#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
long int a[20],i,n,count=0,b[20],c[20],sum=0;
printf("ENter the number in binary form=\t");
scanf("%ld",&n); // Get a binary number from the user
for (i=0;n>=1;i++)
{
a[i]=n%10;
n=n/10; // Loop To reverse the number And put all reversed numbers in arry a[i]
count=count + 1; // count to count the number of times this loop runs
}
for (i=0;i<=count-1;i++) // count -1 condition is used to run the loop till the previous loop run
{
b[i]=pow(2,i); // This is to raise the power of 2 to no of times previous loop runned.
}
for (i=0;i<=count-1;i++)
{
c[i]=a[i] * b[i]; // Multiply a[i] or reveresed binary no with b[i] or increasing pow of 2 to count-1
sum=sum +c[i]; // it is to add the c[i] elements with each other n put into sum variable.
}
printf("Decimal form =%ld",sum); // printing the sum to get the decimal form
getch();
}
// Hope it solves your problem, ANy more assistance honey.gupta13@Yahoo.com ffffff
#include<stdio.h>
#include<conio.h>
void main()
{
char base16[16]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
long num;
long cnum[68];
int base,number,i,j;
clrscr();
printf("\nEnter a decimal number to convert : ");
scanf("%ld",&num);
printf("\nEnter base for convert : 2(for binary),8(for octal),16(for hexadecimal) : ");
scanf("%d",&base);
i=0;
while(num!=0)
{
cnum[i]=num%base;
num=num/base;
i++;
}
i=i--;
printf("\nThe number in your base is : ");
for(i=i;i>=0;i--)
{
number=cnum[i];
printf("%c",base16[number]);
}
getch();
}
write a c++ program to convert binary number to decimal number by using while statement
This is not a question.
include <iostream> using namespace std; int main() { int n; // number to convert to binary while (cin >> n) { if (n > 0) { cout << n << " (decimal) = "; while (n > 0) { cout << n%2; n = n/2; } cout << " (binary) in reverse order" << endl; } else { cout << "Please enter a number greater than zero." << endl; } } return 0; }//end main
Suppose your binary number is stored in a series of bits in the unsigned long type named bits.Then the fragment of a C program to convert this number to a decimal value would be ..double decimal_value = 0.0;for ( unsigned long i = 0; i < sizeof(unsigned long); ++i){decimal_value += pow(2,i) * ( bits & 1 );bits >> 1; // shift all the bits to the right one spot.} // end for iDoing this work is generally unnecessary as functions such as these are built in to most programing languages.Another method for this: double decimal_value= bits;
Binary is a base 2 number system, while octal is base 8. This happens to make conversion between binary and octal fairly trivial, although more complex than conversion to hexadecimal. To convert to octal from binary, take each three bits, starting from the least significant bit, and convert them to their octal equivalent. Examples: 25510 = 111111112 = 11 111 111 = 3778 17410 = 101011102 = 10 101 110 = 2568 You can repeat this process for as many bits as you need. A 24-bit number should translate into 8 octal numbers, for reference.
write a c++ program to convert binary number to decimal number by using while statement
Example Binary 00111000 Convert to Decimal 56 Convert to BCD by using groups of four binary numbers for each digit 5 6 0101 0110
The conversion of octal number to binary can be obtained by using two methods. First, it can be converted into decimal and then obtained decimal is converted into binary. In the second method
becomes heavy because the ang decimal number ay marami kay sa sa stack ng tsinelas
111100002 equals 24010 using unsigned notation. It equals -1610 using signed notation.
0.9028
The binary equivalent is 101110000. If you're using Windows 7, the built-in calculator will convert numbers between base 10, 8, 2 & hex
IF you are asking what that binary number is in decimal form... it would be 7. The question though seems to be asking waht that decimal number is in binary. You want to know what 111 is in binary? 1101111. Try using google. "111 in binary" as a search phrase gives you the answer.
#include<stdio.h> #include<stdlib.h> main() { int number,binary[10000],b=0; printf("Enter decimal number "); scanf("%d",&number); printf("\nBinary: "); for(;number;number/=2,b++) binary[b]=number%2; for(b--;b>-1;b--) printf("%d ",binary[b]); }
This is actually a question in my Digital Circuits text. Are they kidding? Is there a way to tell that a discrete decimal will have an endless binary equivalent?
To convert decimal to binary, divide the decimal number you want to convert by 2 and write down the remainder. Repeat this until the final result is zero. The remainders you wrote down, written from the last one you wrote to the first (so the opposite order from which you derived them) is the binary equivalent.So using this method with the number 23 we get:23/2 = 11 remainder 111/2 = 5 remainder 15/2 = 2 remainder 12/2 = 1 remainder 01/2 = 0 remainder 1So the binary equivalent is 10111
BCD (Binary Coded Decimal) output can be generated using decimal-to-BCD conversion algorithms. One common method involves dividing the decimal number by 10 and storing the remainder as the Binary Coded Decimal digit. This process is repeated until all decimal digits are converted into BCD form. Alternatively, some microcontrollers have built-in instructions to directly convert decimal numbers to BCD format.