user: 下面的汇编指令对应的C源代码是什么:

```asm
<breakUpLongWordsIn>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xc48,%rsp
  mov    %rdi,%r12
  mov    %si,0xa(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x1c38(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  lea    0x20(%rsp),%rdi
  mov    $0x382,%ecx
  rep stos %rax,%es:(%rdi)
  movzbl (%r12),%eax
  test   %al,%al
  je     L70
  mov    %esi,%r8d
  mov    %edx,%r14d
  mov    %r12,%rsi
  mov    $0x0,%r13d
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  mov    $0x0,%r15d
  movswl %r8w,%edx
  lea    -0x1(%rdx),%edi
  mov    %edi,0xc(%rsp)
  lea    0x10(%rsp),%rcx
  mov    %rcx,(%rsp)
  jmp    L10
L89:
  movswq %bx,%rdi
  add    (%rsp),%rdi
  mov    $0x4,%edx
  call   <strncpy@plt>
  add    $0x4,%ebp
  add    $0x4,%ebx
  jmp    L18
L139:
  add    $0x1,%ebp
  movswq %bx,%rdx
  mov    %al,0x10(%rsp,%rdx,1)
  add    $0x1,%ebx
  mov    %r15d,%r13d
L18:
  movswq %bp,%rsi
  add    %r12,%rsi
  movzbl (%rsi),%eax
  test   %al,%al
  je     L91
L10:
  cmp    $0x19,%al
  je     L89
  cmp    $0x20,%al
  je     L139
  cmp    $0xa,%al
  je     L139
  test   %r14b,%r14b
  jne    L145
  cmp    %r13w,0xa(%rsp)
  jg     L145
  lea    0x1(%rbx),%edx
  movswq %bx,%rbx
  movb   $0xa,0x10(%rsp,%rbx,1)
  mov    %r15d,%r13d
L35:
  add    $0x1,%ebp
  lea    0x1(%rdx),%ebx
  movswq %dx,%rdx
  mov    %al,0x10(%rsp,%rdx,1)
  add    $0x1,%r13d
  jmp    L18
L145:
  mov    %ebx,%edx
  test   %r14b,%r14b
  je     L35
  movswl %r13w,%ecx
  cmp    0xc(%rsp),%ecx
  jl     L35
  lea    0x1(%rbp),%edx
  movswq %dx,%rcx
  movzbl (%r12,%rcx,1),%ecx
  cmp    $0x19,%cl
  jne    L12
L52:
  add    $0x4,%edx
  movswq %dx,%rcx
  movzbl (%r12,%rcx,1),%ecx
  cmp    $0x19,%cl
  je     L52
L12:
  test   $0xdf,%cl
  je     L151
  cmp    $0xa,%cl
  jne    L184
L151:
  mov    %ebx,%edx
  jmp    L35
L184:
  movswq %bx,%rdx
  movb   $0x2d,0x10(%rsp,%rdx,1)
  lea    0x2(%rbx),%edx
  lea    0x1(%rbx),%ecx
  movswq %cx,%rcx
  movb   $0xa,0x10(%rsp,%rcx,1)
  mov    %r15d,%r13d
  jmp    L35
L70:
  mov    $0x0,%ebx
L91:
  movswq %bx,%rbx
  movb   $0x0,0x10(%rsp,%rbx,1)
  lea    0x10(%rsp),%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    0x1c38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L114
  add    $0x1c48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L114:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void breakUpLongWordsIn(char *sourceText, short width, char useHyphens) {
  char buf[100 * (31 + 5) * 2] = "";
  short i, m, nextChar, wordWidth;
  wordWidth = 0;
  for (i = 0, m = 0; sourceText[i] != 0;) {
    if (sourceText[i] == 25) {
      strncpy(&(buf[m]), &(sourceText[i]), 4);
      i += 4;
      m += 4;
    } else if (sourceText[i] == ' ' || sourceText[i] == '\n') {
      wordWidth = 0;
      buf[m++] = sourceText[i++];
    } else {
      if (!useHyphens && wordWidth >= width) {
        buf[m++] = '\n';
        wordWidth = 0;
      } else if (useHyphens && wordWidth >= width - 1) {
        nextChar = i + 1;
        while (sourceText[nextChar] == 25) {
          nextChar += 4;
        }
        if (sourceText[nextChar] && sourceText[nextChar] != ' ' && sourceText[nextChar] != '\n') {
          buf[m++] = '-';
          buf[m++] = '\n';
          wordWidth = 0;
        }
      }
      buf[m++] = sourceText[i++];
      wordWidth++;
    }
  }
  buf[m] = '\0';
  strcpy(sourceText, buf);
}
```