$x == 1 ? print "x was '1'" : print "x was something else";

While that's technically correct it doesn't take advantage of the fact that the ternary operator has a return value; whatever expression is evaluate after the ?.

See another adaptation that takes advantage of this feature.

print( "x was " . $x == 1 ? "'1'" : "someting else" );

[download]

Dave

In reply to Re^2: "x ? y : z" notation by davido
in thread "x ? y : z" notation by apok

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