Why is the output of this code '1'?

Question

cout << 1&&0||1||0;

Answer 1

The truth table for AND (&&) logical operator for two operands.

a b output
-------------------
0 0 0
0 1 0
1 0 0
1 1 1

The truth table for OR (||) logical operator for two operands.

a b output
-------------------
0 0 0
0 1 1
1 0 1
1 1 1

// original expression
cout << 1 && 0 || 1 || 0;

// their precedence
cout << (((1 && 0) || 1) || 0);

cout << ((0 || 1) || 0);
cout << (1 || 0);
cout << 1;

Answer 2

Break it down:
Anything AND with 0 will be 0
1&&0 =0
Entire expression would be 0||1||0.
Anything OR with 1 will be 1
Thats why whole expression = 1

Answer 3

1 && 0 || 1 1||0
I | |
true. && true. && true. = true = 1.

Answer 4

yess babak is right .