[blutrane]

all i get are symbols if change i to 48 and i<=58 i get an illegal operation

 for(r=0; r<=3; r++)
{for(i=0; i<=10; i++)
 {cout<<(rou[(r*10)+i]=(r*10)+i);}}

[supersloth]

Quote

Quote: from blutrane on 12:15 pm on Sep. 29, 2001
all i get are symbols if change i to 48 and i<=58 i get an illegal operation

 for(r=0; r<=3; r++)
{for(i=0; i<=10; i++)
 {cout<<(rou[(r*10)+i]=(r*10)+i);}}

ok, i dont think you can do the parentheses inside the bracket part of the array, and on you post i see a ] after the semicolon, put in the box im typing in, i see 2 }'s so maybe that might have something to do with it also.

[blutrane]

it don't work the 2 parenthesis are for the loops

[supersloth]

{cout<<(rou[(r*10)+i]=(r*10)+i);}}

the parentheses tthat the r*10 are in... maybe, they cant work inside the array.

[blutrane]

still only get symbols

[runtime error]

can you post the declaration of your array. Something doesn't seem right with this multiplication thing

[supersloth]

yea, im getting the same vibe.... dont know though..

[blutrane]

its just a regular array
char rou[37];

[malkiri]

Ok, forgive me if I'm being dense...but what's the point of this code block? It seems to me that all you're doing is trying to set all the indices of rou[] to the value of each index, i.e. rou[0] = 0
In addition to this, if your array is only of length 37, then why, at their maximum values, does r*10 + i = 40?
And one last thing...what does "all i get are symbols" mean, exactly? If you mean, you get stuff other than the alphanumeric characters, then yes, I can see that. Say you set a character variable equal to an integer, as in:
char theChar = 10;
cout << theChar;
Your output won't be:
10
It'll actually treat the integer "10" as an ASCII code. This is typecasting - when possible, C++ will cast a value that is not the "correct" type to the "correct" one. In this case, it does this my using the value as the ASCII code -- not by "translating" the value into its ASCII code, and storing it. That's how chars are stored -- as ASCII codes.

Let me know how this goes.

-Malkiri

[Teckwiz]

Add  47,  or 48   to (r*10)+i,   to get the correct decimal numbers,  because doing this meathod starts at decimal symbol  0 and goes to 37,  but  you need to add 48  to start at the  deciamal  characters for number symbols 48-57

((r*10)+i)+48)

[blutrane]

Quote

Quote: from malkiri on 8:35 pm on Sep. 30, 2001
Ok, forgive me if I'm being dense...but what's the point of this code block? It seems to me that all you're doing is trying to set all the indices of rou[] to the value of each index, i.e. rou[0] = 0
In addition to this, if your array is only of length 37, then why, at their maximum values, does r*10 + i = 40?
And one last thing...what does "all i get are symbols" mean, exactly? If you mean, you get stuff other than the alphanumeric characters, then yes, I can see that. Say you set a character variable equal to an integer, as in:
char theChar = 10;
cout << theChar;
Your output won't be:
10
It'll actually treat the integer "10" as an ASCII code. This is typecasting - when possible, C++ will cast a value that is not the "correct" type to the "correct" one. In this case, it does this my using the value as the ASCII code -- not by "translating" the value into its ASCII code, and storing it. That's how chars are stored -- as ASCII codes.

Let me know how this goes.

-Malkiri

i know that. that's the *10 is for ascii numbers go 0-9 and then can multiply them so 0-36