Is there a way to do a parity check using bit shifting, and without using xor of a string with bits?

#include <stdio.h>
#include <string.h>

int main(int argc, char *argv[]) {
    if (argc == 2) {
        int iParity = 0;
        int bitmask = 0;
        for (int i = 1; i < strlen(argv[1]); i++) {
            switch (argv[1][i]) {
              case '0':
                if (iParity == 0)
                    iParity = 0;
                else
                    iParity = 1;
                break;
              case '1':
                if (iParity == 0)
                    iParity = 1;
                else
                    iParity = 0;
                break;
              default:
                break;
            }
        }
        printf("The parity is: %d", iParity);
    }
}

Basically I put the input directly into the execute line, like ./check 10010, check is the name of the program, and afterwards I need to put binary number, and I need to parity check the number using bit shifting ( << or >> ) and I SHOULD NOT use "xor" operator, is there a way to do that without really long code?

#include <stdio.h>
#include <string.h>

int main(int argc, char *argv[]) {
      if(argc == 2){
          int iParity = 0;
          char c;
          int iSel = 0;
          for (int i = 0; i < strlen(argv[1]); i++){
              c = argv[1][i] - '0';
              iSel = (iParity<<1) + c;
              switch(iSel){
                case 1:
                case 2:
                    iParity = 1;
                    break;
                case 0:
                case 3:
                    iParity = 0;
                    break;
                default:
                  break;
              }
          }
        printf("Parity is: %d\n", iParity);
      }
  return 0;
}

Basically this is the solution my teacher told us in the out lesson, and yes, you had to convert a string index into a integer, and create a bit mask that works similar to a XOR operator, and that's pretty much it.

Addendum (by a commenter, scs):

The way the expression iSel = (iParity<<1) + c and the following switch statement work is that they implement the truth table for the XOR operator, like this:

In the end, the effect is the same as the much simpler

iParity = iParity ^ c;