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Is `widget w(gadget(), doodad());` a Variable Declaration in C ?

Linda Hamilton
Release: 2024-11-19 07:53:02
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Is `widget w(gadget(), doodad());` a Variable Declaration in C  ?

C 's Most Vexing Parse Resolved

The issue arises with code like the following:

widget w(gadget(), doodad());  // Pitfall: Not a Variable Declaration
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One might assume this code is declaring a variable named w of type widget. However, this is not the case.

In C , arguments of type array decay into pointers to the first element, and arguments of type function decay into a function pointer. This means the signature of the function being declared is:

widget w(gadget(*)(), doodad(*)());
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This function takes two arguments: one is a pointer to a function taking no arguments and returning a gadget, and the other is a pointer to a function taking no arguments and returning a doodad. The function itself returns a widget.

Even more confusing cases arise when extra parentheses are added to function arguments, as in:

widget w(gadget(x));
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This looks like it should declare a variable named x of type gadget, but it actually declares a function that takes a first argument named x of type gadget and returns a widget.

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