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} };

是否有一种简洁的方法来初始化指向指针的动态分配“双数组”？

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

你需要一个C99'复合文字'，它看起来像一个演员，然后是大括号中的初始化器。

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";
}

Answer 2

您只能在(struct edge[])之前以C99模式初始化整个结构。否则，您必须自己初始化每个成员：

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

或者：

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;

尝试初始化结构数组数组时在'{'标记之前的预期表达式

2 个答案: