C Programming: Explanation of Code with Increment and Decrement Operators

#include <stdio.h> main() { int a=3,b=2,c=1; c-=++b; b*=a+c; { int b=5,c=12; c/=b*2; a-=c; printf("%d,%d,%d,", a,b,c); a+=--c; printf("%d,%d,%d\n", a, b, c); } return 0; } The code provided is incomplete and contains syntax errors. Here is the corrected version: c #include <stdio.h> int main() { int a = 3, b = 2, c = 1; c -= ++b; b *= a + c; { int b = 5, c = 12; c /= b * 2; a -= c; printf("%d,%d,%d,", a, b, c); a += --c; printf("%d,%d,%d\n", a, b, c); } return 0; } Explanation: 1. c -= ++b; - ++b increments the value of b by 1 and then assigns it to c. So, c becomes 3 and b becomes 3. 2. b *= a + c; - a + c evaluates to 6. The value of b is multiplied by 6, so b becomes 18. 3. Inside the inner block, int b = 5, c = 12; declares and initializes new variables b and c with values 5 and 12 respectively. 4. c /= b * 2; - b * 2 evaluates to 10. The value of c is divided by 10, so c becomes 1. 5. a -= c; - c is subtracted from a, so a becomes 2. 6. printf("%d,%d,%d,", a, b, c); prints the values of a, b, and c which are 2, 18, and 1 respectively. 7. a += --c; - --c decrements the value of c by 1 and then assigns it to a. So, c becomes 0 and a becomes 2. 8. printf("%d,%d,%d\n", a, b, c); prints the updated values of a, b, and c which are 2, 18, and 0 respectively. Output: 2,18,1,2,18,0