{"id":11570,"date":"2022-09-27T22:33:02","date_gmt":"2022-09-27T17:03:02","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-doubling-a-number\/"},"modified":"2022-09-27T22:33:02","modified_gmt":"2022-09-27T17:03:02","slug":"solved-doubling-a-number","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-doubling-a-number\/","title":{"rendered":"[Solved] Doubling a number"},"content":{"rendered":"

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

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<\/div>\n
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Well, in your description you say that you need to take one<\/strong> input. However, your program reads two<\/strong> inputs.<\/p>\n

Maybe this is what you need:<\/p>\n

#include <stdio.h>\nint sum(int x, int y)\n{\n return y == 0 ? x : sum(x+1, y-1);\n}\nint main(void)\n{\n int x;\n if (scanf(\"%d\", &x) != 1)  \/\/ Only 1 input\n {\n    \/\/ Input error\n    return -1;\n }\n int z = sum(x, x);    \/\/ Use x for both arguments\n printf(\"%d\\n\", z);\n return 0;\n} \n<\/code><\/pre>\n
This works because the sum<\/code> functions increments one argument and decrements the other in the recursive call and continues doing that until one of the arguments reach zero. In this way the end result will be 2 * x<\/code><\/p>\n
Let’s say the input is 3. Then you’ll have these calls:<\/p>\n
sum(3, 3)\nsum(4, 2)\nsum(5, 1)\nsum(6, 0) \/\/ Which will end the recursion and return 6 all the way up\n<\/code><\/pre>\n<\/p><\/div>\n
<\/div>\n<\/div>\n
            1<\/span> <\/p><\/div>\n<\/div>\n
[ad_2]<\/p>\n
solved Doubling a number <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] Well, in your description you say that you need to take one input. However, your program reads two inputs. Maybe this is what you need: #include <stdio.h> int sum(int x, int y) { return y == 0 ? x : sum(x+1, y-1); } int main(void) { int x; if (scanf(“%d”, &x) != 1) \/\/ … 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-11570","post","type-post","status-publish","format-standard","hentry","category-solved","tag-c"],"yoast_head":"\n[Solved] Doubling a number - 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-doubling-a-number\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Doubling a number - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] Well, in your description you say that you need to take one input. However, your program reads two inputs. Maybe this is what you need: #include <stdio.h> int sum(int x, int y) { return y == 0 ? x : sum(x+1, y-1); } int main(void) { int x; if (scanf("%d", &x) != 1) \/\/ ... 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However, your program reads two inputs. Maybe this is what you need: #include <stdio.h> int sum(int x, int y) { return y == 0 ? x : sum(x+1, y-1); } int main(void) { int x; if (scanf(\"%d\", &x) != 1) \/\/ ... 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