Multiple increment operators in single statement [duplicate]

Answer 1

The behavior will be undefined as told by others. The output depends upon the implementation of compiler.

But as per the standard it should be undefined.

Answer 2

The behavior here is undefined. See this question

Relevant standard quote:

§5/4.1 Between the previous and next sequence point a scalar object shall have its stored value modified at most once by the evaluation of an expression.

The most common sequence point is the end of a statement.

Also worth noting from the standard:

§5.2.2/8 The order of evaluation of arguments is unspecified.

Answer 3

AS this is undefined behaviour, one can't tell the end result. The result depends on the implementation.

Answer 4

The standard says this is undefined. The compiler is free to evaluate the statements in any order is sees fit as long as it follows the operator precedence rules. This results in UB:

b++ * ++b; // b is modified more than once

Answer 5

Undefined behavior, Compiler is free to evaluate this expression in any order because of the same precedence of the operators. Consider using

(b++)*(++b)

instead