1. The following program fragments illustrate the short-circuit behavior of logical expressions. Show the output produced by each, assuming that i, j, and k are int variables.

(a) i = 3; j = 4; k = 5; printf("%d ", i++ < j || ++j < k); printf("%d %d %d", i, j, k);

(b) i = 3; j = 4; k = 5; printf("%d ", i - 7 && j++ < k); printf("%d %d %d", i, j, k);

(c) i = 7; j = 0; k = 9; printf("%d ", (i = j) || (j = k)); printf("%d %d %d", i, j, k);

(d) i = -1; j = 2; k = 4; printf("%d ", ++i || ++j || ++k); printf("%d %d %d", i, j, k);

2. What output does each of the following program fragment produce? Translate (a) and (b) into a single for statement and (c) to a single while statement.

(a)

i = 1;

while(i<=136){ printf("%d*",i); i *= 3;

}

(b)

i = 3698; do {

printf("%d*",i%10);

i /= 10;

} while (i > 0);

(c)

i = -5;

j = -i;

for (;i>0||j>0;--i,j--)

printf("(%d,%d) ",i,j);

} while (i > 0);

3. The value of the mathematical constant e can be expressed as an infinite series : e = 1 + 1/1! + 1/2! + 1/3! + ...

Write a C program that approximate e by computing the value of e = 1 + 1/1! + 1/2! + 1/3! + ... + 1/n!

where n is an integer entered by the user. Save the program in a file named as epsilon.c,and submit the file online as your solution to this question.