user: Demonstrate the C code that generates the assembly output below:

```asm
<challenge9>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  lea    D26(%rip),%rdi
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  mov    %r12d,%ebx
  sub    $0x8,%rsp
  call   <puts@plt>
  mov    %r13,%rsi
  lea    D33(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12d,%esi
  lea    D58(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %eax,%ebp
  mov    %r12d,%eax
  cltd
  idiv   %ebp
  test   %edx,%edx
  cmovne %edx,%ebx
  lea    (%rbx,%rbp,1),%edi
  shl    $0x2,%edi
  call   <malloc@plt>
  mov    %ebp,%edx
  mov    %r13,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memcpy@plt>
  test   %ebx,%ebx
  jle    L2
  lea    -0x1(%rbx),%r14d
  cmp    $0x2,%r14d
  jbe    L70
  mov    %ebx,%edx
  movslq %ebp,%r15
  movd   %ebx,%xmm1
  shr    $0x2,%edx
  lea    (%r12,%r15,4),%rax
  pshufd $0x0,%xmm1,%xmm0
  sub    $0x1,%edx
  lea    0x10(%rax),%rcx
  shl    $0x4,%rdx
  add    %rcx,%rdx
  jmp    L45
  nopl   0x0(%rax)
L36:
  add    $0x10,%rcx
L45:
  movups %xmm0,(%rax)
  mov    %rcx,%rax
  cmp    %rdx,%rcx
  jne    L36
  mov    %ebx,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%bl
  je     L59
L34:
  lea    0x0(%rbp,%rax,1),%edx
  movslq %edx,%rdx
  mov    %ebx,(%r12,%rdx,4)
  lea    0x1(%rax),%edx
  cmp    %edx,%ebx
  jle    L59
  add    %ebp,%edx
  add    $0x2,%eax
  movslq %edx,%rdx
  mov    %ebx,(%r12,%rdx,4)
  cmp    %eax,%ebx
  jle    L59
  add    %eax,%ebp
  movslq %ebp,%rbp
  mov    %ebx,(%r12,%rbp,4)
L59:
  mov    %r13,%rsi
  lea    D24(%rip),%rdi
  xor    %eax,%eax
  add    %r15,%r14
  call   <printf@plt>
  lea    (%r12,%r15,4),%rbx
  lea    0x4(%r12,%r14,4),%r13
  lea    D43(%rip),%rbp
L40:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r13,%rbx
  jne    L40
L47:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  mov    %r12,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    10b0 <free@plt>
  nopl   (%rax)
L2:
  mov    %r13,%rsi
  lea    D24(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L47
L70:
  xor    %eax,%eax
  movslq %ebp,%r15
  jmp    L34
```

tool_calls: name: retrieve_data
arguments: data_tag: D26
type_infered: string
name: retrieve_data
arguments: data_tag: D33
type_infered: string
name: retrieve_data
arguments: data_tag: D58
type_infered: string
name: retrieve_data
arguments: data_tag: D24
type_infered: string
name: retrieve_data
arguments: data_tag: D43
type_infered: string

tool: D26: *** Challenge 9: Implement PKCS#7 padding

tool: D33: Input: %s


tool: D58: Block size: %d


tool: D24: Result: %s

tool: D43: \x%02d

assistant: ```c
void challenge9(char *str, int blk_size) {
  int len;
  int padding_len;
  uint8_t *result;
  int i;
  printf("*** Challenge 9: Implement PKCS#7 padding\n");
  printf("Input: %s\n", str);
  printf("Block size: %d\n", blk_size);
  len = strlen(str);
  padding_len = blk_size % len;
  if (padding_len == 0)
    padding_len += blk_size;
  result = malloc((len + padding_len) * sizeof(uint8_t));
  memcpy(result, str, len);
  for (i = 0; i < padding_len; i++)
    result[len + i] = padding_len;
  printf("Result: %s", str);
  for (i = 0; i < padding_len; i++)
    printf("\\x%02d", result[len + i]);
  printf("\n");
  free(result);
}
```