notes/education/software development/ECE1400/Chapter.md

2. The following program fragments illustrate the logical operators. Show the output produced by each, assuming that i, j, and k are int variables.

a. i = 10; j = 5;

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

// Expected output: `1`, because `!i` evaluates to 0, and 0 is less than 5, so that expression evaluates to true, or 1.

b. i = 2; j = 1;

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

// Expected output: `1`, because !!2 evaluates to 1, and !j evaluates to 0

c. i = 5; j = 0; k = -5;

printf("%d", i && j || k);

// Expected output: `1`, because i && j should evaluate to 0, but `0 || 1` should evalulate to true.

d. i = 1; j = 2; k = 3;

printf("%d", i < j || k);

// Expected output: `1`

4. Write a single expression whose value is either -1, 0, or 1 depending on whether i is less than, equal to, or greater than j, respectively.

/*
	If i < j, the output should be -1.
	If i == j, the output should be zero
	If i > j, the output should be 1.
*/
(i > j) - (i < j)

6. Is the following if statement legal?

if (n == 1-10)
	printf("n is between 1 and 10\n");

Yes the statement is legal, but it does not produce the intended effect. It would not produce an output when n = 5, because 1-10 evaluates to -9, and -9 != 5.

10. What output does the following program fragment produce? (Assume that i is an integer variable.)

int i = 1;
switch (i % 3) {
	case 0: printf("zero");
	case 1: printf("one");
	case 2: printf("two");
}

The program would print onetwo because each case is missing a break statement.

11. The following table shows the telephone ar