Expected expression before ‘{’ token in trying to initialize array of array of structs

Question

My code is below. On the lines marked with *'s, I get:

error: expected expression before ‘{’ token
rval.adj_list[0] = { {"B","C",3},{"B","A",2} };

Is there a compact way to initialize a dynamically allocated "double array" of pointers to pointers?

struct node;

struct edge {
    char *from;
    char *to;
    int weight;
};

struct digraph {
    char **vertices;
    struct edge **adj_list;
};

int main( int argc, char *argv[] ) {
    struct digraph rval;
    int size = 5;
    rval.vertices = malloc( size * sizeof(char*));
    rval.adj_list = malloc( size * sizeof(struct edge*));
    rval.vertices[0] = "A";
    rval.adj_list[0] = { {"A","B",2},{"A","E",1} };  //********
    rval.vertices[1] = "B";
    rval.adj_list[1] = { {"B","C",3},{"B","A",2} };  //********
    rval.vertices[2] = "C";
    rval.vertices[3] = "D";
    rval.vertices[4] = "E";
}

What's rval.vertices = malloc( size * sizeof(char*));? char* here is just a pointer, not the length you need. — herohuyongtao
– herohuyongtao, Commented Jan 31, 2014 at 0:34
char * is a pointer, not an array. arrays can be initialized with = {data}; , but not assigned. You're attempting to assign = {data} to a pointer, which is violates both of those. — Mooing Duck
– Mooing Duck, Commented Jan 31, 2014 at 0:42

Jonathan Leffler · Accepted Answer · 2014-01-31 00:43:58Z

4

You'd need a C99 'compound literal', which looks like a cast followed by an initializer in braces.

rval.adj_list[0] = (struct edge []){ {"A","B",2},{"A","E",1} };

This leads to:

#include <stdlib.h>

struct node;

struct edge {
    char *from;
    char *to;
    int weight;
};

struct digraph {
    char **vertices;
    struct edge **adj_list;
};

int main(void)
{
    struct digraph rval;
    int size = 5;
    rval.vertices = malloc( size * sizeof(char*));
    rval.adj_list = malloc( size * sizeof(struct edge*));
    rval.vertices[0] = "A";
    rval.adj_list[0] = (struct edge[]){ {"A","B",2}, {"A","E",1} };
    rval.vertices[1] = "B";
    rval.adj_list[1] = (struct edge[]){ {"B","C",3}, {"B","A",2} };
    rval.vertices[2] = "C";
    rval.vertices[3] = "D";
    rval.vertices[4] = "E";
}

answered Jan 31, 2014 at 0:43

Jonathan Leffler

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3 Comments

WhozCraig Over a year ago

+1 I wish I could up vote this again. This comes up more often than people admit, and usually those that don't know how to do it often hack up middle-man solutions.

WhozCraig Over a year ago

One thing eludes me. Both the types in the compound literal and the structure definition should be const, shouldn't they ? I would think it mattered (and I always do them that way) but i noticed they aren't here. Just curious.

Jonathan Leffler Over a year ago

If you can work out the correct mix of const qualifiers, be my guest to take the answer over. It isn't trivial to add them. The adjlist element itself must be writable; the array it points to, and the items that each element in the array point to, are not modifiable, strictly.

rullof · Accepted Answer · 2014-01-31 00:46:48Z

2

You could only initialize a whole structure in C99 mode preceding it by (struct edge[]). Otherwise you will have to initialize each member by itself:

rval.adj_list[0] = (struct edge[]){ {"A","B",2}, {"A","E",1} };
...
rval.adj_list[1] = (struct edge[]){ {"B","C",3}, {"B","A",2} };

Or:

rval.vertices[0] = "A";
rval.adj_list[0][0].from = "A";
rval.adj_list[0][0].to = "B";
rval.adj_list[0][0].weight = 2;
rval.adj_list[0][1].from = "A";
rval.adj_list[0][1].to = "E";
rval.adj_list[0][1].weight = 1;
rval.vertices[1] = "B";
rval.adj_list[1][0].from = "B";
rval.adj_list[1][0].to = "C";
rval.adj_list[1][0].weight = 3;
rval.adj_list[1][1].from = "B";
rval.adj_list[1][1].to = "A";
rval.adj_list[1][1].weight = 2;

answered Jan 31, 2014 at 0:46

rullof

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