{"id":32925,"date":"2023-02-03T02:29:43","date_gmt":"2023-02-02T20:59:43","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-why-does-operator-work-as-operator-in-c-when-compared-with-0-9-duplicate\/"},"modified":"2023-02-03T02:29:43","modified_gmt":"2023-02-02T20:59:43","slug":"solved-why-does-operator-work-as-operator-in-c-when-compared-with-0-9-duplicate","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-why-does-operator-work-as-operator-in-c-when-compared-with-0-9-duplicate\/","title":{"rendered":"[Solved] Why does >= operator work as > operator in C when compared with 0.9 [duplicate]"},"content":{"rendered":"

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Your code is equivalent to:<\/p>\n

double x1 = 0.9;\nfloat xf = (float)x1; \/\/ discard some precision\ndouble x2 = (double)xf;\n\nif (x2 >= x1) {\n    \/\/ ...\n}\n<\/code><\/pre>\n
The problem is that, after you discarded some precision from x1<\/code> by converting it to float<\/code>, it can never be recovered; casting it back to double<\/code> doesn’t regain that precision. So x2<\/code> is still not the same as 0.9<\/code>, and due to the rounding direction, it happens to be a bit less. (Note: if you try the same thing with other values, you will sometimes find that it’s a bit more<\/em>.)<\/p>\n<\/p><\/div>\n
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            <\/span> <\/p><\/div>\n<\/div>\n
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solved Why does >= operator work as > operator in C when compared with 0.9 [duplicate] <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] Your code is equivalent to: double x1 = 0.9; float xf = (float)x1; \/\/ discard some precision double x2 = (double)xf; if (x2 >= x1) { \/\/ … } The problem is that, after you discarded some precision from x1 by converting it to float, it can never be recovered; casting it back 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":[324,359,639,1299],"class_list":["post-32925","post","type-post","status-publish","format-standard","hentry","category-solved","tag-c","tag-floating-point","tag-if-statement","tag-precision"],"yoast_head":"\n[Solved] Why does >= operator work as > operator in C when compared with 0.9 [duplicate] - 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-why-does-operator-work-as-operator-in-c-when-compared-with-0-9-duplicate\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Why does >= operator work as > operator in C when compared with 0.9 [duplicate] - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] Your code is equivalent to: double x1 = 0.9; float xf = (float)x1; \/\/ discard some precision double x2 = (double)xf; if (x2 >= x1) { \/\/ ... } The problem is that, after you discarded some precision from x1 by converting it to float, it can never be recovered; casting it back to ... 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