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cchardigits

Increment a char in C

I have a small not understanding with increment a char in c. let suppose:

char c = 'A';

now if I do c+1, the value in c will be 'B'. This is fine. But why when I do the following:

c = c + argv[1][2]

then I got 'r' in c why?

please suppose that the argument is "in12345", then argv[1][2] is equal to 1.

What I am trying to do is for some input, lets say ABCDEF is to receive BDFHJG when I am do it in cyclic manner from the arguments. but I don't understand why the above is not working for me.

why it is not working and what can I do to fix it?

Solution

  • please suppose that the argument is in12345, then argv[1][2] is equal to 1.

    No it isn't. It's equal to '1', or (assuming ASCII) 49.

    'A' (65) + '3' (49) = 'r' (114)

    If you want to get the integer 1 from the character '1', you need to convert it. A cast won't help; one way to do this is with some arithmetic magic:

    char c = 'A';

const int valToAdd = argv[1][2] - '0';
c = c + valToAdd;

// c is now 66, or `'B'`

    However, this is a bit of a hack, and it will break if argv[1][2] is not in the range '0' to '9'. You can do further research on the best (better?) ways to get an integer from an ASCII numeric digit.