Salute..
Let's see this example:
int x,y,s;
cin>>x>>y;
s=x+y;
here we have three variables for adding two values..
Can we do this just with one variable?
thanks.
8 Answers
How about zero variables?
#include <numeric>
#include <iterator>
#include <iostream>
int main()
{
std::cout <<
std::accumulate(
std::istream_iterator<int>(std::cin),
std::istream_iterator<int>(),
0) <<
"\n";
}
Comments
struct Accumulator {
int value;
Accumulator(): value(0) {}
friend std::istream& operator>>(std::istream& ss, Accumulator& acc)
{ int x; ss >> x; acc.value += x; return ss; }
};
int main() {
Accumulator a;
std::cin >> a >> a;
std::cout << "Total is " << a.value << "\n";
return 0;
}
See how useful abstraction is?
Comments
You can cut out one by using the extraction operator twice.
int x, s = 0;
cin >> x;
s += x;
cin >> x;
s += x;
You could cut that down even more by using a single variable that's twice the size of int. I can't believe I am typing this:
long long s;
assert(sizeof(int)*2 == sizeof(long long));
cin >> *(int*)(&s);
cin >> *((int*)(&s)+1);
s = (s & 0xffffffff) + ((s >> 32) & 0xffffffff);
You're only allowed to do things like this when you absolutely need to do something like store two 32-bit values in a 64-bit register for arcane performance reasons, or the gods will smite you. In such a case you are likely not using the iostream library anyway, but there you go. I'm going to go take a shower to wash the code smell off. I might need some lye.
Comments
It is possible.
Very important note, int is 16 bit. The code is very, very long. There is a lot of constants. Something like this.
int x;
cin >> x;
if (x == -32768) {
cin >> x;
x = x - 32768;
} else if (x == -32767) {
cin >> x;
x = x - 32767;
} else ...
...
} else if (x == -1) {
cin >> x;
x = x - 1;
} else if (x == 0) {
cin >> x;
x = x;
} else if (x == 1) {
cin >> x;
x = x + 1;
} else ...
...
} else if (x == 32766) {
cin >> x;
x = x + 32766;
} else {
cin >> x;
x = x + 32767;
}
cout << x << endl;
return 0;
1 Comment
flight
Come on, at least use a switch statement.
I know you can cut it down to 2 variables by using x+=y
You're not allowed to do anything like cin<<x<<x+=y which would be the only way I would know how to cut it down to a single variable.
Comments
You can't as long as that variable is an int. But why do you want to anyway?
Of course, you can get rid of s by x += y.
Comments
int main() {
typedef long long int64;
int64 x;
cin >> ((int*)(&x))[0] >> ((int*)(&x))[1];
x = ((int*)(&x))[0] + ((int*)(&x))[1];
cout << x << endl;
return 0;
}
Input:
100 250
Output
350
See yourself at ideone : http://ideone.com/2AmK0
Note: I don't know how portable this solution is. But this works with gcc-4.3.4.
Comments
I think the below one would work.....
#include<stdio.h>
int x=0;
int getno()
{
scanf("%d",&x);
return x;
}
int main()
{
x=getno()+getno();
printf("addition of the number is = %d",x);
}
int x[ 2 ]; cin >> x[ 0 ] >> x[ 1 ]; x[ 0 ] += x[ 1 ];?