{"id":5850,"date":"2022-08-30T23:03:35","date_gmt":"2022-08-30T17:33:35","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-i-want-to-create-a-program-in-c-to-check-for-and-if-typed-in-by-user-closed\/"},"modified":"2022-08-30T23:03:35","modified_gmt":"2022-08-30T17:33:35","slug":"solved-i-want-to-create-a-program-in-c-to-check-for-and-if-typed-in-by-user-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-i-want-to-create-a-program-in-c-to-check-for-and-if-typed-in-by-user-closed\/","title":{"rendered":"[Solved] I want to create a program in C to check for @ and . if typed in by user [closed]"},"content":{"rendered":"

[ad_1]
\n<\/p>\n

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<\/div>\n
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The major problem your code is facing is not about C but about the logical structure of your function. If your emailadd<\/code> contains any character different from @<\/code> and\/or different from .<\/code> you ask for a new email address.
\nThat’s not what you want. <\/p>\n

Let’s structure your problem. Here is what you want to do:<\/p>\n

    \n
  1. Get user input<\/li>\n
  2. Check if the input contains an @<\/code><\/li>\n
  3. If not, ask for new input<\/li>\n
  4. Check if the input contains a .<\/code><\/li>\n
  5. If not, aks for new input<\/li>\n<\/ol>\n

    You don’t want to do this once, but as long the user input is correct. To be even more precise you want to get a .<\/code> after the @<\/code> (otherwise an email address like stack.overflow@so<\/code> would be valid.<\/p>\n

    Consider this approach which implements exactly the behaviour defined above:<\/p>\n

    #include <stdio.h>\n#include <stdlib.h>\n#include <string.h>\n\nint main()\n{\n    char emailadd[50];\n    size_t emailaddL;   \/* Use size_t instead of int *\/\n\n    \/* Get the email address for the first time *\/\n    printf(\"Please enter your email address\\n\");\n    scanf(\"%s\", emailadd);\n    emailaddL=strlen(emailadd);\n\n    \/* Loop while the address has a wrong format *\/\n    while(1){\n        \/* Create flags: 0 character missing, 1 character found *\/\n        int at_flag=0;\n        int dot_flag=0;\n        emailaddL=strlen(emailadd);\n\n        \/* Loop through the char array. You have to loop from 0 to the array lenght -1*\/\n        for(int i=0; i<emailaddL; i++){    \/* You can declare variables in for loops *\/\n\n            \/* If an @ has been found, set the flag *\/\n            if(emailadd[i] == '@'){\n                at_flag=1;\n            }\n            \/* If a . has been found and @ has been found earlier, set the flag *\/\n            if(at_flag && emailadd[i] == '.'){\n                dot_flag=1;\n            }\n            \/* If both flags have been set, you can stop searching *\/\n            if(at_flag && dot_flag){\n                break;\n            }\n        }\n\n        \/* If no @ has been found, ask for it *\/\n        if(!at_flag){\n            printf(\"The entered email address is incorrect. Please re-enter your email address and it must contain the '@' specifier\\n\");\n            scanf(\"%s\", emailadd);\n        \/* If an @ has been found, but no ., ask for it *\/\n        } else if(!dot_flag){\n            printf(\"The entered email address is incorrect. Please re-enter your email address and it must contain the '.' specifier, after the '@' specifier\\n\");\n            scanf(\"%s\", emailadd);\n        \/* Otherwise leave the loop *\/\n        } else{\n            printf(\"Everything is fine!\\n\");\n            break;\n        }\n    }\n}\n<\/code><\/pre>\n
    Let’s try some scenarios:<\/p>\n
    $ Please enter your email address\n$ stack.overflow \n$ The entered email address is incorrect. Please re-enter your email address and it must contain the '@' specifier\n$ stack.overflow@so\n$ The entered email address is incorrect. Please re-enter your email address and it must contain the '.' specifier, after the '@' specifier\n$ stack.overflow@so.com\n$ Everything is fine!\n<\/code><\/pre>\n<\/p><\/div>\n
    <\/div>\n<\/div>\n
                1<\/span> <\/p><\/div>\n<\/div>\n
    [ad_2]<\/p>\n
    solved I want to create a program in C to check for @ and . if typed in by user [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"
    [ad_1] The major problem your code is facing is not about C but about the logical structure of your function. If your emailadd contains any character different from @ and\/or different from . you ask for a new email address. That’s not what you want. Let’s structure your problem. Here is what you want to … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[324],"class_list":["post-5850","post","type-post","status-publish","format-standard","hentry","category-solved","tag-c"],"yoast_head":"\n[Solved] I want to create a program in C to check for @ and . if typed in by user [closed] - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-i-want-to-create-a-program-in-c-to-check-for-and-if-typed-in-by-user-closed\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] I want to create a program in C to check for @ and . if typed in by user [closed] - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] The major problem your code is facing is not about C but about the logical structure of your function. 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