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

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

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

```c
/*
	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?
```c
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.)
```c
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