Question

#include<stdio.h>
#define FNC(x, y) ((x) > (y)) ? (* * y) : (y - x);
int main() {
int p = 7, 9 = 6, r = 4;
r += FNC (++p, q++);
printf("%d%d %d", p, q, r);
return 0;
0}​

Answer 1

Answer:

67

Explanation:

Here ,there is small correction in question."#define FNC(x, y) ((x) > (y)) ? (* * y) : (y - x);"==>This statement should actually be#define FNC(x, y) ((x) > (y)) ? (x * y) : (y - x);

When FNC(++p,q++) is called===>FNC(++p>q++)?(++p*q++):(q++-++p)

                      p=7 q=6                                8>6           9*7

                      p=9   q=7

                      r=4+63=67

Answer 2

Answer:

The output of the given code is 9 8 67.

Explanation:

• A function FNC is defined at the start of the program. This function takes two arguments.
• The FNC function compares which of the two arguments is greater. If the first argument is greater than the second, the first expression after the question mark is executed.
• If the second argument is greater than the first, the second expression after the question mark (or the expression after the colon) is executed.
• The corresponding result is passed back to the main function.
• In the main function, three variables p, q, and r are initialized to values 7, 6, and 4 respectively.
• Then the FNC function is called with arguments as ++p and q++. This means that the value of p is incremented by 1 before it is passed as a function argument and the value of q is incremented by 1 after the function is run.
• So, the function call may look like this r+=FNC ( 8, 6)
• The function compares the two arguments ++p (8) and q++ (6) and the value of q is incremented by 1 to q=7. Since ++p is greater than q++, the first expression after the question mark is executed.
• Here too, ++p causes the value of p to be incremented by 1 to 9 before the execution of the statement, and the value of q++ is incremented after the execution.
• (++p * q++) will be (9 * 7) which results in 63. The value is q is incremented by 1 to 8.
• The result of the function execution is then passed back to the main function.
• In the main function, this result is added to the value of r which was initialized to 4. So, the new value of r becomes r=4+63=67.
• The final values of p, q, and r are printed and the program is terminated.

Therefore, the program prints the final values of the variables p, q, and r which are 9, 8, and 67.

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