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// Binary to Decimal By C

#include<stdio.h>
#include<math.h>
#include<stdlib.h>

int main()
{
    int rem,num,i=0,Deci=0;
    char Oct[50];
    
    printf("Enter Binary no: ");
    scanf("%d", &num); //Taking num
    
    int j,numb =num;
    while(numb>0) { // Check binary
        j=numb%10;
        if( j!=0 && j!=1 ) {
            printf("number is not binary");
            exit(0);
        } 
        numb=numb/10;}
    
    if(num!=0){
    while(num>0){  // Oct to Dec Func
    rem = num % 10;
    Deci=Deci+rem*pow(2,i++);
    num=num/10; }
    
    }
    else{
    printf("Decimal and Octal Value is 0");
    exit (0);
    }
    
    printf("Decimal Number is %d",Deci);
    rem=0;
    i=0;
    
    while(Deci>0){ // Dec to Oct
    Oct[i++]=Deci%8;
    Deci =Deci/8;
    }

    printf("\nOctal Number is ");
    for(j=i-1;j>=0;j--){
    printf("%d",Oct[j]);
    }
    return 0;
}
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// Checking binary Number
#include<stdio.h>
#include<conio.h>
int main() {
    int j,num;
    printf("Please enter a number :");
    scanf("%d",&num);
    while(num>0) {
        j=num%10;
        if( j!=0 && j!=1 ) {
            printf("num is not binary");
            break;
        }
        num=num/10;
        if(num==0) {
            printf("num is binary");
        }
    }
    getch();
}

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// Complex Numbers adding with friend function
#include<iostream>
using namespace std;
class Complex{
    int a,b;
    public:
    friend Complex sumComplex(Complex o1, Complex o2);
    void setNumber(int n1, int n2){
        a=n1;
        b=n2;
        }
    void printNumber(){
    cout<<"Your number is "<<a<<" + "<<b<<"i"<<endl;
}};

Complex sumComplex(Complex o1, Complex o2){
    Complex o3;
    o3.setNumber((o1.a + o2.a), (o1.b + o2.b));
    return o3;
    }
int main(){
    Complex c1,c2, sum;
    c1.setNumber(1, 4);
    c1.printNumber();
    
    c2.setNumber(3, 6);
    c2.printNumber();
    
    
    
    sum = sumComplex(c1, c2);
    sum.printNumber();
    
    return 0;
    }

/*
Output: 
    Your number is 1 + 4i
    Your number is 3 + 6i
    Your number is 4 + 10i   
*/

// 2 Numbers adder
#include<iostream>
using namespace std;

int main (){
    int num1,num2, res;
    cout<<"Enter first number : ";
    cin>>num1;
    cout<<"Enter second number : ";
    cin>>num2;
    res=num1+num2;
    cout<<"Total: "<<res;
    return 0;
    }

Which of the following output can be possible for given line in C Programming:- int a = 5; printf("%d %d %d\n", a, ++a , a++);
Anonymous voting

// Binary to Decimal By C
#include<stdio.h>
#include<math.h>

int main()
{
    int rem,num,i=0,Deci=0;
    
    printf("Enter Binary no: ");
    scanf("%d", &num);
    
    if(num!=0){
    
    while(num>0){
    rem = num % 10;
    Deci=Deci+rem*pow(2,i++);
    num=num/10; }
    
    }
    else{
    printf("0");
    }
    
    printf("Decimal Number is %d",Deci);
    
    return 0;
}

Output of above code 🙄
Anonymous voting

What will be the output of the following C code?
void main()
{
    int x = 10;
    int y = x++ + 20;
    
    printf("%d,%d",x,y);
    
    return 0;
}

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// Decimal to Binary By C
#include<stdio.h>

int main()
{
    int rem,num,i=1,j;
    char Bin[50];
    
    printf("Enter Decimal no: ");
    scanf("%d", &num);
    
    if(num!=0){
    
    while(num!=0){
    rem = num % 2;
    Bin[i++]= rem+48;
    num=num/2; }
    
    }
    else{
    printf("0");
    }
    
    printf("Binary Number is "); 
    for(j=i+1;0<j;j--){
    printf("%c",Bin[j]);
    }
    
    return 0;
}

// Decimal to Octal By C
#include<stdio.h>

int main()
{
    int rem,num,i=1,j;
    char Oct[50];
    
    printf("Enter Decimal no: ");
    scanf("%d", &num);
    
    if(num!=0){
    
    while(num!=0){
    rem = num % 8;
    Oct[i++]= rem+48;
    num=num/8; }
    
    }
    else{
    printf("0");
    }
    
    printf("\nOctal Number is "); 
    for(j=i+1;0<j;j--){
    printf("%c",Oct[j]);
    }
    return 0;
}

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