Statements and expressions for making decisions, or choices, are similar to those in Java, with a few small differences.
If-statements
Statement
if (condition)
{
statements1
}
else
{
statements2
}
starts by evaluating condition, an expression that normally yields
either true or false. If the condition is true, then
statements1 are performed. If the condition is
false, then statements2 are performed. For example,
if(x > 0)
{
y = x;
}
else
{
y = -x;
}
ends with y equal to the absolute value of x.
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Omitting the else-part
If you omit the word else and the
compound statement that
follows it, then nothing is done when the condition is false.
For example,
if(x < 0)
{
x = -x;
}
will always end with x ≥ 0, since it does nothing if
x ≥ 0.
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Multiway tests
If you have more than two cases, use the following
indentation scheme.
if(test1)
{
…
}
else if(test2)
{
…
}
else if(test3)
{
…
}
else
{
…
}
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Details on if-statements
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The parentheses around the condition in an if-statement are
required. Do not omit them.
The condition is not required (by C++) to be a boolean value. It can be an integer or anything that is represented internally as an integer, such as a character or a member of an enumerated type or a pointer. In all cases, 0 means false and anything other than 0 means true. However, for this course the coding standards require (with very minor exceptions) that conditions be boolean values. (optional) The statements following the condition and following else are shown as compound statements (enclosed in braces) above. C++ does not require that. You can write any statement at all. However, the coding standards for this course do require a compound statement. (optional) |
Watch out: equality tests
If you are doing an equality test, be sure to use
==, not =. Statement
if(n = 0)
{
m = 1;
}
else {
m = 2;
}
always makes m = 2. Expression
n = 0 is not a test, it is an assignment.
It stores 0 into n, and yields result 0.
Since 0 is treated as false, the else-part is performed.
If you ask to compiler to give common warnings, it will
let you know about this mistake.
Do not use = in a test even if it is what you mean. The coding standards do not allow that. |
Watch out: semicolons
Be careful about semicolons. Statement
if(n == 0);
{
printf("n is zero\n");
}
always writes "n is zero", regardless of the value of n.
The semicolon at the end of the first line is an empty statement, so
the if-statement says, if n equals 0, do nothing. The statement after
that is just a
compound statement.
This really does two statements in a row.
The coding standards require you not to use a semicolon as an empty statement. |
Making choices in expressions
Expression a ? b : c starts by evaluating expression a.
If a is true (or nonzero), then it yields the value of
expression b (and does not
compute c at all). If a is false (or 0), it yields the value
of expression c (and does not compute b at all).
For example,
int m = x > 0 ? x : -x;creates variable m and stores the absolute value of x into m. Only use this for expressions, not for statements. The coding standards require that. |
Assume that integer variable n already has a value. Write a statement that makes m equal to 3n+1 if n is odd and to n/2 if n is even. Answer
The coding standards forbid the following.
if(n == 0){}
else
{
z = 1;
}
Rewrite that into an acceptable form that does the same thing.
Answer
The coding standards disallow the following. Explain what is going on and how to avoid this.
if(x == 0)
{
y = 1;
}
else if(x != 0)
{
y = 2;
}
Answer
Suppose that you want to set y equal to
Explain why the following does not do the job.
if(x < 0)
{
y = -1;
}
if(x == 0)
{
y = 0;
}
else
{
y = 1;
}
Answer