user: Translate the below assembly code into C programming language:

```asm
<replace>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r13
  call   <countStrStr@plt>
  mov    %eax,0x4(%rsp)
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%r15d
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%r12d
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %eax,0x8(%rsp)
  mov    %rbp,%r14
  cmp    %eax,%r12d
  jne    L98
L145:
  mov    $0x0,%r8d
  mov    $0x0,%edi
  mov    %r15d,%r9d
  sub    %r12d,%r9d
  js     L20
  mov    %r12d,%esi
  lea    -0x1(%r12),%r11d
  mov    0x8(%rsp),%eax
  lea    -0x1(%rax),%ecx
  mov    %ecx,0xc(%rsp)
  mov    %eax,%r10d
  jmp    L57
L98:
  mov    %eax,%esi
  sub    %r12d,%esi
  mov    %esi,%eax
  imul   0x4(%rsp),%eax
  add    %r15d,%eax
  mov    %eax,0x4(%rsp)
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L115
  mov    0x4(%rsp),%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  jmp    L145
L20:
  mov    $0x0,%r8d
L82:
  lea    0x1(%r9),%eax
  cmp    %eax,%r15d
  jle    L29
  cltq
  movslq %r8d,%rcx
  movslq %r9d,%rdx
  sub    %rdx,%rcx
  add    %r14,%rcx
L78:
  movzbl 0x0(%rbp,%rax,1),%edx
  mov    %dl,-0x1(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%r15d
  jg     L78
  lea    -0x1(%r15),%eax
  sub    %r9d,%eax
  add    %eax,%r8d
L29:
  movslq %r8d,%r8
  movb   $0x0,(%r14,%r8,1)
L115:
  mov    %r14,%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L26:
  mov    0x4(%rsp),%edi
L77:
  cmp    %edx,%r12d
  je     L47
  movslq %edi,%rax
  movzbl 0x0(%rbp,%rax,1),%edx
  movslq %r8d,%rax
  mov    %dl,(%r14,%rax,1)
L90:
  add    $0x1,%edi
  add    $0x1,%r8d
  cmp    %r9d,%edi
  jg     L82
L57:
  mov    $0x0,%eax
  movslq %edi,%rcx
  add    %rbp,%rcx
  mov    $0x0,%edx
  test   %r12d,%r12d
  jle    L77
  mov    %edi,0x4(%rsp)
L141:
  mov    %eax,%edx
  movzbl (%rbx,%rax,1),%edi
  cmp    %dil,(%rcx,%rax,1)
  jne    L26
  lea    0x1(%rax),%edx
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L141
  mov    0x4(%rsp),%edi
  jmp    L77
L47:
  cmpl   $0x0,0x8(%rsp)
  jle    L58
  mov    $0x0,%eax
  movslq %r8d,%rcx
  add    %r14,%rcx
L122:
  movzbl 0x0(%r13,%rax,1),%edx
  mov    %dl,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %r10,%rax
  jne    L122
L58:
  add    %r11d,%edi
  add    0xc(%rsp),%r8d
  jmp    L90
```

assistant: ```c
char *replace(char *line, char *pattern, char *replace) {
  int i, j, k, llen, plen, rlen, count;
  char *newline;
  count = countStrStr(line, pattern);
  llen = strlen(line);
  plen = strlen(pattern);
  rlen = strlen(replace);
  if (rlen != plen) {
    newline = malloc(llen + count * (rlen - plen));
    if (!newline) {
      return NULL;
    }
    memset(newline, 0, llen + count * (rlen - plen));
  } else {
    newline = line;
  }
  for (i = 0, k = 0; i <= llen - plen; i++, k++) {
    for (j = 0; j < plen; j++) {
      if (line[i + j] != pattern[j]) {
        break;
      }
    }
    if (j == plen) {
      for (j = 0; j < rlen; j++) {
        newline[k + j] = replace[j];
      }
      i += plen - 1;
      k += rlen - 1;
      count++;
    } else {
      newline[k] = line[i];
    }
  }
  for (i = llen - plen + 1; i < llen; i++, k++) {
    newline[k] = line[i];
  }
  newline[k] = '\0';
  return newline;
}
```