What is the output from this program? #include void do_something (int * thisp, int that) { int the_other; the_other = 5; that = 2 + the_other; * thisp = the_other * that; } int main (void) { int first, second; first = 1; second = 2; do_something (&second, first); printf ("%4d%4d/n", first, second); return (0); }

Question

Moshe James

Computers & Technology

26 January, 10:46

What is the output from this program? #include void do_something (int * thisp, int that) { int the_other; the_other = 5; that = 2 + the_other; * thisp = the_other * that; } int main (void) { int first, second; first = 1; second = 2; do_something (&second, first); printf ("%4d%4d/n", first, second); return (0); }

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Nikolai Garrison · Answer 1

1 35

Explanation:

* There is a little typo in printf. It should be "/n".

Initially, the value of the first is 1, and the value of the second is 2. Then, do_something (&second, first) is called. The value of the first will still be 1. However, there is a call by reference for second variable. That means the change made by the function do_something will affect the value of the second variable.

When you look at the calculation inside the do_something function, you may see that value of the second will be 35.