獲取錯誤：無法比擬的類型：字符和字符串

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有沒有辦法將字符串數組轉換爲字符數組。目前，我的代碼：獲取錯誤：無法比擬的類型：字符和字符串

String[] alphabet = {"a","A","b","B","c","C","d","D","e","E","f","F","g","G","h","H","i","I","j","J","k","K","l","L","m","M","n","N","o","O","p","P","q","Q","r","R","s","S","t","T","u","U","v","V","w","W","x","X","y","Y","z","Z"}; 
    char[] charArray = alphabet.toCharArray();

目前這給我的錯誤信息：

error: cannot find symbol 
     char[] charArray = alphabet.toCharArray();

來源

2017-10-21 Auroch0

爲什麼你需要在所有使用'String'陣列？ –

你需要一個字符串而不是一個字符串數組。 –

此代碼將轉換字符串數組，字符數組

String[] alphabet = {"a","A","b","B","c","C","d","D","e","E","f","F","g","G","h","H","i","I","j","J","k","K","l","L","m","M","n","N","o","O","p","P","q","Q","r","R","s","S","t","T","u","U","v","V","w","W","x","X","y","Y","z","Z"}; 
String s = ""; 
for (String n:alphabet) 
    s+= n; 
char[] c = s.toCharArray();

來源

2017-10-21 12:00:09

爲什麼要做這樣的事情...... –

@Aominè其實，問題是從字符串數組中獲得一個char數組。我知道這種類型轉換可以避免，但這是個問題 –

你需要使用String對象，因爲toCharArray是String類的方法

String alphabet = "aAbBcCdDeEfFgGhHiIjJkKlLmMnNoOpPqQrRsStTuUvVwWxXyYzZ"; 
char[] charArray = alphabet.toCharArray();

剛如果你仍然想要一個字符串數組，只需分割空字符串

String [] arr = alphabet.split(""); // your String array

來源

2017-10-21 11:58:35

獲取錯誤：無法比擬的類型：字符和字符串

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