Conditional Operator Used in Cout Statement

Conditional operator used in cout statement

The ?: operator has lower precedence than the << operator i.e., the compiler interprets your last statement as:

(std::cout << (a != 0)) ? 42.0f : -42.0f;

Which will first stream the boolean value of (a!=0) to cout. Then the result of that expression (i.e., a reference to cout) will be cast to an appropriate type for use in the ?: operator (namely void*: see cplusplus.com), and depending on whether that value is true (i.e., whether cout has no error flags set), it will grab either the value 42 or the value -42. Finally, it will throw that value away (since nothing uses it).

C++, ternary operator, std::cout

Depends on what type is result1, result2 etc.

expressionC ? expression1 : expression2 isn't valid for all types of expression1 and expression2. They must necessarily be convertible to a common type, roughly speaking (exact rules and exceptions can be read in the standard). Now, if results are strings, then you do it like this:

std::cout << ( condition1 ? result1 : "Error" ) 
                         ^^^
          << ( condition2 ? result2 : "Error") 
                         ^^^
          << etc.

But if results are integers, for example, you can't do it.

HTH

Cout String and int in ternary operator

Ugly:

i%3==0 ? cout<< "Hello\n" : cout<<i;

Nice:

if ( i%3 == 0 )
   cout << "Hello\n";
else
   cout << i;

Your version doesn't work because the result types of the expressions on each side of : need to be compatible.

C++, ternary operator and cout

cout << 5 ? (5 ? 0 : 2) : 5;

is parsed as

(cout << 5) ? (5 ? 0 : 2) : 5;

what is the output of conditional operator with unary operator

The expression on the right hand side is evaluated like an if-statement.

if (Delay == true)
   return 1;
else
   return -1;

The result is then used for the += assignment.

In the
C++20 draft standard that's

7.6.19 (6) (Assignment and compound assignment operators)

The behavior of an expression of the form E1 op= E2 is equivalent to E1 = E1 op E2 except that E1 is
evaluated only once. [...]

Since Delay == false, the return value of the ternary operator is -1. The fact that you're operating on a boolean instead of an int can make it look like you got the +1 back.

Note that you get a compiler warning C4804:

warning C4804: '+=': unsafe use of type 'bool' in operation

Is it undefined behavior? No.

7.6.19 (6) (Assignment and compound assignment operators)

[...] For += and
-=, E1 shall either have arithmetic type or be a pointer to a possibly cv-qualified completely-defined object
type. In all other cases, E1 shall have arithmetic type.

and

7.3.8 (2) (Integral conversions)

If the destination type is bool, see 7.3.14.

which says

7.3.14 (1) (Boolean conversions)

A prvalue of arithmetic, unscoped enumeration, pointer, or pointer-to-member type can be converted to a
prvalue of type bool. A zero value, null pointer value, or null member pointer value is converted to false;
any other value is converted to true.

So the -1 is converted to true and true prints as 1.

How do I use the conditional (ternary) operator?

It works like this:

(condition) ? true-clause : false-clause

It's most commonly used in assignment operations, although it has other uses as well. The ternary operator ? is a way of shortening an if-else clause, and is also called an immediate-if statement in other languages (IIf(condition,true-clause,false-clause) in VB, for example).

For example:

bool Three = SOME_VALUE;
int x = Three ? 3 : 0;

is the same as

bool Three = SOME_VALUE;
int x;
if (Three)
    x = 3;
else
    x = 0;

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