array initializing

Question

I came across this code construct in Linux and would like to understand it

struct mystruct {
  int x;
  int b[40];
};


/*later */

static struct mystruct e = { .x = 5,
                             .b = {-1},   
                           };

What does .b = {-1} do ? Does it initialize only the first or all elements of array b ? How does it work ?

Designated Initializers: gcc.gnu.org/onlinedocs/gcc/Designated-Inits.html — P.P
– P.P, Commented May 9, 2012 at 12:56

Aftnix · Accepted Answer · 2012-05-09 12:46:48Z

3

static struct mystruct e = {
    .x = 5,
    .b = {-1},   
             };

here it initializes b[0] to -1. other elements are initialized to 0.

answered May 9, 2012 at 12:46

Aftnix

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Comments

MByD · Accepted Answer · 2012-05-09 12:41:35Z

1

It means initialize the member b of the struct with an array that starts with -1 and followed by 0

The .b = is a GCC extensions. (as commented, from C99 this is part of the standard as well)
The {-1} is standard array initialization.

edited May 9, 2012 at 12:41

answered May 9, 2012 at 12:35

MByD

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2 Comments

wildplasser Over a year ago

Designated initialisers were introduced in C99 (but IIRC gcc is the only platform that implements c99) So, calling it a gcc extension would be unfair.

unwind Over a year ago

Considering the positioning of the code (inside the braces of a struct initialization), I don't think it's totally correct to call the = part of the standard notation. Before C99, you would have written this as e = { 5, { -1 } };, without the equals.

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