C++, Integer and Char array conversion trouble

Question

I have the char array:

char* chararray = new char[33];

and the int:

int exponent = 11111111;

What I want to do, but am confused as to how, is: input the values of exponent into chararray. With the restrictions that exponent has to take up the 2nd to 9th values of chararray. chararray will be all 32 0s,and I want it to become 0xxxxxxxx0000....00, the x's being the 8 digits in exponent.

Furthermore, no build-in conversion functions like atof or atoi. I also want to avoid using Floats or doubles not that you'd really need to.

Note, this is for making IEEE754 32bit values to get some understanding.

Will edit for additional details or clarification if needed.

also note that x+'0' is equal to 'x' for all 1-digit integers — Sam I am says Reinstate Monica
– Sam I am says Reinstate Monica, Commented Jul 3, 2013 at 4:35

raj raj · Accepted Answer · 2013-07-03 04:53:17Z

1

Try this after initializing the array with '0':

for(int i=9; i>=2; i--) {
        chararray[i] = (exponent%10) + '0';
        exponent = exponent/10;
    }
chararray[32] = '\0';

edited Jul 3, 2013 at 4:53

answered Jul 3, 2013 at 4:41

raj raj

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2 Comments

Tony Delroy Over a year ago

You're printing the digits into chararray backwards ;-). Easiest to start at the right and work back towards the left (i.e. chararray[10-i] = exponent % 10 + '0';)....

raj raj Over a year ago

Yes, you are right. Edited the post! Checked the test code with 11111111, that's why couldn't catch it :).

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