{"id":5476,"date":"2022-08-08T00:01:19","date_gmt":"2022-08-08T07:01:19","guid":{"rendered":"https:\/\/c-for-dummies.com\/blog\/?p=5476"},"modified":"2022-08-06T11:11:25","modified_gmt":"2022-08-06T18:11:25","slug":"from-a-and-a-to-z-and-z-solution","status":"publish","type":"post","link":"https:\/\/c-for-dummies.com\/blog\/?p=5476","title":{"rendered":"From ‘A’ and ‘a’ to ‘Z’ and ‘z’ – Solution"},"content":{"rendered":"

I don’t believe one definitive solution exists for this month’s Exercise<\/a>: Output a string of upper- and lowercase letters. No, the challenge is more to discover different insights a coder has to a specific problem.
\n
\nHere is the desired output string:<\/p>\n

AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWwXxYyZz<\/code><\/p>\n

My first approach is one I consider the most obvious:<\/p>\n

2022_08-Exercise-a.c<\/a><\/h3>\n
\r\n#include <stdio.h>\r\n\r\nint main()\r\n{\r\n    int x;\r\n\r\n    for( x='A'; x<='Z'; x++ )\r\n        printf(\"%c%c\",x,x+32);\r\n    putchar('\\n');\r\n\r\n    return(0);\r\n}<\/pre>\n
A for<\/em> loop plows through characters 'A'<\/code> through 'Z'<\/code> at Line 7. The character is output at Line 8, along with its companion lowercase letter, which is x+32<\/code> (thank you ASCII table). This solution was my first stab, though it could also be done by using two variables instead:<\/p>\n
2022_08-Exercise-b.c<\/a><\/h3>\n
\r\n#include <stdio.h>\r\n\r\nint main()\r\n{\r\n    int x,y;\r\n\r\n    for( x='A',y='a'; x<='Z'; x++,y++ )\r\n        printf(\"%c%c\",x,y);\r\n    putchar('\\n');\r\n\r\n    return(0);\r\n}<\/pre>\n
Here both x<\/code> and y<\/code> are initialized in the for<\/em> loop. Only one terminating condition needs to be specified, x<='Z'<\/code>. Then both x<\/code> and y<\/code> are incremented each turn. This solution avoids math in the printf()<\/em> statement, but it requires a knowledge that you can perform multiple actions within a single for<\/em> loop statement.<\/p>\n
This next solution uses a while<\/em> loop and a single char<\/em> variable alphabet<\/code>:<\/p>\n
2022_08-Exercise-c.c<\/a><\/h3>\n
\r\n#include <stdio.h>\r\n\r\nint main()\r\n{\r\n    char alphabet = 'A';\r\n\r\n    while( alphabet!='Z'+1)\r\n    {\r\n        putchar(alphabet);\r\n        putchar(alphabet+32);\r\n        alphabet++;\r\n    }\r\n    putchar('\\n');\r\n\r\n    return(0);\r\n}<\/pre>\n
The while<\/em> loop spins as long as the value of alphabet<\/code> doesn't equal 'Z'+1<\/code>. This expression allows letters Z and z to be output, because the loop stops after Z is output. The putchar()<\/em> at Line 10 adds 32 to the value of alphabet<\/code>, generating lowercase letter output. Most importantly, alphabet<\/code> is incremented at Line 11, which keeps the loop moving.<\/p>\n
Now I get bizarre:<\/p>\n
2022_08-Exercise-d.c<\/a><\/h3>\n
\r\n#include <stdio.h>\r\n\r\nint main()\r\n{\r\n    \/* size the array for two alphabets (26)\r\n       and the null character *\/<\/span>\r\n    char a[26*2+1];\r\n    char alpha;\r\n    int x;\r\n\r\n    \/* do uppercase *\/<\/span>\r\n    alpha = 'A';\r\n    for( x=0; x<26*2; x+=2 )\r\n    {\r\n        a[x] = alpha;\r\n        alpha++;\r\n    }\r\n\r\n    \/* do lowercase *\/<\/span>\r\n    alpha = 'a';\r\n    for( x=0; x<26*2; x+=2 )\r\n    {\r\n        a[x+1] = alpha;\r\n        alpha++;\r\n    }\r\n\r\n    \/* append the null char *\/<\/span>\r\n    a[26+26] = '\\0';\r\n\r\n    \/* output result *\/<\/span>\r\n    puts(a);\r\n\r\n    return(0);\r\n}<\/pre>\n
Two for<\/em> loops fill array a[]<\/code>. The first sets uppercase letters to every other position in the array. The second for<\/em> loop works similarly, but for lowercase and using a[x+1]<\/code> as its index value. The +1<\/code> shifts the lowercase letters over, squeezing them between uppercase letters already set in the array. It's kinda crazy, but it doesn't attain the level of insanity as this version:<\/p>\n
2022_08-Exercise-e.c<\/a><\/h3>\n
\r\n#include <stdio.h>\r\n#include <stdlib.h>\r\n\r\nint main()\r\n{\r\n    const int size = 26*2+1;\r\n    char *a;\r\n    int x,y;\r\n    \r\n    \/* allocate storage for two alphabets plus\r\n       a null character *\/<\/span>\r\n    a = malloc( sizeof(char) * size );\r\n    if( a==NULL )\r\n    {\r\n        fprintf(stderr,\"Unable to allocate memory\\n\");\r\n        exit(1);\r\n    }\r\n\r\n    \/* initialize the buffer to all As *\/<\/span>\r\n    for( x=0; x<size; x++ )\r\n        *(a+x) = 'A';\r\n\r\n\r\n    \/* manipulate contents *\/<\/span>\r\n    for( x=0,y=0; x<26*2; x+=2,y++ )\r\n    {\r\n        *(a+x) = *(a+x) + y;\r\n        *(a+x+1) = *(a+x+1) + y + 32;\r\n    }\r\n\r\n    \/* output the results *\/<\/span>\r\n    *(a+size-1) = '\\0';\r\n    puts(a);\r\n\r\n    return(0);\r\n}<\/pre>\n
My nuttiest solution uses a pointer, allocates storage, then fills the buffer with uppercase letter 'A'<\/code>. The action takes place at Line 25, where a for<\/em> loop uses two variables to process the buffer. The first looping variable x<\/code> is increased by two each turn; variable y<\/code> incremented by one.<\/p>\n
The statements at Lines 27 and 28 adjust the value of character 'A'<\/code> (already in the buffer) to create the mixed upper- lowercase string. It took a few tries for me to code something so convoluted, but it works.<\/p>\n
I hope you came up with a working solution, easy or hard. If you're an experienced coder, I hope you wove a tapestry of statements that inspires and delights. That's the point of the exercise.<\/p>\n","protected":false},"excerpt":{"rendered":"
I don’t believe one definitive solution exists for this month’s Exercise: Output a string of upper- and lowercase letters. No, the challenge is more to discover different insights a coder has to a specific problem.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"class_list":["post-5476","post","type-post","status-publish","format-standard","hentry","category-solution"],"_links":{"self":[{"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/5476","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=5476"}],"version-history":[{"count":3,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/5476\/revisions"}],"predecessor-version":[{"id":5483,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/5476\/revisions\/5483"}],"wp:attachment":[{"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5476"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5476"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5476"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}