Just as it’s possible to rotate elements in an array, shuffling them all up a notch and moving the first element to the end, it’s also possible to flip an array, reversing the order of its elements.
\n
\nFor example, you start with this array:<\/p>\n
After the operation, the array’s values are:<\/p>\n As shown in last week’s Lesson<\/a>, it’s possible to perform this operation without requiring a second array. Here’s how I do it:<\/p>\n After the array is output, a for<\/em> loop reverses the elements. Variable This operation repeats as variable This code is designed to handle an array of any size, which is why the statement I modified the value of Because the This type of manipulation is demonstrated with an integer array. In next week’s Lesson<\/a> I cover messing with an array of strings.<\/p>\n","protected":false},"excerpt":{"rendered":" Rather than just shuffle the elements, this time the code flips first for last all values from element zero through element n<\/em>. Continue reading { 1, 2, 3, 4, 5, 6, 7, 8 }<\/code><\/p>\n{ 8, 7, 6, 5, 4, 3, 2, 1 }<\/code><\/p>\n2025_03_01-Lesson.c<\/a><\/h3>\n
\r\n#include <stdio.h>\r\n\r\nint main()\r\n{\r\n int a[] = { 1, 2, 3, 4, 5, 6, 7, 8 };\r\n int size,x,save;\r\n\r\n \/* obtain array size *\/<\/span>\r\n size = sizeof(a)\/sizeof(a[0]);\r\n\r\n \/* output the array *\/<\/span>\r\n for( x=0; x<size; x++ )\r\n printf(\" %d\",a[x]);\r\n putchar('\\n');\r\n\r\n \/* reverse the elements *\/<\/span>\r\n for( x=0; x<(size\/2); x++ )\r\n {\r\n save = a[x];\r\n a[x] = a[size-x-1];\r\n a[size-x-1] = save;\r\n }\r\n\r\n \/* output the array *\/<\/span>\r\n for( x=0; x<size; x++ )\r\n printf(\" %d\",a[x]);\r\n putchar('\\n');\r\n\r\n return 0;\r\n}<\/pre>\nsave<\/code> holds the value of the first element, which is then replaced with the value of the last element. Then the last element is set to the value of int<\/em> variable save<\/code>:<\/p>\nfor( x=0; x<(size\/2); x++ )
\n{
\n save = a[x];
\n a[x] = a[size-x-1];
\n a[size-x-1] = save;
\n}<\/code><\/p>\nx<\/code> increments to half the size of the array. A final for<\/em> loop outputs the results:<\/p>\n 1 2 3 4 5 6 7 8
\n 8 7 6 5 4 3 2 1<\/code><\/p>\nsize = sizeof(a)\/sizeof(a[0]);<\/code> is used to obtain the array’s size. But what about an array with an odd number of elements?<\/p>\na[]<\/code> to add three more elements: 9, 10, and 11. Here’s the program’s updated output:<\/p>\n 1 2 3 4 5 6 7 8 9 10 11
\n 11 10 9 8 7 6 5 4 3 2 1<\/code><\/p>\nsize\/2<\/code> expression works with integers, the “half” value is rounded down, which means the middle element — no matter what the array size — isn’t swapped. It doesn’t need to be!<\/p>\n