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:^(
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阿蘭·達瓦卓瑪
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想向專少少發展
想知Oracle Database Administration Certified Professional 考完有冇用?
心思思想報systematics
本人淨係得個degree同uipath RPA既cert
唔知點開始好
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學左新既黑魔法
The Fortran I compiler would expand each operator with a sequence of parentheses. In a simplified form of the algorithm, it would
replace + and – with ))+(( and ))-((, respectively;
replace * and / with )*( and )/(, respectively;
add (( at the beginning of each expression and after each left parenthesis in the original expression; and
add )) at the end of the expression and before each right parenthesis in the original expression.
Although not obvious, the algorithm was correct, and, in the words of Knuth, “The resulting formula is properly parenthesized, believe it or not.”
:^(
:^(
The Fortran I compiler would expand each operator with a sequence of parentheses. In a simplified form of the algorithm, it would
replace + and – with ))+(( and ))-((, respectively;
replace * and / with )*( and )/(, respectively;
add (( at the beginning of each expression and after each left parenthesis in the original expression; and
add )) at the end of the expression and before each right parenthesis in the original expression.
Although not obvious, the algorithm was correct, and, in the words of Knuth, “The resulting formula is properly parenthesized, believe it or not.”
Code4Food
Yes, you can generalize it to + - * / and ^.
+ -> )))+((( ditto for -
* -> ))*((
^ -> )^(
enclose the whole expression with ((( and )))
If you have n different precedence level, use one level of () for the operators with the highest precedence and add one () for each level going down.
+ -> )))+((( ditto for -
* -> ))*((
^ -> )^(
enclose the whole expression with ((( and )))
If you have n different precedence level, use one level of () for the operators with the highest precedence and add one () for each level going down.