Consider the following C program. #include int f1(void); int f2(void); int f3(void); int x=10; int…

Question

Consider the following C program.
#include<stdio.h>

int f1(void);

int f2(void);

int f3(void);

int x=10;

int main()

{

int x=1;

x += f1() + f2 () + f3() + f2();

printf("%d", x);

return 0;

}

int f1() { int x = 25; x++; return x;}

int f2() { static int x = 50; x++; return x;}

int f3() { x *= 10; return x;}

The output of the program is ________.

Accepted Answer

Correct answer: 230 — Answer: 230 Explanation: Call to f1: it declares a local x = 25, then increments and returns 26. First call to f2: static x starts at 50, is incremented to 51 and returns 51. Call to f3: it multiplies the global x (initially 10) by 10, changing the global x to 100 and returning 100. Second call to f2: the static x is incremented again from 51 to 52 and returns 52. Compute the sum returned by the functions: 26 + 51 + 100 + 52 = 229. Add this to the local x in main (initially 1): 1 + 229 = 230. Therefore, the program prints 230. Note: The C standard does not guarantee the order of evaluation of the operands of +, but in this program the final result is the same regardless of evaluation order because the functions' side effects do not interfere in a way that changes the computed sum.

Consider the following C program. #include<stdio.h> int f1(void); int…