In C does the logical operator end if one side of a || is true? [duplicate]

Answer 1

No, it won't. This is called "short-circuiting" and it is a common flow-control mechanism:

With a && b, b is only evaluated if a is true; if a is false, the whole expression must necessarily be false.

With a || b, b is only evaluated if a is false; if a is false, the whole expression may still be true.

Answer 2

abc(2) will be called only if abc(1) is false

According to C99 specification, logical OR operator says

The || operator guarantees left-to-right evaluation; there is a sequence point after the evaluation of the first operand. If the first operand compares unequal to 0, the second operand is not evaluated.

Answer 3

|| (logical comparison) breaks further checking, while | (bitwise comparison) doesn't.

You might also read: Difference between | and || or & and && for comparison

Answer 4

nop, the second abc(2) is called only if the left statement is false

Answer 5

No. This is actually non trivial and defined in standard that logical operators are evaluated from left to right. Evaluation is stopped when the value can be determined with out further evaluation of operands. At least I am 100% sure for AND and OR.

This is non trivial problem, because therefore the evaluation of operands cannot be implicitly parallelized or optimized by reorganization of order, as the expected outcome could differ.

E.g. run-time troubles in wide-use cases such as if (*ptr && (ptr->number > other_number) )

Answer 6

If abc(1==1) returns true will then call abc(2) ?

No, it won't. This behavior is known as short-circuiting. It is guaranteed by C and C++ standards.

C11(n1570), § 6.5.13 Logical AND operator

Unlike the bitwise binary & operator, the && operator guarantees left-to-right evaluation; if the second operand is evaluated, there is a sequence point between the evaluations of the first and second operands. If the first operand compares equal to 0, the second operand is not evaluated.

(Emphasis is mine.)

The same applies to the || operator.