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
<ssds_cgi_new_class>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x290,%rsp
  mov    %edi,%r12d
  mov    %rdx,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x288(%rsp)
  xor    %eax,%eax
  lea    0x180(%rsp),%rdi
  mov    %rsi,%rdx
  lea    D7(%rip),%rsi
  call   <sprintf@plt>
  mov    %eax,%ebp
  mov    %rbx,%rdi
  call   <strlen@plt>
  test   %rax,%rax
  jg     L6
L4:
  lea    0x80(%rsp),%rbp
  lea    0x180(%rsp),%rsi
  mov    %rbp,%rdx
  mov    %r12d,%edi
  call   <ssds_send@plt>
  mov    %eax,%ebx
  mov    %rsp,%rsi
  mov    $0x10,%edx
  mov    %rbp,%rdi
  call   <get_argc@plt>
  mov    0x288(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  mov    %ebx,%eax
  add    $0x290,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  movslq %ebp,%rax
  lea    0x180(%rsp,%rax,1),%rdi
  mov    %rbx,%rdx
  lea    D13(%rip),%rsi
  call   <sprintf@plt>
  jmp    L4
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
type_infered: string
name: extract_symbolic_data
arguments: tag: D13
type_infered: string

tool: D7: msg 5678:class:%s:action:create_class:reply:yes:

tool: D13: id:%s:

assistant: ```c
int ssds_cgi_new_class(int fd, char *classsp, char *idsp) {
  int ret;
  char **argv;
  char *myargv[16];
  int argc;
  char rbuf[256];
  char sbuf[256];
  argv = myargv;
  ret = sprintf(sbuf, "msg 5678:class:%s:action:create_class:reply:yes:", classsp);
  if (strlen(idsp) > 0) {
    ret += sprintf(&sbuf[ret], "id:%s:", idsp);
  }
  ret = ssds_send(fd, sbuf, rbuf);
  argc = get_argc(rbuf, argv, 16);
  if (0 && argc < 0)
    printf(" Argc error %d\n", argc);
  return ret;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ssds_cgi_new_class>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rdx,%r12
  mov    %rsi,%rdx
  lea    D16(%rip),%rsi
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x290,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x288(%rsp)
  xor    %eax,%eax
  lea    0x180(%rsp),%r13
  mov    %r13,%rdi
  call   <sprintf@plt>
  mov    %r12,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  test   %rax,%rax
  jne    L10
L21:
  lea    0x80(%rsp),%r14
  mov    %r13,%rsi
  mov    %ebp,%edi
  mov    %r14,%rdx
  call   <ssds_send@plt>
  mov    %rsp,%rsi
  mov    $0x10,%edx
  mov    %r14,%rdi
  mov    %eax,%r12d
  call   <get_argc@plt>
  mov    0x288(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x290,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L10:
  movslq %ebx,%rax
  mov    %r12,%rdx
  lea    D5(%rip),%rsi
  lea    0x0(%r13,%rax,1),%rdi
  call   <sprintf@plt>
  jmp    L21
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D16
type: string
name: retrieve_data
arguments: data_tag: D5
type: string

tool: D16: msg 5678:class:%s:action:create_class:reply:yes:

tool: D5: id:%s:

assistant: ```c
int ssds_cgi_new_class(int fd, char *classsp, char *idsp) {
  int ret;
  char **argv;
  char *myargv[16];
  int argc;
  char rbuf[256];
  char sbuf[256];
  argv = myargv;
  ret = sprintf(sbuf, "msg 5678:class:%s:action:create_class:reply:yes:", classsp);
  if (strlen(idsp) > 0) {
    ret += sprintf(&sbuf[ret], "id:%s:", idsp);
  }
  ret = ssds_send(fd, sbuf, rbuf);
  argc = get_argc(rbuf, argv, 16);
  if (0 && argc < 0)
    printf(" Argc error %d\n", argc);
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ssds_cgi_new_class>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rdx,%r12
  mov    %rsi,%rdx
  lea    D4(%rip),%rsi
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x290,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x288(%rsp)
  xor    %eax,%eax
  lea    0x180(%rsp),%r13
  mov    %r13,%rdi
  call   <sprintf@plt>
  mov    %r12,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  test   %rax,%rax
  jne    L14
L27:
  lea    0x80(%rsp),%r14
  mov    %r13,%rsi
  mov    %ebp,%edi
  mov    %r14,%rdx
  call   <ssds_send@plt>
  mov    %rsp,%rsi
  mov    $0x10,%edx
  mov    %r14,%rdi
  mov    %eax,%r12d
  call   <get_argc@plt>
  mov    0x288(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x290,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  movslq %ebx,%rax
  mov    %r12,%rdx
  lea    D5(%rip),%rsi
  lea    0x0(%r13,%rax,1),%rdi
  call   <sprintf@plt>
  jmp    L27
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D4
type_guessed: string
name: retrieve_data
arguments: tag: D5
type_guessed: string

tool: D4: msg 5678:class:%s:action:create_class:reply:yes:

tool: D5: id:%s:

assistant: ```c
int ssds_cgi_new_class(int fd, char *classsp, char *idsp) {
  int ret;
  char **argv;
  char *myargv[16];
  int argc;
  char rbuf[256];
  char sbuf[256];
  argv = myargv;
  ret = sprintf(sbuf, "msg 5678:class:%s:action:create_class:reply:yes:", classsp);
  if (strlen(idsp) > 0) {
    ret += sprintf(&sbuf[ret], "id:%s:", idsp);
  }
  ret = ssds_send(fd, sbuf, rbuf);
  argc = get_argc(rbuf, argv, 16);
  if (0 && argc < 0)
    printf(" Argc error %d\n", argc);
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x66647361,-0xd(%rbp)
  movb   $0x0,-0x9(%rbp)
  movzbl -0xd(%rbp),%eax
  movsbl %al,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int main() {
  char t[] = "asdf";
  return t[0];
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  mov    $0x61,%eax
  ret
```

assistant: ```c
int main() {
  char t[] = "asdf";
  return t[0];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  mov    $0x61,%eax
  ret
```

assistant: ```c
int main() {
  char t[] = "asdf";
  return t[0];
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  mov    $0x61,%eax
  ret
```

assistant: ```c
int main() {
  char t[] = "asdf";
  return t[0];
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<advbubblesort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x1,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L49
L53:
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L56
L8:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L14
  movl   $0x1,-0x8(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
L14:
  addl   $0x1,-0x10(%rbp)
L56:
  mov    -0x1c(%rbp),%eax
  sub    -0xc(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x10(%rbp)
  jl     L8
  addl   $0x1,-0xc(%rbp)
L49:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jge    L1
  cmpl   $0x0,-0x8(%rbp)
  jne    L53
L1:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void advbubblesort(int a[], int n) {
  int j, i, temp, flag = 1;
  for (i = 0; i < n - 1 && flag; i++) {
    flag = 0;
    for (j = 0; j < n - i - 1; j++) {
      if (a[j] > a[j + 1]) {
        flag = 1;
        temp = a[j];
        a[j] = a[j + 1];
        a[j + 1] = temp;
      }
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<advbubblesort>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x1(%rsi),%ebp
  test   %ebp,%ebp
  jle    L26
  mov    %rdi,%r8
  mov    %esi,%ebx
  mov    %ebp,%r9d
  mov    $0x0,%r11d
  mov    $0x0,%r13d
  mov    $0x0,%edx
  mov    $0x1,%r10d
  mov    $0x0,%r12d
  jmp    L51
L22:
  add    $0x1,%edx
  cmp    %r9d,%edx
  jge    L57
L51:
  movslq %edx,%rax
  shl    $0x2,%rax
  lea    (%r8,%rax,1),%rsi
  mov    (%rsi),%ecx
  lea    0x4(%r8,%rax,1),%rax
  mov    (%rax),%edi
  cmp    %edi,%ecx
  jle    L22
  mov    %edi,(%rsi)
  mov    %ecx,(%rax)
  mov    %r10d,%r13d
  jmp    L22
L57:
  add    $0x1,%r11d
  test   %r13d,%r13d
  je     L26
  cmp    %ebp,%r11d
  jge    L26
  mov    %ebx,%r9d
  sub    %r11d,%r9d
  sub    $0x1,%r9d
  test   %r9d,%r9d
  jle    L26
  mov    %r12d,%r13d
  mov    $0x0,%edx
  jmp    L51
L26:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void advbubblesort(int a[], int n) {
  int j, i, temp, flag = 1;
  for (i = 0; i < n - 1 && flag; i++) {
    flag = 0;
    for (j = 0; j < n - i - 1; j++) {
      if (a[j] > a[j + 1]) {
        flag = 1;
        temp = a[j];
        a[j] = a[j + 1];
        a[j + 1] = temp;
      }
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<advbubblesort>:
  endbr64
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  test   %ebp,%ebp
  jle    L26
  mov    %esi,%ebx
  mov    %rdi,%r9
  mov    %ebp,%r10d
  xor    %esi,%esi
  nopw   0x0(%rax,%rax,1)
L17:
  xor    %r11d,%r11d
  xor    %edx,%edx
  nopl   (%rax)
L37:
  movslq %edx,%rax
  shl    $0x2,%rax
  lea    (%r9,%rax,1),%rdi
  lea    0x4(%r9,%rax,1),%rax
  mov    (%rdi),%ecx
  mov    (%rax),%r8d
  cmp    %r8d,%ecx
  jle    L43
  mov    %r8d,(%rdi)
  mov    $0x1,%r11d
  mov    %ecx,(%rax)
L43:
  add    $0x1,%edx
  cmp    %r10d,%edx
  jl     L37
  add    $0x1,%esi
  cmp    %esi,%ebp
  jle    L26
  test   %r11d,%r11d
  je     L26
  mov    %ebx,%r10d
  sub    %esi,%r10d
  sub    $0x1,%r10d
  test   %r10d,%r10d
  jg     L17
L26:
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void advbubblesort(int a[], int n) {
  int j, i, temp, flag = 1;
  for (i = 0; i < n - 1 && flag; i++) {
    flag = 0;
    for (j = 0; j < n - i - 1; j++) {
      if (a[j] > a[j + 1]) {
        flag = 1;
        temp = a[j];
        a[j] = a[j + 1];
        a[j + 1] = temp;
      }
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<advbubblesort>:
  endbr64
  push   %rbx
  lea    -0x1(%rsi),%ebx
  xor    %r11d,%r11d
  mov    %ebx,%r9d
  test   %ebx,%ebx
  jle    L51
  nopw   0x0(%rax,%rax,1)
L42:
  xor    %r10d,%r10d
  xor    %edx,%edx
  nopl   (%rax)
L47:
  movslq %edx,%rax
  lea    (%rdi,%rax,4),%rcx
  movq   (%rcx),%xmm0
  pshufd $0xe5,%xmm0,%xmm2
  movd   %xmm0,%r8d
  movd   %xmm2,%eax
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%r8d
  jle    L19
  movq   %xmm1,(%rcx)
  mov    $0x1,%r10d
L19:
  add    $0x1,%edx
  cmp    %r9d,%edx
  jl     L47
  add    $0x1,%r11d
  cmp    %r11d,%ebx
  jle    L51
  test   %r10d,%r10d
  je     L51
  mov    %esi,%r9d
  sub    %r11d,%r9d
  sub    $0x1,%r9d
  test   %r9d,%r9d
  jg     L42
L51:
  pop    %rbx
  ret
```

assistant: ```c
void advbubblesort(int a[], int n) {
  int j, i, temp, flag = 1;
  for (i = 0; i < n - 1 && flag; i++) {
    flag = 0;
    for (j = 0; j < n - i - 1; j++) {
      if (a[j] > a[j + 1]) {
        flag = 1;
        temp = a[j];
        a[j] = a[j + 1];
        a[j + 1] = temp;
      }
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fgetsr>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%rsi
  jbe    L70
  mov    %rdi,%rbp
  mov    %rsi,%r14
  mov    %rdx,%r13
  call   <ftello@plt>
  mov    %eax,%ebx
  lea    0x1(%rax),%edx
  mov    $0x0,%eax
  cmp    $0x1,%edx
  jbe    L70
  lea    -0x1(%r14),%r12
  movb   $0x0,-0x1(%r13,%r14,1)
  movl   $0x1,0xc(%rsp)
  mov    D23(%rip),%r15
  jmp    L93
L128:
  sub    $0x1,%r12
  lea    0x14(%rsp),%rsi
  lea    0x0(%r13,%r12,1),%rdi
  mov    $0x1,%edx
  call   <memcpy@plt>
L59:
  movl   $0x0,0xc(%rsp)
  test   %ebx,%ebx
  je     L82
L93:
  sub    $0x1,%ebx
  mov    (%r15),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <fseeko@plt>
  test   %rax,%rax
  jne    L53
  mov    %rbp,%rdi
  call   <fgetc@plt>
  mov    D25(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L80
  cmp    $0xa,%eax
  jne    L98
  cmpl   $0x0,0xc(%rsp)
  je     L27
L98:
  cmp    $0xd,%eax
  je     L59
  cmp    $0xa,%eax
  je     L59
  mov    %eax,0x14(%rsp)
  test   %r12,%r12
  jne    L128
  lea    -0x2(%r14),%rdx
  lea    0x1(%r13),%rdi
  mov    %r13,%rsi
  call   <memmove@plt>
  mov    $0x1,%r12d
  jmp    L128
L82:
  mov    D23(%rip),%rax
  mov    (%rax),%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <fseeko@plt>
L27:
  sub    %r12,%r14
  lea    0x0(%r13,%r12,1),%rsi
  mov    %r14,%rdx
  mov    %r13,%rdi
  call   <memmove@plt>
  lea    -0x1(%r14),%rax
L70:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L25
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L53:
  mov    $0x0,%eax
  jmp    L70
L80:
  mov    $0x0,%eax
  jmp    L70
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t fgetsr(FILE *fs, size_t n, char *buf) {
  off_t fpos;
  size_t bufpos;
  bool first = true;
  if (n <= 1 || (fpos = ftello(fs)) == -1 || fpos == 0) {
    return 0;
  }
  bufpos = n - 1;
  buf[bufpos] = '\0';
  while (true) {
    int c;
    if (fseeko(fs, --fpos, SEEK_SET) != 0 || (c = fgetc(fs)) == EOF) {
      return 0;
    }
    if (c == '\n' && !first) {
      break;
    }
    first = false;
    if (c != '\r' && c != '\n') {
      uint8_t ch = (uint8_t)c;
      if (bufpos == 0) {
        memmove(buf + 1, buf, n - 2);
        ++bufpos;
      }
      memcpy(buf + --bufpos, &ch, 1);
    }
    if (fpos == 0) {
      fseeko(fs, 0, SEEK_SET);
      break;
    }
  }
  memmove(buf, buf + bufpos, n - bufpos);
  return n - bufpos - 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<fgetsr>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%rsi
  jbe    L14
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%r13
  call   <ftello@plt>
  mov    %eax,%ebx
  lea    0x1(%rax),%eax
  cmp    $0x1,%eax
  jbe    L14
  lea    0x14(%rsp),%rax
  movb   $0x0,-0x1(%r13,%r12,1)
  lea    -0x1(%r12),%r14
  mov    $0x1,%r15d
  mov    %rax,0x8(%rsp)
  jmp    L119
  nopl   (%rax)
L86:
  sub    $0x1,%r14
  lea    0x0(%r13,%r14,1),%rdi
L80:
  mov    0x8(%rsp),%rsi
  mov    $0x1,%edx
  call   <memcpy@plt>
L79:
  xor    %r15d,%r15d
  test   %ebx,%ebx
  je     L135
L119:
  mov    D26(%rip),%rax
  sub    $0x1,%ebx
  mov    %rbp,%rdi
  mov    %ebx,%esi
  mov    (%rax),%edx
  call   <fseeko@plt>
  test   %rax,%rax
  jne    L14
  mov    %rbp,%rdi
  call   <fgetc@plt>
  mov    D11(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L14
  cmp    $0xa,%eax
  jne    L126
  test   %r15d,%r15d
  je     L6
L126:
  cmp    $0xd,%eax
  je     L79
  cmp    $0xa,%eax
  je     L79
  mov    %eax,0x14(%rsp)
  test   %r14,%r14
  jne    L86
  lea    0x1(%r13),%rdi
  lea    -0x2(%r12),%rdx
  mov    %r13,%rsi
  call   <memmove@plt>
  mov    %r13,%rdi
  jmp    L80
  nopl   0x0(%rax)
L14:
  xor    %eax,%eax
L21:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L135:
  mov    D26(%rip),%rax
  xor    %esi,%esi
  mov    %rbp,%rdi
  mov    (%rax),%edx
  call   <fseeko@plt>
L6:
  sub    %r14,%r12
  lea    0x0(%r13,%r14,1),%rsi
  mov    %r13,%rdi
  mov    %r12,%rdx
  call   <memmove@plt>
  lea    -0x1(%r12),%rax
  jmp    L21
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t fgetsr(FILE *fs, size_t n, char *buf) {
  off_t fpos;
  size_t bufpos;
  bool first = true;
  if (n <= 1 || (fpos = ftello(fs)) == -1 || fpos == 0) {
    return 0;
  }
  bufpos = n - 1;
  buf[bufpos] = '\0';
  while (true) {
    int c;
    if (fseeko(fs, --fpos, SEEK_SET) != 0 || (c = fgetc(fs)) == EOF) {
      return 0;
    }
    if (c == '\n' && !first) {
      break;
    }
    first = false;
    if (c != '\r' && c != '\n') {
      uint8_t ch = (uint8_t)c;
      if (bufpos == 0) {
        memmove(buf + 1, buf, n - 2);
        ++bufpos;
      }
      memcpy(buf + --bufpos, &ch, 1);
    }
    if (fpos == 0) {
      fseeko(fs, 0, SEEK_SET);
      break;
    }
  }
  memmove(buf, buf + bufpos, n - bufpos);
  return n - bufpos - 1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<fgetsr>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%rsi
  jbe    L33
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%r13
  call   <ftello@plt>
  mov    %eax,%ebx
  lea    0x1(%rax),%eax
  cmp    $0x1,%eax
  jbe    L33
  lea    0x14(%rsp),%rax
  movb   $0x0,-0x1(%r13,%r12,1)
  lea    -0x1(%r12),%r14
  mov    $0x1,%r15d
  mov    %rax,0x8(%rsp)
  jmp    L66
  nopl   (%rax)
L119:
  sub    $0x1,%r14
  lea    0x0(%r13,%r14,1),%rdi
L126:
  mov    0x8(%rsp),%rsi
  mov    $0x1,%edx
  call   <memcpy@plt>
L89:
  xor    %r15d,%r15d
  test   %ebx,%ebx
  je     L129
L66:
  mov    D6(%rip),%rax
  sub    $0x1,%ebx
  mov    %rbp,%rdi
  mov    %ebx,%esi
  mov    (%rax),%edx
  call   <fseeko@plt>
  test   %rax,%rax
  jne    L33
  mov    %rbp,%rdi
  call   <fgetc@plt>
  mov    D16(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L33
  cmp    $0xa,%eax
  jne    L90
  test   %r15d,%r15d
  je     L80
L90:
  cmp    $0xd,%eax
  je     L89
  cmp    $0xa,%eax
  je     L89
  mov    %eax,0x14(%rsp)
  test   %r14,%r14
  jne    L119
  lea    0x1(%r13),%rdi
  lea    -0x2(%r12),%rdx
  mov    %r13,%rsi
  call   <memmove@plt>
  mov    %r13,%rdi
  jmp    L126
  nopl   0x0(%rax)
L33:
  xor    %eax,%eax
L84:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L121
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L129:
  mov    D6(%rip),%rax
  xor    %esi,%esi
  mov    %rbp,%rdi
  mov    (%rax),%edx
  call   <fseeko@plt>
L80:
  sub    %r14,%r12
  lea    0x0(%r13,%r14,1),%rsi
  mov    %r13,%rdi
  mov    %r12,%rdx
  call   <memmove@plt>
  lea    -0x1(%r12),%rax
  jmp    L84
L121:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t fgetsr(FILE *fs, size_t n, char *buf) {
  off_t fpos;
  size_t bufpos;
  bool first = true;
  if (n <= 1 || (fpos = ftello(fs)) == -1 || fpos == 0) {
    return 0;
  }
  bufpos = n - 1;
  buf[bufpos] = '\0';
  while (true) {
    int c;
    if (fseeko(fs, --fpos, SEEK_SET) != 0 || (c = fgetc(fs)) == EOF) {
      return 0;
    }
    if (c == '\n' && !first) {
      break;
    }
    first = false;
    if (c != '\r' && c != '\n') {
      uint8_t ch = (uint8_t)c;
      if (bufpos == 0) {
        memmove(buf + 1, buf, n - 2);
        ++bufpos;
      }
      memcpy(buf + --bufpos, &ch, 1);
    }
    if (fpos == 0) {
      fseeko(fs, 0, SEEK_SET);
      break;
    }
  }
  memmove(buf, buf + bufpos, n - bufpos);
  return n - bufpos - 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<count_hash_results>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L12
L1:
  mov    -0x8(%rbp),%eax
  add    $0x1,%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  div    %rcx
  mov    %rdx,%rax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x5,%rax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%edx
  lea    0x1(%rdx),%ecx
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  mov    %ecx,(%rax,%rdx,4)
  addl   $0x1,-0x8(%rbp)
L12:
  cmpl   $0x7,-0x8(%rbp)
  jle    L1
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void count_hash_results(unsigned long hash, int counts[8][8]) {
  int srv, nsrv;
  for (nsrv = 0; nsrv < 8; nsrv++) {
    srv = hash % (nsrv + 1);
    counts[nsrv][srv]++;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<count_hash_results>:
  endbr64
  mov    $0x1,%ecx
L3:
  mov    %rdi,%rax
  mov    $0x0,%edx
  div    %rcx
  movslq %edx,%rdx
  addl   $0x1,(%rsi,%rdx,4)
  add    $0x1,%rcx
  add    $0x20,%rsi
  cmp    $0x9,%rcx
  jne    L3
  ret
```

assistant: ```c
void count_hash_results(unsigned long hash, int counts[8][8]) {
  int srv, nsrv;
  for (nsrv = 0; nsrv < 8; nsrv++) {
    srv = hash % (nsrv + 1);
    counts[nsrv][srv]++;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<count_hash_results>:
  endbr64
  mov    $0x1,%ecx
  nopl   0x0(%rax)
L9:
  mov    %rdi,%rax
  xor    %edx,%edx
  div    %rcx
  add    $0x1,%rcx
  movslq %edx,%rdx
  addl   $0x1,(%rsi,%rdx,4)
  add    $0x20,%rsi
  cmp    $0x9,%rcx
  jne    L9
  ret
```

assistant: ```c
void count_hash_results(unsigned long hash, int counts[8][8]) {
  int srv, nsrv;
  for (nsrv = 0; nsrv < 8; nsrv++) {
    srv = hash % (nsrv + 1);
    counts[nsrv][srv]++;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<count_hash_results>:
  endbr64
  mov    %rdi,%rax
  addl   $0x1,(%rsi)
  mov    %rsi,%rcx
  and    $0x1,%eax
  addl   $0x1,0x20(%rsi,%rax,4)
  mov    %rdi,%rax
  movabs $0xaaaaaaaaaaaaaaab,%rsi
  mul    %rsi
  mov    %rdx,%rax
  mov    %rdx,%rsi
  and    $0xfffffffffffffffe,%rdx
  shr    %rax
  shr    $0x2,%rsi
  add    %rax,%rdx
  mov    %rdi,%rax
  sub    %rdx,%rax
  movabs $0xcccccccccccccccd,%rdx
  cltq
  addl   $0x1,0x40(%rcx,%rax,4)
  mov    %rdi,%rax
  and    $0x3,%eax
  addl   $0x1,0x60(%rcx,%rax,4)
  mov    %rdi,%rax
  mul    %rdx
  mov    %rdx,%rax
  and    $0xfffffffffffffffc,%rdx
  shr    $0x2,%rax
  add    %rax,%rdx
  mov    %rdi,%rax
  sub    %rdx,%rax
  lea    (%rsi,%rsi,2),%rdx
  cltq
  add    %rdx,%rdx
  addl   $0x1,0x80(%rcx,%rax,4)
  mov    %rdi,%rax
  sub    %rdx,%rax
  movabs $0x2492492492492493,%rdx
  cltq
  addl   $0x1,0xa0(%rcx,%rax,4)
  mov    %rdi,%rax
  mul    %rdx
  mov    %rdi,%rax
  sub    %rdx,%rax
  shr    %rax
  add    %rax,%rdx
  mov    %rdi,%rax
  and    $0x7,%edi
  shr    $0x2,%rdx
  lea    0x0(,%rdx,8),%rsi
  sub    %rdx,%rsi
  sub    %rsi,%rax
  cltq
  addl   $0x1,0xc0(%rcx,%rax,4)
  addl   $0x1,0xe0(%rcx,%rdi,4)
  ret
```

assistant: ```c
void count_hash_results(unsigned long hash, int counts[8][8]) {
  int srv, nsrv;
  for (nsrv = 0; nsrv < 8; nsrv++) {
    srv = hash % (nsrv + 1);
    counts[nsrv][srv]++;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<test_most_a>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x61617a6161616161,%rax
  movabs $0x6161616161616161,%rdx
  mov    %rax,-0x50(%rbp)
  mov    %rdx,-0x48(%rbp)
  movabs $0x6161616161616161,%rax
  movabs $0x6161616161616161,%rdx
  mov    %rax,-0x40(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
  movabs $0x6161616161616161,%rax
  mov    %rax,-0x20(%rbp)
  movl   $0x61616161,-0x18(%rbp)
  movw   $0x6161,-0x14(%rbp)
  movb   $0x0,-0x12(%rbp)
  lea    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <use@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
void test_most_a() {
  char a[] = "aaaaazaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa";
  use(a);
}
```