Is there a representation with logical operators that returns the same result for ternary operator?

user_5698517 0
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I am curious about if a logical representation exists for ternary operator that gives the same result for each condition combination. I tried below but it doesn't cover all situations. Is there a correct representation for this problem?

Ternary Condition : a ? b : c
Logical Version : (a && b) || c

Logical version fails in one condition compared to ternary results when the conditions below are given:

(true  && true)  || true  = true 
(true  && true)  || false = true 
(true  && false) || true  = true //must be false
(true  && false) || false = false 
(false && true)  || true  = true 
(false && true)  || false = false
(false && false) || true  = true
(false && false) || false = false
logical-operators conditional-operator c#
1 Answers

If we talk about pre-evaluated boolean variables or constants, a logical equivalent to the ternary operator can be written like Lee commented

(a && b) || (!a && c)

Or alternatively as

(!a || b) && (a || c)

However, since this question is flagged as C#, I think it's important to discuss the case of a, b or c actually being boolean expressions with side effects (like a function call or a property getter with backing logic)

So lets compare the approaches with regard to, how often a, b and c are evaluated:

Case 1: the ternary operator a ? b : c, a is evaluated exactly once, and depending on a, b or c is evaluated once.

Case 2: (a && b) || (!a && c) The number of evaluations of a for a=1 depend on the value of b. Note that #a is the number of evaluations for a expression.

a   b   c   #a  #b  #c

0   0   0   2   0   1
0   0   1   2   0   1
0   1   0   2   0   1
0   1   1   2   0   1
1   0   0   2   1   0
1   0   1   2   1   0
1   1   0   1   1   0
1   1   1   1   1   0

Case 3: (!a || b) && (a || c) again, the number of evaluations of a for a=1 depend on b, but in reverse, compared to case 2

a   b   c   #a  #b  #c

0   0   0   2   0   1
0   0   1   2   0   1
0   1   0   2   0   1
0   1   1   2   0   1
1   0   0   1   1   0
1   0   1   1   1   0
1   1   0   2   1   0
1   1   1   2   1   0

I don't consider the expression evaluation side effect, where the evaluation of a changes the value of subsequent evaluations, since it would probably be off topic.

user_52652920
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