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>> DAVID J. MALAN: Let's write a program
that prompts the user for a string and

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then proceed to print that string
character for character one per line.

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Now in the past, we would have done so
probably with square bracket notation,

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effectively treating a string
is an array of characters.

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>> But this time, let's instead treat
a string for what it really is, a

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pointer or an address.

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Specifically, the address of a
character, really the address of the

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first character, in a sequence of
characters that we collectively know

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as a string.

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>> Let's first declare a string for
what it really is, char*.

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And we'll call it s.

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And then assign it the return
value of get string.

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>> Let's next do some error checking.

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If s is null, let's immediately return
so that we don't accidentally

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dereference that null pointer.

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>> Next, let's iterate over the
characters in s as follows.

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For int, i gets 0.

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n equals the string length of s.

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Do this so long as i is less than n.

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And on each iteration, increment i.

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>> And what do we want to
do on each iteration?

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Let's now print out on each iteration
a single character

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followed by a new line.

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Well, what character do
we want to print?

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I propose that we go to the address
that equals the sum of s plus i.

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>> Now, why that expression?

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Well, recall that stored in s is the
address of the first character

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in our string, s.

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Meanwhile, i is being incremented on
each iteration so that it starts at 0,

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then goes to 1, then goes to 2.

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>> So in other words, s plus i effectively
represents the address of

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the i-th character in s.

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So if we go to that address by way of
the * operator, we'll be going to the

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i-th character in the string.

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And that's the value that will be
substituted for our placeholder,

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percent C.

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>> Let's confirm as much.

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Let's save, compile, and
run this program.

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Make pointers, dot slash pointers.

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And now I'll give it a
string like hello.

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Enter.

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>> And indeed, I see H-E-L-L-O, with
each char on its own line.

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