Palindromes revisited

Here's a simpler program for checking palindromes: int nums[100]; int i = 0, a; cin >> a; while(a > 0) { nums[i++] = a; cin >> a; } for(int j=0; j<i; j++) if(nums[j] != nums[i-1-j]) { cout << "no" << endl; return 0; } cout << "yes" << endl;

Text Palindromes

Instead of using numbers, let's use letters, or chars.

Also, we will read in until the end of the user input,

rather than reading in until a sentinel value. The

expression: cin >> a; (where a is a char)

will return a value equivalent to true if a character

is read in and a value equivalent to false if

nothing is read in. We can use that in a test.

Palindrome Code char lets[100], a; int i = 0; while(cin >> a) lets[i++] = a; for(int j=0; j<i; j++) if(lets[j] != lets[i-1-j]) { cout << "no" << endl; return 0; } cout << "yes" << endl;

To indicate the end of input, the user would type ^Z

(in Windows) or ^D (in Unix).

Another Example

Here's the problem: Let's read in characters, just like

in the last program, but then print out a histogram

showing the number of a's, b's, c's,, ... read in. For

example, if the program read in the text:This is a test. This is only a test. In the event of an

actual emergency, you would be told where to tune on

your radio dial.

Example (cont'd)

a: *******b: *c: **d: ****e: ************f: *g: *h: ****i: ******j:

Example Code

We will need a counter for each letter, a – z.The counters should be initialized to 0.Every time a char is read in, if it is a letter (a-z, or

A-Z), the appropriate counter should be

incremented. This requires a read-to-end-of-input

loop.Once everything is read in, print out the histogram.

Converting chars to ints

We need to convert the chars read in to be ints, since

arrays are indexed by ints. This is easy to do in C+

+. Any arithmetic operation applied to a char

automatically converts it to an int. The value of the

char is given by the ASCII (American Standard

Code for Information Interchange) code. For

example, 'A' is 65, and 'a' is 97, but we don't need to

know that.

Converting chars to ints (cont'd)

What we need to know is how to convert 'a' to 0, and

'b' to 1, and 'c' to 3, ... (and likewise 'A' to 0 and 'B'

to 1, ...). The expression c – 'a' will

automatically convert the value of c to its ASCII

value, and also 'a' to its ASCII and subtract the

two values. This does exactly what we want for

lower case letters. c – 'A' works for upper case

letters.

Example Code

int count[26]; char c; for(int i=0; i<26; i++) count[i] = 0; while(cin >> c) if('a' <= c && c <= 'z') count[c - 'a']++; else if('A' <= c && c <= 'Z') count[c - 'Z']++; for(int i=0; i<26; i++) { cout << (char)(i+'a') << ": "; printStars(count[i]); }

More About Arrays

Arrays must be declared to be a fixed size (in most

compilers – in some, arrays which are local

variables, can be declared with a variable)Usually named constants are used to denote the

size: const int NUMCHARS = 26; int a[NUMCHARS]; Array always start at 0, not 1.Arrays may be initialized:

int primes[4] = {2, 3, 5, 7};

Using Indexed Variables as Args.

Index variables (array elements) may be used as

arguments in function calls.They may correspond to either call-by-value or call-

by-reference parameters.If the indexed variable has a variable subscript, say

a[i], the subscript is evaluated when the function

is called, and that data object is used.

Using Whole Arrays as Arguments

A whole array may be used an argument in a

function call.Whole arrays are always call-by-reference

parameters, even without writing the “&” in the

parameter list.A parameter which is an array variable doesn't

cause a data object to be created – it becomes

another name of the array argument. No need to give the size of the array parameter.

Example

int foo(int x[], int size);

int main(void) { int a[10], s = 10; // read in array a; foo(a,s);}

Exercise

Write a function, maxElement, that is passed an

array of ints, a, and the number of elements in that

array, size, and returns the maximum element of

the array.