ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<chop_words>:
  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
  movq   $0x0,-0x28(%rbp)
  movl   $0x2,-0x34(%rbp)
  cmpq   $0x0,-0x48(%rbp)
  jne    L25
  mov    $0x0,%eax
  jmp    L10
L25:
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <calloc@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x34(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    -0x20(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x48(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x34(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  lea    -0x30(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  lea    D9(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtok_r@plt>
  mov    %rax,(%rbx)
L30:
  addl   $0x1,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x28(%rbp)
  mov    -0x34(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdx
  lea    D9(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok_r@plt>
  mov    %rax,(%rbx)
  mov    (%rbx),%rax
  test   %rax,%rax
  jne    L30
  mov    -0x28(%rbp),%rax
L10:
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L31
  call   <__stack_chk_fail@plt>
L31:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D9
type_guessed: string

tool: D9:  	


assistant: ```c
char **chop_words(const char *src) {
  char **v = NULL, *str, *r;
  int count = 2;
  if (!src)
    return NULL;
  str = calloc(1 + strlen(src), sizeof(char));
  *(v = malloc(count * sizeof(char **))) = str;
  strcpy(str, src);
  *(v + count - 1) = strtok_r(str, " \t\r\n", &r);
  do
    v = realloc(v, ++count * sizeof(char **));
  while ((*(v + count - 1) = strtok_r(NULL, " \t\r\n", &r)));
  return v;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<chop_words>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L39
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,%r12
  mov    $0x10,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %r12,(%rax)
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %rsp,%rdx
  lea    D5(%rip),%rsi
  mov    %r12,%rdi
  call   <strtok_r@plt>
  mov    %rax,0x8(%rbx)
  mov    $0x10,%ebp
  lea    D5(%rip),%r13
L23:
  lea    0x8(%rbp),%esi
  mov    %rbx,%rdi
  call   <realloc@plt>
  mov    %rax,%rbx
  lea    (%rax,%rbp,1),%r12
  mov    %rsp,%rdx
  mov    %r13,%rsi
  mov    $0x0,%edi
  call   <strtok_r@plt>
  mov    %rax,(%r12)
  add    $0x8,%rbp
  test   %rax,%rax
  jne    L23
L7:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  mov    %rbx,%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L39:
  mov    %rdi,%rbx
  jmp    L7
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D5
datatype: string

tool: D5:  	


assistant: ```c
char **chop_words(const char *src) {
  char **v = NULL, *str, *r;
  int count = 2;
  if (!src)
    return NULL;
  str = calloc(1 + strlen(src), sizeof(char));
  *(v = malloc(count * sizeof(char **))) = str;
  strcpy(str, src);
  *(v + count - 1) = strtok_r(str, " \t\r\n", &r);
  do
    v = realloc(v, ++count * sizeof(char **));
  while ((*(v + count - 1) = strtok_r(NULL, " \t\r\n", &r)));
  return v;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<chop_words>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L39
  mov    %rdi,%r12
  mov    %rsp,%r13
  mov    $0x10,%ebx
  call   <strlen@plt>
  mov    $0x1,%esi
  lea    0x1(%rax),%rdi
  call   <calloc@plt>
  mov    $0x10,%edi
  mov    %rax,%r14
  call   <malloc@plt>
  mov    %r12,%rsi
  mov    %r14,%rdi
  lea    D4(%rip),%r12
  mov    %r14,(%rax)
  mov    %rax,%rbp
  call   <strcpy@plt>
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <strtok_r@plt>
  mov    %rax,0x8(%rbp)
  xchg   %ax,%ax
L19:
  lea    0x8(%rbx),%esi
  mov    %rbp,%rdi
  call   <realloc@plt>
  mov    %r13,%rdx
  mov    %r12,%rsi
  xor    %edi,%edi
  lea    (%rax,%rbx,1),%r14
  mov    %rax,%rbp
  add    $0x8,%rbx
  call   <strtok_r@plt>
  mov    %rax,(%r14)
  test   %rax,%rax
  jne    L19
L17:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x10,%rsp
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L39:
  xor    %ebp,%ebp
  jmp    L17
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D4
datatype: string

tool: D4:  	


assistant: ```c
char **chop_words(const char *src) {
  char **v = NULL, *str, *r;
  int count = 2;
  if (!src)
    return NULL;
  str = calloc(1 + strlen(src), sizeof(char));
  *(v = malloc(count * sizeof(char **))) = str;
  strcpy(str, src);
  *(v + count - 1) = strtok_r(str, " \t\r\n", &r);
  do
    v = realloc(v, ++count * sizeof(char **));
  while ((*(v + count - 1) = strtok_r(NULL, " \t\r\n", &r)));
  return v;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<chop_words>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L9
  mov    %rdi,%r12
  mov    %rsp,%r13
  mov    $0x10,%ebx
  call   <strlen@plt>
  mov    $0x1,%esi
  lea    0x1(%rax),%rdi
  call   <calloc@plt>
  mov    $0x10,%edi
  mov    %rax,%r14
  call   <malloc@plt>
  mov    %r12,%rsi
  mov    %r14,%rdi
  lea    D7(%rip),%r12
  mov    %r14,(%rax)
  mov    %rax,%rbp
  call   <strcpy@plt>
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <strtok_r@plt>
  mov    %rax,0x8(%rbp)
  xchg   %ax,%ax
L20:
  lea    0x8(%rbx),%esi
  mov    %rbp,%rdi
  call   <realloc@plt>
  mov    %r13,%rdx
  mov    %r12,%rsi
  xor    %edi,%edi
  lea    (%rax,%rbx,1),%r14
  mov    %rax,%rbp
  add    $0x8,%rbx
  call   <strtok_r@plt>
  mov    %rax,(%r14)
  test   %rax,%rax
  jne    L20
L11:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x10,%rsp
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  xor    %ebp,%ebp
  jmp    L11
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D7
datatype: string

tool: D7:  	


assistant: ```c
char **chop_words(const char *src) {
  char **v = NULL, *str, *r;
  int count = 2;
  if (!src)
    return NULL;
  str = calloc(1 + strlen(src), sizeof(char));
  *(v = malloc(count * sizeof(char **))) = str;
  strcpy(str, src);
  *(v + count - 1) = strtok_r(str, " \t\r\n", &r);
  do
    v = realloc(v, ++count * sizeof(char **));
  while ((*(v + count - 1) = strtok_r(NULL, " \t\r\n", &r)));
  return v;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  movq   $0x1,-0x8(%rbp)
  mov    $0x0,%eax
  call   <fun@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  cmpl   $0x48,-0x10(%rbp)
  jne    L4
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  je     L27
L4:
  mov    $0x1,%eax
  jmp    L13
L27:
  mov    $0x0,%eax
L13:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D10
type: string
name: disassemble_data
arguments: data_tag: D5
type: string

tool: D10: NR OF TRAPS %d


tool: D5: RETURNED %d


assistant: ```c
int main(int argc, const char *argv[]) {
  int ret = 0;
  int nr_of_traps = 0;
  long nr_of_reductions = 1;
  ret = fun();
  printf("NR OF TRAPS %d\n", nr_of_traps);
  printf("RETURNED %d\n", ret);
  if (ret != 72 || nr_of_traps != ret) {
    return 1;
  } else {
    return 0;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <fun@plt>
  mov    %eax,%ebx
  mov    $0x0,%esi
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,%esi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: label: D12
type: string
name: parse_data
arguments: label: D16
type: string

tool: D12: NR OF TRAPS %d


tool: D16: RETURNED %d


assistant: ```c
int main(int argc, const char *argv[]) {
  int ret = 0;
  int nr_of_traps = 0;
  long nr_of_reductions = 1;
  ret = fun();
  printf("NR OF TRAPS %d\n", nr_of_traps);
  printf("RETURNED %d\n", ret);
  if (ret != 72 || nr_of_traps != ret) {
    return 1;
  } else {
    return 0;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <fun@plt>
  xor    %esi,%esi
  lea    D13(%rip),%rdi
  mov    %eax,%ebp
  call   <printf@plt>
  mov    %ebp,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D13
data_type: string
name: retrieve_data
arguments: data_label: D7
data_type: string

tool: D13: NR OF TRAPS %d


tool: D7: RETURNED %d


assistant: ```c
int main(int argc, const char *argv[]) {
  int ret = 0;
  int nr_of_traps = 0;
  long nr_of_reductions = 1;
  ret = fun();
  printf("NR OF TRAPS %d\n", nr_of_traps);
  printf("RETURNED %d\n", ret);
  if (ret != 72 || nr_of_traps != ret) {
    return 1;
  } else {
    return 0;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <fun@plt>
  xor    %esi,%esi
  lea    D14(%rip),%rdi
  mov    %eax,%ebp
  call   <printf@plt>
  mov    %ebp,%esi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D14
type_guessed: string
name: disassemble_data
arguments: label: D10
type_guessed: string

tool: D14: NR OF TRAPS %d


tool: D10: RETURNED %d


assistant: ```c
int main(int argc, const char *argv[]) {
  int ret = 0;
  int nr_of_traps = 0;
  long nr_of_reductions = 1;
  ret = fun();
  printf("NR OF TRAPS %d\n", nr_of_traps);
  printf("RETURNED %d\n", ret);
  if (ret != 72 || nr_of_traps != ret) {
    return 1;
  } else {
    return 0;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L13
L9:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x4(%rbp)
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
L13:
  cmpl   $0x0,-0x14(%rbp)
  jg     L9
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int reverse(int N) {
  int a, rev = 0;
  while (N > 0) {
    a = N % 10;
    rev = rev * 10 + a;
    N /= 10;
  }
  return rev;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<reverse>:
  endbr64
  test   %edi,%edi
  jle    L13
  mov    $0x0,%edx
L6:
  lea    (%rdx,%rdx,4),%esi
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  mov    %edi,%ecx
  sub    %edx,%ecx
  lea    (%rcx,%rsi,2),%edx
  mov    %edi,%ecx
  mov    %eax,%edi
  cmp    $0x9,%ecx
  jg     L6
L16:
  mov    %edx,%eax
  ret
L13:
  mov    $0x0,%edx
  jmp    L16
```

assistant: ```c
int reverse(int N) {
  int a, rev = 0;
  while (N > 0) {
    a = N % 10;
    rev = rev * 10 + a;
    N /= 10;
  }
  return rev;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<reverse>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  jle    L8
  mov    $0xcccccccd,%r9d
  nopl   0x0(%rax)
L16:
  mov    %edi,%eax
  mov    %edi,%ecx
  lea    (%r8,%r8,4),%esi
  imul   %r9,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%ecx
  mov    %edi,%edx
  mov    %eax,%edi
  lea    (%rcx,%rsi,2),%r8d
  cmp    $0x9,%edx
  jg     L16
L8:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int reverse(int N) {
  int a, rev = 0;
  while (N > 0) {
    a = N % 10;
    rev = rev * 10 + a;
    N /= 10;
  }
  return rev;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<reverse>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  jle    L12
  mov    $0xcccccccd,%r9d
  nopl   0x0(%rax)
L17:
  mov    %edi,%eax
  mov    %edi,%ecx
  lea    (%r8,%r8,4),%esi
  imul   %r9,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%ecx
  mov    %edi,%edx
  mov    %eax,%edi
  lea    (%rcx,%rsi,2),%r8d
  cmp    $0x9,%edx
  jg     L17
L12:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int reverse(int N) {
  int a, rev = 0;
  while (N > 0) {
    a = N % 10;
    rev = rev * 10 + a;
    N /= 10;
  }
  return rev;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x430,%rsp
  mov    %edi,-0x424(%rbp)
  mov    %rsi,-0x430(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xffffffff,-0x414(%rbp)
  jmp    L45
L24:
  lea    -0x410(%rbp),%rax
  mov    %rax,%rdi
  call   <cleanup@plt>
  movzbl -0x410(%rbp),%eax
  cmp    $0xa,%al
  je     L16
  cmpl   $0x0,-0x414(%rbp)
  jle    L36
  mov    $0xa,%edi
  call   <putchar@plt>
L36:
  movl   $0x0,-0x414(%rbp)
  mov    D9(%rip),%rax
  mov    (%rax),%edx
  lea    -0x410(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <fputs@plt>
  jmp    L45
L16:
  cmpl   $0x0,-0x414(%rbp)
  jns    L41
  jmp    L45
L41:
  addl   $0x1,-0x414(%rbp)
L45:
  mov    D1(%rip),%rax
  mov    (%rax),%edx
  lea    -0x410(%rbp),%rax
  mov    $0x400,%esi
  mov    %rax,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L24
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L40
  call   <__stack_chk_fail@plt>
L40:
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int empties = -1;
  char line[1024];
  while (fgets(line, sizeof(line), stdin)) {
    cleanup(line);
    if (line[0] != '\n') {
      if (empties > 0)
        putchar('\n');
      empties = 0;
      fputs(line, stdout);
      continue;
    }
    if (empties < 0)
      continue;
    empties++;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    $0xffffffff,%ebx
  mov    D15(%rip),%rbp
  jmp    L21
L65:
  test   %ebx,%ebx
  jg     L8
L62:
  mov    %rsp,%rdi
  mov    D14(%rip),%rax
  mov    (%rax),%esi
  call   <fputs@plt>
  mov    $0x0,%ebx
L21:
  mov    %rsp,%rdi
  mov    0x0(%rbp),%edx
  mov    $0x400,%esi
  call   <fgets@plt>
  test   %rax,%rax
  je     L41
  mov    %rsp,%rdi
  call   <cleanup@plt>
  cmpb   $0xa,(%rsp)
  jne    L65
  cmp    $0x80000000,%ebx
  adc    $0x0,%ebx
  jmp    L21
L8:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L62
L41:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  mov    $0x0,%eax
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  ret
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  int empties = -1;
  char line[1024];
  while (fgets(line, sizeof(line), stdin)) {
    cleanup(line);
    if (line[0] != '\n') {
      if (empties > 0)
        putchar('\n');
      empties = 0;
      fputs(line, stdout);
      continue;
    }
    if (empties < 0)
      continue;
    empties++;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  mov    $0xffffffff,%ebp
  push   %rbx
  sub    $0x410,%rsp
  mov    D1(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  nopl   0x0(%rax)
L39:
  mov    (%r12),%edx
  mov    $0x400,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  je     L0
  mov    %rbx,%rdi
  call   <cleanup@plt>
  cmpb   $0xa,(%rsp)
  je     L1
  test   %ebp,%ebp
  jle    L4
  mov    $0xa,%edi
  call   <putchar@plt>
L4:
  mov    D0(%rip),%rax
  mov    %rbx,%rdi
  xor    %ebp,%ebp
  mov    (%rax),%esi
  call   <fputs@plt>
  jmp    L39
  nop
L1:
  xor    %eax,%eax
  cmp    $0xffffffff,%ebp
  setne  %al
  add    %eax,%ebp
  jmp    L39
  nopl   0x0(%rax)
L0:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x410,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  int empties = -1;
  char line[1024];
  while (fgets(line, sizeof(line), stdin)) {
    cleanup(line);
    if (line[0] != '\n') {
      if (empties > 0)
        putchar('\n');
      empties = 0;
      fputs(line, stdout);
      continue;
    }
    if (empties < 0)
      continue;
    empties++;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  mov    $0xffffffff,%ebp
  push   %rbx
  sub    $0x410,%rsp
  mov    D11(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  nopl   0x0(%rax)
L25:
  mov    (%r12),%edx
  mov    $0x400,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  je     L45
  mov    %rbx,%rdi
  call   <cleanup@plt>
  cmpb   $0xa,(%rsp)
  je     L28
  test   %ebp,%ebp
  jle    L11
  mov    $0xa,%edi
  call   <putchar@plt>
L11:
  mov    D14(%rip),%rax
  mov    %rbx,%rdi
  xor    %ebp,%ebp
  mov    (%rax),%esi
  call   <fputs@plt>
  jmp    L25
  nop
L28:
  xor    %eax,%eax
  cmp    $0xffffffff,%ebp
  setne  %al
  add    %eax,%ebp
  jmp    L25
  nopl   0x0(%rax)
L45:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x410,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  int empties = -1;
  char line[1024];
  while (fgets(line, sizeof(line), stdin)) {
    cleanup(line);
    if (line[0] != '\n') {
      if (empties > 0)
        putchar('\n');
      empties = 0;
      fputs(line, stdout);
      continue;
    }
    if (empties < 0)
      continue;
    empties++;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<calc_xi_sum_dev>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  movss  %xmm0,-0x3c(%rbp)
  mov    %r9d,-0x40(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L21
L10:
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L25:
  mov    -0x4(%rbp),%eax
  imul   -0x40(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm1
  mov    0x18(%rbp),%eax
  imul   -0x40(%rbp),%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm2
  mov    -0x4(%rbp),%eax
  imul   -0x40(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  mulss  %xmm0,%xmm2
  mov    0x10(%rbp),%eax
  imul   -0x40(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  mulss  %xmm0,%xmm2
  mov    0x18(%rbp),%eax
  add    $0x1,%eax
  imul   -0x40(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  mulss  %xmm2,%xmm0
  divss  -0x3c(%rbp),%xmm0
  mov    -0x4(%rbp),%eax
  imul   -0x40(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L11:
  mov    -0x4(%rbp),%eax
  cmp    -0x40(%rbp),%eax
  jl     L25
  addl   $0x1,-0x8(%rbp)
L21:
  mov    -0x8(%rbp),%eax
  cmp    -0x40(%rbp),%eax
  jl     L10
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void calc_xi_sum_dev(float *xi_sum_d, float *a_d, float *b_d, float *alpha_d, float *beta_d, float sum_ab, int nstates, int obs_t, int t) {
  int i, j;
  for (i = 0; i < nstates; ++i) {
    for (j = 0; j < nstates; ++j) {
      xi_sum_d[(j * nstates) + i] += alpha_d[(t * nstates) + j] * a_d[(j * nstates) + i] * b_d[(obs_t * nstates) + i] * beta_d[((t + 1) * nstates) + i] / sum_ab;
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<calc_xi_sum_dev>:
  endbr64
  test   %r9d,%r9d
  jle    L5
  push   %rbp
  push   %rbx
  mov    %rsi,%r10
  mov    %rcx,%r11
  mov    %r9d,%eax
  imul   0x20(%rsp),%eax
  mov    %r9d,%ecx
  imul   0x18(%rsp),%ecx
  movslq %ecx,%rcx
  lea    (%rdx,%rcx,4),%rsi
  lea    (%rax,%r9,1),%edx
  movslq %edx,%rdx
  lea    (%r8,%rdx,4),%rcx
  mov    %r9d,%ebp
  cltq
  lea    (%r11,%rax,4),%rbx
  movslq %r9d,%r8
  shl    $0x2,%r8
  lea    -0x1(%r9),%edx
  add    %rdx,%rax
  lea    0x4(%r11,%rax,4),%r9
  mov    $0x0,%r11d
L13:
  lea    0x0(,%r11,4),%rax
  mov    %rbx,%rdx
L26:
  movss  (%rdx),%xmm1
  mulss  (%r10,%rax,1),%xmm1
  mulss  (%rsi),%xmm1
  mulss  (%rcx),%xmm1
  divss  %xmm0,%xmm1
  addss  (%rdi,%rax,1),%xmm1
  movss  %xmm1,(%rdi,%rax,1)
  add    $0x4,%rdx
  add    %r8,%rax
  cmp    %r9,%rdx
  jne    L26
  add    $0x1,%r11
  add    $0x4,%rsi
  add    $0x4,%rcx
  cmp    %rbp,%r11
  jne    L13
  pop    %rbx
  pop    %rbp
  ret
L5:
  ret
```

assistant: ```c
void calc_xi_sum_dev(float *xi_sum_d, float *a_d, float *b_d, float *alpha_d, float *beta_d, float sum_ab, int nstates, int obs_t, int t) {
  int i, j;
  for (i = 0; i < nstates; ++i) {
    for (j = 0; j < nstates; ++j) {
      xi_sum_d[(j * nstates) + i] += alpha_d[(t * nstates) + j] * a_d[(j * nstates) + i] * b_d[(obs_t * nstates) + i] * beta_d[((t + 1) * nstates) + i] / sum_ab;
    }
  }
}
```