How to declare variable in function arguments

Question

I need to declare a variable inside function arguments. Please advice the syntax to use?

I've got something like this:

#include <stdio.h>

int foo (int *a)
{
   printf ("%d\n", *a);
}

int main (void)
{
   foo (&(int){int a=1});
   return 0;
}

And GCC fails with the message:

$ gcc a.c
a.c: In function 'main':
a.c:10: error: expected expression before '{' token

As an option I can put use not named variable like this (same question at russian version of Stack Overflow):

foo(&(int) { 1 });

and it works, but it is interesting why compiler accept {1} but does not accept {int a=1}

What do you mean by "declare it as a variable but not as a constant"? — StoryTeller - Unslander Monica
– StoryTeller - Unslander Monica, Commented Jan 31, 2018 at 7:52
'as a variable' means something like {int a=1}, not like just {1}. Sorry if I ask stupid questions. — Peter K.
– Peter K., Commented Jan 31, 2018 at 7:56
What is the purpose of doing this? I fail to see what advantage this has over int a = 1; foo(&a); — UnholySheep
– UnholySheep, Commented Jan 31, 2018 at 7:57
Since you know about the compound literal notation, and your &(int){int a = 1} notation seems to be trying to use it too, it isn't clear what you're hoping for. You can't define a variable name inside the compound literal. It really isn't clear what you're asking for. — Jonathan Leffler
– Jonathan Leffler, Commented Jan 31, 2018 at 8:16

Lundin · Accepted Answer · 2018-01-31 12:53:15Z

3

You can use compound literal - I suspect that's what you tried, you almost got it right:

foo (&(int){1});

This is by no means a "constant", I don't know how you got that idea.

Note that a compound literal only have local ("automatic") storage duration - if the calling block goes out of scope, so does the compound literal.

answered Jan 31, 2018 at 12:53

Lundin

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3 Comments

Peter K. Over a year ago

Yes, this variant works (as in description). And sorry - I've used wrong terms. You are correct - {1} here is not a "constant" - as I understoot it is just a variable on stack that has no name. My question was how to give a name to this variable. If I try to set {int a=1} instead of just {1} the gcc prints error. I just do not understand how compiler treat this construction and why does it suppose it as an error.

Lundin Over a year ago

@PeterK. Why on earth do you need to name a variable, when in this case the name is only useful from the point where the variable is created until the function is called? Once inside a function, the variable name is that of the pointer parameter. Your explanation about writing all of this in comments doesn't make sense. This is an XY problem if I ever saw one.

Peter K. Over a year ago

ok, accept this answer (even having it in my initial comment) because it actually resolve my request 'to have declaration inside argument'. Again, I would like to understand why compiler suppose that {int a=1} and just {1} are different.

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