ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<addForward>:
  endbr64
  cmp    %esi,%edi
  jg     L1
  movsd  D7(%rip),%xmm3
  add    $0x1,%esi
  pxor   %xmm0,%xmm0
  nopw   0x0(%rax,%rax,1)
L4:
  pxor   %xmm2,%xmm2
  movapd %xmm3,%xmm1
  cvtss2sd %xmm0,%xmm0
  cvtsi2ss %edi,%xmm2
  add    $0x1,%edi
  cvtss2sd %xmm2,%xmm2
  divsd  %xmm2,%xmm1
  addsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  cmp    %esi,%edi
  jne    L4
  ret
  nopl   0x0(%rax,%rax,1)
L1:
  pxor   %xmm0,%xmm0
  ret
```

tool_calls: name: read_data
arguments: label: D7
type: f64

tool: D7: 1

assistant: ```c
float addForward(int from, int to) {
  float sum = 0;
  int i;
  for (i = from; i <= to; i++) {
    sum += 1.0 / (float)i;
  }
  return sum;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<addForward>:
  endbr64
  cmp    %esi,%edi
  jg     L19
  movsd  D4(%rip),%xmm3
  add    $0x1,%esi
  pxor   %xmm0,%xmm0
  nopw   0x0(%rax,%rax,1)
L4:
  pxor   %xmm2,%xmm2
  movapd %xmm3,%xmm1
  cvtss2sd %xmm0,%xmm0
  cvtsi2ss %edi,%xmm2
  add    $0x1,%edi
  cvtss2sd %xmm2,%xmm2
  divsd  %xmm2,%xmm1
  addsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  cmp    %esi,%edi
  jne    L4
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  pxor   %xmm0,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D4
type_guessed: f64

tool: D4: 1

assistant: ```c
float addForward(int from, int to) {
  float sum = 0;
  int i;
  for (i = from; i <= to; i++) {
    sum += 1.0 / (float)i;
  }
  return sum;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L29
L7:
  mov    -0x10(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  mov    %rcx,%rsi
  sar    $0x3f,%rsi
  sub    %rsi,%rax
  mov    %rax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  test   %rdx,%rdx
  jne    L31
  addl   $0x1,-0x14(%rbp)
  mov    -0x10(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x10(%rbp)
L29:
  mov    -0x10(%rbp),%rax
  test   %rax,%rax
  jne    L7
  jmp    L30
L31:
  nop
L30:
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D25
data_type: string
name: retrieve_data
arguments: label: D23
data_type: string
name: retrieve_data
arguments: label: D4
data_type: string

tool: D25: Enter an integer: 

tool: D23: %lld

tool: D4: Number of digits: %d

assistant: ```c
int main() {
  long long n;
  int count = 0;
  printf("Enter an integer: ");
  scanf("%lld", &n);
  while (n != 0) {
    if (n % 10 == 0) {
      ++count;
      n /= 10;
    } else
      break;
  }
  printf("Number of digits: %d", count);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rcx
  test   %rcx,%rcx
  je     L14
  mov    $0x0,%r9d
  mov    $0x0,%esi
  movabs $0x6666666666666667,%rdi
  mov    $0x1,%r8d
  jmp    L33
L37:
  mov    %rdx,%rbx
L33:
  mov    %rcx,%rax
  imul   %rdi
  sar    $0x2,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  cmp    %rax,%rcx
  jne    L69
  add    $0x1,%esi
  mov    %rcx,%rax
  imul   %rdi
  sar    $0x2,%rdx
  sar    $0x3f,%rcx
  sub    %rcx,%rdx
  mov    %rdx,%rcx
  mov    %r8d,%r9d
  jne    L37
  movq   $0x0,(%rsp)
L68:
  lea    D19(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L69:
  test   %r9b,%r9b
  je     L68
  mov    %rbx,(%rsp)
  jmp    L68
L14:
  mov    $0x0,%esi
  jmp    L68
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D5
type_guessed: string
name: read_data
arguments: label: D24
type_guessed: string
name: read_data
arguments: label: D19
type_guessed: string

tool: D5: Enter an integer: 

tool: D24: %lld

tool: D19: Number of digits: %d

assistant: ```c
int main() {
  long long n;
  int count = 0;
  printf("Enter an integer: ");
  scanf("%lld", &n);
  while (n != 0) {
    if (n % 10 == 0) {
      ++count;
      n /= 10;
    } else
      break;
  }
  printf("Number of digits: %d", count);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbx
  lea    D15(%rip),%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D1(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rcx
  test   %rcx,%rcx
  je     L73
  movabs $0xcccccccccccccccd,%r9
  xor    %r8d,%r8d
  xor    %esi,%esi
  movabs $0x1999999999999998,%rdi
  movabs $0x6666666666666667,%r10
  jmp    L10
  cs nopw 0x0(%rax,%rax,1)
L7:
  mov    %rcx,%rax
  add    $0x1,%esi
  mov    $0x1,%r8d
  imul   %r10
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  sub    %rax,%rdx
  mov    %rdx,%rcx
  je     L12
  mov    %rcx,%rbx
L10:
  mov    %rcx,%rax
  imul   %r9,%rax
  add    %rdi,%rax
  ror    %rax
  cmp    %rdi,%rax
  jbe    L7
  test   %r8b,%r8b
  je     L1
  mov    %rbx,(%rsp)
L1:
  xor    %eax,%eax
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L75
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L12:
  movq   $0x0,(%rsp)
  jmp    L1
L73:
  xor    %esi,%esi
  jmp    L1
L75:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D15
type_infered: string
name: read_data
arguments: data_label: D1
type_infered: string
name: read_data
arguments: data_label: D5
type_infered: string

tool: D15: Enter an integer: 

tool: D1: %lld

tool: D5: Number of digits: %d

assistant: ```c
int main() {
  long long n;
  int count = 0;
  printf("Enter an integer: ");
  scanf("%lld", &n);
  while (n != 0) {
    if (n % 10 == 0) {
      ++count;
      n /= 10;
    } else
      break;
  }
  printf("Number of digits: %d", count);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D11(%rip),%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rcx
  test   %rcx,%rcx
  je     L25
  movabs $0xcccccccccccccccd,%r9
  xor    %r8d,%r8d
  xor    %esi,%esi
  movabs $0x1999999999999998,%rdi
  movabs $0x6666666666666667,%r10
  jmp    L52
  cs nopw 0x0(%rax,%rax,1)
L14:
  mov    %rcx,%rax
  add    $0x1,%esi
  mov    $0x1,%r8d
  imul   %r10
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  sub    %rax,%rdx
  mov    %rdx,%rcx
  je     L30
  mov    %rcx,%rbx
L52:
  mov    %rcx,%rax
  imul   %r9,%rax
  add    %rdi,%rax
  ror    %rax
  cmp    %rdi,%rax
  jbe    L14
  test   %r8b,%r8b
  je     L9
  mov    %rbx,(%rsp)
L9:
  xor    %eax,%eax
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L30:
  movq   $0x0,(%rsp)
  jmp    L9
L25:
  xor    %esi,%esi
  jmp    L9
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D11
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D14
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D4
type_guessed: string

tool: D11: Enter an integer: 

tool: D14: %lld

tool: D4: Number of digits: %d

assistant: ```c
int main() {
  long long n;
  int count = 0;
  printf("Enter an integer: ");
  scanf("%lld", &n);
  while (n != 0) {
    if (n % 10 == 0) {
      ++count;
      n /= 10;
    } else
      break;
  }
  printf("Number of digits: %d", count);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fgets2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x30,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    D268(%rip),%rax
  mov    %rax,-0x20(%rbp)
L86:
  mov    D80(%rip),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <find_lf@plt>
  mov    %rax,-0x20(%rbp)
  mov    D80(%rip),%rax
  cmp    %rax,-0x20(%rbp)
  jae    L98
  mov    D268(%rip),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,D166(%rip)
  mov    -0x18(%rbp),%rax
  jmp    L103
L98:
  mov    D166(%rip),%rax
  lea    D166(%rip),%rdx
  cmp    %rdx,%rax
  jbe    L60
  mov    D80(%rip),%rdx
  mov    D166(%rip),%rax
  cmp    %rax,%rdx
  je     L59
  mov    D80(%rip),%rax
  mov    D166(%rip),%rdx
  sub    %rdx,%rax
  mov    %eax,%edx
  mov    D166(%rip),%rax
  mov    %rax,%rsi
  lea    D166(%rip),%rax
  mov    %rax,%rdi
  call   <memmove@plt>
L59:
  mov    D80(%rip),%rax
  mov    D166(%rip),%rdx
  sub    %rdx,%rax
  mov    %rax,%rdx
  lea    D166(%rip),%rax
  add    %rdx,%rax
  mov    %rax,D23(%rip)
  mov    D23(%rip),%rax
  mov    %rax,-0x20(%rbp)
  lea    D166(%rip),%rax
  mov    %rax,D23(%rip)
  jmp    L117
L60:
  mov    D23(%rip),%rax
  lea    D23(%rip),%rdx
  cmp    %rdx,%rax
  jne    L117
  mov    $0x0,%eax
  jmp    L103
L117:
  lea    D23(%rip),%rax
  mov    D23(%rip),%rdx
  sub    %rdx,%rax
  mov    %eax,%r12d
  mov    D23(%rip),%rbx
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <fileno@plt>
  mov    %r12d,%edx
  mov    %rbx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jg     L119
  mov    D23(%rip),%rdx
  mov    D23(%rip),%rax
  cmp    %rax,%rdx
  jne    L50
  mov    $0x0,%eax
  jmp    L103
L50:
  mov    D23(%rip),%rax
  movb   $0x0,(%rax)
  mov    D23(%rip),%rax
  mov    %rax,D117(%rip)
  mov    D23(%rip),%rax
  jmp    L103
L119:
  mov    D117(%rip),%rdx
  mov    -0x24(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    %rax,D117(%rip)
  mov    D117(%rip),%rax
  movb   $0xa,(%rax)
  jmp    L86
L103:
  add    $0x30,%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
const char *fgets2(FILE *stream) {
  static char buffer[(256 * 1024) + 68];
  static char *end = buffer;
  static char *line = buffer;
  char *next;
  int ret;
  next = line;
  while (1) {
    next = find_lf(next, end);
    if (next < end) {
      const char *start = line;
      *next = '\0';
      line = next + 1;
      return start;
    }
    if (line > buffer) {
      if (end != line)
        memmove(buffer, line, end - line);
      end = buffer + (end - line);
      next = end;
      line = buffer;
    } else {
      if (end == buffer + (256 * 1024))
        return NULL;
    }
    ret = read(fileno(stream), end, buffer + (256 * 1024) - end);
    if (ret <= 0) {
      if (end == line)
        return NULL;
      *end = '\0';
      end = line;
      return line;
    }
    end += ret;
    *end = '\n';
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fgets2>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    D31(%rip),%rbx
  lea    D139(%rip),%rbp
  lea    0x40000(%rbp),%r12
  jmp    L98
L74:
  mov    D31(%rip),%rax
  movb   $0x0,(%rbx)
  add    $0x1,%rbx
  mov    %rbx,D73(%rip)
L3:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L60:
  mov    %rbp,%rbx
  add    D73(%rip),%rbx
  sub    D73(%rip),%rbx
  mov    %rbx,D78(%rip)
  mov    %rbp,D78(%rip)
L73:
  mov    D78(%rip),%r14
  mov    %r13,%rdi
  call   <fileno@plt>
  mov    %eax,%edi
  mov    %r12d,%edx
  sub    %r14d,%edx
  mov    %r14,%rsi
  call   <read@plt>
  test   %eax,%eax
  jle    L19
  cltq
  add    D78(%rip),%rax
  mov    %rax,D78(%rip)
  movb   $0xa,(%rax)
L98:
  mov    D78(%rip),%rsi
  mov    %rbx,%rdi
  call   <find_lf@plt>
  mov    %rax,%rbx
  mov    D78(%rip),%rdx
  cmp    %rax,%rdx
  ja     L74
  mov    D78(%rip),%rsi
  cmp    %rbp,%rsi
  jbe    L25
  cmp    %rsi,%rdx
  je     L60
  sub    %esi,%edx
  mov    %rbp,%rdi
  call   <memmove@plt>
  jmp    L60
L25:
  cmp    %r12,%rdx
  jne    L73
  mov    $0x0,%eax
  jmp    L3
L19:
  mov    D78(%rip),%rdx
  mov    D78(%rip),%rax
  cmp    %rax,%rdx
  je     L31
  movb   $0x0,(%rdx)
  mov    %rax,D78(%rip)
  jmp    L3
L31:
  mov    $0x0,%eax
  jmp    L3
```

assistant: ```c
const char *fgets2(FILE *stream) {
  static char buffer[(256 * 1024) + 68];
  static char *end = buffer;
  static char *line = buffer;
  char *next;
  int ret;
  next = line;
  while (1) {
    next = find_lf(next, end);
    if (next < end) {
      const char *start = line;
      *next = '\0';
      line = next + 1;
      return start;
    }
    if (line > buffer) {
      if (end != line)
        memmove(buffer, line, end - line);
      end = buffer + (end - line);
      next = end;
      line = buffer;
    } else {
      if (end == buffer + (256 * 1024))
        return NULL;
    }
    ret = read(fileno(stream), end, buffer + (256 * 1024) - end);
    if (ret <= 0) {
      if (end == line)
        return NULL;
      *end = '\0';
      end = line;
      return line;
    }
    end += ret;
    *end = '\n';
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<fgets2>:
  endbr64
  push   %r14
  mov    D143(%rip),%rsi
  push   %r13
  mov    D92(%rip),%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  lea    D80(%rip),%rbx
  lea    0x40000(%rbx),%rbp
  jmp    L21
  nopl   (%rax)
L17:
  mov    %rsi,%rax
  cmp    %rsi,%r14
  je     L33
  mov    %r14d,%edx
  mov    %rbx,%rdi
  sub    %esi,%edx
  call   <memmove@plt>
  mov    D143(%rip),%rsi
  mov    D92(%rip),%rax
L33:
  sub    %rax,%rsi
  mov    %rbx,D40(%rip)
  lea    (%rbx,%rsi,1),%r13
  mov    %r13,D40(%rip)
  mov    %r13,%r14
L57:
  mov    %r12,%rdi
  call   <fileno@plt>
  mov    %ebp,%edx
  mov    %r14,%rsi
  sub    %r14d,%edx
  mov    %eax,%edi
  call   <read@plt>
  test   %eax,%eax
  jle    L63
  movslq %eax,%rsi
  add    D40(%rip),%rsi
  mov    %rsi,D40(%rip)
  movb   $0xa,(%rsi)
L21:
  mov    %r13,%rdi
  call   <find_lf@plt>
  mov    D40(%rip),%r14
  mov    %rax,%r13
  cmp    %rax,%r14
  ja     L62
  mov    D40(%rip),%rsi
  cmp    %rbx,%rsi
  ja     L17
  cmp    %rbp,%r14
  jne    L57
L31:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L62:
  movb   $0x0,0x0(%r13)
  add    $0x1,%r13
  pop    %rbx
  mov    D40(%rip),%rax
  pop    %rbp
  mov    %r13,D40(%rip)
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L63:
  mov    D40(%rip),%rdx
  mov    D40(%rip),%rax
  cmp    %rax,%rdx
  je     L31
  movb   $0x0,(%rdx)
  pop    %rbx
  mov    %rax,D40(%rip)
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
const char *fgets2(FILE *stream) {
  static char buffer[(256 * 1024) + 68];
  static char *end = buffer;
  static char *line = buffer;
  char *next;
  int ret;
  next = line;
  while (1) {
    next = find_lf(next, end);
    if (next < end) {
      const char *start = line;
      *next = '\0';
      line = next + 1;
      return start;
    }
    if (line > buffer) {
      if (end != line)
        memmove(buffer, line, end - line);
      end = buffer + (end - line);
      next = end;
      line = buffer;
    } else {
      if (end == buffer + (256 * 1024))
        return NULL;
    }
    ret = read(fileno(stream), end, buffer + (256 * 1024) - end);
    if (ret <= 0) {
      if (end == line)
        return NULL;
      *end = '\0';
      end = line;
      return line;
    }
    end += ret;
    *end = '\n';
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fgets2>:
  endbr64
  push   %r14
  mov    D88(%rip),%rsi
  push   %r13
  mov    D54(%rip),%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  lea    D15(%rip),%rbx
  lea    0x40000(%rbx),%rbp
  jmp    L44
  nopl   (%rax)
L35:
  mov    %rsi,%rax
  cmp    %rsi,%r14
  je     L20
  mov    %r14d,%edx
  mov    %rbx,%rdi
  sub    %esi,%edx
  call   <memmove@plt>
  mov    D88(%rip),%rsi
  mov    D54(%rip),%rax
L20:
  sub    %rax,%rsi
  mov    %rbx,D75(%rip)
  lea    (%rbx,%rsi,1),%r13
  mov    %r13,D75(%rip)
  mov    %r13,%r14
L63:
  mov    %r12,%rdi
  call   <fileno@plt>
  mov    %ebp,%edx
  mov    %r14,%rsi
  sub    %r14d,%edx
  mov    %eax,%edi
  call   <read@plt>
  test   %eax,%eax
  jle    L57
  movslq %eax,%rsi
  add    D75(%rip),%rsi
  mov    %rsi,D75(%rip)
  movb   $0xa,(%rsi)
L44:
  mov    %r13,%rdi
  call   <find_lf@plt>
  mov    D75(%rip),%r14
  mov    %rax,%r13
  cmp    %rax,%r14
  ja     L60
  mov    D75(%rip),%rsi
  cmp    %rbx,%rsi
  ja     L35
  cmp    %rbp,%r14
  jne    L63
L58:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L60:
  movb   $0x0,0x0(%r13)
  add    $0x1,%r13
  pop    %rbx
  mov    D75(%rip),%rax
  pop    %rbp
  mov    %r13,D75(%rip)
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L57:
  mov    D75(%rip),%rdx
  mov    D75(%rip),%rax
  cmp    %rax,%rdx
  je     L58
  movb   $0x0,(%rdx)
  pop    %rbx
  mov    %rax,D75(%rip)
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
const char *fgets2(FILE *stream) {
  static char buffer[(256 * 1024) + 68];
  static char *end = buffer;
  static char *line = buffer;
  char *next;
  int ret;
  next = line;
  while (1) {
    next = find_lf(next, end);
    if (next < end) {
      const char *start = line;
      *next = '\0';
      line = next + 1;
      return start;
    }
    if (line > buffer) {
      if (end != line)
        memmove(buffer, line, end - line);
      end = buffer + (end - line);
      next = end;
      line = buffer;
    } else {
      if (end == buffer + (256 * 1024))
        return NULL;
    }
    ret = read(fileno(stream), end, buffer + (256 * 1024) - end);
    if (ret <= 0) {
      if (end == line)
        return NULL;
      *end = '\0';
      end = line;
      return line;
    }
    end += ret;
    *end = '\n';
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<dynmacro>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    -0x38(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isalpha@plt>
  test   %eax,%eax
  jne    L74
  mov    $0x0,%eax
  jmp    L85
L74:
  mov    -0x38(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x20(%rbp)
  jmp    L93
L72:
  addq   $0x1,-0x20(%rbp)
L93:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L17
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isalpha@plt>
  mov    %eax,%ebx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  or     %ebx,%eax
  test   %eax,%eax
  jne    L72
L17:
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L15
L63:
  addq   $0x1,-0x20(%rbp)
L15:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L63
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x3a,%al
  jne    L79
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x3d,%al
  je     L52
L79:
  mov    $0x0,%eax
  jmp    L85
L52:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x21(%rbp)
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x2,%rax
  mov    %rax,%rdi
  call   <copys@plt>
  mov    %eax,%ecx
  mov    -0x38(%rbp),%rax
  mov    $0x1,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <setvar@plt>
  mov    -0x18(%rbp),%rax
  movzbl -0x21(%rbp),%edx
  mov    %dl,(%rax)
  mov    $0x1,%eax
L85:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int dynmacro(char *line) {
  char *s;
  char endc, *endp;
  if (!isalpha(line[0]))
    return 0;
  for (s = line + 1; *s && (isalpha(*s) | isdigit(*s)); ++s)
    ;
  endp = s;
  while (isspace(*s))
    ++s;
  if (s[0] != ':' || s[1] != '=')
    return 0;
  endc = *endp;
  *endp = '\0';
  setvar(line, copys(s + 2), 1);
  *endp = endc;
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<dynmacro>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  movsbl (%rdi),%edi
  call   <isalpha@plt>
  test   %eax,%eax
  je     L23
  lea    0x1(%r12),%rbp
  movzbl 0x1(%r12),%edi
  test   %dil,%dil
  je     L32
L34:
  movsbl %dil,%edi
  call   <isalpha@plt>
  mov    %eax,%ebx
  movsbl 0x0(%rbp),%edi
  call   <isdigit@plt>
  or     %eax,%ebx
  je     L32
  add    $0x1,%rbp
  movzbl 0x0(%rbp),%edi
  test   %dil,%dil
  jne    L34
L32:
  mov    %rbp,%rbx
  jmp    L61
L13:
  add    $0x1,%rbx
L61:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L13
  cmpb   $0x3a,(%rbx)
  jne    L23
  cmpb   $0x3d,0x1(%rbx)
  je     L54
L23:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L54:
  movzbl 0x0(%rbp),%r13d
  movb   $0x0,0x0(%rbp)
  lea    0x2(%rbx),%rdi
  call   <copys@plt>
  mov    %eax,%esi
  mov    $0x1,%edx
  mov    %r12,%rdi
  call   <setvar@plt>
  mov    %r13b,0x0(%rbp)
  mov    $0x1,%eax
  jmp    L23
```

assistant: ```c
int dynmacro(char *line) {
  char *s;
  char endc, *endp;
  if (!isalpha(line[0]))
    return 0;
  for (s = line + 1; *s && (isalpha(*s) | isdigit(*s)); ++s)
    ;
  endp = s;
  while (isspace(*s))
    ++s;
  if (s[0] != ':' || s[1] != '=')
    return 0;
  endc = *endp;
  *endp = '\0';
  setvar(line, copys(s + 2), 1);
  *endp = endc;
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<dynmacro>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  call   <isalpha@plt>
  test   %eax,%eax
  je     L84
  movsbl 0x1(%r12),%edi
  lea    0x1(%r12),%rbp
  test   %dil,%dil
  jne    L41
  jmp    L73
  nop
L12:
  movsbl 0x1(%rbp),%edi
  add    $0x1,%rbp
  test   %dil,%dil
  je     L73
L41:
  call   <isalpha@plt>
  movsbl 0x0(%rbp),%edi
  mov    %eax,%ebx
  call   <isdigit@plt>
  or     %eax,%ebx
  jne    L12
  movsbl 0x0(%rbp),%edi
L73:
  mov    %rbp,%rbx
  jmp    L2
  nopw   0x0(%rax,%rax,1)
L39:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
L2:
  call   <isspace@plt>
  test   %rax,%rax
  jne    L39
  cmpb   $0x3a,(%rbx)
  je     L51
L84:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L51:
  cmpb   $0x3d,0x1(%rbx)
  jne    L84
  movzbl 0x0(%rbp),%r13d
  movb   $0x0,0x0(%rbp)
  lea    0x2(%rbx),%rdi
  call   <copys@plt>
  mov    %r12,%rdi
  mov    $0x1,%edx
  mov    %eax,%esi
  call   <setvar@plt>
  mov    %r13b,0x0(%rbp)
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int dynmacro(char *line) {
  char *s;
  char endc, *endp;
  if (!isalpha(line[0]))
    return 0;
  for (s = line + 1; *s && (isalpha(*s) | isdigit(*s)); ++s)
    ;
  endp = s;
  while (isspace(*s))
    ++s;
  if (s[0] != ':' || s[1] != '=')
    return 0;
  endc = *endp;
  *endp = '\0';
  setvar(line, copys(s + 2), 1);
  *endp = endc;
  return 1;
}
```