user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r15
  lea    D60(%rip),%rdi
  lea    D81(%rip),%r15
  push   %r14
  push   %r13
  lea    D64(%rip),%r13
  push   %r12
  lea    D78(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D66(%rip),%rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x1c(%rsp),%r14
  lea    0x20(%rsp),%rbp
  call   <printf@plt>
  lea    D55(%rip),%rdi
  call   <puts@plt>
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x24(%rsp),%rax
  mov    %rax,0x8(%rsp)
  nopl   0x0(%rax,%rax,1)
L17:
  mov    %r15,%rdi
  call   <puts@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %r12,%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x20(%rsp),%esi
  mov    0x1c(%rsp),%edi
  call   <fnRightRot@plt>
  mov    0x20(%rsp),%edx
  mov    0x1c(%rsp),%esi
  lea    D14(%rip),%rdi
  mov    %eax,%ecx
  xor    %eax,%eax
  call   <printf@plt>
  lea    D48(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%rsi
  xor    %eax,%eax
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmpl   $0x1,0x24(%rsp)
  je     L17
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D60
data_type: string
name: unpack_data
arguments: data_tag: D81
data_type: string
name: unpack_data
arguments: data_tag: D64
data_type: string
name: unpack_data
arguments: data_tag: D78
data_type: string
name: unpack_data
arguments: data_tag: D66
data_type: string
name: unpack_data
arguments: data_tag: D55
data_type: string
name: unpack_data
arguments: data_tag: D38
data_type: string
name: unpack_data
arguments: data_tag: D14
data_type: string
name: unpack_data
arguments: data_tag: D48
data_type: string

tool: D60: 
*********************************************************

tool: D81: 
Enter the value to be rotated

tool: D64: %u

tool: D78: 
Enter the number of positions by which the value has to be rotated

tool: D66: %d

tool: D55: 
*	PROGRAM TO IMPLEMENT CIRCULAR RIGHT SHIFT	*

tool: D38: *********************************************************

tool: D14: 
The value %u after right rotation by %d bits = %u


tool: D48: 
Press 1 to continue or any other key to exit....

assistant: ```c
int main(void) {
  unsigned int iVal, iNewVal;
  int iNum, iChoice;
  printf("\n*********************************************************");
  printf("\n*\tPROGRAM TO IMPLEMENT CIRCULAR RIGHT SHIFT\t*\n");
  printf("*********************************************************");
  do {
    printf("\nEnter the value to be rotated\n");
    scanf("%u", &iVal);
    printf("\nEnter the number of positions by which the value has to be rotated\n");
    scanf("%d", &iNum);
    iNewVal = fnRightRot(iVal, iNum);
    printf("\nThe value %u after right rotation by %d bits = %u\n", iVal, iNum, iNewVal);
    printf("\nPress 1 to continue or any other key to exit....\n");
    scanf("%d", &iChoice);
  } while (1 == iChoice);
  return 0;
}
```