Why cast pointer into void

Question

I came across some code. Right after the construction, the author casts the pointer into a void type. What is the purpose of this line? Is it like assert( m_some_class_ptr != 0) to check the integrity of the pointer?

Some_class * m_some_class_ptr = new Some_class();

(void)m_some_class_ptr;

Thank you.

can't be sure without seeing more code, but that's usually done to get rid of the "variable set but not used" warning from the compiler. — user26347
– user26347, Commented Mar 19, 2014 at 23:32
you know what, i think you are right. These pointers are all global variables (evil, i know), which are not used in the context. — Frank Liu
– Frank Liu, Commented Mar 19, 2014 at 23:35

gnasher729 · Accepted Answer · 2014-03-19 23:35:08Z

1

Cast to void is a common idiom to avoid compiler warnings that a variable is unused. If the initialisation was the only use of m_some_class_ptr, you would quite possible get a warning. So you want to add a second use of the variable, but without doing anything.

(void)m_some_class_ptr; does nothing. It evaluates the expression m_some_class_ptr and throws the result away.

answered Mar 19, 2014 at 23:35

gnasher729

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