ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  lea    0xe(%rsp),%r13
  jmp    L65
  nopl   0x0(%rax)
L13:
  cmp    %ebx,%ebp
  jl     L50
L65:
  mov    $0xa,%esi
  mov    %r12,%rdi
  call   <getLine@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jg     L13
  xor    %eax,%eax
  mov    %ebp,%esi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  test   %ebp,%ebp
  jne    L34
L69:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L50:
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %ebx,%ebp
  call   <copy@plt>
  jmp    L65
L34:
  lea    0xe(%rsp),%rsi
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L69
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D11
datatype: string
name: parse_data
arguments: data_tag: D13
datatype: string

tool: D11: Longest line was: %d characters


tool: D13: %s

assistant: ```c
int main(void) {
  int len;
  int max;
  char line[10];
  char longest[10];
  max = 0;
  while ((len = getLine(line, 10)) > 0) {
    if (len > max) {
      max = len;
      copy(longest, line);
    }
  }
  printf("Longest line was: %d characters\n", max);
  if (max > 0) {
    printf("%s", longest);
  }
  printf("\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<die>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  lea    D14(%rip),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <error@plt>
  mov    $0x0,%eax
  call   <exithal@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D14
type: string
name: unpack_data
arguments: data_tag: D1
type: string

tool: D14: MegaHAL received signal %d

tool: D1: die

assistant: ```c
void die(int sig) {
  error("die", "MegaHAL received signal %d", sig);
  exithal();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<die>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%edx
  lea    D15(%rip),%rsi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <error@plt>
  mov    $0x0,%eax
  call   <exithal@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D15
datatype: string
name: extract_symbolic_data
arguments: tag: D4
datatype: string

tool: D15: MegaHAL received signal %d

tool: D4: die

assistant: ```c
void die(int sig) {
  error("die", "MegaHAL received signal %d", sig);
  exithal();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<die>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%edx
  lea    D10(%rip),%rsi
  xor    %eax,%eax
  lea    D1(%rip),%rdi
  call   <error@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <exithal@plt>
```

tool_calls: name: parse_data
arguments: tag: D10
type_infered: string
name: parse_data
arguments: tag: D1
type_infered: string

tool: D10: MegaHAL received signal %d

tool: D1: die

assistant: ```c
void die(int sig) {
  error("die", "MegaHAL received signal %d", sig);
  exithal();
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<die>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%edx
  lea    D11(%rip),%rsi
  xor    %eax,%eax
  lea    D17(%rip),%rdi
  call   <error@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <exithal@plt>
```

tool_calls: name: read_data
arguments: tag: D11
type: string
name: read_data
arguments: tag: D17
type: string

tool: D11: MegaHAL received signal %d

tool: D17: die

assistant: ```c
void die(int sig) {
  error("die", "MegaHAL received signal %d", sig);
  exithal();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<aligned_block_copy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %edx,-0x44(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  jmp    L3
L14:
  mov    -0x40(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x40(%rbp)
  mov    (%rax),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x40(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x40(%rbp)
  mov    (%rax),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x40(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x40(%rbp)
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x40(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x40(%rbp)
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x28(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x28(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x28(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x40(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x40(%rbp)
  mov    (%rax),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x40(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x40(%rbp)
  mov    (%rax),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x40(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x40(%rbp)
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x40(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x40(%rbp)
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x28(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x28(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x28(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
L3:
  subl   $0x40,-0x44(%rbp)
  cmpl   $0x0,-0x44(%rbp)
  jns    L14
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void aligned_block_copy(int64_t *__restrict dst_, int64_t *__restrict src, int size) {
  volatile int64_t *dst = dst_;
  int64_t t1, t2, t3, t4;
  while ((size -= 64) >= 0) {
    t1 = *src++;
    t2 = *src++;
    t3 = *src++;
    t4 = *src++;
    *dst++ = t1;
    *dst++ = t2;
    *dst++ = t3;
    *dst++ = t4;
    t1 = *src++;
    t2 = *src++;
    t3 = *src++;
    t4 = *src++;
    *dst++ = t1;
    *dst++ = t2;
    *dst++ = t3;
    *dst++ = t4;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<aligned_block_copy>:
  endbr64
  mov    %rdi,%rcx
  mov    %rsi,%rax
  cmp    $0x3f,%edx
  jle    L16
  sub    $0x40,%edx
  shr    $0x6,%edx
  add    $0x1,%edx
  mov    %edx,%edx
  shl    $0x6,%rdx
  add    %rsi,%rdx
L15:
  mov    (%rax),%r9
  mov    0x8(%rax),%r8
  mov    0x10(%rax),%rdi
  mov    0x18(%rax),%rsi
  mov    %r9,(%rcx)
  mov    %r8,0x8(%rcx)
  mov    %rdi,0x10(%rcx)
  mov    %rsi,0x18(%rcx)
  mov    0x20(%rax),%r9
  mov    0x28(%rax),%r8
  mov    0x30(%rax),%rsi
  add    $0x40,%rax
  mov    -0x8(%rax),%rdi
  mov    %r9,0x20(%rcx)
  mov    %r8,0x28(%rcx)
  mov    %rsi,0x30(%rcx)
  mov    %rcx,%rsi
  add    $0x40,%rcx
  mov    %rdi,0x38(%rsi)
  cmp    %rax,%rdx
  jne    L15
L16:
  ret
```

assistant: ```c
void aligned_block_copy(int64_t *__restrict dst_, int64_t *__restrict src, int size) {
  volatile int64_t *dst = dst_;
  int64_t t1, t2, t3, t4;
  while ((size -= 64) >= 0) {
    t1 = *src++;
    t2 = *src++;
    t3 = *src++;
    t4 = *src++;
    *dst++ = t1;
    *dst++ = t2;
    *dst++ = t3;
    *dst++ = t4;
    t1 = *src++;
    t2 = *src++;
    t3 = *src++;
    t4 = *src++;
    *dst++ = t1;
    *dst++ = t2;
    *dst++ = t3;
    *dst++ = t4;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<aligned_block_copy>:
  endbr64
  mov    %rdi,%rcx
  mov    %rsi,%rax
  cmp    $0x3f,%edx
  jle    L2
  sub    $0x40,%edx
  shr    $0x6,%edx
  add    $0x1,%edx
  shl    $0x6,%rdx
  add    %rsi,%rdx
  nop
L11:
  mov    (%rax),%r9
  mov    0x8(%rax),%r8
  add    $0x40,%rax
  mov    -0x28(%rax),%rsi
  mov    -0x30(%rax),%rdi
  mov    %r9,(%rcx)
  mov    -0x20(%rax),%r9
  mov    %r8,0x8(%rcx)
  mov    -0x18(%rax),%r8
  mov    %rdi,0x10(%rcx)
  mov    %rsi,0x18(%rcx)
  mov    -0x10(%rax),%rsi
  mov    -0x8(%rax),%rdi
  mov    %r9,0x20(%rcx)
  mov    %r8,0x28(%rcx)
  mov    %rsi,0x30(%rcx)
  mov    %rcx,%rsi
  add    $0x40,%rcx
  mov    %rdi,0x38(%rsi)
  cmp    %rax,%rdx
  jne    L11
L2:
  ret
```

assistant: ```c
void aligned_block_copy(int64_t *__restrict dst_, int64_t *__restrict src, int size) {
  volatile int64_t *dst = dst_;
  int64_t t1, t2, t3, t4;
  while ((size -= 64) >= 0) {
    t1 = *src++;
    t2 = *src++;
    t3 = *src++;
    t4 = *src++;
    *dst++ = t1;
    *dst++ = t2;
    *dst++ = t3;
    *dst++ = t4;
    t1 = *src++;
    t2 = *src++;
    t3 = *src++;
    t4 = *src++;
    *dst++ = t1;
    *dst++ = t2;
    *dst++ = t3;
    *dst++ = t4;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<aligned_block_copy>:
  endbr64
  mov    %rdi,%rcx
  mov    %rsi,%rax
  cmp    $0x3f,%edx
  jle    L8
  sub    $0x40,%edx
  shr    $0x6,%edx
  add    $0x1,%edx
  shl    $0x6,%rdx
  add    %rsi,%rdx
  nop
L11:
  mov    (%rax),%r9
  mov    0x8(%rax),%r8
  add    $0x40,%rax
  mov    -0x28(%rax),%rsi
  mov    -0x30(%rax),%rdi
  mov    %r9,(%rcx)
  mov    -0x20(%rax),%r9
  mov    %r8,0x8(%rcx)
  mov    -0x18(%rax),%r8
  mov    %rdi,0x10(%rcx)
  mov    %rsi,0x18(%rcx)
  mov    -0x10(%rax),%rsi
  mov    -0x8(%rax),%rdi
  mov    %r9,0x20(%rcx)
  mov    %r8,0x28(%rcx)
  mov    %rsi,0x30(%rcx)
  mov    %rcx,%rsi
  add    $0x40,%rcx
  mov    %rdi,0x38(%rsi)
  cmp    %rax,%rdx
  jne    L11
L8:
  ret
```

assistant: ```c
void aligned_block_copy(int64_t *__restrict dst_, int64_t *__restrict src, int size) {
  volatile int64_t *dst = dst_;
  int64_t t1, t2, t3, t4;
  while ((size -= 64) >= 0) {
    t1 = *src++;
    t2 = *src++;
    t3 = *src++;
    t4 = *src++;
    *dst++ = t1;
    *dst++ = t2;
    *dst++ = t3;
    *dst++ = t4;
    t1 = *src++;
    t2 = *src++;
    t3 = *src++;
    t4 = *src++;
    *dst++ = t1;
    *dst++ = t2;
    *dst++ = t3;
    *dst++ = t4;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<client>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x430,%rsp
  mov    %rdi,-0x428(%rbp)
  mov    %esi,-0x42c(%rbp)
  mov    %edx,-0x430(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x428(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x420(%rbp)
  mov    -0x420(%rbp),%rax
  sub    $0x1,%rax
  movzbl -0x410(%rbp,%rax,1),%eax
  cmp    $0xa,%al
  jne    L9
  subq   $0x1,-0x420(%rbp)
L9:
  mov    -0x420(%rbp),%rdx
  mov    -0x428(%rbp),%rcx
  mov    -0x430(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  jmp    L6
L38:
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  mov    -0x418(%rbp),%rdx
  lea    -0x410(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
L6:
  lea    -0x410(%rbp),%rcx
  mov    -0x42c(%rbp),%eax
  mov    $0x400,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %rax,-0x418(%rbp)
  cmpq   $0x0,-0x418(%rbp)
  jne    L38
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L31
  call   <__stack_chk_fail@plt>
L31:
  leave
  ret
```

assistant: ```c
void client(const char *pathname, int read_fd, int write_fd) {
  size_t len;
  ssize_t n;
  char buff[1024];
  len = strlen(pathname);
  if (buff[len - 1] == '\n') {
    len--;
  }
  write(write_fd, pathname, len);
  while ((n = read(read_fd, buff, 1024)) > 0) {
    write(STDOUT_FILENO, buff, n);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<client>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    %edx,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  lea    -0x1(%rax),%rdx
  cmpb   $0xa,-0x1(%rsp,%rax,1)
  cmovne %rax,%rdx
  mov    %rbx,%rsi
  mov    %r12d,%edi
  call   <write@plt>
  mov    %rsp,%rbx
  jmp    L10
L21:
  mov    %rbx,%rsi
  mov    D8(%rip),%rax
  mov    (%rax),%edi
  call   <write@plt>
L10:
  mov    $0x400,%edx
  mov    %rbx,%rsi
  mov    %ebp,%edi
  call   <read@plt>
  mov    %rax,%rdx
  test   %rax,%rax
  jne    L21
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void client(const char *pathname, int read_fd, int write_fd) {
  size_t len;
  ssize_t n;
  char buff[1024];
  len = strlen(pathname);
  if (buff[len - 1] == '\n') {
    len--;
  }
  write(write_fd, pathname, len);
  while ((n = read(read_fd, buff, 1024)) > 0) {
    write(STDOUT_FILENO, buff, n);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<client>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <strlen@plt>
  mov    %r12,%rsi
  mov    %r13d,%edi
  mov    %rax,%rdx
  sub    $0x1,%rax
  cmpb   $0xa,-0x1(%rsp,%rdx,1)
  cmove  %rax,%rdx
  call   <write@plt>
  jmp    L22
  nopl   0x0(%rax)
L8:
  mov    D3(%rip),%rax
  mov    %rbx,%rsi
  mov    (%rax),%edi
  call   <write@plt>
L22:
  mov    $0x400,%edx
  mov    %rbx,%rsi
  mov    %ebp,%edi
  call   <read@plt>
  mov    %rax,%rdx
  test   %rax,%rax
  jne    L8
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void client(const char *pathname, int read_fd, int write_fd) {
  size_t len;
  ssize_t n;
  char buff[1024];
  len = strlen(pathname);
  if (buff[len - 1] == '\n') {
    len--;
  }
  write(write_fd, pathname, len);
  while ((n = read(read_fd, buff, 1024)) > 0) {
    write(STDOUT_FILENO, buff, n);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<client>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <strlen@plt>
  mov    %r12,%rsi
  mov    %r13d,%edi
  mov    %rax,%rdx
  sub    $0x1,%rax
  cmpb   $0xa,-0x1(%rsp,%rdx,1)
  cmove  %rax,%rdx
  call   <write@plt>
  jmp    L22
  nopl   0x0(%rax)
L28:
  mov    D4(%rip),%rax
  mov    %rbx,%rsi
  mov    (%rax),%edi
  call   <write@plt>
L22:
  mov    $0x400,%edx
  mov    %rbx,%rsi
  mov    %ebp,%edi
  call   <read@plt>
  mov    %rax,%rdx
  test   %rax,%rax
  jne    L28
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void client(const char *pathname, int read_fd, int write_fd) {
  size_t len;
  ssize_t n;
  char buff[1024];
  len = strlen(pathname);
  if (buff[len - 1] == '\n') {
    len--;
  }
  write(write_fd, pathname, len);
  while ((n = read(read_fd, buff, 1024)) > 0) {
    write(STDOUT_FILENO, buff, n);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_strupcase>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L39
L25:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x60,%al
  jle    L14
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x7a,%al
  jg     L14
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  lea    -0x20(%rax),%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L14:
  addl   $0x1,-0x4(%rbp)
L39:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L25
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *ft_strupcase(char *str) {
  int it;
  it = 0;
  while (str[it]) {
    if (str[it] >= 'a' && str[it] <= 'z')
      str[it] = str[it] - 32;
    it++;
  }
  return (str);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_strupcase>:
  endbr64
  mov    %rdi,%rax
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L36
  mov    %rdi,%rcx
  jmp    L29
L19:
  add    $0x1,%rcx
  movzbl (%rcx),%edx
  test   %dl,%dl
  je     L36
L29:
  lea    -0x61(%rdx),%esi
  cmp    $0x19,%sil
  ja     L19
  sub    $0x20,%edx
  mov    %dl,(%rcx)
  jmp    L19
L36:
  ret
```

assistant: ```c
char *ft_strupcase(char *str) {
  int it;
  it = 0;
  while (str[it]) {
    if (str[it] >= 'a' && str[it] <= 'z')
      str[it] = str[it] - 32;
    it++;
  }
  return (str);
}
```
user: What is the c source code of the assembly code below:

```asm
<ft_strupcase>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rdi,%rax
  test   %dl,%dl
  je     L25
  mov    %rdi,%rcx
  nopl   0x0(%rax)
L1:
  lea    -0x61(%rdx),%esi
  cmp    $0x19,%sil
  ja     L24
  sub    $0x20,%edx
  mov    %dl,(%rcx)
L24:
  movzbl 0x1(%rcx),%edx
  add    $0x1,%rcx
  test   %dl,%dl
  jne    L1
L25:
  ret
```

assistant: ```c
char *ft_strupcase(char *str) {
  int it;
  it = 0;
  while (str[it]) {
    if (str[it] >= 'a' && str[it] <= 'z')
      str[it] = str[it] - 32;
    it++;
  }
  return (str);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ft_strupcase>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rdi,%rax
  test   %dl,%dl
  je     L34
  mov    %rdi,%rcx
  nopl   0x0(%rax)
L29:
  lea    -0x61(%rdx),%esi
  cmp    $0x19,%sil
  ja     L22
  sub    $0x20,%edx
  add    $0x1,%rcx
  mov    %dl,-0x1(%rcx)
  movzbl (%rcx),%edx
  test   %dl,%dl
  jne    L29
  ret
  nopl   0x0(%rax,%rax,1)
L22:
  movzbl 0x1(%rcx),%edx
  add    $0x1,%rcx
  test   %dl,%dl
  jne    L29
L34:
  ret
```

assistant: ```c
char *ft_strupcase(char *str) {
  int it;
  it = 0;
  while (str[it]) {
    if (str[it] >= 'a' && str[it] <= 'z')
      str[it] = str[it] - 32;
    it++;
  }
  return (str);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<computer_move>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x108,%rsp
  mov    %rdi,0x10(%rsp)
  mov    %edx,0x2c(%rsp)
  mov    %dl,0x23(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0xf8(%rsp)
  xor    %eax,%eax
  cmp    $0x4f,%dl
  mov    $0x40,%edx
  mov    $0x4f,%eax
  cmove  %edx,%eax
  mov    %al,0x22(%rsp)
  mov    %rsi,0x18(%rsp)
  movl   $0x64,0xc(%rsp)
  movl   $0x0,0x28(%rsp)
  movl   $0x0,0x24(%rsp)
  movl   $0x0,0x8(%rsp)
  lea    0x54(%rsp),%r13
  jmp    L86
L72:
  add    $0x1,%rbx
  cmp    $0x6,%rbx
  je     L32
L85:
  mov    %ebx,0x4(%rsp)
  cmpl   $0x0,(%r14,%rbx,4)
  je     L72
  mov    0x10(%rsp),%rdi
  lea    0x30(%rsp),%rcx
L0:
  mov    %rdi,%rsi
  mov    $0x0,%eax
L36:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    $0x6,%rax
  jne    L36
  add    $0x6,%rdi
  add    $0x6,%rcx
  cmp    %r13,%rcx
  jne    L0
  movsbl 0x23(%rsp),%ecx
  lea    0x30(%rsp),%r12
  mov    0x4(%rsp),%edx
  mov    0x8(%rsp),%esi
  mov    %r12,%rdi
  call   <make_move@plt>
  movsbl 0x22(%rsp),%ebp
  lea    0x60(%rsp),%r15
  mov    %ebp,%edx
  mov    %r15,%rsi
  mov    %r12,%rdi
  call   <valid_moves@plt>
  mov    %ebp,%edx
  mov    %r15,%rsi
  mov    %r12,%rdi
  call   <best_move@plt>
  cmp    0xc(%rsp),%eax
  jge    L72
  mov    %eax,0xc(%rsp)
  mov    0x4(%rsp),%eax
  mov    %eax,0x28(%rsp)
  mov    0x8(%rsp),%eax
  mov    %eax,0x24(%rsp)
  jmp    L72
L32:
  addl   $0x1,0x8(%rsp)
  mov    0x8(%rsp),%eax
  addq   $0x18,0x18(%rsp)
  cmp    $0x6,%eax
  je     L68
L86:
  mov    0x18(%rsp),%r14
  mov    $0x0,%ebx
  jmp    L85
L68:
  movsbl 0x2c(%rsp),%ecx
  mov    0x28(%rsp),%edx
  mov    0x24(%rsp),%esi
  mov    0x10(%rsp),%rdi
  call   <make_move@plt>
  mov    0xf8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x108,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void computer_move(char board[][6], int moves[][6], char player) {
  int row = 0;
  int col = 0;
  int best_row = 0;
  int best_col = 0;
  int i = 0;
  int j = 0;
  int new_score = 0;
  int score = 100;
  char temp_board[6][6];
  int temp_moves[6][6];
  char opponent = (player == 'O') ? '@' : 'O';
  for (row = 0; row < 6; row++)
    for (col = 0; col < 6; col++) {
      if (moves[row][col] == 0)
        continue;
      for (i = 0; i < 6; i++)
        for (j = 0; j < 6; j++)
          temp_board[i][j] = board[i][j];
      make_move(temp_board, row, col, player);
      valid_moves(temp_board, temp_moves, opponent);
      new_score = best_move(temp_board, temp_moves, opponent);
      if (new_score < score) {
        score = new_score;
        best_row = row;
        best_col = col;
      }
    }
  make_move(board, best_row, best_col, player);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<computer_move>:
  endbr64
  push   %r15
  movsbl %dl,%r15d
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    $0x40,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  sub    $0xf8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xe8(%rsp)
  xor    %eax,%eax
  cmp    $0x4f,%dl
  mov    $0x4f,%eax
  lea    0x20(%rsp),%rbp
  cmove  %r13d,%eax
  movl   $0x64,0x10(%rsp)
  xor    %r13d,%r13d
  movl   $0x0,0x1c(%rsp)
  mov    %al,0x17(%rsp)
  movl   $0x0,0x18(%rsp)
L51:
  xor    %ebx,%ebx
  jmp    L54
  nopl   0x0(%rax)
L47:
  add    $0x1,%rbx
  cmp    $0x6,%rbx
  je     L34
L54:
  mov    (%r12,%rbx,4),%eax
  mov    %ebx,%edx
  test   %eax,%eax
  je     L47
  movdqu (%r14),%xmm0
  mov    0x20(%r14),%eax
  mov    %r15d,%ecx
  mov    %r13d,%esi
  movdqu 0x10(%r14),%xmm1
  mov    %rbp,%rdi
  movaps %xmm0,0x20(%rsp)
  movaps %xmm1,0x30(%rsp)
  mov    %eax,0x40(%rsp)
  call   <make_move@plt>
  movsbl 0x17(%rsp),%edx
  lea    0x50(%rsp),%rsi
  mov    %rbp,%rdi
  mov    %rsi,(%rsp)
  mov    %edx,0xc(%rsp)
  call   <valid_moves@plt>
  mov    0xc(%rsp),%edx
  mov    (%rsp),%rsi
  mov    %rbp,%rdi
  call   <best_move@plt>
  cmp    0x10(%rsp),%eax
  jge    L47
  mov    %ebx,0x1c(%rsp)
  add    $0x1,%rbx
  mov    %eax,0x10(%rsp)
  mov    %r13d,0x18(%rsp)
  cmp    $0x6,%rbx
  jne    L54
L34:
  add    $0x1,%r13d
  add    $0x18,%r12
  cmp    $0x6,%r13d
  jne    L51
  mov    0x1c(%rsp),%edx
  mov    0x18(%rsp),%esi
  mov    %r15d,%ecx
  mov    %r14,%rdi
  call   <make_move@plt>
  mov    0xe8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  add    $0xf8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L30:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void computer_move(char board[][6], int moves[][6], char player) {
  int row = 0;
  int col = 0;
  int best_row = 0;
  int best_col = 0;
  int i = 0;
  int j = 0;
  int new_score = 0;
  int score = 100;
  char temp_board[6][6];
  int temp_moves[6][6];
  char opponent = (player == 'O') ? '@' : 'O';
  for (row = 0; row < 6; row++)
    for (col = 0; col < 6; col++) {
      if (moves[row][col] == 0)
        continue;
      for (i = 0; i < 6; i++)
        for (j = 0; j < 6; j++)
          temp_board[i][j] = board[i][j];
      make_move(temp_board, row, col, player);
      valid_moves(temp_board, temp_moves, opponent);
      new_score = best_move(temp_board, temp_moves, opponent);
      if (new_score < score) {
        score = new_score;
        best_row = row;
        best_col = col;
      }
    }
  make_move(board, best_row, best_col, player);
}
```