#include<stdio.h>
#include<conio.h>
long long int add(int,int);
void main()
{
long long int a,b;
printf("Enter the two Numbers: ");
scanf("%lld%lld",&a,&b);
printf("Addition of two num. is : %lld",add(a,b));
getch();
}
long long int add(int a, int b)
{
if (!a)
return b;
else
return add((a & b) << 1, a ^ b);
}
Not possible. Let's not forget than in C the followings are all operators:+, -+=, -=++, --=&, *, []function-call
int main() { int a=10,b=20; while(a--) b++; printf("%d",b); } or: a -= -b;
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You cannot compare 2 numbers without using relational operators. Certainly, you could subtract them, but you still need to test the result, and that is a relational operator in itself.
Without any additional plugins, this can be done using the command "/f add".
They are two large numbers, without any operator between them.
No, you cannot play Uno without using any cards.
You cannot perform any operation by giving any operator. The operation must be valid in the domain and range.For example, you often cannot perform the square rootoperation using the square root operator if your domain and range are integers. At the level of maths that I guess you are at (from this question), the square root of a negative number is not a operation that is defined.
Java does not have the sizeOf() operator or any operator that gives an equivalent result.
(0! + 0! +0! + 0! + 0!)!
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9876543210, without using any mathemeatical operations.