void main()

{

int a=10,b=20;

while(b--) a++;

printf("Sum is :%d",a);

}

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int i,j;

if(i<j)

{

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

j++;

cout<<j;

}

else

{

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

i++;

cout<<i;

}

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#include

int main()

{

unsigned int t = 0xffffffff;

unsigned int z = 1;

int a = 10, b = 10, sum = 0, r = 0;

while (t) {

int t1, t2;

t1 = a & 0x00000001;

t2 = b & 0x00000001;

if ((t1 ^ t2 ^ r) == 1) {

if (t1 == 1 && t2 == 1 && r == 1)

r = 1;

else

r = 0;

sum = sum | z;

} else {

if (t1 == 0 && t2 == 0 && r == 0)

r = 0;

else

r = 1;

}

a >>= 1;

b >>= 1;

z <<= 1;

t >>= 1;

}

printf("sum: %dn", sum);

return 0;

}

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Arithmetic operators |

Operator Name | Syntax | Overloadable | Included in C |

Unary Plus | **+**a | Yes | Yes |

Addition (Sum) | a **+** b | Yes | Yes |

Prefix Increment | **++**a | Yes | Yes |

Postfix Increment | a**++** | Yes | Yes |

Assignment by Addition | a **+=** b | Yes | Yes |

Unary Minus (Negation) | **-**a | Yes | Yes |

Subtraction (Difference) | a **-** b | Yes | Yes |

Prefix Decrement | **--**a | Yes | Yes |

Postfix Decrement | a**--** | Yes | Yes |

Assignment by Subtraction | a **-=** b | Yes | Yes |

Multiplication (Product) | a ***** b | Yes | Yes |

Assignment by Multiplication | a ***=** b | Yes | Yes |

Division (Quotient) | a **/** b | Yes | Yes |

Assignment by Division | a **/=** b | Yes | Yes |

Modulus (Remainder) | a **%** b | Yes | Yes |

Assignment by Modulus | a **%=** b | Yes | Yes |

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Use minus operator :-

int a=10,b=20,c;

c=a-(-b);

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void main()

{

int a=10,b=10;

a=a<<1;

b=b<<1;

printf("Sum of a&b is %dn",(a&b));

}

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#include add(int a,int b)

{

if(!a) return b;

else return add((a&b)<<1,a^b);

}

void main()

{ int a=2, b=5,c; c= add(a,b);

cout<<c;

}

hope it 'll help you right...........

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void main(){

int a=5,b=7,i;

for(i=1;i<=a;i++)

b++;

printf("sum value:%d",b);

}

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Code

for (i = 0; i < a; i++)

{

count++;

}

for (j = 0; j < b; j++)

{

count++;

}

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Code

#include<stdio.h>

int main()

{

int i = 1, j = 20, c;

c = (i ^ j);

return 0;

}

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Code

main()<br />{<br />cout<<"Enter the numbers to be added";<br /> int a,b,x,y;<br /> cin>>a>>b;<br /> <br />do{<br /> x=a&b;<br /> y=a^b;<br /> x=a<<1;<br /> y=b;<br />}while(a);<br />cotu<<"The sum is:"<<y;<br />}

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Code

#include<stdio.h>

main()

{

int a, b, c, d, sum;

printf("n Please enter a,b values:");
scanf("%d %d", &a, &b);

if (b > a) {

c = b - a;

} else {

c = a - b; // using minus operator

}

d = square(b) - square(a); // using square function and - minus operator

sum = d / c; // using division operator

printf("The sum of a and b is:%d", sum

);
return 0;

}

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Code

#include<stdio.h>

int result = 0X0;

int a, b, C = 0X0, aB, bB;

int main()

{

int val = 0X1;

scanf("%d%d", &a, &b);

while (val) {

aB = !!(a & val);

bB = !!(b & val);

if (aB ^ bB ^ C)

result |= val;

C = (aB & bB) | ((aB ^ bB) & C);

val <<= 1;

}

return 0;

}

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Code

class Test {

public static void main(String ... z) {

System.out.println(4 || 8); //use logical or symbol u can add it

}}

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**sravan**
Answered On : Aug 20th, 2011

Code

#include<iostream.h>

void main()

{

int i=20,j=30,m,n;

m=i,n=j;

if(i<j)

{

for(i=1;i<=m;i++)

j++;

cout<<j;

}

else

{

for(j=1;j<=n;j++)

i++;

cout<<i;

}

}

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**mukesh kumar **
Answered On : Mar 25th, 2012

Code

int main(){

int a,b;

printf("Enter the two numbers:
");

scanf("%d",&a);

scanf("%d",&b);

printf("Sum is: %d",add

(a

,b

));
}

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**mahamad**
Answered On : Jun 14th, 2012

Code

#include<stdio.h>

#include<conio.h>

main()

{

int a=20,b=10,c;

clrscr();

c=a-~b-1; //it will change the sign of operator truly magic

getch();

return 0;

}

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**And**
Answered On : Aug 28th, 2012

For two positive numbers:

int main(){

int a = 10;

int b = 10;

printf("%d",a ^ b | ((a & b)<< 1));

return 0;

}

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