ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<ieee_nextafter>:
  endbr64
  movq   %xmm0,%rax
  movd   %xmm0,%r8d
  movq   %xmm0,%rdi
  sar    $0x20,%rdi
  movq   %xmm1,%rdx
  movq   %xmm1,%r9
  sar    $0x20,%r9
  movd   %xmm0,%esi
  and    $0x7fffffff,%esi
  movd   %xmm1,%ecx
  and    $0x7fffffff,%ecx
  cmp    $0x7fefffff,%esi
  jle    L162
  lea    -0x7ff00000(%rsi),%r10d
  or     %edi,%r10d
  jne    L179
L162:
  cmp    $0x7fefffff,%ecx
  jle    L146
  sub    $0x7ff00000,%ecx
  or     %r9d,%ecx
  jne    L179
L146:
  movapd %xmm0,%xmm2
  ucomisd %xmm1,%xmm0
  jp     L214
  je     L131
L214:
  or     %edi,%esi
  jne    L228
  and    $0x80000000,%edx
  bts    $0x20,%rdx
  movq   %rdx,%xmm0
  mov    %rdx,%rax
  movq   %rdx,%xmm2
  mulsd  %xmm0,%xmm2
  ucomisd %xmm2,%xmm0
  jp     L76
  je     L131
L76:
  movq   %rax,%xmm2
  jmp    L131
L179:
  movapd %xmm0,%xmm2
  addsd  %xmm1,%xmm2
L131:
  movapd %xmm2,%xmm0
  ret
L228:
  test   %eax,%eax
  js     L204
  cmp    %edx,%eax
  jg     L167
  jne    L194
  cmp    %r9d,%edi
  jbe    L194
L167:
  sub    $0x1,%eax
  test   %edi,%edi
  cmove  %eax,%r8d
  sub    $0x1,%edi
L174:
  mov    %r8d,%eax
  and    $0x7ff00000,%eax
  cmp    $0x7fefffff,%eax
  jg     L83
  cmp    $0xfffff,%eax
  jg     L130
  movapd %xmm2,%xmm0
  mulsd  %xmm2,%xmm0
  ucomisd %xmm0,%xmm2
  jp     L68
  jne    L68
L130:
  shl    $0x20,%rdi
  mov    %r8d,%eax
  or     %rdi,%rax
  movq   %rax,%xmm2
  jmp    L131
L194:
  add    $0x1,%eax
  add    $0x1,%edi
  cmove  %eax,%r8d
  jmp    L174
L204:
  test   %edx,%edx
  jns    L111
  cmp    %edx,%eax
  jg     L111
  jne    L63
  cmp    %r9d,%edi
  jbe    L63
L111:
  sub    $0x1,%eax
  test   %edi,%edi
  cmove  %eax,%r8d
  sub    $0x1,%edi
  jmp    L174
L63:
  add    $0x1,%eax
  add    $0x1,%edi
  cmove  %eax,%r8d
  jmp    L174
L83:
  addsd  %xmm2,%xmm2
  jmp    L131
L68:
  shl    $0x20,%rdi
  mov    %r8d,%eax
  or     %rdi,%rax
  movq   %rax,%xmm2
  jmp    L131
```

assistant: ```c
double ieee_nextafter(double x, double y) {
  int hx, hy, ix, iy;
  unsigned lx, ly;
  hx = *(int *)&x;
  lx = *(1 + (int *)&x);
  hy = *(int *)&y;
  ly = *(1 + (int *)&y);
  ix = hx & 0x7fffffff;
  iy = hy & 0x7fffffff;
  if (((ix >= 0x7ff00000) && ((ix - 0x7ff00000) | lx) != 0) || ((iy >= 0x7ff00000) && ((iy - 0x7ff00000) | ly) != 0))
    return x + y;
  if (x == y)
    return x;
  if ((ix | lx) == 0) {
    *(int *)&x = hy & 0x80000000;
    *(1 + (int *)&x) = 1;
    y = x * x;
    if (y == x)
      return y;
    else
      return x;
  }
  if (hx >= 0) {
    if (hx > hy || ((hx == hy) && (lx > ly))) {
      if (lx == 0)
        hx -= 1;
      lx -= 1;
    } else {
      lx += 1;
      if (lx == 0)
        hx += 1;
    }
  } else {
    if (hy >= 0 || hx > hy || ((hx == hy) && (lx > ly))) {
      if (lx == 0)
        hx -= 1;
      lx -= 1;
    } else {
      lx += 1;
      if (lx == 0)
        hx += 1;
    }
  }
  hy = hx & 0x7ff00000;
  if (hy >= 0x7ff00000)
    return x + x;
  if (hy < 0x00100000) {
    y = x * x;
    if (y != x) {
      *(int *)&y = hx;
      *(1 + (int *)&y) = lx;
      return y;
    }
  }
  *(int *)&x = hx;
  *(1 + (int *)&x) = lx;
  return x;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ieee_nextafter>:
  endbr64
  movd   %xmm0,%esi
  movq   %xmm0,%rdi
  movq   %xmm1,%r9
  movd   %xmm1,%ecx
  and    $0x7fffffff,%esi
  movq   %xmm0,%rax
  sar    $0x20,%rdi
  movd   %xmm0,%r8d
  movq   %xmm1,%rdx
  sar    $0x20,%r9
  and    $0x7fffffff,%ecx
  cmp    $0x7fefffff,%esi
  jle    L83
  lea    -0x7ff00000(%rsi),%r10d
  or     %edi,%r10d
  jne    L151
L83:
  cmp    $0x7fefffff,%ecx
  jg     L178
L211:
  ucomisd %xmm1,%xmm0
  movapd %xmm0,%xmm2
  jp     L131
  je     L36
L131:
  or     %edi,%esi
  jne    L129
  and    $0x80000000,%edx
  bts    $0x20,%rdx
  movq   %rdx,%xmm0
  movq   %rdx,%xmm2
  mulsd  %xmm0,%xmm2
  ucomisd %xmm2,%xmm0
  jp     L115
  jne    L115
L36:
  movapd %xmm2,%xmm0
  ret
  nopl   0x0(%rax,%rax,1)
L178:
  sub    $0x7ff00000,%ecx
  or     %r9d,%ecx
  je     L211
L151:
  movapd %xmm0,%xmm2
  addsd  %xmm1,%xmm2
  movapd %xmm2,%xmm0
  ret
  nopl   0x0(%rax,%rax,1)
L129:
  test   %eax,%eax
  js     L100
  cmp    %edx,%eax
  jg     L114
  jne    L123
L145:
  cmp    %r9d,%edi
  jbe    L123
L114:
  sub    $0x1,%eax
  test   %edi,%edi
  cmove  %eax,%r8d
  sub    $0x1,%edi
L49:
  mov    %r8d,%eax
  and    $0x7ff00000,%eax
  cmp    $0x7ff00000,%eax
  je     L135
  cmp    $0xfffff,%eax
  jg     L27
  movapd %xmm0,%xmm1
  mulsd  %xmm0,%xmm1
  ucomisd %xmm1,%xmm0
  jp     L55
  jne    L55
L27:
  shl    $0x20,%rdi
  mov    %r8d,%eax
  or     %rdi,%rax
  movq   %rax,%xmm0
L115:
  movapd %xmm0,%xmm2
  movapd %xmm2,%xmm0
  ret
  nopl   0x0(%rax)
L100:
  test   %edx,%edx
  jns    L114
  cmp    %edx,%eax
  jg     L114
  je     L145
L123:
  add    $0x1,%eax
  add    $0x1,%edi
  cmove  %eax,%r8d
  jmp    L49
  cs nopw 0x0(%rax,%rax,1)
L135:
  movapd %xmm0,%xmm2
  addsd  %xmm0,%xmm2
  jmp    L36
  nopl   (%rax)
L55:
  shl    $0x20,%rdi
  mov    %r8d,%eax
  or     %rdi,%rax
  movq   %rax,%xmm2
  jmp    L36
```

assistant: ```c
double ieee_nextafter(double x, double y) {
  int hx, hy, ix, iy;
  unsigned lx, ly;
  hx = *(int *)&x;
  lx = *(1 + (int *)&x);
  hy = *(int *)&y;
  ly = *(1 + (int *)&y);
  ix = hx & 0x7fffffff;
  iy = hy & 0x7fffffff;
  if (((ix >= 0x7ff00000) && ((ix - 0x7ff00000) | lx) != 0) || ((iy >= 0x7ff00000) && ((iy - 0x7ff00000) | ly) != 0))
    return x + y;
  if (x == y)
    return x;
  if ((ix | lx) == 0) {
    *(int *)&x = hy & 0x80000000;
    *(1 + (int *)&x) = 1;
    y = x * x;
    if (y == x)
      return y;
    else
      return x;
  }
  if (hx >= 0) {
    if (hx > hy || ((hx == hy) && (lx > ly))) {
      if (lx == 0)
        hx -= 1;
      lx -= 1;
    } else {
      lx += 1;
      if (lx == 0)
        hx += 1;
    }
  } else {
    if (hy >= 0 || hx > hy || ((hx == hy) && (lx > ly))) {
      if (lx == 0)
        hx -= 1;
      lx -= 1;
    } else {
      lx += 1;
      if (lx == 0)
        hx += 1;
    }
  }
  hy = hx & 0x7ff00000;
  if (hy >= 0x7ff00000)
    return x + x;
  if (hy < 0x00100000) {
    y = x * x;
    if (y != x) {
      *(int *)&y = hx;
      *(1 + (int *)&y) = lx;
      return y;
    }
  }
  *(int *)&x = hx;
  *(1 + (int *)&x) = lx;
  return x;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ieee_nextafter>:
  endbr64
  movd   %xmm0,%esi
  movq   %xmm0,%r8
  movq   %xmm1,%r9
  movd   %xmm1,%ecx
  and    $0x7fffffff,%esi
  movd   %xmm0,%edi
  sar    $0x20,%r8
  movq   %xmm0,%rax
  movq   %xmm1,%rdx
  sar    $0x20,%r9
  and    $0x7fffffff,%ecx
  cmp    $0x7fefffff,%esi
  jle    L122
  lea    -0x7ff00000(%rsi),%r10d
  or     %r8d,%r10d
  jne    L189
L122:
  cmp    $0x7fefffff,%ecx
  jg     L193
L165:
  ucomisd %xmm1,%xmm0
  movapd %xmm0,%xmm2
  jp     L50
  je     L8
L50:
  or     %r8d,%esi
  jne    L66
  and    $0x80000000,%edx
  bts    $0x20,%rdx
  movq   %rdx,%xmm0
  movq   %rdx,%xmm2
  mulsd  %xmm0,%xmm2
  ucomisd %xmm2,%xmm0
  jp     L26
  jne    L26
L8:
  movapd %xmm2,%xmm0
  ret
  nopl   0x0(%rax,%rax,1)
L193:
  sub    $0x7ff00000,%ecx
  or     %r9d,%ecx
  je     L165
L189:
  movapd %xmm0,%xmm2
  addsd  %xmm1,%xmm2
  movapd %xmm2,%xmm0
  ret
L66:
  test   %eax,%eax
  js     L39
  cmp    %edx,%eax
  jg     L153
  jne    L7
L148:
  cmp    %r9d,%r8d
  jbe    L7
L153:
  sub    $0x1,%eax
  test   %r8d,%r8d
  cmove  %eax,%edi
  sub    $0x1,%r8d
L204:
  mov    %edi,%eax
  and    $0x7ff00000,%eax
  cmp    $0x7ff00000,%eax
  je     L31
  movd   %edi,%xmm2
  movd   %r8d,%xmm3
  punpckldq %xmm3,%xmm2
  cmp    $0xfffff,%eax
  jg     L8
  movapd %xmm0,%xmm1
  mulsd  %xmm0,%xmm1
  ucomisd %xmm1,%xmm0
  jp     L164
  je     L8
L164:
  shl    $0x20,%r8
  mov    %edi,%eax
  or     %r8,%rax
  movq   %rax,%xmm2
  jmp    L8
  xchg   %ax,%ax
L26:
  movapd %xmm0,%xmm2
  movapd %xmm2,%xmm0
  ret
  nopl   0x0(%rax)
L39:
  test   %edx,%edx
  jns    L153
  cmp    %edx,%eax
  jg     L153
  je     L148
L7:
  add    $0x1,%eax
  add    $0x1,%r8d
  cmove  %eax,%edi
  jmp    L204
  cs nopw 0x0(%rax,%rax,1)
L31:
  movapd %xmm0,%xmm2
  addsd  %xmm0,%xmm2
  jmp    L8
```

assistant: ```c
double ieee_nextafter(double x, double y) {
  int hx, hy, ix, iy;
  unsigned lx, ly;
  hx = *(int *)&x;
  lx = *(1 + (int *)&x);
  hy = *(int *)&y;
  ly = *(1 + (int *)&y);
  ix = hx & 0x7fffffff;
  iy = hy & 0x7fffffff;
  if (((ix >= 0x7ff00000) && ((ix - 0x7ff00000) | lx) != 0) || ((iy >= 0x7ff00000) && ((iy - 0x7ff00000) | ly) != 0))
    return x + y;
  if (x == y)
    return x;
  if ((ix | lx) == 0) {
    *(int *)&x = hy & 0x80000000;
    *(1 + (int *)&x) = 1;
    y = x * x;
    if (y == x)
      return y;
    else
      return x;
  }
  if (hx >= 0) {
    if (hx > hy || ((hx == hy) && (lx > ly))) {
      if (lx == 0)
        hx -= 1;
      lx -= 1;
    } else {
      lx += 1;
      if (lx == 0)
        hx += 1;
    }
  } else {
    if (hy >= 0 || hx > hy || ((hx == hy) && (lx > ly))) {
      if (lx == 0)
        hx -= 1;
      lx -= 1;
    } else {
      lx += 1;
      if (lx == 0)
        hx += 1;
    }
  }
  hy = hx & 0x7ff00000;
  if (hy >= 0x7ff00000)
    return x + x;
  if (hy < 0x00100000) {
    y = x * x;
    if (y != x) {
      *(int *)&y = hx;
      *(1 + (int *)&y) = lx;
      return y;
    }
  }
  *(int *)&x = hx;
  *(1 + (int *)&x) = lx;
  return x;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<snis_format_hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %ecx,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x3736353433323130,%rax
  movabs $0x6665646362613938,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  movb   $0x0,-0x10(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L25
L7:
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  shr    $0x4,%al
  movzbl %al,%eax
  cltq
  movzbl -0x20(%rbp,%rax,1),%edx
  mov    -0x24(%rbp),%eax
  add    %eax,%eax
  movslq %eax,%rcx
  mov    -0x48(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  and    $0xf,%eax
  cltq
  movzbl -0x20(%rbp,%rax,1),%edx
  mov    -0x24(%rbp),%eax
  add    %eax,%eax
  cltq
  lea    0x1(%rax),%rcx
  mov    -0x48(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  addl   $0x1,-0x24(%rbp)
L25:
  mov    -0x24(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jge    L24
  mov    -0x24(%rbp),%eax
  add    %eax,%eax
  cmp    %eax,-0x40(%rbp)
  jg     L7
L24:
  mov    -0x24(%rbp),%eax
  add    %eax,%eax
  movslq %eax,%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L39
  call   <__stack_chk_fail@plt>
L39:
  leave
  ret
```

assistant: ```c
void snis_format_hash(unsigned char *hash, int hashlen, unsigned char *buffer, int len) {
  int i;
  const char hexdigit[] = "0123456789abcdef";
  for (i = 0; i < hashlen && i * 2 < len; i++) {
    buffer[i * 2] = hexdigit[((hash[i] & 0xf0) >> 4)];
    buffer[i * 2 + 1] = hexdigit[(hash[i] & 0x0f)];
  }
  buffer[i * 2] = '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<snis_format_hash>:
  endbr64
  sub    $0x28,%rsp
  mov    %rdi,%r8
  mov    %rdx,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movabs $0x3736353433323130,%rax
  movabs $0x6665646362613938,%rdx
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  movb   $0x0,0x10(%rsp)
  test   %esi,%esi
  jle    L67
  test   %ecx,%ecx
  jle    L35
  mov    %esi,%esi
  sub    $0x1,%ecx
  shr    %ecx
  lea    0x2(%rcx),%ecx
  mov    $0x1,%eax
L38:
  movzbl -0x1(%r8,%rax,1),%edx
  shr    $0x4,%dl
  and    $0xf,%edx
  movzbl (%rsp,%rdx,1),%edx
  mov    %dl,-0x2(%rdi,%rax,2)
  movzbl -0x1(%r8,%rax,1),%edx
  and    $0xf,%edx
  movzbl (%rsp,%rdx,1),%edx
  mov    %dl,-0x1(%rdi,%rax,2)
  mov    %eax,%edx
  cmp    %rsi,%rax
  je     L63
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L38
L63:
  add    %edx,%edx
  movslq %edx,%rdx
  movb   $0x0,(%rdi,%rdx,1)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L46
  add    $0x28,%rsp
  ret
L67:
  mov    $0x0,%edx
  jmp    L63
L35:
  mov    $0x0,%edx
  jmp    L63
L46:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void snis_format_hash(unsigned char *hash, int hashlen, unsigned char *buffer, int len) {
  int i;
  const char hexdigit[] = "0123456789abcdef";
  for (i = 0; i < hashlen && i * 2 < len; i++) {
    buffer[i * 2] = hexdigit[((hash[i] & 0xf0) >> 4)];
    buffer[i * 2 + 1] = hexdigit[(hash[i] & 0x0f)];
  }
  buffer[i * 2] = '\0';
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<snis_format_hash>:
  endbr64
  sub    $0x28,%rsp
  movdqa D6(%rip),%xmm0
  mov    %rdi,%r8
  mov    %rdx,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movb   $0x0,0x10(%rsp)
  movaps %xmm0,(%rsp)
  test   %esi,%esi
  jle    L14
  test   %ecx,%ecx
  jle    L14
  sub    $0x1,%ecx
  movslq %esi,%rsi
  mov    $0x1,%eax
  shr    %ecx
  add    $0x2,%ecx
  jmp    L32
  nopl   0x0(%rax)
L15:
  add    $0x1,%rax
  cmp    %rax,%rcx
  je     L5
L32:
  movzbl -0x1(%r8,%rax,1),%edx
  shr    $0x4,%dl
  and    $0xf,%edx
  movzbl (%rsp,%rdx,1),%edx
  mov    %dl,-0x2(%rdi,%rax,2)
  movzbl -0x1(%r8,%rax,1),%edx
  and    $0xf,%edx
  movzbl (%rsp,%rdx,1),%edx
  mov    %dl,-0x1(%rdi,%rax,2)
  lea    (%rax,%rax,1),%edx
  cmp    %rax,%rsi
  jne    L15
L5:
  movslq %edx,%rdx
  add    %rdx,%rdi
L14:
  movb   $0x0,(%rdi)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  add    $0x28,%rsp
  ret
L36:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D6
type_guessed: byte[16]

tool: {"D6": [0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x61,0x62,0x63,0x64,0x65,0x66]}

assistant: ```c
void snis_format_hash(unsigned char *hash, int hashlen, unsigned char *buffer, int len) {
  int i;
  const char hexdigit[] = "0123456789abcdef";
  for (i = 0; i < hashlen && i * 2 < len; i++) {
    buffer[i * 2] = hexdigit[((hash[i] & 0xf0) >> 4)];
    buffer[i * 2 + 1] = hexdigit[(hash[i] & 0x0f)];
  }
  buffer[i * 2] = '\0';
}
```
user: Translate the below assembly code into C programming language:

```asm
<snis_format_hash>:
  endbr64
  sub    $0x28,%rsp
  movdqa D9(%rip),%xmm0
  mov    %rdi,%r8
  mov    %rdx,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movb   $0x0,0x10(%rsp)
  movaps %xmm0,(%rsp)
  test   %esi,%esi
  jle    L44
  test   %ecx,%ecx
  jle    L44
  sub    $0x1,%ecx
  movslq %esi,%rsi
  mov    $0x1,%eax
  shr    %ecx
  add    $0x2,%ecx
  jmp    L50
  nopl   0x0(%rax)
L25:
  add    $0x1,%rax
  cmp    %rax,%rcx
  je     L29
L50:
  movzbl -0x1(%r8,%rax,1),%edx
  shr    $0x4,%dl
  and    $0xf,%edx
  movzbl (%rsp,%rdx,1),%edx
  mov    %dl,-0x2(%rdi,%rax,2)
  movzbl -0x1(%r8,%rax,1),%edx
  and    $0xf,%edx
  movzbl (%rsp,%rdx,1),%edx
  mov    %dl,-0x1(%rdi,%rax,2)
  lea    (%rax,%rax,1),%edx
  cmp    %rax,%rsi
  jne    L25
L29:
  movslq %edx,%rdx
  add    %rdx,%rdi
L44:
  movb   $0x0,(%rdi)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  add    $0x28,%rsp
  ret
L38:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D9
type: byte[16]

tool: {"D9": [0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x61,0x62,0x63,0x64,0x65,0x66]}

assistant: ```c
void snis_format_hash(unsigned char *hash, int hashlen, unsigned char *buffer, int len) {
  int i;
  const char hexdigit[] = "0123456789abcdef";
  for (i = 0; i < hashlen && i * 2 < len; i++) {
    buffer[i * 2] = hexdigit[((hash[i] & 0xf0) >> 4)];
    buffer[i * 2 + 1] = hexdigit[(hash[i] & 0x0f)];
  }
  buffer[i * 2] = '\0';
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<VarNameEnd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rbx
  movzbl (%rbx),%eax
  cmp    $0x7b,%al
  jne    L78
  add    $0x1,%rbx
  jmp    L50
L79:
  add    $0x1,%rbx
L50:
  movzbl (%rbx),%eax
  cmp    $0x7d,%al
  je     L17
  cmp    -0x20(%rbp),%rbx
  jne    L79
L17:
  mov    %rbx,%rax
  jmp    L95
L8:
  add    $0x1,%rbx
L78:
  movzbl (%rbx),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <isalnum@plt>
  test   %eax,%eax
  jne    L8
  movzbl (%rbx),%eax
  cmp    $0x5f,%al
  je     L8
  movzbl (%rbx),%eax
  cmp    $0x28,%al
  jne    L28
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  cmp    %rax,%rbx
  je     L28
  lea    0x1(%rbx),%rcx
  mov    -0x20(%rbp),%rax
  mov    $0x29,%edx
  mov    %rax,%rsi
  mov    %rcx,%rdi
  mov    $0x0,%eax
  call   <QuoteEnd@plt>
  cltq
  jmp    L95
L28:
  lea    -0x1(%rbx),%rax
L95:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *VarNameEnd(string, lastChar)
char *string;
char *lastChar;
{
  register char *p = string + 1;
  if (*p == '{') {
    for (p++; (*p != '}') && (p != lastChar); p++) {
    }
    return p;
  }
  while (isalnum(((unsigned char)(*p))) || (*p == '_')) {
    p++;
  }
  if ((*p == '(') && (p != string + 1)) {
    return QuoteEnd(p + 1, lastChar, ')');
  }
  return p - 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<VarNameEnd>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%r12
  lea    0x1(%rdi),%r13
  mov    %r13,%rbx
  cmpb   $0x7b,0x1(%rdi)
  jne    L48
  lea    0x2(%rdi),%rax
  cmpb   $0x7d,0x2(%rdi)
  je     L1
  cmp    %rax,%rsi
  je     L1
L65:
  add    $0x1,%rax
  cmpb   $0x7d,(%rax)
  je     L1
  cmp    %rax,%r12
  jne    L65
  jmp    L1
L75:
  add    $0x1,%rbx
L48:
  movzbl (%rbx),%ebp
  movzbl %bpl,%edi
  call   <isalnum@plt>
  cmp    $0x5f,%bpl
  je     L75
  test   %eax,%eax
  jne    L75
  cmp    %r13,%rbx
  je     L82
  cmp    $0x28,%bpl
  je     L35
L82:
  lea    -0x1(%rbx),%rax
L1:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L35:
  lea    0x1(%rbx),%rdi
  mov    $0x29,%edx
  mov    %r12,%rsi
  call   <QuoteEnd@plt>
  cltq
  jmp    L1
```

assistant: ```c
char *VarNameEnd(string, lastChar)
char *string;
char *lastChar;
{
  register char *p = string + 1;
  if (*p == '{') {
    for (p++; (*p != '}') && (p != lastChar); p++) {
    }
    return p;
  }
  while (isalnum(((unsigned char)(*p))) || (*p == '_')) {
    p++;
  }
  if ((*p == '(') && (p != string + 1)) {
    return QuoteEnd(p + 1, lastChar, ')');
  }
  return p - 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<VarNameEnd>:
  endbr64
  push   %r13
  lea    0x1(%rdi),%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %r13,%rbx
  sub    $0x8,%rsp
  movzbl 0x1(%rdi),%ebp
  cmp    $0x7b,%bpl
  jne    L39
  jmp    L97
  nopl   0x0(%rax)
L90:
  movzbl 0x1(%rbx),%ebp
  add    $0x1,%rbx
L39:
  movzbl %bpl,%edi
  call   <isalnum@plt>
  test   %eax,%eax
  jne    L90
  cmp    $0x5f,%bpl
  je     L90
  cmp    $0x28,%bpl
  jne    L106
  cmp    %r13,%rbx
  jne    L31
L106:
  lea    -0x1(%rbx),%rax
L32:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L97:
  cmpb   $0x7d,0x2(%rdi)
  lea    0x2(%rdi),%rax
  je     L32
  cmp    %rax,%r12
  je     L32
  nop
L82:
  add    $0x1,%rax
  cmpb   $0x7d,(%rax)
  je     L32
  cmp    %rax,%r12
  jne    L82
  jmp    L32
L31:
  lea    0x1(%rbx),%rdi
  mov    $0x29,%edx
  mov    %r12,%rsi
  call   <QuoteEnd@plt>
  cltq
  jmp    L32
```

assistant: ```c
char *VarNameEnd(string, lastChar)
char *string;
char *lastChar;
{
  register char *p = string + 1;
  if (*p == '{') {
    for (p++; (*p != '}') && (p != lastChar); p++) {
    }
    return p;
  }
  while (isalnum(((unsigned char)(*p))) || (*p == '_')) {
    p++;
  }
  if ((*p == '(') && (p != string + 1)) {
    return QuoteEnd(p + 1, lastChar, ')');
  }
  return p - 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<VarNameEnd>:
  endbr64
  push   %r13
  lea    0x1(%rdi),%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %r13,%rbx
  sub    $0x8,%rsp
  cmpb   $0x7b,0x1(%rdi)
  jne    L83
  jmp    L102
L39:
  add    $0x1,%rbx
L83:
  movzbl (%rbx),%edi
  mov    %edi,%ebp
  call   <isalnum@plt>
  test   %eax,%eax
  jne    L39
  cmp    $0x5f,%bpl
  je     L39
  cmp    $0x28,%bpl
  jne    L76
  cmp    %r13,%rbx
  jne    L42
L76:
  lea    -0x1(%rbx),%rax
L34:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L102:
  cmpb   $0x7d,0x2(%rdi)
  lea    0x2(%rdi),%rax
  je     L34
  cmp    %rax,%r12
  je     L34
  nopw   0x0(%rax,%rax,1)
L38:
  add    $0x1,%rax
  cmpb   $0x7d,(%rax)
  je     L34
  cmp    %rax,%r12
  jne    L38
  jmp    L34
L42:
  lea    0x1(%rbx),%rdi
  mov    $0x29,%edx
  mov    %r12,%rsi
  call   <QuoteEnd@plt>
  cltq
  jmp    L34
```

assistant: ```c
char *VarNameEnd(string, lastChar)
char *string;
char *lastChar;
{
  register char *p = string + 1;
  if (*p == '{') {
    for (p++; (*p != '}') && (p != lastChar); p++) {
    }
    return p;
  }
  while (isalnum(((unsigned char)(*p))) || (*p == '_')) {
    p++;
  }
  if ((*p == '(') && (p != string + 1)) {
    return QuoteEnd(p + 1, lastChar, ')');
  }
  return p - 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Section2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <Section2Fourth@plt>
  mov    %eax,%r13d
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <Section2Third@plt>
  mov    %eax,%r12d
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <Section2Second@plt>
  mov    %eax,%ebx
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <Section2First@plt>
  mov    %r13d,%r8d
  mov    %r12d,%ecx
  mov    %ebx,%edx
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D6
type_infered: string

tool: D6: 
2.- | %d | %d | %d | %d |

assistant: ```c
int Section2(int number) {
  printf("\n2.- | %d | %d | %d | %d |", Section2First(number), Section2Second(number), Section2Third(number), Section2Fourth(number));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Section2>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  call   <Section2Fourth@plt>
  mov    %eax,%r13d
  mov    %ebx,%edi
  call   <Section2Third@plt>
  mov    %eax,%r12d
  mov    %ebx,%edi
  call   <Section2Second@plt>
  mov    %eax,%ebp
  mov    %ebx,%edi
  call   <Section2First@plt>
  mov    %eax,%esi
  mov    %r13d,%r8d
  mov    %r12d,%ecx
  mov    %ebp,%edx
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
data_type: string

tool: D6: 
2.- | %d | %d | %d | %d |

assistant: ```c
int Section2(int number) {
  printf("\n2.- | %d | %d | %d | %d |", Section2First(number), Section2Second(number), Section2Third(number), Section2Fourth(number));
  return 0;
}
```