Programming
October 10, 2023Programming
October 10, 2023Programming
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Question 4
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What is the output of the following program?
#include <stdio.h>
int funcf (int x);
int funcg (int y);
main()
{
int x = 5, y = 10, count;
for (count = 1; count <= 2; ++count)
{
y += funcf(x) + funcg(x);
printf ("%d ", y);
}
}
funcf(int x)
{
int y;
y = funcg(x);
return (y);
}
funcg(int x)
{
static int y = 10;
y += 1;
return (y+x);
}
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43 80
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42 74
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33 37
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32 32
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Question 4 Explanation:
In first iteration:
In first case of funcf, which in turn calls funcg, y becomes 11 and it returns 5+11 = 16.
In second call of funcg, y becomes 12 and it returns 5+12 = 17.
So, in main y is incremented by 16+17 = 33 to become 10+33 = 43.
In second iteration:
y will be incremented by 18+19 = 37 to give 43+37 = 80.
In first case of funcf, which in turn calls funcg, y becomes 11 and it returns 5+11 = 16.
In second call of funcg, y becomes 12 and it returns 5+12 = 17.
So, in main y is incremented by 16+17 = 33 to become 10+33 = 43.
In second iteration:
y will be incremented by 18+19 = 37 to give 43+37 = 80.
Correct Answer: A
Question 4 Explanation:
In first iteration:
In first case of funcf, which in turn calls funcg, y becomes 11 and it returns 5+11 = 16.
In second call of funcg, y becomes 12 and it returns 5+12 = 17.
So, in main y is incremented by 16+17 = 33 to become 10+33 = 43.
In second iteration:
y will be incremented by 18+19 = 37 to give 43+37 = 80.
In first case of funcf, which in turn calls funcg, y becomes 11 and it returns 5+11 = 16.
In second call of funcg, y becomes 12 and it returns 5+12 = 17.
So, in main y is incremented by 16+17 = 33 to become 10+33 = 43.
In second iteration:
y will be incremented by 18+19 = 37 to give 43+37 = 80.
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