WAP to convert octal to binary

Question

In a C++ program to convert octal numbers to binary, I was able to run the program, but the problem is the output is coming,

Enter number in octal: 1
1 in octal:1

But as we know octal is a 3 bit number so what to do to bring the result as 001? For information, I declared a function, and used a return value of 'binary'(a variable I declared) in the function definition. Will I declare the binary variable as binary [3]? Or if not, what is the correct way?

You didn't post any of your code. There's nothing we can do for you without seeing your code. — Carey Gregory, Jan 03 '16 at 06:56

score 1 · Accepted Answer · answered Jan 03 '16 at 06:58

1

#include <iomanip>

cout << setfill('0') << setw(3) << binary

answered Jan 03 '16 at 06:58

amallard

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Exactly. I did tried it out few minutes back and it worked out. Thanks! – Aneek Jan 03 '16 at 07:07

score 0 · Answer 2 · answered Jan 03 '16 at 07:15

If leading zeros matter to you... this could sort of be a dup. See "How can I pad an int with leading zeros when using cout << operator?" ( How can I pad an int with leading zeros when using cout << operator? ).

Anyway, aside from (in spite of?) the lack of example code in the question... :-)

I would suggest looking at the answer from that post about printf(), which I find to be much more concise than cout formatting.

Try this (assuming your number is stored in a variable named 'i', adjust accordingly):

int i = 0113; // should yield decimal=75, hex=4B, and octal=113.
// important: note that we give printf 3 arguments after the string: i, i, i.
printf("decimal: %06d  hex: %06x  octal: %06o\n", i, i, i); // zero filled, all 6 long.
printf("decimal: %6d  hex: %6x  octal: %6o\n", i, i, i); // note lack of leading zeros, still 6 long.
printf("decimal: %d  hex: %x  octal: %o\n", i, i, i); // just as long as they need to be (1 to many).

This is one place (of many) to get some printf formatting info: http://www.lix.polytechnique.fr/~liberti/public/computing/prog/c/C/FUNCTIONS/format.html

fyi - Internally your integers are already in binary, and are already probably using 32 (or maybe 64) bits each... depending on whether they're shot, int, long and so on. So I suspect there isn't much to change your variable declarations.

The trick here is to understand that printf (or cout, if you prefer) generates the characters to represent the binary number (i, above) on your output. This is why the compiler wants you to use 123 for decimal, 0123 for octal (not the leading zero) and 0x123 for hex so it can convert them from characters in your source code (or input at run time) into the proper binary format in a given variable's memory.

WAP to convert octal to binary

2 Answers2