Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 3715-2•Know how to declare arrays, access elements of an array.•Understand how character arrays can represent strings.•Use compact C notation such as assignment operators or the increment and decrement operators.•Two dimensional arrays •Related Chapters: ABC Chapter 6.1, 6.2, 6.1215-3A one dimensional array is a list of data values, with all values having the same data type(the base type), such as:• integer• float• double• charTechnically, an array is a uniform data structure. Individual array elements are referenced using the array name and a subscript that identifies the element position in the array.15-4For a one dimensional array, we specify the array name, its base data type, and the number of storage locations required using declarations such asint a[25];float x[100], y[100];which specifies 25 integer locations for a and 100 floating-point locations for arrays x and y. Storage for array elements are in contiguous locations in memory, referenced by subscript(or index) values starting at 0. Thus for array a above, the storage is. . .a[0] a[1] a[24]RAM15-5We can initialize array elements in declaration statements; e.g., int counts[5] = {1, 2, 3, 4, 5};int evens[] = {2, 4, 6, 8};The array size could also be specified using a symbolic constant:#define ARRAY_SIZE 25int a[ARRAY_SIZE];.../* evens has 4 elements */We cannot specify more initial values than there are array elements, but we can specify fewer initial values, and the remaining elements will be initialized to 0. e.g.,int b[10] = {2};double x[250] = {0};15-6Note: when referencing a particular element of an array use square brackets, not parenthesis or curly braces.Given that the array x is declared as:int x[250];To initialize a large array to a nonzero value - say 10, we can use a loop. e.g., for (k = 0; k <= 249; k = k+1) x[k] = 10;k is a subscript15-7A subscript value for an array element can be specified as any integer expression. For example, given the following declarations:double y[4] = {-1.0, 12.0, 3.5, 3.2e-2};int c = 2;the following statement would assign the value of 5.5 to y[1]y[2 * c - 3] = 5.5;15-81. Problem DefinitionWrite a program “fib” that prompts the user for an integer n. The program prints the first n Fibonacci numbers. 2. Refine, Generalize, Decompose the problem definition(i.e., identify sub-problems, I/O, etc.)Input = n - the count of Fibonacci Numbers to be printedOutput= Fib Numbers printed to screen. 3. Develop AlgorithmUse the formula: fib0= 1, fib1 =1 and for k > 1fibk = fibk-1 + fibk-2#include <stdio.h>/* C program to print the first n Fibonacci Numbers. */void main(void){ int k,n; int fib[100] = {1,1}; /* note: n must be <= 100 , why ? */ printf("How many Fibonacci number do you want listed? "); scanf("%i",&n); for (k = 2; k < n; k++) /* the ++ is the increment operator */ fib[k] = fib[k-1] + fib[k-2]; /* print the results, four per line */ for (k = 0; k < n;++k) { if (k % 4 == 0)printf(“\n”); printf("%8i", fib[k]); /* that’s 8 ints */ } /* end of for loop */ printf("\n");} /* end of main */(For large values of parameter n, we would need more space for each printed column. Also, the calculated values may become too large for the allocated storage space for a datatype int.)15-10: .Since incrementing and decrementing (negative increments)are common programming operations, the C languageprovides shortcut Increment / Decrement operators.++k Increment k by 1 and use incremented value in expression containing ++k.k++ Use the current value of k then increment by 1.--k Decrement k by 1 and use decremented value in expression containing --k. k-- Use current value of k then decrement by 1.++k;15-11++k; /*C statement k = k + 1; */int x,z;int y = 1;int a = 2;x = y + a++;/* now x = 3 , y = 1 , a = 3 */z = x + --a;/* now a = 2, z = 5, x = 3, y = 1 */ z = --x * ++y;/* now x = 2, y = 2, z = 4, a = 2 */x = a++ / --y;/* now y = 1, x = 2, a = 3, z = 4 */Note: The increment/decrement operators can be applied only to single variables. They cannot be applied to expressions.15-121. Problem DefinitionWrite a program that inputs a list of numbers into an array. The program then calculates the average value, and then prints a list of differences. The differences are computed by taking the original values in the list minus the average. 2. Refine, Generalize, Decompose the problem definition(i.e., identify sub-problems, I/O, etc.)Input = list of reals in a file “input.dat” . We will use Unix redirection to read values from this file.Output= The average and the list of differences.#include <stdio.h>void main(void){ int counter,k; double datAve, sum = 0.0, datVal[500]; /* 500 element max */ /* read the file and compute the average */ counter = 0; while (EOF != scanf("%lf", &datVal[counter])) { sum += datVal[counter]; /* += is an assignment operator */ counter++; } /* compute and print the average */ datAve = sum /counter; printf("The average is:%lf \n", datAve); /* compute and print the diff list */ for(k=0;k<counter;++k) printf("%lf\n", datVal[k]-datAve);}15-14- a shortcut method for writing assignment statements of the formvar1 = var1 op expression; Using an "assignment operator", we can rewrite the above asvar1 op= expression; where op can be any one of the arithmetic binary operators:+ - * / %sum += datVal[counter];15-15sum += datVal[counter]; or sum = sum + datVal[counter];k += 5; or k = k +5;n -= 4; or n = n -4;a *= 2.0; or a = a*2.0;x /= b; or x = x/b;remainder %= 6; orremainder = remainder % 6;newValue += (a-b)/(2*c); or newValue = newValue + (a-b)/(2*c);15-16Unlike many other languages, there is no data type for character strings in C. We will implement strings in C by using a one dimensional array of characters where the last character is the “NULL” character ‘\0’.15-17char aString[5] = "Zip!";char atomic[ ] = "hydrogen";Declaring character arrays with a string constantIMPORTANT!!!In C, you must always terminate a character array with the NULL character, ‘\0’ . Therefore, the array size of your character array should be one plus the maximum length of the string you want to store.