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Carole Cox
429 PointsI can't get past varl not being defined. it looks like I am following the lecture to the "t".
It looks like I am not getting the challenge question again. I am going back and forth with the lecture-this looks right....but :-(
varl = "smart"
varl.index('t')
2 Answers
Rick Buffington
8,146 PointsYou are super close. All they are looking for is a variable named var1 being assigned a string where var1[5] is the letter "t". Where you are a smidgen off is your index start point. Remember, the indexes on strings and arrays start at 0. So, if they want your var1[5] to be "t", that means it would be 0, 1, 2, 3, 4, [5] - the 6th character in the string. Add 1 more letter before the "t" and you will be golden! You do not need the var1.index piece - only declaring var1 is required.
Carole Cox
429 PointsOh my! How "duh" of me. I also saw another mistake that was creating the defined issue. the type in the challenge mad var1 look like varl. I was using varl which the challenge would not see! Once I saw your help, I saw another problem. Finally a pass!. Thanks Rick!
Rick Buffington
8,146 PointsNo problem. Glad I could help.