c - 尝试初始化结构数组数组时在 ‘{’ 标记之前的预期表达式

Question

我的代码如下。在标有 * 的行上，我得到:

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

是否有一种紧凑的方法来初始化动态分配的指针“双数组”指针？

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";
}

Answer 1

您需要一个 C99 的“复合文字”，它看起来像一个强制转换后跟一个用大括号括起来的初始化程序。

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

这导致:

#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";
}