{"id":57753,"date":"2022-12-29T03:01:43","date_gmt":"2022-12-28T17:01:43","guid":{"rendered":"http:\/\/www.rjmprogramming.com.au\/ITblog\/?p=57753"},"modified":"2022-12-27T15:01:30","modified_gmt":"2022-12-27T05:01:30","slug":"c-recursion-fibonacci-series-tutorial","status":"publish","type":"post","link":"https:\/\/www.rjmprogramming.com.au\/ITblog\/c-recursion-fibonacci-series-tutorial\/","title":{"rendered":"C++ Recursion Fibonacci Series Tutorial"},"content":{"rendered":"
\"C++<\/a>

C++ Recursion Fibonacci Series Tutorial<\/p><\/div>\n

We’re revisiting the Fibonacci Series code of C Recursion Find Revisit Tutorial<\/a> and compiling it, plus some extended functionality, via the g++ c++ compiler on a macOS command line, as per …<\/p>\n


\ng++ fibonacci_series.cpp -o fibonacci_series.out
\n<\/code><\/p>\n

… regarding the changed<\/a> fibonacci_series.cpp<\/a> code. Today’s tutorial picture<\/a> displays the execution run involving the interactive input<\/font> as per …<\/p>\n


\n.\/fibonacci_series.out<\/font>
\nHow many numbers are there in your fibonacci sequence (-ve for in a row): 23<\/font>
\n0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657
\n
\n 0, 1
\n 0 + 1 = 1
\n 0, 1, 1
\n 0 + 1 = 1
\n 0, 1, 1, 2
\n 1 + 1 = 2
\n 0, 1, 1, 2, 3
\n 1 + 2 = 3
\n 0, 1, 1, 2, 3, 5
\n 2 + 3 = 5
\n 0, 1, 1, 2, 3, 5, 8
\n 3 + 5 = 8
\n 0, 1, 1, 2, 3, 5, 8, 13
\n 5 + 8 = 13
\n 0, 1, 1, 2, 3, 5, 8, 13, 21
\n 8 + 13 = 21
\n 0, 1, 1, 2, 3, 5, 8, 13, 21, 34
\n 13 + 21 = 34
\n 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55
\n 21 + 34 = 55
\n 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89
\n 34 + 55 = 89
\n 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144
\n 55 + 89 = 144
\n 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233
\n 89 + 144 = 233
\n 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377
\n 144 + 233 = 377
\n 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610
\n 233 + 377 = 610
\n 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987
\n 377 + 610 = 987
\n 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597
\n 610 + 987 = 1597
\n 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584
\n 987 + 1597 = 2584
\n 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181
\n 1597 + 2584 = 4181
\n 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765
\n 2584 + 4181 = 6765
\n 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946
\n 4181 + 6765 = 10946
\n 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711
\n 6765 + 10946 = 17711
\n 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657
\n 10946 + 17711 = 28657
\n<\/code><\/p>\n

… using the C++ command line code …<\/p>\n


\n\/\/ Fibonacci sequence
\n\/\/ Number in series interactively or via command line argument
\n\/\/ fibonacci_series.cpp
\n\/\/ RJM Programming - December, 2022
\n
\n#include <stdio.h>
\n#include <stdlib.h>
\n#include <string.h>
\n
\nchar aline[500];
\nint done=0;
\nint last=-1;
\nchar oneline[500];
\nchar lblanks[500];
\n\/\/int nums[]={0,1};
\nint lastf=0;
\nint lastlastf=0;
\n
\nint countFibonacci(int i) { \/\/ Recursive function
\n int fibonacci;
\n if (i == 0) { return 0; } else if (i == 1) { if (done == 0) { done=1; sprintf((aline + strlen(aline)), \"1%s\", \" \"); } return 1; }
\n

\n fibonacci = countFibonacci(i-1) + countFibonacci(i-2);
\n if (fibonacci > last) {
\n strcpy(aline, lblanks);
\n strcat(aline, oneline);
\n \/\/sprintf((aline + strlen(aline)), \"%d%s\", fibonacci, \", \");
\n sprintf((aline + strlen(aline)), \"%d%s\", fibonacci, \" \");
\n \/\/nums[1]=fibonacci;
\n }
\n last=fibonacci;
\n return fibonacci;
\n}
\n
\nint main(int argc, char *args[]) {
\n char delim[3]=\", \";
\n char explanation[5000];
\n int nsize = 0, fibonacci, i, j, k;
\n char catend[20];
\n

\n \/\/ Check for argument 1 number in series
\n if (argc > 1) {
\n for(i=0; i<strlen(args[1]); i++) {
\n if (args[1][i] == '-' || (args[1][i] >= '0' && args[1][i] <= '9')) {
\n nsize = nsize;
\n } else if (args[0][i] != '-') {
\n nsize--;
\n }
\n }
\n if (nsize == 0) sscanf(args[1], \"%d\", &nsize);
\n }
\n strcpy(explanation, \"\\n\\n\");
\n strcpy(aline, \"0, \");
\n oneline[0]=0;
\n lblanks[0]=0;
\n if (argc <= 1) { \/\/nsize <= 0) {
\n \/\/printf(\"How many numbers are there in your fibonacci sequence (-ve for in a row): \");
\n printf(\"How many numbers are there in your fibonacci sequence: \");
\n scanf(\" %d\", &nsize);
\n }
\n \/\/if (nsize > 0) { strcpy(delim, \"\\n\\0\"); }
\n

\n for (i=-5; i<=(int) abs(nsize * 3); i++) {
\n sprintf((lblanks + strlen(lblanks)), \"%s\", \" \");
\n }
\n k=strlen(lblanks);
\n strcpy(aline, lblanks);
\n strcat(aline, \"0, \");
\n

\n for (i=0; i<=(int) abs(nsize); i++) {
\n fibonacci = countFibonacci(i);
\n if (i == (int) abs(nsize)) {
\n printf(\"%d\", fibonacci);
\n sprintf((oneline + strlen(oneline)), \"%d%s\", fibonacci, \"\\0\");
\n if (strstr(aline, \"1\")) {
\n if (lastf != 0) {
\n sprintf(catend, \"%d%s%d%s%d%s\", lastlastf, \" + \", lastf, \" = \", fibonacci, \"\\n\");
\n } else {
\n sprintf(catend, \"%d%s%d%s%d%s\", lastlastf, \" + \", 1, \" = \", fibonacci, \"\\n\");
\n }
\n strcat(explanation, strcat(strcat(aline, \"\\n \"), catend));
\n }
\n \/\/oneline[0]=0;
\n aline[0]=0;
\n k-=2;
\n sprintf(lblanks, \"%s\", \"\\0\");
\n for (j=0; j<k; j++) {
\n sprintf((lblanks + strlen(lblanks)), \"%s\", \" \");
\n }
\n sprintf(aline, \"%s%s\", lblanks, \"\\0\");
\n strcat(aline, \"0, \");
\n lastlastf=lastf;
\n lastf=fibonacci;
\n } else {
\n printf(\"%d%s\", fibonacci, delim);
\n sprintf((oneline + strlen(oneline)), \"%d%s%s\", fibonacci, delim, \"\\0\");
\n if (strstr(aline, \"1\")) {
\n if (lastf != 0) {
\n sprintf(catend, \"%d%s%d%s%d%s\", lastlastf, \" + \", lastf, \" = \", fibonacci, \"\\n\");
\n } else {
\n sprintf(catend, \"%d%s%d%s%d%s\", lastlastf, \" + \", 1, \" = \", fibonacci, \"\\n\");
\n }
\n strcat(explanation, strcat(strcat(aline, \"\\n \"), catend));
\n }
\n \/\/oneline[0]=0;
\n aline[0]=0;
\n k-=2;
\n sprintf(lblanks, \"%s\", \"\\0\");
\n for (j=0; j<k; j++) {
\n sprintf((lblanks + strlen(lblanks)), \"%s\", \" \");
\n }
\n sprintf(aline, \"%s%s\", lblanks, \"\\0\");
\n strcat(aline, \"0, \");
\n lastlastf=lastf;
\n lastf=fibonacci;
\n }
\n }
\n \/\/strcat(explanation, strcat(oneline, \" a new line\\n\"));
\n printf(\"\\n%s\\n\", explanation);
\n}
\n<\/code><\/p>\n

\n

\n

Previous relevant C Recursion Find Revisit Tutorial<\/a> is shown below.<\/p>\n

\"C<\/a>

C Recursion Find Revisit Tutorial<\/p><\/div>\n

Yesterday’s C Recursion Revisit Tutorial<\/a> had us finding a macOS Xcode produced executable via the command line “find” command. We find this excellent “find” command trades off<\/i> the time it takes for the convenience to the user. And this came into play when we noted, from yesterday’s …<\/p>\n


\nfind \/ -name 'Fibonacci' 2> \/dev\/null
\n<\/code><\/p>\n

… a mix of executable and non-executable files, causing a degree of confusion, and yet online published alternatives such as find \/ -name ‘Fibonacci’ -type f -executable 2> \/dev\/null<\/i> just didn’t work, and some solutions, such as an xargs<\/i> one might need to wait for the whole list to be produced before producing any (hence our trades off<\/i> assertion). So, what else? Well, macOS “find”<\/a> has this useful “-ls” switch as per …<\/p>\n

\n -ls This primary always evaluates to true. The following information
\n for the current file is written to standard output: its inode
\n number, size in 512-byte blocks, file permissions, number of hard
\n links, owner, group, size in bytes, last modification time, and
\n pathname. If the file is a block or character special file, the
\n major and minor numbers will be displayed instead of the size in
\n bytes. If the file is a symbolic link, the pathname of the
\n linked-to file will be displayed preceded by ‘->’. The format
\n is identical to that produced by ls -dgils.\n<\/p><\/blockquote>\n

… that brings along information at each record that could help, and means we can pipe<\/i> some extra (thanks to this useful link<\/a> for ideas) …<\/p>\n


\nfind \/ -name 'Fibonacci' -ls<\/b> 2> \/dev\/null | grep 'rwx' | grep -v 'drwx' | rev | cut -d' ' -f 1 | rev<\/i>
\n<\/code><\/p>\n

… to narrow the result set down to just executables for the longer “find” command line “parented” command set. And the result is? Yes, both files found were Xcode executables which worked, as you can see at today’s tutorial picture<\/a>.<\/p>\n

A word of warning here is that we do not know the behaviour, with methodology above, should filenames contain any blank characters.<\/p>\n

\n

\n

Previous relevant C Recursion Revisit Tutorial<\/a> is shown below.<\/p>\n

\"C<\/a>

C Recursion Revisit Tutorial<\/p><\/div>\n

Back in October, 2013 we presented C Recursion Primer Tutorial<\/a> as a C++ ( but really just straight C code) Xcode (IDE<\/a>) project on Mac OS X. That’s a long time ago, looking at it from late in 2021. So what’s the go now with …<\/p>\n