Precedence of assignment

Question

In c ,

main()   {
 int a = (1,2,3,4);
  printf("%d",a); 
}

yields an output of

4

This is because comma(,)operator has a right to left precedence . But

main()   {
int a = {1,2,3,4};
 printf("%d",a);
}

yields an output

1

anyone pls explain the logic behind this. Thanks

I think that this is a [constraint violation](http://port70.net/~nsz/c/c11/n1570.html#6.7.9p2), and hence undefined behavior. — ad absurdum, Jun 26 '17 at 14:05
None of this looks remotely standard, nor correct as of 2017, or the past 30 years. — DeiDei, Jun 26 '17 at 14:05
It is actually perfectly fine C. Nearly identical duplicate: https://stackoverflow.com/questions/35266987/int-nums-5-2-1-4-causes-a-segmentation-fault — Lundin, Jun 26 '17 at 14:14
@Lundin (replaced comment) -- I see it's trying to do this to a pointer, but this question doesn't discuss whether this is allowed. See my answer, is my interpretation of §6.7.9 -- 2 wrong? — , Jun 26 '17 at 14:21

score 5 · Answer 1 · 2017-06-26T14:19:30.157

{1,2,3,4} is the syntax for an initializer, used to initialize something that has more than one value, like an array or a struct. So the commas inside it are not operators, they are just part of the initializer syntax.

When initializing, C uses the values from the initializer from the left. As you initialize a single scalar variable, only one element is needed.

Your compiler is supposed to tell you that this code doesn't make sense:

x.c:2:12: warning: excess elements in scalar initializer
 int a = {1,2,3,4};
            ^

I should add that this code violates a constraint of the standard, see C11 draft N1570, § 6.7.9 -- 2:

No initializer shall attempt to provide a value for an object not contained within the entity being initialized.

This requires the compiler to emit a diagnostic when compiling such broken code.

From the same paragraph is the following rule (number 17):

Each brace-enclosed initializer list has an associated current object. When no designations are present, subobjects of the current object are initialized in order according to the type of the current object: array elements in increasing subscript order, structure members in declaration order, and the first named member of a union.1

So your compiler decides to do the "next closest" thing to the standard and just use the first value you provide.

@Jean-FrançoisFabre it is required to do so (constraint violation), I just updated my answer. — , Jun 26 '17 at 14:20

score 2 · Answer 2 · answered Jun 26 '17 at 14:05

That's an initializer (usually used for arrays), but you are not using it wisely, didn't your compiler tell you?

Georgioss-MacBook-Pro:~ gsamaras$ gcc -Wall main.c 
main.c:2:13: warning: excess elements in scalar initializer
 int a = {1,2,3,4};
            ^

The precedence here is from left to right, and since you have only one element to initialize, one element is chosen from the initializer (that is 1 in this case).

score 1 · Answer 3 · answered Jun 26 '17 at 14:04

The curly braces mean initialization of the variable, mostly useful for arrays.

In your case, compiling with gcc yields:

test.c:6:12: warning: excess elements in scalar initializer
 int a = {1,2,3,4};
            ^

(same for 3 & 4)

Means that only value 1 is useful for your case (it's a scalar)

Precedence of assignment

3 Answers3