{"id":4490,"date":"2020-12-08T00:01:33","date_gmt":"2020-12-08T08:01:33","guid":{"rendered":"https:\/\/c-for-dummies.com\/blog\/?p=4490"},"modified":"2021-01-08T09:00:34","modified_gmt":"2021-01-08T17:00:34","slug":"some-weighty-conversions-solution","status":"publish","type":"post","link":"https:\/\/c-for-dummies.com\/blog\/?p=4490","title":{"rendered":"Some Weighty Conversions – Solution"},"content":{"rendered":"

This month’s Exercise<\/a> is to create a program that converts weight measurements by using this format:<\/p>\n

nnnFT<\/em><\/code><\/p>\n

Where nnn<\/em><\/code> is a value, F<\/em><\/code> is the unit to convert from, and T<\/em><\/code> is the unit to convert to. This problem is more of an input-parsing exercise than a straight conversion.
\n
\nFor my solution, I use an if-else if-else<\/em> structure to sift through the various permutations of K (kilos), P (pounds), and S (stone). The first step, however, is to pull out the relevant information.<\/p>\n

To process the string I use the strtof()<\/em> function. This function is far superior to the old atof()<\/em> function; strtof()<\/em> uses two pointers, illustrated in Figure 1.<\/p>\n

\"\"

Figure 1. How the strtof()<\/em> function works<\/p><\/div>\n

The first pointer, input<\/code>, references the string containing the value — and the value must come first. The second pointer, which is the address of a pointer (convert<\/code> in my example), will hold the location of the first non-value character in the string, which comes in handy for plucking out the conversion characters.<\/p>\n

After input is read, this statement converts the value:<\/p>\n

w = strtof(input,&convert);<\/code><\/p>\n

Variable w<\/code> holds the floating point value stored in string input<\/code>, with pointer variable convert<\/code> holding the location of the next character after the value. To ensure that the input was valid, a test is performed on w:<\/p>\n

\tif( w==0.0 && convert==input )
\n    error_exit();<\/code><\/p>\n

When the value returned by strtof()<\/em> is zero and both pointers are equal, the input contained no value. The code exits with an error message (error_exit()<\/em>) when true.<\/p>\n

After converting and checking the value, another test is performed to ensure that two characters are typed after the value, where convert<\/code> is a pointer referencing the first character:<\/p>\n

if( *convert=='\\0' ||
\n    *convert=='\\n' ||
\n    *(convert+1)=='\\0' ||
\n    *(convert+1)=='\\n')
\n{
\n    error_exit();
\n}<\/code><\/p>\n

At this point, I’m assured that two characters are found after the value, so a series of tests are performed to examine these characters and output the proper results. If anything goes awry during these tests, control is thrust to the error_exit()<\/em> function, which terminates the code.<\/p>\n

Here is my solution:<\/p>\n

2020_12-Exercise.c<\/a><\/h3>\n
\r\n#include <stdio.h>\r\n#include <stdlib.h>\r\n#include <ctype.h>\r\n\r\nvoid error_exit(void)\r\n{\r\n    puts(\"Invalid input\");\r\n    exit(1);\r\n}\r\n\r\nint main()\r\n{\r\n    float w;\r\n    char input[16];\r\n    char *r,*convert;\r\n\r\n    \/* intro text *\/<\/span>\r\n    puts(\"Type a value to convert, suffixed by conversion types:\");\r\n    puts(\"K - kilos, P - pounds, S - stones\");\r\n    puts(\"Example: 2.0PK (convert 2.0 pounds to kilograms)\");\r\n    printf(\"Convert: \");\r\n\r\n    \/* gather input *\/<\/span>\r\n    r = fgets(input,16,stdin);\r\n    if( r==NULL )\r\n        error_exit();\r\n\r\n    \/* process input *\/<\/span>\r\n    w = strtof(input,&convert);\r\n        \/* text for nothing input *\/<\/span>\r\n    if( w==0.0 && convert==input )\r\n        error_exit();\r\n        \/* confirm that conversion characters exist *\/<\/span>\r\n    if( *convert=='\\0' ||\r\n        *convert=='\\n' ||\r\n        *(convert+1)=='\\0' ||\r\n        *(convert+1)=='\\n')\r\n    {\r\n        error_exit();\r\n    }\r\n        \/* kilo conversions *\/<\/span>\r\n    if( toupper(*convert)=='K' )\r\n    {\r\n        if( toupper( *(convert+1))=='P' )\r\n            printf(\"%.2f kilos is %.2f pounds\\n\",w,w*2.20462);\r\n        else if( toupper( *(convert+1))=='S' )\r\n            printf(\"%.2f kilos is %.2f stone\\n\",w,w*0.157473);\r\n        else\r\n            error_exit();\r\n    }\r\n        \/* pounds conversions *\/<\/span>\r\n    else if( toupper(*convert)=='P' )\r\n    {\r\n        if( toupper( *(convert+1))=='K' )\r\n            printf(\"%.2f pounds is %.2f kilos\\n\",w,w*0.453592);\r\n        else if( toupper( *(convert+1))=='S' )\r\n            printf(\"%.2f pounds is %.2f stone\\n\",w,w*0.0714286);\r\n        else\r\n            error_exit();\r\n    }\r\n        \/* stone conversions *\/<\/span>\r\n    else if( toupper(*convert)=='S' )\r\n    {\r\n        if( toupper( *(convert+1))=='K' )\r\n            printf(\"%.2f stone is %.2f kilos\\n\",w,w*6.35029);\r\n        else if( toupper( *(convert+1))=='P' )\r\n            printf(\"%.2f stone is %.2f pounds\\n\",w,w*14.0);\r\n        else\r\n            error_exit();\r\n    }\r\n    else\r\n        error_exit();\r\n\r\n    return(0);\r\n}<\/pre>\n
The if-else if-else<\/em> monster scours for permutations of K, P, and S. Then a nested if-else if-else<\/em> structure scans for the other letter and handles bogus input. Conversions are made within the printf()<\/em> statements that output the results. Various toupper()<\/em> functions translate lowercase letters to uppercase.<\/p>\n
Type a value to convert, suffixed by conversion types:
\nK - kilos, P - pounds, S - stones
\nExample: 2.0PK (convert 2.0 pounds to kilograms)
\nConvert: 103.5kp
\n103.50 kilos is 228.18 pounds<\/code><\/p>\n
In the sample run above, 103.5 kilograms is converted into pounds: 228.18.<\/p>\n
I hope your solution works and that proper output is generated.<\/p>\n","protected":false},"excerpt":{"rendered":"
This month’s Exercise is to create a program that converts weight measurements by using this format: nnnFT Where nnn is a value, F is the unit to convert from, and T is the unit to convert to. This problem is … Continue reading →<\/span><\/a><\/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-4490","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\/4490","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=4490"}],"version-history":[{"count":9,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/4490\/revisions"}],"predecessor-version":[{"id":4528,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/4490\/revisions\/4528"}],"wp:attachment":[{"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4490"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=4490"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=4490"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

(I don’t cover the strtof()<\/em> function specifically in this blog, though I do cover its relative strtod()<\/em> in this blog post<\/a>.)<\/p>\n