Using ternary operator to initialize a reference variable?

Question

Putting all the maintainability and reading issues aside, can these lines of code generate undefined behavior?

float  a = 0, b = 0;
float& x = some_condition()? a : b;
x = 5;
cout << a << ", " << b;

could be even simpler: `( some_condition() ? a : b ) = 5;` – Slava Oct 20 '16 at 18:10 — Slava, Oct 20 '16 at 18:10

score 13 · Accepted Answer · answered Jul 15 '12 at 19:39

13

No, it's just fine. It would not create undefined behavior in this code. You will just change value of a or b to 5, according to condition.

answered Jul 15 '12 at 19:39

Blood

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score 9 · Answer 2 · answered Jul 15 '12 at 19:42

9

This is absolutely fine, as long as both sides of the conditional are expressions that can be used to initialize a reference (e.g. variables, pointer dereferences, etc)

float& x = some_condition()? a : *(&b); // This is OK - it is the same as your code
float& x = some_condition()? a : b+1;   // This will not compile, because you cannot take reference of b+1

answered Jul 15 '12 at 19:42

Sergey Kalinichenko

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Well, inability to compile is a perfectly defined behavior, isn't it? – Desmond Hume Jul 15 '12 at 19:45
Arguably, yes- an ill-formed program does exhibit defined behaviour- nothing. – Puppy Jul 15 '12 at 19:46

Using ternary operator to initialize a reference variable?

2 Answers2

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