I just started learning C programming and we just finished learning about function calls and base conversion. Can you give me a simple program that converts a decimal based integer (user input) to binary, octal, and hexadecimal bases? It can't have printf %o (octal) or %x (hexidecimal). Just a simple program for someone who just started learning about programming and learning about calling a function.

I'd really appreciate it. Thanks alot!

I'd really appreciate it. Thanks alot!

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#define OCT_MASK 0x07 //largest possible number represented by a single octal-digit(octet)

unsigned int num=0xffffffff; // this is a sample 32 bit number

int nBits=3; //number of bits to skip each time.. for octal it is 3, for hex it is 4, for binary it is 1.

int max_bits=8*sizeof(int); // number of bits in the integer on this platform

char *digit_char="0123456789abc

int digit;

for(i=0;i<max_bits;i+=nBit

digit=num & OCT_MASK; // take the number in the right most 3 bits

printf("%c",digit_char[dig

num= num>>nBits; //drop of the last 3 bits

}

above loop will print the octets of num from right to left, to print them in the left-to-right order, u may keep storing the octets in the array...and print it later on

for hexadecimal representation use

#define HEX_MASK 0x0f

and nBits=4

for binary representation use

#define BIN_MASK 0x01

and nBits=1

change the other part of the codes accrodingly, try to bring together all this in a program, better will be to make a function which takes in the argument to choose from oct/hex/bin and return the appropriate character string

good luck

akshay

Is there another simpler way?

Again, thanks alot.

Another method to convert to another base is to find the remainder when the number is divided by the base i.e. by 2, 8, 16 for binary, octal and hexadecimal.

sample code for binary would be somewhat like:

while (num > 0)

{

x = num % 2; //gets the remainder when number is divided by 2

num = num / 2;

array[i] = x + '0'; //array is character array so we have to add ascii value of 0 to store correctly

i++;

}

//Now reverse the array

for(j=0;j < i/2 ;j++)

{

temp = array[j];

array[j] = array[i-j-1];

array[i-j-1] = temp;

}

Similarly you can do for octal and hex. For octal you have to divide by 8 and for hex by 16. For hex you have to check whether remainder is > 9. If it is you have to convert it to A,B,C,D,E,F and then store.

Dhyanesh

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Start your 7-day free trialBut got two questions:

Can you do this without using arrays? I know it might be a longer way but how do you do it? I tried to figure it out.

Second, when I tried to run your (Dhyanesh) program, I got errors saying that I need pointers for the arrays. That's the reason I don't really want to use arrays. My question is: I'm learning functions calls and if I add your program, would it be the function at the top and I would just need to call it at my void main function?

Again, I really appreciate your help.

-cooldean

if you wish to do without arrays it will be somewhat like:

binary = 0;

mult = 1;

while (num > 0)

{

x = num % 2; //gets the remainder when number is divided by 2

num = num / 2;

binary = mult *x + binary; //binary is integer variable

mult = mult * 10;

}

return binary;

However this will not work for hexadecimal as it involves A,B,C,D,E,F which require a char array to store.

>>would it be the function at the top and I would just need to call it at my void main function?

You can create a function on top of main or even after the end of main. If you create it before main you just need to call the function in main. However if you write function after closing brace of main you require a function prototype in the main before you call it i.e.

void main()

{

int func1(int); //function prototype

.....

abc = func1(xyz);

}

int func1(int t)

{

................

}

Dhyanesh

another way to do it would be by using recursion.(function calling itself)

void binary(int n)

{

if (n>0)

{

binary(n/2);

printf("%d",n%2);

}

}

this function is called from the main.

similar for octal and hex. and for hex again you will have to check if remainder > 9 etc..

Dhyanesh

#include<stdio.h>

#include<conio.h>

typedef struct byte

{

unsigned int fi : 1; //First bit denoted by fi.

unsigned int se : 1; //Second bit denoted by se.

unsigned int th : 1; //Third bit denoted by th.

unsigned int fo : 1; //Fourth bit denoted by fo.

unsigned int ff : 1; //Fifth bit denoted by ff.

unsigned int si : 1; //Sixth bit denoted by si.

unsigned int sv : 1; //Seventh bit denoted by sv.

unsigned int ei : 1; //Eighth bit denoted by ei.

} byte;

byte test;

void byte_print(byte);

void convert_byte(int);

void main()

{

int a;

// struct byte test;

printf("Enter the number < 255 to be converted to binary");

scanf("%d", &a);

convert_byte(a);

byte_print(test);

getch();

}

void convert_byte(int x)

{

int i;

// extern byte test;

test.fi = x & 1;

x >>= 1;

test.se = x & 1;

x >>= 1;

test.th = x & 1;

x >>= 1;

test.fo = x & 1;

x >>= 1;

test.ff = x & 1;

x >>= 1;

test.si = x & 1;

x >>= 1;

test.sv = x & 1;

x >>= 1;

test.ei = x & 1;

x >>= 1;

}

void byte_print(byte x)

{

printf("\n%d",x.ei);

printf("%d",x.sv);

printf("%d",x.si);

printf("%d",x.ff);

printf("%d",x.fo);

printf("%d",x.th);

printf("%d",x.se);

printf("%d\n",x.fi);

}

C

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1. all numbers are stored in binary format.

2. octal and hexadecimal are powers of 2.

3. to get bit by bit of a number, u can use the 'right-shift'( >> ) operation in C.

4. to divide a number by 2^n, right shift the number by n bits.

so here is the procedure to get the binary/octal/hexadecimal representation of the number.

keep this string of bits/octets/hex-bits

char *nbits="0123456789abcdef";

1. to get the binary representation of the number..

right-shift bit by bit and keep storing/printing the bits one by one.

(for each shifted bit print---> nbits[value_of_shifted_bit

2. to get the octal representation of the number..

right-shift the number by 3-bits at a time, and each time checking the number in the thus got 3 shifted bits, and print the appropriate octet.

(for each shifted bit print---> nbits[value_of_shifted_bit

3. for hexadecimal, shift the bits 4 at a time

(for each shifted bit print---> nbits[value_of_shifted_bit

the number that u'll print this way will be in the right-to-left direction ( reversed from normal left-to-right)

try implementing code on this and we will help you out to make it work.

akshay