Bit mask with 0's in C - c

Question

I need to build a method in C that will return an int, take 3 ints as parameters. The first and second int are the starting and ending bit position. The third int is a 0 or 1 to determine the type of mask.
For example,
getMask(2, 6, 1);
//Set bits 2 to 6 to 1, set all others to zero
should set the bits 2 through 6 to a 1 and all other bits to zero.
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0
So getMask(2, 6, 1) should return the integer 124.
And getMask(11, 31, 0) (set bits 11 to 31 to 0) should return 2047.
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1
This is what I have currently:
#include <stdio.h>
int getM(int start, int end, int choice) {
if (choice == 1){
return ~(~0 << (end - start + 1)) << (start);
}
else if (choice == 0){
return ~(~1 << (end - start + 1)) << (start);
}
else{
return 0;
}
}
It works when I the choice is 1, but for 0 I am completely lost.
I currently get -2048 for getMask(11, 31, 0).
I know I can use ands and ors, but I cannot figure out how to use them the way I am doing this.

Answer 1

#AnttiHaapala is correct: choice==0 is just the bitwise negation of choice==1 for the same start and end. Therefore (as an MCVE):
#include <stdio.h>
int getM(int start, int end, int choice) {
if (choice == 1){
return ~(~0 << (end - start + 1)) << (start);
}
else if (choice == 0){
return ~getM(start, end, 1); /* Just use what you have, but ~ it */
}
else{
return 0;
}
}
int main() {
printf("2 6 1 %d\n", getM(2,6,1));
printf("11 31 0 %d\n", getM(11,31,0));
}