Final exercise JS basic algorithms#48
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ta-web-mad
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Oct 31, 2017
| for (var i = 0; i < halfString; i++) { | ||
| if (inputUppercase[i] !== inputUppercase[inputUppercase.length - i - 1]) { | ||
| console.log("The word is not a palindrome"); | ||
| break; |
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intenta no utilizar los break si no son un switch. Esto lo puedes resolver utilizando un return del console.log(). Ten en cuenta que el return es lo último que ejecuta una función
ta-web-mad
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ta-web-mad
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Buen ejercicio Bart, está muy bien, te he dejado un comentario en el código.
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