Java 我不知道为什么这个代码不能正常工作，因为它'；s supossed

我不确定如何解释这段代码的作用，但我希望你能得到我想要的。它应该取hash的值behinde index 7，并用字母表的值behinde index“var”替换它

所以最后它应该在cosole中打印如下内容：“0123456789abcdefghijklmnopqrstuvwxyzabefghijklmnopqrstuvxyz”

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我不确定错误在哪里。这可能是一些非常基本的东西，但我对编程一无所知……

你看到的是“成功”，不是吗？这可能是因为abort变量。var从不为61，因为在执行for块之前将检查此条件。 这是必须的

public String createHash()
{
String[] ALPHABET={"0","1","2","3","4","5","6","7","8","9","a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"};
String[] hash={"0","1","2","3","4","5","6","7"};
for(int var=0; var==61; var++)
{
   hash[7]=ALPHABET[var];
   System.out.print(hash[7]);
}
return "Success";
}

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for（int-var=0；var）恐怕我不明白你想做什么。如果你只是想打印字母表中的所有字符，你需要
hash
做什么？错误是什么？“（int-var=0；var==61；var++”-你需要重新阅读for循环章节。@OldProgrammer没有错误。它返回“Success”应该如此，但没有错误。for循环的中间部分是一个连续条件，只有当
var==61
时，它才会进入循环的下一步-如果
var=0
，这种情况发生的频率是多少？“var永远不会是61，因为在执行for块之前会检查此条件”这是我的错误。我认为它首先将var设置为0，然后遍历整个块，将var增加1，并检查它是否为61，因为它是1，所以一直持续到var为61。感谢alotYup。在do while循环中就是这种情况！：）
for(int var = 0; var <= 61; var++)