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Compound assignment in C++

I would like to know the execution flow of compound assignments in C++. I came across a CodeChef question, where I am calculating NCR mod p values and adding them together to get the final answer:

// correct
for(int i=min1; i<=max1; i+=2){
     ans = (ans+ncr_mod_p(n,i))%mod;
}
// incorrect
for(int i=min1; i<=max1; i+=2){
     ans+=ncr_mod_p(n,i)%mod;
}

This is happening because of integer overflow.

So, what is the execution sequence of compound assignment?

Let's say, if we have an equation a+=b%c then what would be the execution sequence:

a = (a+b)%c
// OR
a = a+(b)%c;
like image 467
Rahul Avatar asked Sep 12 '26 08:09

Rahul


1 Answers

The compound assignment operators are in the second lowest precedence group of all in C++ (taking priority over only the comma operator). Thus, your a += b % c case would be equivalent to a += ( b % c ), or a = a + ( b % c ).

This explains why your two code snippets are different. The second:

    ans+=ncr_mod_p(n,i)%mod;

is equivalent to:

    ans = ans + ( ncr_mod_p(n,i) % mod );

Which is clearly different from the first (correct) expression:

    ans = ( ans + ncr_mod_p(n,i) ) % mod;
like image 91
Adrian Mole Avatar answered Sep 14 '26 23:09

Adrian Mole



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