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- C Conversion programs
- C Decimal to Binary
- C Decimal to Octal
- C Decimal to Hexadecimal
- C Binary to Decimal
- C Binary to Octal
- C Binary to Hexadecimal
- C Octal to Decimal
- C Octal to Binary
- C Octal to Hexadecimal
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- C Hexadecimal to Octal
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# C Program to Convert Octal to Hexadecimal

In this tutorial, we will learn about how to create a program in C that converts any given octal number into its equivalent hexadecimal
value. At last we have also created a program using a user-defined function **OctToHex()** that also
does the same job.

But before going through the program, if you are not aware of

- Octal Number
- Hexadecimal Number
- Octal to Hexadecimal Conversion Process

then refer to Octal to Hexadecimal conversion step by step process. Now let's move on to the program.

## Octal to Hexadecimal in C

To convert octal number to hexadecimal number in C programming, you have to ask from user to enter any octal number. Then convert it
into hexadecimal and display **hex** value as output. The question is, **Write a program in C that converts any given octal number by
user (at run-time) to its equivalent hexadecimal value.** The answer to this question is given below:

#include<stdio.h> #include<conio.h> #include<string.h> int main() { int octnum, rev=0, rem, count=0, hex=0, mul=1, i=0, k=0; char binnum[40] = "", hexnum[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; 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; } while(binnum[k]!='\0') k++; count=1; k--; while(k>=0) { if(binnum[k]=='0') rem = 0; else rem = 1; hex = hex + (rem*mul); if(count%4==0) { if(hex<10) hexnum[i] = hex+48; else hexnum[i] = hex+55; mul = 1; hex = 0; count = 1; i++; } else { mul = mul*2; count++; } k--; } if(count!=1) hexnum[i] = hex+48; if(count==1) i--; printf("\nEquivalent Hexadecimal Value = "); count = 0; for(i=i; i>=0; i--) { if(hexnum[i]=='0' && count==0) { count++; continue; } else printf("%c", hexnum[i]); } } else printf("\nInvalid Octal Digit %d", rem); getch(); return 0; }

As the above program was written under **Code::Blocks** IDE, therefore after successful build and run, here is the first
snapshot of the sample run:

Now supply any octal number say **1452** as input and press `ENTER`

key to see its hexadecimal equivalent as shown in the second
snapshot of the sample run:

Here is the final snapshot of another sample run:

#### Program Explained

- Receive any octal number from user at run-time say
**1452**as input - Now reverse the given octal number and initialize its reverse to the original number say
**octnum**itself - Here we have reverse the number because using modulous (%) or remainder operator
we will get the last digit. For example, while
applying modulous operator in a way like
**1452%10**, then we will get 2 as result, and 2 is the last digit, but after reversing the number, it will become 2541. And then after applying modulous operator in same way like**2541%10**, we will get 1 as result, and 1 is the first digit of original number say**1452** - While reversing the number, we have checked whether any digit is greater than 7 or not. As in octal number system, all the digit
must be less than or equal to 7. But if any digit is greater than 7, then we have incremented a
variable say
**count**(it is initialized with 0 at start of the program), so that we can check after exiting from the while loop that it holds its original value (0) or not, if it holds, then we have to proceed to convert the number from octal to hexadecimal, otherwise we have to print a message like**you have entered an invalid octal digit** - Now if
**count**holds its original value, then program flow goes inside the if block - And
**rev**(reverse of octal number) or**2541**gets initialized to**octnum** - Then we have created a
**while**loop to convert octal number to binary - The condition of the
**while**loop**octnum!=0**or**2541!=0**evaluates to be true, and program flow goes inside the loop - There,
**octnum%10**or**2541%10**or**1**gets initialized to**rem** - Using switch case, we have checked, what value present inside the variable
**rem** - Whatever the value in
**rem**, concatenate it to**binnum**character array - Here we have used
**strcat()**function of**string.h**library, so that the binary equivalent (in string form) of each octal digit gets concatenated into**binnum**character array one by one - After concatenation,
**octnum/10**or**2541/10**or**254**gets initialized to**octnum** - And the program flow goes back to the condition of the loop, there
**octnum!=0**or**254!=0**evaluates to be true, and program flow again goes inside the loop and do the same operation as told above until the value of**octnum**becomes 0 - After exiting from the
**while**loop, we have**binnum**character array that holds**001100101010** - Here
**001**,**100**,**101**, and**010**are the binary equivalent of**1**,**4**,**5**, and**2** - Now we have used
**while**loop, to count the length of**binnum**character array, that how many character present inside**binnum**array - The length of array
**binnum**initialized to a variable say**k** - The value of
**k**gets decremented as indexing in array starts with 0 - Now we have binary equivalent of each octal digit
- It's time to make 4-4 binary pair from right side and convert it into its hexadecimal equivalent
- To do this, we have starts from last index of
**binnum**array - At first run of the
**while**loop, the condition**k>=0**or**8>=0**evaluates to be true, therefore program flow goes inside the loop - Inside the loop, we have checked what the character present at current index of the array
**binnum**, if it is 0, then initialized 0 to**rem**, otherwise initialized 1 to**rem** - And do the same procedure as told in binary to hexadecimal in C tutorial
- After doing all the above steps, we only have to print all the hexadecimal digit one by one
- Here we have used
**if**statement inside the**for**loop in printing hexadecimal digit, so that any zero (0) present at beginning of the hexadecimal number gets skipped

**Note - **You can also approach the second option, that is octal to decimal
and then decimal to hexadecimal. Because the second option
is little easier to create and understand than the first one. Therefore, the first option is used here that
converts octal to binary and then
binary to hexadecimal.

You can follow both the article (used for second option) to create the program with yourself. I'm sure, that will be very easy for you after understanding the first option as created above.

### Octal to Hexadecimal in C using User-defined Function

Let's create the same purpose program, but this time using user-defined function. The question is, **Write a program in C that
converts Octal to Hexadecimal using user-defined function named OctToHex()**. The answer to this question is:

#include<stdio.h> #include<conio.h> #include<string.h> void OctToHex(int oct); static int count, i; char hexnum[40]; int main() { int octnum; printf("Enter any Octal Number: "); scanf("%d", &octnum); OctToHex(octnum); if(count==0) printf("\nYou've entered an invalid octal digit"); else { printf("\nEquivalent Hexadecimal Value = "); count = 0; for(i=i; i>=0; i--) { if(hexnum[i]=='0' && count==0) { count++; continue; } else printf("%c", hexnum[i]); } } getch(); return 0; } void OctToHex(int oct) { int rev=0, rem, hex=0, mul=1, k=0; char binnum[40] = ""; while(oct!=0) { rem = oct%10; if(rem>7) { count++; break; } rev = (rev*10) + rem; oct = oct/10; } if(count==0) { oct = rev; while(oct!=0) { rem = oct%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; } oct = oct/10; } while(binnum[k]!='\0') k++; count=1; k--; while(k>=0) { if(binnum[k]=='0') rem = 0; else rem = 1; hex = hex + (rem*mul); if(count%4==0) { if(hex<10) hexnum[i] = hex+48; else hexnum[i] = hex+55; mul = 1; hex = 0; count = 1; i++; } else { mul = mul*2; count++; } k--; } if(count!=1) hexnum[i] = hex+48; if(count==1) i--; } else count=0; }

Here is the final snapshot of the sample run:

Here we have declared the variable **i** and **count** as static variable, so that it can hold or remembers its previous
values. We have not initialized both the static variable with 0, because by default static variable gets initialized 0 as its
initial value automatically. We have declared both static variables and a character array **hexnum[]** as global variables to make
all the three variables known to both the function say **main()** and **OctToHex()**.

#### Same Program in Other Languages

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