The output of the following C program is_________.void f1(int a, int b) { int c; c=a; a=b; b=c; } void f2(int *a, int *b) { int c; c=*a; *a=*b; *b=c; } int main(){ int a=4, b=5, c=6; f1(a,b); f2(&b, &c); printf("%d",c-a-b); }

Let us trace the execution of the program step by step:

1. In main(), local variables are initialized:

   • a = 4

   • b = 5

   • c = 6

2. The function f1(a, b) is called with parameters passed by value. Any modifications made to a and b inside f1 are local to f1 and do not affect the variables in main(). Thus, after f1(a, b) returns, the values in main() remain:

   • a = 4, b = 5, c = 6

3. The function f2(&b, &c) is called with the addresses of b and c (pass-by-reference using pointers):

   • Inside f2, the values pointed to by a (which is &b) and b (which is &c) are swapped.

   • Hence, in main(), the value of b becomes $6$ and the value of c becomes $5$.

4. Now, the printf statement computes c - a - b:

   $$\text{c} - \text{a} - \text{b} = 5 - 4 - 6 = -5$$

Therefore, the output of the program is -5.