Thread: Compound literals

Code:

 #include <stdio.h>
int sum_array(int * , int );

int main(void)
{	
	
int total = sum_array( (int []) { 1 , 2 , 3 , 4 } , 4 );

printf(" Total is: %d" , total);

return 0;
}
int sum_array( int arr[] , int n)
{
	int i , sum=0 ;
	
	for(i=0; i<n; i++)
	sum += arr[i];
	
	return sum;
}

We create an unnamed array "on the fly" that it means array has no permament storage during the program execution?

Thank you in advance

Originally Posted by grumpy

As far as your program is concerned, the array does not exist in any statement of main() after the call of sum_array(), if that's what you mean.

As to how (or if) that array will exist in memory for as long as your program is running, that is compiler-dependent. That array is essentially a compile-time constant, and it (or the population of it) has to be represented somehow in executable code.

It is not a technique you should use if your goal is to minimise memory usage - as measured, for example, by a host operating system - of your program at some point after calling sum_array().

Ok. I was wondering what is the meaning of creation an array "on the fly" . That was my basic doubt.

It's an array of integers, it's pretty clear regarding the type (perhaps you are asking something else).

Originally Posted by migf1

It's an array of integers, it's pretty clear regarding the type (perhaps you are asking something else).

I don't know maybe is a silly question... but I can't understand what we mean when we say "type name" in a compound literal using structures is more clear to me.... (struct tag_name) yes that is a type name.... but here ? I don't see any name in (int [])

sorry if the question is silly.

I didn't imply your question was silly, only that perhaps I didn't understand it.

So to make it clear, your question is "why int[] specifies a type"? If so, what would you think as a better alternative for describing the type of an array of integers?

Originally Posted by migf1

I didn't imply your question was silly, only that perhaps I didn't understand it.

So to make it clear, your question is "why int[] specifies a type"? If so, what would you think as a better alternative for describing the type of an array of integers?

No. I am just wondering what does it means with "type name" I know that the type is int [] the name of type where is? maybe my question is bad or weird.... I don't want to tire you.

Originally Posted by Mr.Lnx

No. I am just wondering what does it means with "type name"

"type name" is defined in the C standard (section 6.7.6 in C99, 6.7.7 in C11).

Originally Posted by Mr.Lnx

I know that the type is int [] the name of type where is?

The type name of the type int [] is int [] :-).

Bye, Andreas

Originally Posted by C11 standard

In several contexts, it is necessary to specify a type. This is accomplished using a type
name, which is syntactically a declaration for a function or an object of that type that
omits the identifier.

So

Code:

int i; // type name "int"
int arr[5]; // type name "int[5]"
void func(int i, int j); //type name "void (int,int)"

So yes, the int[] between the brackets is the type name. It's not a stupid question -- you might have expected it to be "array" or something. But nope, it's easier than that.

Thank you very much guys. Your explanations is what I wanted