Ternary operator with no condition

Question

I found a section of code that shows the following:

int A = 4;
int Z;

Z = (A ? 55 : 3);

Why does the result for Z give 55?

`A` gets implicitly converted to `bool` and any numeric value different than `0` is evaluated as `true`. — Yksisarvinen, Mar 15 '19 at 23:51
Possible doplicate: [bool to int conversion](https://stackoverflow.com/questions/5369770/bool-to-int-conversion) — Yksisarvinen, Mar 15 '19 at 23:53

Matteo Italia · Answer 1 · 2019-03-16T00:10:13.003

You seem to have a common misconception about the fact that expressions in conditional statements (if, while, ...) and ternary operations must "look like" a condition, so they should contain a relational/equality/logical operator.

It's not like that. Commonly used relational/equality/... operators don't have any particular relationship with conditional statements/expressions; they can live on their own

bool foo = 5 > 4;
std::cout<<foo<<"\n"; // prints 1

and conditional statements/expressions don't care particularly for them

if(5) std::cout << "hello\n"; // prints hello

if/?/while/... just evaluate the expression, check if the result, converted to bool, is true or false, and act accordingly. If the expression doesn't "looks like" a condition is irrelevant, as long as the result can be converted to bool you can use it in a conditional.

Now, in this particular case A evaluates to 4, which is not zero, so when converted to bool is true, hence the ternary expression evaluates to its second expression, so 55.

Ternary operator with no condition

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