嵌套"for"循环n次 [英] Nesting "for" Loop n Times

问题描述

我正在编写一个查找密码的程序.当我看到必须替换所选密码长度的变量的"for"循环来替换部分密码时，我遇到了一个问题.该程序的目标是生成并检查任何字符数组的密码，从"0"开始并经过?(n次)"，其中"0"是第一个字符，?"是最后一个字符.

I'm writing a program that finds passwords. I ran into a problem when I saw that the "for" loops to replace parts of the password would have to be repeated for the variable of the chosen password length. The goal of this program is to generate and check a password of any character array, starting with "0" and going through "? (n times)", where '0' is the first character and '?' is the last character.

有没有办法重复for循环可变的次数 而无需分别对其进行编码?

Is there any way to repeat a for loop a variable number of times without coding them in individually?

注意，基于广为接受的评论:
重复一个for循环"可能更正确地表示为嵌套多个for循环".

Note, based on well-received comments:
The "repeat a for loop" is probably more correctly expressed as "nest several for loops".

int maxLength = 32; //Length of the password, changes via input
char output[maxLength] = "";
for (int currentLength = 1; currentLength < maxLength + 1; currentLength++) {
    for (int characterNumber = 0; characterNumber < 96 /*characters found with switch/case, 95 total*/; characterNumber++) {
        for (int position = currentLength /*position in string*/; position > 0; position--) {
            //For loops of character and position currentLength times
            char newCharacter = '\0'; //Avoiding not initialized error, could be and character
            output[position - 1] = getChar(characterNumber, newCharacter);
        }
    }
}

输出示例为: ... 01，02，03 ...，10，11 ...，a0，a1 ...，???? 4afuh7yfzdn_)aj901 ...

Example of an output would be: ...01, 02, 03..., 10, 11..., a0, a1..., ????4afuh7yfzdn_)aj901...

推荐答案

您无需重复for循环.相反，您需要将其嵌套.
实现此目的最吸引人的解决方案是递归构造.

You do not need to repeat the for loop. Instead you need to nest it.
The most inviting solution to implement this is a recursive construct.

伪代码:

void nestloop(int depth, int width)
{
    if(depth>0)
    {
        int i;
        for (i=0; i<width; i++)
        {
            nestloop(depth-1, width);
        }
    } else
    {
        /* do whatever you need done inside the innermost loop */
    }
}

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