user: 这段汇编代码用C语言表示是什么样的:

```asm
<trim>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    -0x1(%rax),%ebx
  nop
L20:
  movsbl (%r12,%rbp,1),%edi
  mov    %ebp,%r14d
  mov    %ebp,%r13d
  add    $0x1,%rbp
  call   <isspace@plt>
  test   %rax,%rax
  jne    L20
  cmp    %ebx,%r14d
  jg     L101
  movslq %ebx,%rbx
  jmp    L114
  nopw   0x0(%rax,%rax,1)
L0:
  sub    $0x1,%rbx
  cmp    %ebx,%r13d
  jg     L101
L114:
  movsbl (%r12,%rbx,1),%edi
  mov    %ebx,%ebp
  mov    %ebx,%r15d
  call   <isspace@plt>
  test   %rax,%rax
  jne    L0
  cmp    %ebx,%r13d
  jg     L101
  mov    %ebx,%esi
  movslq %r13d,%rdx
  sub    %r14d,%esi
  cmp    $0x6,%esi
  jbe    L72
  lea    0x1(%rsi),%edi
  cmp    $0xe,%esi
  jbe    L162
  mov    %edi,%ecx
  add    %r12,%rdx
  shr    $0x4,%ecx
  shl    $0x4,%rcx
  nopl   0x0(%rax,%rax,1)
L183:
  movdqu (%rdx,%rax,1),%xmm0
  movups %xmm0,(%r12,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rcx
  jne    L183
  mov    %edi,%ecx
  and    $0xfffffff0,%ecx
  lea    (%rcx,%r13,1),%eax
  cmp    %ecx,%edi
  je     L23
  sub    %ecx,%esi
  lea    0x1(%rsi),%edi
  cmp    $0x6,%esi
  jbe    L121
L46:
  mov    (%rdx,%rcx,1),%rdx
  mov    %rdx,(%r12,%rcx,1)
  mov    %edi,%edx
  and    $0xfffffff8,%edx
  add    %edx,%eax
  cmp    %edx,%edi
  je     L23
L121:
  movslq %eax,%rdx
  movzbl (%r12,%rdx,1),%ecx
  mov    %eax,%edx
  sub    %r13d,%edx
  movslq %edx,%rdx
  mov    %cl,(%r12,%rdx,1)
  lea    0x1(%rax),%edx
  cmp    %edx,%r15d
  jl     L23
  movslq %edx,%rcx
  sub    %r13d,%edx
  movzbl (%r12,%rcx,1),%ecx
  movslq %edx,%rdx
  mov    %cl,(%r12,%rdx,1)
  lea    0x2(%rax),%edx
  cmp    %edx,%r15d
  jl     L23
  movslq %edx,%rcx
  sub    %r13d,%edx
  movzbl (%r12,%rcx,1),%ecx
  movslq %edx,%rdx
  mov    %cl,(%r12,%rdx,1)
  lea    0x3(%rax),%edx
  cmp    %edx,%r15d
  jl     L23
  movslq %edx,%rcx
  sub    %r13d,%edx
  movzbl (%r12,%rcx,1),%ecx
  movslq %edx,%rdx
  mov    %cl,(%r12,%rdx,1)
  lea    0x4(%rax),%edx
  cmp    %edx,%r15d
  jl     L23
  movslq %edx,%rcx
  sub    %r13d,%edx
  movzbl (%r12,%rcx,1),%ecx
  movslq %edx,%rdx
  mov    %cl,(%r12,%rdx,1)
  lea    0x5(%rax),%edx
  cmp    %edx,%r15d
  jl     L23
  movslq %edx,%rcx
  sub    %r13d,%edx
  add    $0x6,%eax
  movzbl (%r12,%rcx,1),%ecx
  movslq %edx,%rdx
  mov    %cl,(%r12,%rdx,1)
  cmp    %eax,%r15d
  jl     L23
  movslq %eax,%rdx
  sub    %r13d,%eax
  movzbl (%r12,%rdx,1),%edx
  cltq
  mov    %dl,(%r12,%rax,1)
L23:
  add    $0x1,%ebp
  sub    %r13d,%ebp
  movslq %ebp,%rax
  add    %rax,%r12
L101:
  movb   $0x0,(%r12)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L72:
  mov    %r12,%rax
  lea    0x1(%r12,%rsi,1),%rsi
  nopl   0x0(%rax,%rax,1)
L55:
  movzbl (%rax,%rdx,1),%ecx
  add    $0x1,%rax
  mov    %cl,-0x1(%rax)
  cmp    %rsi,%rax
  jne    L55
  jmp    L23
L162:
  mov    %r13d,%eax
  xor    %ecx,%ecx
  add    %r12,%rdx
  jmp    L46
```

assistant: ```c
void trim(char *str) {
  int i;
  int begin = 0;
  int end = strlen(str) - 1;
  while (isspace(str[begin]))
    begin++;
  while ((end >= begin) && isspace(str[end]))
    end--;
  for (i = begin; i <= end; i++)
    str[i - begin] = str[i];
  str[i - begin] = '\0';
}
```