Why return gives error when used with ? :

Ruby
1

Hi all

in my program I am using a lot of true false conditions. So, I want to
shorten my code by using the conditional operator ? : but when I use the
word return it gives an error which i do not really know why?

e.g. x == 5 ? return true : return false
SyntaxError: compile error
(irb):19: syntax error, unexpected kTRUE

but if i do not use return it works

x == 5 ? true : false

any idea

2

Oops I forgot to ask
which one is faster
if x == 5
return true
else
return false
end

or the conditional operator ? :
x == 5 ? true : false

3

Shuaib Z. wrote:

but if i do not use return it works

x == 5 ? true : false

any idea

That’s the format of the ?: operator.

you could do:
z = (x==5)?true: false
return z

Cheers,
Mohit.
11/6/2007 | 6:11 PM.

4

Shuaib Z. wrote:

but if i do not use return it works

x == 5 ? true : false

The ternary operator ? : evaluates to a value, so it expects values (or
expressions evaluating to values, but not statements as in your example)
after ? and : . You would also not write

x = return 5

because during an assignment, Ruby expects a value on the right side of
the equation mark.

Cheers,
Peter

http://www.rubyrailways.com
http://scrubyt.org

5

On 11/6/07, Shuaib Z. [email protected] wrote:

Hi all

in my program I am using a lot of true false conditions. So, I want to
shorten my code by using the conditional operator ? : but when I use the
word return it gives an error which i do not really know why?

e.g. x == 5 ? return true : return false

you may write it as return x == 5 ? true : false
and in this case you can omit the “? true : false”, because the value
of x == 5 is just that.
so in your case write just
return x == 5
and if it is the lase statement, you can event omit the return

6

Shuaib Z. wrote:

Hi all

in my program I am using a lot of true false conditions. So, I want to
shorten my code by using the conditional operator ? : but when I use the
word return it gives an error which i do not really know why?

e.g. x == 5 ? return true : return false
SyntaxError: compile error
(irb):19: syntax error, unexpected kTRUE

but if i do not use return it works

x == 5 ? true : false

any idea

def f
x = 4
x==5 ? (return true) : (return false)
end

puts f
–>false

7

Hi all,

On Tue, 2007-11-06 at 19:33 +0900, 7stud – wrote:

end

puts f
–>false

How about:

return (x == 5)

:slight_smile:

Cheers,
Arlen

8

Peter S. wrote:

The ternary operator ? : evaluates to a value, so it expects values (or
expressions evaluating to values, but not statements as in your example)
after ? and : . You would also not write

x = return 5

because during an assignment, Ruby expects a value on the right side of
the equation mark.

The ternary expression don’t require value/experession but the
assignment does so the problem is precedence rather than value
requirement. A code like this will work becouse parentheses solves the
problem:

def bla(x)
x == 5 ? (return true) : (return false)
end

but this won’t becouse of the assignment that requires a
value/expression:

x = (x == 5) ? (return true) : (return false)

cheers
MoisesMachado

9

On Nov 6, 3:11 am, Shuaib Z. [email protected] wrote:

Oops I forgot to ask
which one is faster
if x == 5
return true
else
return false
end

or the conditional operator ? :
x == 5 ? true : false

They’re about the same speed. Where you will see a small performance
gain, however, is leaving off the call to the return method
altogether. The value of the last expression in a method is the return
value. The only time to use return if you need to exit a method early.
(And this practice is considered a bad idea by some people in many
situations.)

require ‘benchmark’

def ternary_return1( x )
return ( x==5 ? true : false )
end

def ternary_return2( x )
x==5 ? (return true) : (return false)
end

def if_return1( x )
return (if x == 5
true
else
false
end)
end

def if_return2( x )
if x == 5
return true
else
return false
end
end

def ternary_expression( x )
x == 5 ? true : false
end

def if_expression( x )
if x == 5
true
else
false
end
end

values = (1…8).to_a * 500_000
method_names = %w| ternary_return1 ternary_return2
if_return1 if_return2
ternary_expression if_expression |

Benchmark.bmbm{ |x|
method_names.each{ |name|
meth = method( name )
x.report( name ){
values.each{ |n|
meth.call( n )
}
}
}
}

Rehearsal ------------------------------------------------------
ternary_return1 6.125000 0.000000 6.125000 ( 6.157000)
ternary_return2 5.859000 0.000000 5.859000 ( 5.906000)
if_return1 6.016000 0.000000 6.016000 ( 6.047000)
if_return2 5.906000 0.000000 5.906000 ( 5.953000)
ternary_expression 5.453000 0.000000 5.453000 ( 5.484000)
if_expression 5.672000 0.000000 5.672000 ( 5.687000)
-------------------------------------------- total: 35.031000sec

                     user     system      total        real

ternary_return1 6.047000 0.000000 6.047000 ( 6.078000)
ternary_return2 6.250000 0.000000 6.250000 ( 6.282000)
if_return1 6.032000 0.000000 6.032000 ( 6.063000)
if_return2 6.157000 0.000000 6.157000 ( 6.187000)
ternary_expression 5.656000 0.000000 5.656000 ( 5.703000)
if_expression 5.750000 0.000000 5.750000 ( 5.781000)

10

Shuaib Z. wrote:

Oops I forgot to ask
which one is faster

there not really a difference in functionality between the two forms, i
think that it’s the same underlying implementation so same speed
(correct me if i’m wrong didn’t run the benchs).

for instance I can do a multi command conditional with ?:

cond ? (command1; command2; command3) : command4

and one line with if:

if cond then command1 else command2 end

cheers
MoisesMachado

11

Others have pointed out that (in this particular example) it’s foolish
to branch on a boolean value only to return that particular value as a
new literal, so here’s a modified test showing the fastest way - just
let the boolean value return itself.

require ‘benchmark’

def ternary_return1( x )
return ( x==5 ? true : false )
end

def ternary_return2( x )
x==5 ? (return true) : (return false)
end

def if_return1( x )
return (if x == 5
true
else
false
end)
end

def if_return2( x )
if x == 5
return true
else
return false
end
end

def ternary_expression( x )
x == 5 ? true : false
end

def if_expression( x )
if x == 5
true
else
false
end
end

def boolean_expression( x )
x == 5
end

def return_boolean( x )
return x == 5
end

values = (1…8).to_a * 1_000_000
method_names = %w| ternary_return1 ternary_return2
if_return1 if_return2
ternary_expression if_expression
boolean_expression return_boolean |

Benchmark.bmbm{ |x|
method_names.each{ |name|
meth = method( name )
x.report( name ){
values.each{ |n|
meth.call( n )
}
}
}
}

Rehearsal ------------------------------------------------------
ternary_return1 12.125000 0.000000 12.125000 ( 12.172000)
ternary_return2 11.750000 0.000000 11.750000 ( 11.812000)
if_return1 11.797000 0.000000 11.797000 ( 11.813000)
if_return2 11.453000 0.000000 11.453000 ( 11.484000)
ternary_expression 10.516000 0.000000 10.516000 ( 10.578000)
if_expression 11.218000 0.000000 11.218000 ( 11.250000)
boolean_expression 9.719000 0.000000 9.719000 ( 9.782000)
return_boolean 10.797000 0.000000 10.797000 ( 10.812000)
-------------------------------------------- total: 89.375000sec

                     user     system      total        real

ternary_return1 11.578000 0.000000 11.578000 ( 11.625000)
ternary_return2 12.141000 0.000000 12.141000 ( 12.188000)
if_return1 11.765000 0.000000 11.765000 ( 11.812000)
if_return2 12.031000 0.000000 12.031000 ( 12.078000)
ternary_expression 10.969000 0.000000 10.969000 ( 10.999000)
if_expression 11.078000 0.000000 11.078000 ( 11.125000)
boolean_expression 10.093000 0.000000 10.093000 ( 10.125000)
return_boolean 11.141000 0.000000 11.141000 ( 11.188000)