Question

1. Identify the directives and statements in the following program. What is the output of the program? [2 points]

#include <stdio.h>

int main (void) {

printf(“Parkinson’s Law: \n Work expands so as to \t”);

1. If i and j are positive integers, does (-i)/j always have the same value as –(i/j)? Justify your answer. [2 points]

printf(“fill the time \n”);

1. Supply parenthesis to show how a C compiler would interpret the following expressions: [2 points]
   1. a * b – c *d + e;
   2. a / b % c / d;
   3. – a – b + c - +d;
   4. a * - b / c – d;

printf(“\t available for its completion.”);

return 0;

}

Answer 1

Solution 1)

//Directive -stdio.h
#include <stdio.h>
int main(void)
{
//Statement-printf
printf("\nParkinson's Law:\nWork expands as to ");
//Statement-printf
printf("fill the time\n");
//Statement-printf
printf("available for its completion.\n\n");
return 0;
}

Output of the program:

Solution 2)

(a) a * b - c * d + e
((a*b)-((c*d))+e)

(b) a / b % c / d
(a/b)%(c/d)

(c) - a - b + c - + d
Incorrect expression -error

(d) a * - b / c - d
((a* -(b/c))-d)

Solution 3)

if i and j are positive integers, does (-i)/j always have the same value as –(i/j)?

Yes they both will always have the answer .

Why?
Because c compiler acts just like a normal scientific calculator for this case due to the Operator precedence.

Example:
#include <stdio.h>
int main(void)
{
int i,j,x,y;
i =33;
j=3;
x=((-i)/j);
y= (-(i/j));
printf("%d",x);
printf("\n");
printf("%d",y);
return 0;
}

Output:

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