boolean logic Are both boolean logic equal - boolean-logic

Question

A && B || C && D
(A && B) || (C && D)
Are both boolean logic equal in C++? I am confused.

Answer 1

Whether or not they're equal depends entirely on how you define your operator precedence. If && takes precedence over ||, then yes. Otherwise, no.

Answer 2

In the most programming languages you'll find that operator && is of higher priority than ||.
So for example in Java, C#, C, C++, Python, Ruby, etc.
A && B || C && D
is equivalent to
(A && B) || (C && D)
You can even copy-paste the code:
#include <iostream>
using namespace std;
int main() {
bool A = false;
bool B = false;
bool C = true;
bool D = true;
for(int i = 0; i < 2; ++i) {
A = (i == 0);
for(int j = 0; j < 2; ++j) {
B = (j == 0);
for(int k = 0; k < 2; ++k) {
C = (k == 0);
for(int l = 0; l < 2; ++l) {
D = (l == 0);
cout << A << " " << B << " " << C << " " << D << " -> ";
cout << ((A && B || C && D) == ((A && B) || (C && D))) << endl;
}
}
}
}
return 0;
}
to Ideone to find out for yourself. In C++ for example the output is:
1 1 1 1 -> 1
1 1 1 0 -> 1
1 1 0 1 -> 1
1 1 0 0 -> 1
1 0 1 1 -> 1
1 0 1 0 -> 1
1 0 0 1 -> 1
1 0 0 0 -> 1
0 1 1 1 -> 1
0 1 1 0 -> 1
0 1 0 1 -> 1
0 1 0 0 -> 1
0 0 1 1 -> 1
0 0 1 0 -> 1
0 0 0 1 -> 1
0 0 0 0 -> 1
So the ((A && B || C && D) == ((A && B) || (C && D))) is a tautology.

Answer 3

While the final answer goes to the specifics of the C++ language you're asking about, here's some food for thought on why (and possibly how) to remember:
Conjunction (AND, &&) is often associated with multiplication, while disjunction (OR, ||) is often associated with addition (and we generally know the precedence of multiplication over addition).
Here's a quote from http://www.ocf.berkeley.edu/~fricke/projects/quinto/dnf.html:
... As a practical matter, we usually associate conjunction with
multiplication and disjunction with addition. Indeed, if we identify
true with 1 and false with 0, then {0,1} coupled with the usual
definitions of addition and multiplication over the Galois field of
size 2 (eg, arithmetic modulo 2), then addition (+) and disjunction
(or) really are the same, as are multiplication and conjunction (and).
...
Speaking in rather general terms, the computer languages tend to honor the precedence of multiplicative operators over additive operators.
(Further, these associations, e.g. between operators in logic and in algebra reoccur in other areas, such as type systems. For an interesting exposition of that, see http://blog.lab49.com/archives/3011 on the notion of Algebraic Type Systems.)