:
  lea    (%rcx,%rdx,1),%eax
  add    $0x1,%rdx
  cmpb   $0x0,-0x1(%rdx)
  jne    L8
  cltq
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  xor    %eax,%eax
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  register int i = 0;
  while (*str++)
    i++;
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  lea    0x1(%rdi),%rdx
  je     L11
  mov    $0x1,%ecx
  sub    %edx,%ecx
  nopl   0x0(%rax)
L9:
  lea    (%rcx,%rdx,1),%eax
  add    $0x1,%rdx
  cmpb   $0x0,-0x1(%rdx)
  jne    L9
  cltq
  ret
  nopl   0x0(%rax,%rax,1)
L11:
  xor    %eax,%eax
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  register int i = 0;
  while (*str++)
    i++;
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xtc_sizeofints>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xc0,%rsp
  mov    %edi,-0xb4(%rbp)
  mov    %rsi,-0xc0(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0xa0(%rbp)
  movl   $0x1,-0x90(%rbp)
  movl   $0x0,-0x9c(%rbp)
  movl   $0x0,-0xa8(%rbp)
  jmp    L61
L27:
  movl   $0x0,-0x94(%rbp)
  movl   $0x0,-0x98(%rbp)
  jmp    L62
L29:
  mov    -0x98(%rbp),%eax
  mov    -0x90(%rbp,%rax,4),%edx
  mov    -0xa8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0xc0(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  imul   %edx,%eax
  add    %eax,-0x94(%rbp)
  mov    -0x94(%rbp),%eax
  movzbl %al,%edx
  mov    -0x98(%rbp),%eax
  mov    %edx,-0x90(%rbp,%rax,4)
  shrl   $0x8,-0x94(%rbp)
  addl   $0x1,-0x98(%rbp)
L62:
  mov    -0x98(%rbp),%eax
  cmp    -0xa0(%rbp),%eax
  jb     L29
  jmp    L80
L15:
  mov    -0x98(%rbp),%edx
  lea    0x1(%rdx),%eax
  mov    %eax,-0x98(%rbp)
  mov    -0x94(%rbp),%eax
  movzbl %al,%eax
  mov    %edx,%edx
  mov    %eax,-0x90(%rbp,%rdx,4)
  shrl   $0x8,-0x94(%rbp)
L80:
  cmpl   $0x0,-0x94(%rbp)
  jne    L15
  mov    -0x98(%rbp),%eax
  mov    %eax,-0xa0(%rbp)
  addl   $0x1,-0xa8(%rbp)
L61:
  mov    -0xa8(%rbp),%eax
  cmp    -0xb4(%rbp),%eax
  jl     L27
  movl   $0x1,-0xa4(%rbp)
  subl   $0x1,-0xa0(%rbp)
  jmp    L6
L89:
  addl   $0x1,-0x9c(%rbp)
  shll   -0xa4(%rbp)
L6:
  mov    -0xa0(%rbp),%eax
  mov    -0x90(%rbp,%rax,4),%eax
  cmp    %eax,-0xa4(%rbp)
  jbe    L89
  mov    -0xa0(%rbp),%eax
  lea    0x0(,%rax,8),%edx
  mov    -0x9c(%rbp),%eax
  add    %edx,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L41
  call   <__stack_chk_fail@plt>
L41:
  leave
  ret
```

assistant: ```c
int xtc_sizeofints(int nints, unsigned int *sizes) {
  int i;
  unsigned int num;
  unsigned int nbytes, nbits, bytes[32], bytecnt, tmp;
  nbytes = 1;
  bytes[0] = 1;
  nbits = 0;
  for (i = 0; i < nints; i++) {
    tmp = 0;
    for (bytecnt = 0; bytecnt < nbytes; bytecnt++) {
      tmp = bytes[bytecnt] * sizes[i] + tmp;
      bytes[bytecnt] = tmp & 0xff;
      tmp >>= 8;
    }
    while (tmp != 0) {
      bytes[bytecnt++] = tmp & 0xff;
      tmp >>= 8;
    }
    nbytes = bytecnt;
  }
  num = 1;
  nbytes--;
  while (bytes[nbytes] >= num) {
    nbits++;
    num *= 2;
  }
  return nbits + nbytes * 8;
}
```
user: What is the c source code of the assembly code below?

```asm
<xtc_sizeofints>:
  endbr64
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  test   %edi,%edi
  jle    L81
  mov    %rsi,%r9
  lea    -0x1(%rdi),%eax
  lea    0x4(%rsi,%rax,4),%r10
  mov    $0x1,%ecx
  lea    0x4(%rsp),%r11
  jmp    L15
L54:
  mov    (%r9),%r8d
  mov    %rsp,%rdx
  lea    -0x1(%rcx),%eax
  lea    (%r11,%rax,4),%rdi
  mov    $0x0,%eax
L50:
  mov    %r8d,%esi
  imul   (%rdx),%esi
  add    %esi,%eax
  movzbl %al,%esi
  mov    %esi,(%rdx)
  shr    $0x8,%eax
  add    $0x4,%rdx
  cmp    %rdx,%rdi
  jne    L50
  test   %eax,%eax
  je     L58
L80:
  mov    %ecx,%edx
  add    $0x1,%ecx
  mov    %edx,%edx
  movzbl %al,%esi
  mov    %esi,(%rsp,%rdx,4)
  shr    $0x8,%eax
  jne    L80
L58:
  add    $0x4,%r9
  cmp    %r10,%r9
  je     L65
L15:
  test   %ecx,%ecx
  jne    L54
  jmp    L58
L65:
  sub    $0x1,%ecx
  mov    %ecx,%eax
  mov    (%rsp,%rax,4),%esi
  test   %esi,%esi
  je     L7
  mov    $0x0,%edx
  mov    $0x1,%eax
L106:
  add    $0x1,%edx
  add    %eax,%eax
  cmp    %esi,%eax
  jbe    L106
L34:
  lea    (%rdx,%rcx,8),%eax
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L125
  add    $0x98,%rsp
  ret
L81:
  mov    $0x0,%ecx
  mov    $0x1,%edx
  jmp    L34
L7:
  mov    %esi,%edx
  jmp    L34
L125:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int xtc_sizeofints(int nints, unsigned int *sizes) {
  int i;
  unsigned int num;
  unsigned int nbytes, nbits, bytes[32], bytecnt, tmp;
  nbytes = 1;
  bytes[0] = 1;
  nbits = 0;
  for (i = 0; i < nints; i++) {
    tmp = 0;
    for (bytecnt = 0; bytecnt < nbytes; bytecnt++) {
      tmp = bytes[bytecnt] * sizes[i] + tmp;
      bytes[bytecnt] = tmp & 0xff;
      tmp >>= 8;
    }
    while (tmp != 0) {
      bytes[bytecnt++] = tmp & 0xff;
      tmp >>= 8;
    }
    nbytes = bytecnt;
  }
  num = 1;
  nbytes--;
  while (bytes[nbytes] >= num) {
    nbits++;
    num *= 2;
  }
  return nbits + nbytes * 8;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<xtc_sizeofints>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  test   %edi,%edi
  jle    L107
  lea    -0x1(%rdi),%eax
  mov    %rsi,%r9
  mov    $0x1,%ecx
  mov    %rsp,%rbx
  lea    0x4(%rsi,%rax,4),%r10
  xor    %ebp,%ebp
  lea    0x4(%rsp),%r11
  nopl   0x0(%rax)
L70:
  mov    %ebp,%eax
  mov    (%r9),%r8d
  mov    %rbx,%rdx
  lea    (%r11,%rax,4),%rdi
  xor    %eax,%eax
  xchg   %ax,%ax
L78:
  mov    (%rdx),%esi
  add    $0x4,%rdx
  imul   %r8d,%esi
  add    %esi,%eax
  movzbl %al,%esi
  shr    $0x8,%eax
  mov    %esi,-0x4(%rdx)
  cmp    %rdx,%rdi
  jne    L78
  test   %eax,%eax
  je     L10
  xchg   %ax,%ax
L35:
  mov    %ecx,%edx
  movzbl %al,%esi
  add    $0x1,%ecx
  shr    $0x8,%eax
  mov    %esi,(%rsp,%rdx,4)
  jne    L35
  lea    -0x1(%rcx),%ebp
L10:
  add    $0x4,%r9
  cmp    %r9,%r10
  je     L114
  test   %ecx,%ecx
  jne    L70
L101:
  lea    0x4(%r9),%rax
  cmp    %rax,%r10
  je     L114
  add    $0x8,%r9
  cmp    %r9,%r10
  jne    L101
  nopw   0x0(%rax,%rax,1)
L114:
  mov    %ebp,%eax
  lea    0x0(,%rbp,8),%r8d
  mov    (%rsp,%rax,4),%ecx
  test   %ecx,%ecx
  je     L75
  xor    %edx,%edx
  mov    $0x1,%eax
L85:
  add    %eax,%eax
  add    $0x1,%edx
  cmp    %eax,%ecx
  jae    L85
L8:
  add    %edx,%r8d
L75:
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L52
  add    $0x98,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
L107:
  xor    %r8d,%r8d
  mov    $0x1,%edx
  jmp    L8
L52:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int xtc_sizeofints(int nints, unsigned int *sizes) {
  int i;
  unsigned int num;
  unsigned int nbytes, nbits, bytes[32], bytecnt, tmp;
  nbytes = 1;
  bytes[0] = 1;
  nbits = 0;
  for (i = 0; i < nints; i++) {
    tmp = 0;
    for (bytecnt = 0; bytecnt < nbytes; bytecnt++) {
      tmp = bytes[bytecnt] * sizes[i] + tmp;
      bytes[bytecnt] = tmp & 0xff;
      tmp >>= 8;
    }
    while (tmp != 0) {
      bytes[bytecnt++] = tmp & 0xff;
      tmp >>= 8;
    }
    nbytes = bytecnt;
  }
  num = 1;
  nbytes--;
  while (bytes[nbytes] >= num) {
    nbits++;
    num *= 2;
  }
  return nbits + nbytes * 8;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<xtc_sizeofints>:
  endbr64
  push   %rbx
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  test   %edi,%edi
  jle    L52
  lea    -0x1(%rdi),%eax
  mov    %rsi,%r10
  lea    0x4(%rsp),%rbx
  mov    $0x1,%ecx
  lea    0x4(%rsi,%rax,4),%r11
  nopl   0x0(%rax,%rax,1)
L124:
  test   %ecx,%ecx
  je     L76
  lea    -0x1(%rcx),%eax
  mov    (%r10),%r9d
  mov    %rsp,%rdx
  mov    %rax,%rsi
  lea    (%rbx,%rax,4),%r8
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L23:
  mov    (%rdx),%edi
  add    $0x4,%rdx
  imul   %r9d,%edi
  add    %edi,%eax
  movzbl %al,%edi
  shr    $0x8,%eax
  mov    %edi,-0x4(%rdx)
  cmp    %rdx,%r8
  jne    L23
  test   %eax,%eax
  je     L72
  cs nopw 0x0(%rax,%rax,1)
L3:
  mov    %ecx,%edx
  movzbl %al,%edi
  add    $0x1,%ecx
  shr    $0x8,%eax
  mov    %edi,(%rsp,%rdx,4)
  mov    %rdx,%rsi
  jne    L3
L72:
  add    $0x4,%r10
  cmp    %r10,%r11
  jne    L124
L14:
  mov    %esi,%eax
  mov    (%rsp,%rax,4),%edi
  lea    0x0(,%rsi,8),%eax
  test   %edi,%edi
  je     L58
  xor    %ecx,%ecx
  mov    $0x1,%edx
  nopl   0x0(%rax,%rax,1)
L75:
  add    %edx,%edx
  add    $0x1,%ecx
  cmp    %edx,%edi
  jae    L75
L43:
  add    %ecx,%eax
L58:
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x90,%rsp
  pop    %rbx
  ret
L76:
  add    $0x4,%r10
  mov    $0xffffffff,%esi
  cmp    %r10,%r11
  je     L14
  add    $0x4,%r10
  cmp    %r10,%r11
  jne    L124
  jmp    L14
L52:
  xor    %eax,%eax
  mov    $0x1,%ecx
  jmp    L43
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int xtc_sizeofints(int nints, unsigned int *sizes) {
  int i;
  unsigned int num;
  unsigned int nbytes, nbits, bytes[32], bytecnt, tmp;
  nbytes = 1;
  bytes[0] = 1;
  nbits = 0;
  for (i = 0; i < nints; i++) {
    tmp = 0;
    for (bytecnt = 0; bytecnt < nbytes; bytecnt++) {
      tmp = bytes[bytecnt] * sizes[i] + tmp;
      bytes[bytecnt] = tmp & 0xff;
      tmp >>= 8;
    }
    while (tmp != 0) {
      bytes[bytecnt++] = tmp & 0xff;
      tmp >>= 8;
    }
    nbytes = bytecnt;
  }
  num = 1;
  nbytes--;
  while (bytes[nbytes] >= num) {
    nbits++;
    num *= 2;
  }
  return nbits + nbytes * 8;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_strrev>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rsi
  movl   $0x0,-0x3c(%rbp)
  movl   $0x0,-0x34(%rbp)
  jmp    L57
L37:
  addl   $0x1,-0x34(%rbp)
L57:
  mov    -0x34(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L37
  mov    -0x34(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x30(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  movslq %eax,%rdx
  mov    %rdx,%rcx
  mov    $0x0,%ebx
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%ebx
  mov    $0x0,%edx
  div    %rbx
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L15:
  cmp    %rdx,%rsp
  je     L69
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L15
L69:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L26
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L26:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x34(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x38(%rbp)
  jmp    L7
L60:
  mov    -0x38(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%ecx
  mov    -0x28(%rbp),%rdx
  mov    -0x3c(%rbp),%eax
  cltq
  mov    %cl,(%rdx,%rax,1)
  addl   $0x1,-0x3c(%rbp)
  subl   $0x1,-0x38(%rbp)
L7:
  mov    -0x3c(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L60
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rsi,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L53
  call   <__stack_chk_fail@plt>
L53:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *ft_strrev(char *str) {
  int i;
  int j;
  int size;
  char tmp;
  i = 0;
  size = 0;
  while (str[size])
    size++;
  char result[size];
  j = size - 1;
  while (i < size) {
    result[i] = str[j];
    i++;
    j--;
  }
  char *a = result;
  return (a);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ft_strrev>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpb   $0x0,(%rdi)
  je     L11
  mov    $0x1,%eax
L42:
  lea    -0x1(%rax),%edx
  mov    %eax,%esi
  mov    %rax,%rcx
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L42
  add    $0xf,%rcx
  mov    %rcx,%rdi
  and    $0xfffffffffffffff0,%rdi
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rax
  sub    %rcx,%rax
L59:
  cmp    %rax,%rsp
  je     L68
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L59
L68:
  mov    %rdi,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L23
  orq    $0x0,-0x8(%rsp,%rax,1)
L23:
  mov    %rsp,%rdi
  test   %esi,%esi
  jle    L85
  mov    $0x0,%eax
L52:
  mov    %eax,%ecx
  add    $0x1,%eax
  cmp    %edx,%ecx
  jl     L52
L85:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L72
  mov    %rdi,%rax
  leave
  ret
L11:
  lea    -0x9(%rbp),%rdi
  jmp    L85
L72:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ft_strrev(char *str) {
  int i;
  int j;
  int size;
  char tmp;
  i = 0;
  size = 0;
  while (str[size])
    size++;
  char result[size];
  j = size - 1;
  while (i < size) {
    result[i] = str[j];
    i++;
    j--;
  }
  char *a = result;
  return (a);
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
char *ft_strrev(char *str) {
  int i;
  int j;
  int size;
  char tmp;
  i = 0;
  size = 0;
  while (str[size])
    size++;
  char result[size];
  j = size - 1;
  while (i < size) {
    result[i] = str[j];
    i++;
    j--;
  }
  char *a = result;
  return (a);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

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

assistant: ```c
char *ft_strrev(char *str) {
  int i;
  int j;
  int size;
  char tmp;
  i = 0;
  size = 0;
  while (str[size])
    size++;
  char result[size];
  j = size - 1;
  while (i < size) {
    result[i] = str[j];
    i++;
    j--;
  }
  char *a = result;
  return (a);
}
```
user: Translate the below assembly code into C programming language:

```asm
<bubble_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x5,-0x30(%rbp)
  movl   $0x1,-0x2c(%rbp)
  movl   $0x9,-0x28(%rbp)
  movl   $0xc,-0x24(%rbp)
  movl   $0x63,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x9,-0x18(%rbp)
  movl   $0x7,-0x38(%rbp)
  movl   $0x0,-0x40(%rbp)
  jmp    L17
L5:
  movl   $0x0,-0x3c(%rbp)
  jmp    L46
L25:
  mov    -0x3c(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%edx
  mov    -0x3c(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  jle    L12
  mov    -0x3c(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,-0x34(%rbp)
  mov    -0x3c(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%edx
  mov    -0x3c(%rbp),%eax
  cltq
  mov    %edx,-0x30(%rbp,%rax,4)
  mov    -0x3c(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    -0x34(%rbp),%edx
  mov    %edx,-0x30(%rbp,%rax,4)
L12:
  addl   $0x1,-0x3c(%rbp)
L46:
  mov    -0x38(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x3c(%rbp)
  jl     L25
  addl   $0x1,-0x40(%rbp)
L17:
  mov    -0x40(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L5
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L47
  call   <__stack_chk_fail@plt>
L47:
  leave
  ret
```

assistant: ```c
void bubble_sort() {
  int arr[] = {5, 1, 9, 12, 99, 0, 9};
  int noEle = sizeof(arr) / sizeof(arr[0]);
  int i, j, temp;
  for (i = 0; i < noEle; i++) {
    for (j = 0; j < noEle - 1; j++) {
      if (arr[j] > arr[j + 1]) {
        temp = arr[j];
        arr[j] = arr[j + 1];
        arr[j + 1] = temp;
      }
    }
  }
}
```