Lab 03 - Assert

Thanks for my Lab Tutor Zhang Puyu!

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Lab 3 Slides

Exercise 1 Review

Code Style

1. Initialize the variable immediately after the declaration.

For example,

// Not Recommended
double num;
num = cs1010_read_double();

// Recommended
double num = cs1010_read_double();

1. Do not declare the function and then implement it somewhere else (Usually after main()). Implement your function right after the declaration.

No return after else

Code Example

// Not allowed
if (something)
{
    return something;
}
else
{
    return other_things;
}

Out of succient reasons, the else is of no use. So, we can simplify the code to below

// Allowed
if (something)
{
    return something;
}
return other_things;

No nested If

If your code is like below

if (something_1)
{
    if (something_2)
    {
        // do something
    }
}

Since there is only one if statement inside the "outer" if, so there is no need to use the nested if. And we can simplify it into

if (something_1 && something_2)
{
    // do sommething
}

Assert Library

assert.h is a library designed to help with debugging procedures.

Usage

assert(p) where p is a boolean expression. When the condition p fails during program execution, the program will halt with an error message.

Exercise 1 Further Discussion

odd.c

A solution that uses no condition is that $m=n+(n\%2)*(n\%2)+1$, where $m$ $f(x) = x * e^{2 pi i \xi x}$is the smallest odd number that is strictly bigger than $n$.

The Deriving process

odd.c

multiple.c

Modulo is not defined when we divide a non-zero number by 0. That is $m\%0$, where $m\neq0$, is not defined!!!

date.c

There is a method not using conditions. Map each $(m,d)$ date to the integer $100m+d$.

The only criteria for coefficient of $m$ is that the coefficient must be bigger than the maximum days in a month, which is 31. So $m\geq31$ should be okay.

The reason is that we should make sure the priority of the month is higher.

power.c

Ways to optimize

1. Remove the useless work, when the base is 0, -1 or 1.

long compute_power(long x, long y)
{
    if (x == 0)
    {
        return 0;
    }
    if (x == 1)
    {
        return 1;
    }
    if (x == -1)
    {
        return y % 2 == 0 ? 1 : -1;
    }
    // ...
}

1. Half the calculation

Recall that

$$\large x^y=\begin{cases} \left(x^2\right)^{\frac{y}{2}} & \text{if } y \text{ is even} \\ \left(x^2\right)^{\frac{y-1}{2}} \cdot x & \text{if } y \text{ is odd} \end{cases}$$

Then convert it to code will be intuitive

if (y % 2 == 0)
{
    return compute_power(x * x, y / 2);
}
return x * compute_power(x * x, (y - 1) / 2);

taxi.c

A normal way to compute ceil()

long ceil_of_quotient(long n, long m)
{
    if (n % m == 0)
    {
        return n / m;
    }
    return n / m + 1;
}

A smart way to compute ceil()

Suppose we want to calculate $\lceil \frac{n}{m} \rceil$, where $n$ and $m$ are two possitive integers. First, let us write $n=mq+r$, where $q\geq0$ and $0 \leq r \leq m-1$. Now, let's consider

$$\large\frac{n+m-1}{m}=\frac{mq+r+m-1}{m}=\frac{m(q+1)+r-1}{m}$$

If $r=0$, then we will get $q$ as the output. If $0<r \leq m-1$, then the above numerator is at least $m(q+1)$ but strictly less than $m(q+2)$, so the quotient evaluates to $q+1$, which will be our output.

The case $r=0$ is equivalent to the case that $n\%m=0$ The case $0<r \leq m-1$ is equivalent to the case that $n\%m \neq 0$

Thus, convert the result into `C` code ```c long ceil_of_quotient(long n, long m) { return (n + m - 1) / m; } ```