Initialize a struct in C using {}

Question

In the marked line I get an error Error - expected expression

#include <stdlib.h>

struct list_head {
    struct list_head *next, *prev;
};

struct program_struct {
    const char *name;
    struct list_head node;
};
typedef struct program_struct program_t;

struct task_t {
    program_t blocked_list;
};

int main() {

    struct task_t *p = malloc(sizeof(*p));
    p->blocked_list.name = NULL;
    p->blocked_list.node = {&(p->blocked_list.node), &(p->blocked_list.node)}; //error

    return 0;
}

I know I can replace this line with

p->blocked_list.node.next = &(p->blocked_list.node);
p->blocked_list.node.prev = &(p->blocked_list.node);

But can I make it work using {} like I tried in the first piece of code?

Even though you can use a compound literal for this, there is no reason to do so. Instead use the most readable form, which is two rows: p->blocked_list.node.next = &p->blocked_list.node; p->blocked_list.node.prev = &p->blocked_list.node; This is the clearest form and therefore the one preferred. — Lundin
– Lundin, Commented Oct 14, 2016 at 8:33

P.P · Accepted Answer · 2016-10-14 08:19:21Z

2

Initialization is allowed only when you define a variable. So, you can't use initializers in assignment. You can instead use C99's compound literals:

p->blocked_list.node = (struct list_head) {&(p->blocked_list.node), &(p->blocked_list.node)}; //error

answered Oct 14, 2016 at 8:19

P.P

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1 Comment

alk Over a year ago

Please note that the compound literal defined by doing (struct list_head) {..} lives on until the current scope is left. To have it be deallocated immediately after the assignment just wrap the whole assignment statement into a new scope by enclosing it into braces {...}.

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