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.