char.charCodeAt(0) + 1 handles the mathematical shift for non-vowel characters, turning a 'b' into a 'c', a 't' into a 'u', etc. CodeHS 8.3.8 Python Implementation
: Double-check your for loop syntax: (let i = 0; i < str.length; i++) .
根据上面的编码表：
In this part, students are asked to create a simple encoding scheme that replaces each letter with a letter a fixed number of positions down the alphabet.
去掉空格后得到最终的答案串： 83 8 create your own encoding codehs answers exclusive
26个大写字母 + 1个空格 = 27种不同的符号。 2^4 = 16 ，不够（16 < 27）。 2^5 = 32 ，够用（32 ≥ 27）。
Are you having trouble with a in the CodeHS console, or does the logic make sense now? This seems to be about CodeHS exercises, specifically
text.charAt(i) isolates the single character at the current index.
The code above is reversible and deterministic. leaving you ready to explain
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Modulo arithmetic keeps numbers within a reasonable range (0-255). The decoding requires modular inverse or a lookup table.