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"Int" multiplication in c++ with "long long" result [duplicate]

I am trying to find the square of a int. My code looks like below:

long long sqr=0;
int num=77778;
sqr= num*num;

The result should have been 6049417284 But when I check the output it shows 1754449988. What is the mistake I am doing? long long should be able to store the result but why am I getting a different value?

like image 747
srip Avatar asked Jun 10 '16 06:06

srip


Video Answer


2 Answers

Intermediate result type is the same as first argument type in this case. So this code puts wrong value into sqr (because here you have integer overflow). Try this:

long long num = 77778;
long long sqr = num * num;

Another way with casting:

int num = 77778;
long long sqr = (long long) num * num;

To understand it better you can check these lines:

int ia = 1, ib = 2;
double da = 1.0, db = 2.0;
double ic = ia / ib; // c == 0.0, because intermediate result of division is 0
double dc = da / db; // c == 0.5, expected result
like image 74
Ilya Avatar answered Oct 17 '22 22:10

Ilya


If you multiply two integer values, the result will be an integer value. Then, the assignment of integer value to long long is meaningless.

So, to get the desired result the num should be long long also.

like image 42
Monirul Islam Milon Avatar answered Oct 17 '22 23:10

Monirul Islam Milon