In this tutorial, we will learn about how to create a program in C that converts any given octal number by user at run-time into its equivalent binary value. At last we have also created a function-driven program that does the same job. In function-driven program, we have defined a function named OctToBin() that takes octal number as argument and converts it into its equivalent binary value.
But before going through the program, if you are not aware of
then refer to Octal to Binary conversion step by step process. Now let's move on to the program.
To convert octal number to binary number in C programming, you have to ask from user to enter octal value and convert it into binary value, then display the equivalent value in binary as output. The question is, Write a program in C that converts any given number (in octal number system) to its equivalent value in binary number system. The answer to this question is:
// Octal to Binary in C
// ----codescracker.com----
#include<stdio.h>
#include<conio.h>
int main()
{
int octnum, rev=0, rem;
printf("Enter any Octal Number: ");
scanf("%d", &octnum);
while(octnum!=0)
{
rem = octnum%10;
rev = (rev*10) + rem;
octnum = octnum/10;
}
octnum = rev;
printf("\nEquivalent Binary value = ");
while(octnum!=0)
{
rem = octnum%10;
switch(rem)
{
case 0: printf("000");
break;
case 1: printf("001");
break;
case 2: printf("010");
break;
case 3: printf("011");
break;
case 4: printf("100");
break;
case 5: printf("101");
break;
case 6: printf("110");
break;
case 7: printf("111");
break;
default: printf(" InvalidOctalDigit(%d) ", rem);
break;
}
octnum = octnum/10;
}
getch();
return 0;
}
As the above program was written under Code::Blocks IDE, therefore after successful build and run, here is the sample run. This is the first snapshot:
Now supply any number say 734 as input and press ENTER key to see its value in Binary form as shown in the second snapshot of the sample run give here:
As the number 734 has 3 digits, that are 7, 3 and 4. Here 7 in binary is 111, 3 in binary is 011 and 4 in binary is 100. Therefore after combining all numbers in octal and binary, we have 734 (in octal) equals 111011100 (in binary) as shown in above snapshot. You can also write (734)8 = (111011100)2
Let's create the same purpose program in another way. Here we have used the string function named strcat() of string.h library, to concatenate the string one by one. We have stored binary value in string format as shown in the program given below. The question is, Write a program in C that converts octal number to its equivalent binary number using library function strcat(). The answer to this question is:
// Octal to Binary in C using strcat() Function
// ----codescracker.com----
#include<stdio.h>
#include<conio.h>
#include<string.h>
int main()
{
int octnum, rev=0, rem, count=0;
char binnum[40] = "";
printf("Enter any Octal Number: ");
scanf("%d", &octnum);
while(octnum!=0)
{
rem = octnum%10;
if(rem>7)
{
count++;
break;
}
rev = (rev*10) + rem;
octnum = octnum/10;
}
if(count==0)
{
octnum = rev;
printf("\nEquivalent Binary value = ");
while(octnum!=0)
{
rem = octnum%10;
switch(rem)
{
case 0: strcat(binnum, "000");
break;
case 1: strcat(binnum, "001");
break;
case 2: strcat(binnum, "010");
break;
case 3: strcat(binnum, "011");
break;
case 4: strcat(binnum, "100");
break;
case 5: strcat(binnum, "101");
break;
case 6: strcat(binnum, "110");
break;
case 7: strcat(binnum, "111");
break;
}
octnum = octnum/10;
}
printf("%s", binnum);
}
else
printf("\nInvalid Octal Digit %d", rem);
getch();
return 0;
}
Here is the final snapshot of the above program's sample run:
As octal digit can not exceed the value greater than 7, therefore we have used count variable to check whether any digit of the given octal number exceeds the value 7 or not. If it exceeds, then its value gets incremented and then exits from the current loop. And before converting octal to binary, we have checked whether count holds its original value or not. If it holds, then all the octal digit is correct, otherwise print any message like Invalid Octal digit was entered by you.
The question is, Write a program in C that converts octal number to its equivalent binary number in such way that the given octal number first converted into decimal number and then decimal number gets converted into binary number. The answer to this question is:
// Octal to Binary in C using Indirect Conversion
// ----codescracker.com----
#include<stdio.h>
#include<conio.h>
int main()
{
int octnum, decnum=0, binnum[40], rem, mul=1, i=0, count=0;
printf("Enter any Octal Number: ");
scanf("%d", &octnum);
while(octnum!=0)
{
rem = octnum%10;
if(rem>7)
{
count++;
break;
}
decnum = decnum + (rem*mul);
mul = mul*8;
octnum = octnum/10;
}
if(count==0)
{
while(decnum!=0)
{
binnum[i] = decnum%2;
i++;
decnum = decnum/2;
}
printf("\nEquivalent Binary Value = ");
for(i=(i-1); i>=0; i--)
printf("%d", binnum[i]);
}
else
printf("\nInvalid Octal Digit %d", rem);
getch();
return 0;
}
Here is the final snapshot of the above program's sample run:
To convert octal to binary using user-defined function named OctToBin(). Create this function in a way that it takes one argument that will be given octal number by user.
Here we have declared all the common variables (present inside both the function main() and OctToBin()) as global variable to make it known through both the function. The variable i and count gets declared as static variable, so that they can remember their previous value. As we all knows that the static variable holds 0 as its initial value automatically. Therefore here we have not initialized zero to both the static variable i and count.
// Octal to Binary in C using Function
// ----codescracker.com----
#include<stdio.h>
#include<conio.h>
void OctToBin(int oct);
static int i, count;
int bin[40];
int main()
{
int octnum;
printf("Enter any Octal Number: ");
scanf("%d", &octnum);
OctToBin(octnum);
if(count==0)
{
printf("\nEquivalent Binary Value = ");
for(i=(i-1); i>=0; i--)
printf("%d", bin[i]);
}
else
printf("\nYou've entered invalid Octal Digit");
getch();
return 0;
}
void OctToBin(int oct)
{
int dec=0, rem, mul=1;
while(oct!=0)
{
rem = oct%10;
if(rem>7)
{
count++;
break;
}
dec = dec + (rem*mul);
mul = mul*8;
oct = oct/10;
}
if(count==0)
while(dec!=0)
{
bin[i] = dec%2;
i++;
dec = dec/2;
}
}
Here is the final snapshot of the sample run:
