What is the difference between the two declarations in case of foo's arguments? The syntax in the second one is familiar to me and declares a pointer to function. Are both declarations fully equivalent?
void foo(int(int));
void foo(int(*)(int));
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this answer has some useful information about the use of, and lack of, an asterisk in function pointer declarations.Remy Lebeau– Remy Lebeau2018-05-24 00:44:08 +00:00Commented May 24, 2018 at 0:44
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Also see: Is the non-pointer syntax for declaring function pointer parameters not worth mentioning?Remy Lebeau– Remy Lebeau2018-05-24 00:45:00 +00:00Commented May 24, 2018 at 0:45
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gcc thinks that they are equal in this context: ideone.com/snKfj6Slava– Slava2018-05-24 00:47:11 +00:00Commented May 24, 2018 at 0:47
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1 Answer
They are equivalent as long as int(int) and int(*)(int) are used in function parameter lists. In function parameter list the int(int) is automatically adjusted by the language to mean int(*)(int).
It is the same adjustment mechanism that makes int [] parameter declaration equivalent to int * parameter declaration.
Outside of this specific context int(int) and int(*)(int) mean two different things.
answered May 24, 2018 at 0:45
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1 Comment
lukeg
Thanks! Can you elaborate on what are the two different meanings of the delcaration outside of function parameter context? Do you mean that one declares a function type and the other a pointer? I.e. something that has already been pointed out in the comments to my question?