C＃将char转换为int [英] C# convert char to int

问题描述

这是我的代码，我希望有mapfull map *的东西，但它的ASCII（或其他东西），我想把它变成正常。 be1 / 2/3是由81个字符长的数字组成的字符串（例如：79456123 ...）。你能帮助我吗？

This is my code and i want to have the mapfull map* stuff but its in ASCII(or something) and i want to turn it "normal". be1/2/3 are strings that are consist of an 81 char long numbers (example: 79456123...). Can you help me?

int ind1 = 0;
            int ind2 = 0;
            int ind3 = 0;
            //MessageBox.Show(be1);
            for (int i = 0; i <= 8; i++)
            {
                for (int j = 0; j <= 8; j++)
                {
                    mapFull[i, j] = be1[ind1++];
                    mapInitial[i, j] = be2[ind2++];
                    mapCurrent[i, j] = be3[ind3++];
                }
            }

推荐答案

假设mapFull，mapInitial和mapCurrent是 int的数组，那么这应该做：

Assuming mapFull, mapInitial and mapCurrent are arrays of int, then this should do:

mapFull[i, j] = (int)be1[ind1++] - '0';
mapInitial[i, j] = (int)be2[ind2++] - '0';
mapCurrent[i, j] = (int)be3[ind3++] - '0';

编辑：请记住这个解决方案是愚蠢的 - 如果字符串中有非数字字符，则不会出现错误/异常。因此，在使用之前，您必须确保字符串纯粹由数字组成。但你的问题就是这样，所以我建议采用这种方法。

Keep in mind that this solution is "stupid" - if there are non-digit characters in your string, you won't get an error / exception. So you have to ensure that the strings are purely made of digits before using this. But your question reads like that is the case, so I suggested this approach.

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