CODEWITHSUNDEEP
c++long-integer
cosmin-andrei paraschiv
cosmin-andrei paraschiv

Reputation: 35

Multiplying 1 million

Why when you do this:

int a = 1000000 , b = 1000000;
long long product = a * b;
cout<<product;

It gives some random trash value?Why do both a and b need to be long long in order to calculate it?

Upvotes: 2

Views: 808

Answers (2)

Ron
Ron

Reputation: 15511

You are observing the effects of undefined behavior. This:

a * b

is an (arithmetic) expression of type int, as both operands a and b are of type int. Trying to store the value of 1000000000000 into an int results in the so-called signed integer overflow which is undefined behavior.

Either cast one of the operands to long long, thus causing the entire expression to become long long, which is sufficiently large to accept the value of 1000000000000:

#include <iostream>

int main()
{
    int a = 1000000, b = 1000000;
    auto product = static_cast<long long>(a) * b;
    std::cout << product;
}

Or define one of the operands as long long:

#include <iostream>

int main()
{
    long long a = 1000000;
    int b = 1000000;
    auto product = a * b;
    std::cout << product;
}

Optionally, use unsigned long instead.

Upvotes: 6

Sahadat Hossain
Sahadat Hossain

Reputation: 4349

This will work fine. When you do multiplication of int and int if it gets out of limit then to put it in a long long variable you multiply the result with 1ll or 1LL. When you multiply the result with 1ll the result is converted to long long during calculation of product itself.

int a = 1000000 , b = 1000000;
    
long long product = 1ll * a * b;
    
cout << product;

Upvotes: 1

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