console.log(“foo” + + “bar”)
result = fooNaN
Please explain the result
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“foo” + + “bar”
= “foo” + (+ “bar”)
= “foo” + NaN
= fooNaN
Unary + operator has higher precedence over binary + operator.
So + “bar” expression will be executed first. Unary + triggers numeric conversion for string “bar”. As the string does not represent a valid number so the result is NaN.
Hence in the end “foo” + NaN is evaluated.
Few Cases-
- console.log(‘foo’ + ’ ’ + ‘bar’)
result= foo bar
In this case you have to add space between both the strings. - console.log(‘foo’ + ‘bar’)
result= foobar
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There are many methods for type conversion like String(anything)=“Anything” or Number (“1.23”) =1.23, similarly + operator has so many functionalities if it is used with “a”+“b” =“ab”,in string used for concatenation ,2+3=5 just an additional operator,now it is also used for converting string to number if the passed value is Numeric data inside double quotes.so const a=“123”,console.log(+a),output=123,so now the precedence of this type conversion is higher than concatnation operator,so +“bar”=NaN,so “foo”+NaN=“fooNaN”
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