C# convert char to int

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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?

C#

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++];
                }
            }

Posted 3-Apr-15 2:29am

Member 11338785

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Sascha Lefèvre · Accepted Answer · 2015-04-03T02:41:00

Solution 1

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

C#

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

Edit: 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.

Posted 3-Apr-15 2:41am

Sascha Lefèvre

Updated 3-Apr-15 4:17am

v2

Comments

Member 11338785 3-Apr-15 8:48am

THANK YOU SOOOO MUCH!!!!

Sascha Lefèvre 3-Apr-15 8:50am

You're welcome!

Sascha Lefèvre 3-Apr-15 10:17am

I added a comment to the solution, please take a look.

Member 11338785 3-Apr-15 10:22am

Yep, they are ONLY contain numbers.