user: 请从给定的汇编指令推导出C源代码：

```asm
<remove_blanks>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <strlen@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jle    L148
  mov    %r13,%r14
  mov    %r13,%rbp
  xor    %ebx,%ebx
  jmp    L31
  nopl   0x0(%rax)
L138:
  add    $0x1,%ebx
  add    $0x1,%rbp
  cmp    %ebx,%r12d
  je     L127
L31:
  movsbl 0x0(%rbp),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L138
  sub    %ebx,%r12d
  test   %r12d,%r12d
  jle    L0
  lea    -0x1(%r12),%eax
  mov    %r12d,%esi
  movslq %ebx,%rdx
  cmp    $0x6,%eax
  jbe    L35
  cmp    $0xe,%eax
  jbe    L191
  mov    %r12d,%ecx
  add    %r13,%rdx
  xor    %eax,%eax
  shr    $0x4,%ecx
  shl    $0x4,%rcx
  cs nopw 0x0(%rax,%rax,1)
L125:
  movdqu (%rdx,%rax,1),%xmm0
  movups %xmm0,0x0(%r13,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rcx
  jne    L125
  mov    %r12d,%ecx
  and    $0xfffffff0,%ecx
  mov    %ecx,%eax
  cmp    %ecx,%r12d
  je     L0
  mov    %r12d,%esi
  sub    %ecx,%esi
  lea    -0x1(%rsi),%edi
  cmp    $0x6,%edi
  jbe    L154
L98:
  mov    (%rdx,%rcx,1),%rdx
  mov    %rdx,0x0(%r13,%rcx,1)
  mov    %esi,%edx
  and    $0xfffffff8,%edx
  add    %edx,%eax
  cmp    %esi,%edx
  je     L0
L154:
  lea    (%rbx,%rax,1),%edx
  movslq %edx,%rdx
  movzbl 0x0(%r13,%rdx,1),%ecx
  movslq %eax,%rdx
  mov    %cl,0x0(%r13,%rdx,1)
  lea    0x1(%rax),%edx
  cmp    %edx,%r12d
  jle    L0
  lea    (%rbx,%rdx,1),%ecx
  movslq %edx,%rdx
  movslq %ecx,%rcx
  movzbl 0x0(%r13,%rcx,1),%ecx
  mov    %cl,0x0(%r13,%rdx,1)
  lea    0x2(%rax),%edx
  cmp    %edx,%r12d
  jle    L0
  lea    (%rbx,%rdx,1),%ecx
  movslq %edx,%rdx
  movslq %ecx,%rcx
  movzbl 0x0(%r13,%rcx,1),%ecx
  mov    %cl,0x0(%r13,%rdx,1)
  lea    0x3(%rax),%edx
  cmp    %edx,%r12d
  jle    L0
  lea    (%rbx,%rdx,1),%ecx
  movslq %edx,%rdx
  movslq %ecx,%rcx
  movzbl 0x0(%r13,%rcx,1),%ecx
  mov    %cl,0x0(%r13,%rdx,1)
  lea    0x4(%rax),%edx
  cmp    %edx,%r12d
  jle    L0
  lea    (%rbx,%rdx,1),%ecx
  movslq %edx,%rdx
  movslq %ecx,%rcx
  movzbl 0x0(%r13,%rcx,1),%ecx
  mov    %cl,0x0(%r13,%rdx,1)
  lea    0x5(%rax),%edx
  cmp    %edx,%r12d
  jle    L0
  lea    (%rbx,%rdx,1),%ecx
  add    $0x6,%eax
  movslq %edx,%rdx
  movslq %ecx,%rcx
  movzbl 0x0(%r13,%rcx,1),%ecx
  mov    %cl,0x0(%r13,%rdx,1)
  cmp    %eax,%r12d
  jle    L0
  add    %eax,%ebx
  cltq
  movslq %ebx,%rbx
  movzbl 0x0(%r13,%rbx,1),%edx
  mov    %dl,0x0(%r13,%rax,1)
L0:
  movslq %r12d,%r12
  pop    %rbx
  pop    %rbp
  movb   $0x0,0x0(%r13,%r12,1)
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L148:
  jne    L0
L127:
  pop    %rbx
  pop    %rbp
  movb   $0x0,0x0(%r13)
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  xchg   %ax,%ax
L35:
  lea    0x1(%r13,%rax,1),%rcx
  nopl   (%rax)
L25:
  movzbl (%r14,%rdx,1),%eax
  add    $0x1,%r14
  mov    %al,-0x1(%r14)
  cmp    %r14,%rcx
  jne    L25
  jmp    L0
L191:
  xor    %ecx,%ecx
  xor    %eax,%eax
  add    %r13,%rdx
  jmp    L98
```

assistant: ```c
void remove_blanks(char *s1) {
  int i = 0, n = strlen(s1), l;
  int j;
  char ch;
  while (i < n) {
    ch = s1[i];
    if (isspace(ch))
      i++;
    else
      break;
  }
  if (i == n)
    s1[0] = 0;
  else {
    l = n - i;
    for (j = 0; j < l; j++)
      s1[j] = s1[j + i];
    s1[l] = 0;
  }
}
```