\nAs a puzzle, commenter nobugz\nasks,\n“What kind of infinity is \nNow, the division by zero results in IEEE positive infinity,\nwould would normally be printed as\n \nThe Visual C runtime library arbitrarily decided that\nall of the exceptional values have one digit before\nthe decimal (namely, the \nOkay, so now you have one digit before the decimal\n(the \nThe first character is \nThat’s why printing IEEE positive infinity to two places\ngives you the strange-looking \nI doubt this behavior was intended;\nit’s just a consequence of taking a rounding algorithm intended\nfor digits and applying it to non-digits.\n<\/P>\n1.#J<\/CODE><\/A>?”\n<\/P>\n\ndouble z = 0;\nprintf(“%.2f”, 1\/z);\n<\/PRE>\n
1#INF<\/CODE>.\nBut the catch here is that the print format says\n“Display at most two places after the decimal point.”\nBut where is the decimal point in infinity?\n<\/P>\n“1”<\/CODE>).\nActually, it turns out that this puzzle might be an\nanswer to Random832’s question,\n“What’s\nthe 1 for<\/A>?”\nMaybe the 1 is there so that there is a digit at all.\n<\/P>\n“1”<\/CODE>),\nand now you need to show at most two places after the decimal.\nBut “#INF”<\/CODE> is too long to fit into two characters.\nThe C runtime then says,\n“Well, then I’d better round it off to two places, then.”\n<\/P>\n“#”<\/CODE>.\nThe second character is “I”<\/CODE>.\nNow we need to round it.\nThat’s done by inspecting the third character,\nwhich is “N”<\/CODE>.\nWe all learned in grade school that you round up if the next\ndigit is 5 or greater.\nAnd it so happens that the code point for\n“N”<\/CODE> is numerically higher than the code point for\n“5”<\/CODE>,\nso the value is rounded up by incrementing the previous digit.\nIncrementing “I”<\/CODE> gives you\n“J”<\/CODE>.\n<\/P>\n“1#J”<\/CODE>.\nThe J<\/CODE> is an I<\/CODE> that got rounded up.\n<\/P>\n