if(Math.abs(j) + Math.abs(k) + Math.abs(l) == i*2 && j + k + l == 0) this.hexCube.push([j,k,l])
时间: 2024-04-21 13:26:29 浏览: 12
这是一段 JavaScript 代码,它首先通过计算 j、k、l 的绝对值之和是否等于 i*2 来判断它们是否符合一个六边形坐标系中的条件。接着,它还判断 j、k、l 的和是否为 0,以确保它们在六边形坐标系中处于同一平面。如果这两个条件都满足,那么这个三元组 [j,k,l] 就会被添加到 hexCube 数组中。在六边形坐标系中,三个坐标轴的和必须为 0,因为六边形坐标系是一个立方体网格,每个立方体都由六个正方形构成,每个正方形都与另外两个正方形共享一个顶点,因此在六边形坐标系中,每个顶点都有三个相邻的正方形。
相关问题
i <= Math.sqrt(Math.abs(toCheck))
As an AI language model, I cannot provide a complete solution without knowing the context and purpose of the code. However, I can explain what this line of code does.
This line is a condition that checks if the value of "i" is less than or equal to the square root of the absolute value of "toCheck".
The Math.abs() function returns the absolute value of a number, which means it returns a positive value regardless of whether the original number was positive or negative.
The Math.sqrt() function returns the square root of a number.
Therefore, this condition is checking if the current value of "i" is less than or equal to the square root of the absolute value of "toCheck".
char c = (char)(Math.abs(i - j) + 'A');
This line of code is incorrect and will result in a compilation error.
The correct code should be:
```
char c = (char)(Math.abs(i - j) + 'A');
```
This code is converting the absolute difference between the values of i and j to a character using the ASCII code of 'A' as an offset.
For example, if i=2 and j=5, the absolute difference is 3. Adding 'A' (which has an ASCII value of 65) to 3 gives a character with ASCII value 68, which is 'D'.
Therefore, the variable c will hold the character 'D' in this example.