And what's the difference between
void*(*void)(void*)
and
void*(*voi)(void*)
and when to use it?
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1I see no difference, exept a typo (voi) – Chris Maes Jan 10 '16 at 16:40
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Don't you have a typo: 'voi' instead of 'void'? – arainone Jan 10 '16 at 16:40
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5well `void*(void)(void)` doesn't compile, because it tries to declare a function that returns a `void*` named `void` and you can't use keywords of the language as names. – PeterT Jan 10 '16 at 16:42
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The formatting is making the \* get lost – Sami Kuhmonen Jan 10 '16 at 16:53
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@SamiKuhmonen You were putting something more, that wasn't in the OP. That's why I rolled back. – πάντα ῥεῖ Jan 10 '16 at 16:55
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Yes, the typo shouldn't have been fixed, otherwise the formatting is making people miss the point – Sami Kuhmonen Jan 10 '16 at 16:55
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@SamiKuhmonen Found a simpler way to clarify with an edit ... – πάντα ῥεῖ Jan 10 '16 at 17:11
1 Answers
4
well void*(void)(void) doesn't compile, because it tries to declare a function that returns a void* named void and you can't use keywords of the language as names.
you can't declare an int named void either (int void=5; doesn't work of course).
Now voi is a valid identifier and you can name a function voi.
Trying to get at what you're actually asking the difference between.
void(*name)(void); and void(name)(void); is that one declares a function and the other declares a function pointer.
In fact the brackets on the second example don't do anything. void(name)(void); is the same as void name(void);.
However when declaring function pointers the brackets are needed or the * will bind left to the return type.
Anyway, I'm not sure what exactly you're asking, so you're best off searching the site for other questions regarding function pointers. Like this one that also explains why using void(name)(void) can be different from void(name)().
