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The value of j at the end of the execution of the following C programint incr (int i){   static int count = 0;   count = count + i;   return (count);}main (){   int i,j;   for (i = 0; i <=4; i++)      j = incr(i);}41067

Question

The value of j at the end of the execution of the following C programint incr (int i){   static int count = 0;   count = count + i;   return (count);}main (){   int i,j;   for (i = 0; i <=4; i++)      j = incr(i);}41067

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Solution

The function incr is a C function that takes an integer i as an argument and increments a static integer count by i. The static keyword means that count retains its value between function calls.

The main function initializes two integers i and j, and then enters a for loop that iterates from i = 0 to i = 4. On each iteration, it calls incr(i) and assigns the result to j.

Let's break down the execution of the program:

  • On the first iteration, i = 0, so incr(0) is called. This adds 0 to count (which is initially 0), so count remains 0 and j is set to 0.
  • On the second iteration, i = 1, so incr(1) is called. This adds 1 to count (which is currently 0), so count becomes 1 and j is set to 1.
  • On the third iteration, i = 2, so incr(2) is called. This adds 2 to count (which is currently 1), so count becomes 3 and j is set to 3.
  • On the fourth iteration, i = 3, so incr(3) is called. This adds 3 to count (which is currently 3), so count becomes 6 and j is set to 6.
  • On the fifth and final iteration, i = 4, so incr(4) is called. This adds 4 to count (which is currently 6), so count becomes 10 and j is set to 10.

So, at the end of the execution of the program, the value of j is 10.

This problem has been solved

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