Could you explain the output of this code. ?

Question

#include <stdio.h>

int main() {
int x=10;

printf("%d \n",x + x++ );
x=10;
printf("%d \n",x + x++ + x++);
x=10;
printf("%d \n",x + x++ + x);
x=10;
printf("%d \n",x + x++ + x+ x);
x=10;
printf("%d \n",x + x++ + x++ +x);
x=10;
printf("%d \n",x + x++ + x++ +x++);
return 0;
}

Output:

21
32
32
43
44
44

Answer 1

printf("%d \n",x + x++ );
// x = 10, x++ = 11 so sum is 21. (10+(10+1) = 21)

x=10;
printf("%d \n",x + x++ + x++);
// x = 10, x++ = 11 and , sum is 32. You may think why sum isn't 33, cause you add x first after you increment its value. (10+(10+1)+11 = 32) + 1 but after printing so it is not calculated while it displays.

x=10;
printf("%d \n",x + x++ + x);
// x = 10, x++ = 11 and x = 11 again, sum is 32. This statement the same with previous statement.(11+(10+1)+11 = 32)

x=10;
printf("%d \n",x + x++ + x+ x);
// x = 10, x++ = 11 now and the final value of x is 11 so sum is 43. (10+(10+1)+11+11 = 43)

x=10;
printf("%d \n",x + x++ + x++ +x);
// x = 10, x++ = 11 now and final value of x is 12. So, sum is 44. (10+(10+1)+(11)+12 = 44)

x=10;
printf("%d \n",x + x++ + x++ +x++);
// x = 10, x++ = 11 and x++= 12. (10+(10+1)+(11+1)+12 = 44) +1 but after prints this statement so it is not calculated.

Answer 2

In that case value are not changing in last two statements but why?
x and x++ are showing the same result.