In [4]:
minutes = 105
minutes/60
Out[4]:
But we don’t normally write hours with decimal points. Floor division returns the integer number of hours, rounding down:
In [6]:
hours = minutes ÷ 60
Out[6]:
To get the remainder, you could subtract off one hour in minutes:
In [7]:
remainder = minutes - hours * 60
Out[7]:
An alternative is to use the modulus function mod
which divides two numbers and returns the remainder:
In [8]:
remainder = mod(minutes, 60)
Out[8]:
The modulus operator is more useful than it seems. For example, you can check whether one number is divisible by another—if mod(x, y)
is zero, then x
is divisible by y
.
Also, you can extract the right-most digit or digits from a number. For example, mod(x, 10)
yields the right-most digit of x
(in base 10). Similarly mod(x, 100)
yields the last two digits.
In [9]:
5 == 5
Out[9]:
In [10]:
5 == 6
Out[10]:
true
and false
are special values that belong to the type Bool
; they are not strings:
In [11]:
typeof(true)
Out[11]:
In [12]:
typeof(false)
Out[12]:
The ==
operator is one of the relational operators; the others are:
x ≠ y # x is not equal to y
x != y
x > y # x is greater than y
x < y # x is less than y
x ≥ y # x is greater than or equal to y
x >= y
x ≤ y # x is less than or equal to y
x <= y
A common error is to use a single equal sign (=
) instead of a double equal sign (==
). Remember that =
is an assignment operator and ==
is a relational operator. There is no such thing as =<
or =>
.
In [17]:
5 ≥ 6
Out[17]:
There are three main logical operators: and (&
), or (|
), and not (!
). The semantics (meaning) of these operators is similar to their meaning in English. For example, x > 0 & x < 10
is true only if x
is greater than 0 and less than 10.
(mod(n, 2) == 0) | (mod(n, 3) == 0)
is true if either or both of the conditions is true, that is, if the number is divisible by 2 or 3.
In [5]:
n = 8
(mod(n, 2) == 0) | (mod(n, 3) == 0)
Out[5]:
Finally, the not operator negates a boolean expression, so !(x > y)
is true if x > y
is false, that is, if x
is less than or equal to y
.
Strictly speaking, the logical operators in Julia are bitwise operators:
In [24]:
42 & true
Out[24]:
In [25]:
typeof(42 & true)
Out[25]:
This flexibility can be useful, but there are some subtleties to it that might be confusing. You might want to avoid it (unless you know what you are doing).
In [29]:
x = 1
if x ≥ 0
println(x, " is positive")
end
The boolean expression after if
is called the condition. If it is true, the statements before end
run. If not, nothing happens.
if
statements have the same structure as function definitions: a header followed by a body terminated with end
. Statements like this are called compound statements.
There is no limit on the number of statements that can appear in the body. Occasionally, it is useful to have a body with no statements (usually as a place keeper for code you haven’t written yet).
In [30]:
if x < 0
# TODO: need to handle negative values!
end
In [35]:
x = 4
if mod(x, 2) == 0
println(x, " is even")
else
println(x, " is odd")
end
If the remainder when x
is divided by 2 is 0, then we know that x
is even, and the program displays an appropriate message. If the condition is false, the second set of statements runs. Since the condition must be true
or false
, exactly one of the alternatives will run. The alternatives are called branches, because they are branches in the flow of execution.
In [34]:
x = 1
y = 2
if x < y
println(x, " is less than ", y)
elseif x > y
println(x, " is greater than ", y)
else
println(x, " and ", y, " are equal")
end
Again, exactly one branch will run. There is no limit on the number of elseif
statements. If there is an else
clause, it has to be at the end, but there doesn’t have to be one.
Each condition is checked in order. If the first is false, the next is checked, and so on. If one of them is true, the corresponding branch runs and the statement ends. Even if more than one condition is true, only the first true branch runs.
Python:
if x < y:
print(x, 'is less than', y)
elif x > y:
print(x, 'is greater than', y)
else:
print(x, 'and', y, 'are equal')
if
statements have the same structure as function definitions: a header followed by an indented body.
There is no limit on the number of statements that can appear in the body, but there has to be at least one. Occasionally, it is useful to have a body with no statements (usually as a place keeper for code you haven’t written yet). In that case, you can use the pass
statement, which does nothing.
if x < 0:
pass # TODO: need to handle negative values
C++:
if (x < y) {
std::cout << x << " is less than " << y << std::endl;
} else if (x > y) {
std::cout << x << " is greater than " << y << std::endl;
} else {
std::cout << x << " and " << y << " are equal" << std::endl;
}
if
statements have the same structure as function definitions: a header followed by a body started with {
and ended with }
. The condition is surrounded by parentheses.
In [39]:
x = 1
y = 2
if x == y
println(x, " and ", y, " are equal")
else
if x < y
println(x, " is less than ", y)
else
println(x, " is greater than ", y)
end
end
The outer conditional contains two branches. The first branch contains a simple statement. The second branch contains another if statement, which has two branches of its own. Those two branches are both simple statements, although they could have been conditional statements as well.
Although the indentation of the statements makes the structure apparent, nested conditionals become difficult to read very quickly. It is a good idea to avoid them when you can.
Logical operators often provide a way to simplify nested conditional statements. For example, we can rewrite the following code using a single conditional:
In [42]:
x = 1
if 0 < x
if x < 10
println(x, " is a positive single-digit number")
end
end
The print
statement runs only if we make it past both conditionals, so we can get the same effect with the &
operator:
In [2]:
x = 1
if (0 < x) & (x < 10)
println(x, " is a positive single-digit number")
end
For this kind of condition. Julia provides a more concise option:
In [3]:
if 0 < x < 10
println(x, " is a positive single-digit number")
end
Short-circuit evaluation is quite similar to conditional evaluation. The behaviour is found in most imperative programming languages having the &&
and ||
boolean operators: in a series of boolean expressions connected by these operators, only the minimum number of expressions are evaluated as are necessary to determine the final boolean value of the entire chain:
In [51]:
x = 1
if 0 < x && x < 10
println(x, " is a positive single-digit number")
end
If x
is less than 0 only the first conditional 0 < x
is evaluated.
Both &&
and ||
associate to the right, but &&
has higher precedence than ||
does.
In [56]:
function countdown(n)
if n <= 0
println("Blastoff!")
else
print(n, " ")
countdown(n-1)
end
end
countdown(3)
If n
is 0 or negative, it outputs the word, “Blastoff!” Otherwise, it outputs n
and then calls a function named countdown —itself— passing n-1
as an argument.
What happens if we call this function like countdown(3)
?
The execution of countdown
begins with n=3
, and since n
is greater than 0, it outputs the value 3, and then calls itself...
The execution of countdown
begins with n=2
, and since n
is greater than 0, it outputs the value 2, and then calls itself...
The execution of countdown
begins with n=1
, and since n
is greater than 0, it outputs the value 1, and then calls itself...
countdown
begins with n=0
, and since n
is not greater than 0, it outputs the word, “Blastoff!”
and then returns.The countdown
that got n=1
returns.
The countdown
that got n=2
returns.
The countdown
that got n=3
returns.
And then you’re back in __main__
.
A function that calls itself is recursive; the process of executing it is called recursion. As another example, we can write a function that prints a string n times:
In [2]:
function print_n(s, n)
if n <= 0
return
end
println(s)
println("before n=",n)
print_n(s, n-1)
println("after n=",n)
end
print_n("Hello", 3)
If n <= 0
the return statement exits the function. The flow of execution immediately returns to the caller, and the remaining lines of the function don’t run.
The rest of the function is similar to countdown
: it displays s
and then calls itself to display s
n−1
additional times. So the number of lines of output is 1 + (n - 1)
, which adds up to n
.
For simple examples like this, it is probably easier to use a for
loop. But we will see examples later that are hard to write with a for
loop and easy to write with recursion, so it is good to start early.
In lecture 5, we used a stack diagram to represent the state of a program during a function call. The same kind of diagram can help interpret a recursive function.
Every time a function gets called, Julia creates a frame to contain the function’s local variables and parameters. For a recursive function, there might be more than one frame on the stack at the same time.
In [66]:
using TikzPictures
TikzPicture(L"""
\node (main) [draw, fill=lightgray, minimum width=4cm] {$\phantom{3}$};
\node [left of=main, xshift=-2cm] {\textrm{\_\_main\_\_}};
\node (c1) [draw, fill=lightgray, below of=main, yshift=0.25cm, minimum width=4cm] {$n\rightarrow3$};
\node [left of=c1, xshift=-2cm] {$\textrm{countdown}$};
\node (c2) [draw, fill=lightgray, below of=c1, yshift=0.25cm, minimum width=4cm] {$n\rightarrow2$};
\node [left of=c2, xshift=-2cm] {$\textrm{countdown}$};
\node (c3) [draw, fill=lightgray, below of=c2, yshift=0.25cm, minimum width=4cm] {$n\rightarrow1$};
\node [left of=c3, xshift=-2cm] {$\textrm{countdown}$};
\node (c4) [draw, fill=lightgray, below of=c3, yshift=0.25cm, minimum width=4cm] {$n\rightarrow0$};
\node [left of=c4, xshift=-2cm] {$\textrm{countdown}$};
"""; options="very thick, scale=3, transform shape", preamble="""
\\usepackage{newtxmath}
\\renewcommand{\\familydefault}{\\sfdefault}
""")
Out[66]:
As usual, the top of the stack is the frame for __main__
. It is empty because we did not create any variables in __main__
or pass any arguments to it.
The four countdown frames have different values for the parameter n
. The bottom of the stack, where n=0
, is called the base case. It does not make a recursive call, so there are no more frames.
In [68]:
function recurse()
recurse()
end
Out[68]:
In most programming environments, a program with infinite recursion does not really run forever.
Julia reports a StackOverflowError
when the maximum recursion depth is reached:
In [69]:
recurse()
When the error occurs, there are 80000 recurse frames on the stack!
If you encounter an infinite recursion by accident, review your function to confirm that there is a base case that does not make a recursive call. And if there is a base case, check whether you are guaranteed to reach it.
The programs we have written so far accept no input from the user. They just do the same thing every time.
Julia provides a built-in function called readline
that stops the program and waits for the user to type something. When the user presses Return
or Enter
, the program resumes and readline
returns what the user typed as a string.
In [70]:
text = readline()
Out[70]:
In [71]:
text
Out[71]:
Before getting input from the user, it is a good idea to print a prompt telling the user what to type.
In [4]:
print("What is your name? ")
name = readline()
Out[4]:
If you expect the user to type an integer, you can try to convert the return value to Int
:
In [8]:
println("What is the airspeed velocity of an unladen swallow?")
speed = readline()
parse(Int, speed)
We will see how to handle this kind of error later.
When a syntax or runtime error occurs, the error message contains a lot of information, but it can be overwhelming. The most useful parts are usually:
Syntax errors are usually easy to find. You should take the time to read error messages carefully, but don’t assume that everything they say is correct.
Runtime errors in Julia throw an Exception
and an error message hints to what is wrong.
Semantic errors are more difficult to troubleshoot. Suppose you are trying to compute a signal-to-noise ratio in decibels. The formula is $$ SNR_{db} = 10 \log_{10} \left(\frac{P_{signal}}{P_{noise}}\right) $$ In Julia, you might write something like this:
In [7]:
signal_power = 9
noise_power = 10
ratio = signal_power ÷ noise_power
decibels = 10 * log10(ratio)
println(decibels)
This result is not what we expected. To find the real error, it might be useful to print the value of ratio
, which turns out to be 0. The problem is in line 3, which uses floor division instead of floating-point division.