initializing array of structure with array elements in it

Question

#include <stdio.h>

int main()
{
    typedef struct s
    {
        int a;
        int b[5];
        char c[2];
    }st;

    st vs[1];

    vs[0] = {1,{1,2,3,4,5},{'c','d'}};

    printf("%d:a\n",vs[1].a);
    printf("%d:b[0]\t %d:b[4]\n",vs[0].b[0],vs[0].b[4]);
    printf("%c:c[0]\t %c:c[1]\n",vs[0].c[0],vs[0].c[1]);

    return 0;
}

why does this doesn't work?

on

gcc -o main *.c

I get this error

main.c: In function 'main': main.c:15:12: error: expected expression before '{' token vs[0] ={1,{1,2,3,4,5},{'c','d'}};

But if I have this:

#include <stdio.h>

int main()
{
    typedef struct s
    {
        int a;
        int b[5];
        char c[2];
    }st;

    st vs[] = {
                {1,{1,2,3,4,5},{'c','d'}}
              };

    printf("%d:a\n",vs[0].a);
    printf("%d:b[0]\t %d:b[4]\n",vs[0].b[0],vs[0].b[4]);
    printf("%c:c[0]\t %c:c[1]\n",vs[0].c[0],vs[0].c[1]);

    return 0;
}

it works. What is the logic in this.

How can I make it work using st vs[1] method?

P.P · Accepted Answer · 2016-01-21 08:36:57Z

8

You can only do braced initialization when you declare a variable. So,

st vs[] = {
            {1,{1,2,3,4,5},{'c','d'}}
          };

is allowed. But

vs[0] = {1,{1,2,3,4,5},{'c','d'}};

is not. Because this is not a initialization but assignment.

However, you can use C99's compound literal, see C11, 6.5.2.5:

vs[0] = (struct s){1,{1,2,3,4,5},{'c','d'}};

edited Jan 21, 2016 at 8:36

answered Jan 21, 2016 at 8:31

P.P

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Comments

M.M · Accepted Answer · 2016-01-21 08:36:11Z

Initialization is when you declare a variable and provide initial values for it as part of the declaration. For example this is legal:

st vs[1] = { {1,{1,2,3,4,5},{'c','d'}} };

Your code is actually attempting assignment. Assignment is when an existing variable has a value assigned to it. The reason your code doesn't work is that {1,{1,2,3,4,5},{'c','d'}} isn't a value.

In a statement (not a declaration), each expression must be readable by the compiler on its own merit, and a more complicated statement is made up of various expressions joined by operators. So the compiler doesn't know what to do with {1,{1,2,3,4,5},{'c','d'}} - at this stage it has no idea that that is supposed to be a st.

Since C99 there is a new language construct you can use here, called compound literal:

vs[0] = (const st){1,{1,2,3,4,5},{'c','d'}};

Note that this is not a cast operator being applied to some sort of braced expression; it is a single syntactic construct (Typename){ initializers } .

My use of const is a micro-optimization, it may help the compiler store the literal in a read-only block of the executable and allow constant folding.

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