C declaration

[Bibi]

What does the following C declaration means?

void *(disp)(int))

I can't even begin to understand it!

Thank you in advance!

--B

 

[Allan Bruce]

What does the following C declaration means?

void *(disp)(int))

I can't even begin to understand it!

Thank you in advance!

It is a pointer to a function that returns a void and takes an int as an
arguement.
Allan

 

[Victor Nazarov]

What does the following C declaration means?

void *(disp)(int))

I can't even begin to understand it!

Thank you in advance!

--B

It is incorrect. What about the brace at the end of the line? If you
have meant
void *(disp)(int);
then this is an equevalent for
void *disp (int);
which is the declaration of function with single argument of type int
which returns pointer to void (nothing).

vir

 

[Bibi]

Thanks!

--B

It is a pointer to a function that returns a void and takes an int as an
arguement.
Allan

 

[Thomas stegen]

It is a pointer to a function that returns a void and takes an int as an
arguement.

Err, nope... Assuming
void *(disp)(int);

Then it is a function prototype for a function called disp
which returns a pointer to void and takes an int as the argument.

 

[osmium]

What does the following C declaration means?

void *(disp)(int))

I can't even begin to understand it!

It is a fragment of something. There is no semi colon and there are
mismatched parens. Try to type more carefully.

 

[Allan Bruce]

Err, nope... Assuming
void *(disp)(int);

Then it is a function prototype for a function called disp
which returns a pointer to void and takes an int as the argument.

so it is, I should have read more carefully, if the OP had asked about

void (*disp)(int);

then that is a pointer to a function. The OPs function can be rewritten as:

void * disp(int);

which is a 'normal' function called disp returning a void pointer.

Allan

 

[Bibi]

so it is, I should have read more carefully, if the OP had asked about

void (*disp)(int);

then that is a pointer to a function. The OPs function can be rewritten as:

void * disp(int);

which is a 'normal' function called disp returning a void pointer.

Allan

Well sorry for the mistake. it was in fact
void *(disp)(int)
It was in the argument of the function sigset(2).
void(*sigset(int sig, void (*disp)(int)))(int);
Now it makes sense to me.
Thank you all!

--B

 

[Emmanuel Delahaye]

Well sorry for the mistake. it was in fact
void *(disp)(int)

Of course, you meant

void (*disp)(int)

 

[James McIninch]

<posted & mailed>

What does the following C declaration means?

void *(disp)(int))

I can't even begin to understand it!

Thank you in advance!

--B

You are missing a parenthesis, but it says 'disp is a pointer to a function
that takes and integer argument'.