Question

我是编程新手。我刚刚学习了c ++和一些数据结构的概念。我正在尝试使用邻接列表编写图表表示。我使用codeblocks作为我的编译器。但是，每当我尝试编译我的程序时，我都会收到以下错误...

C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 22 | error：变量或字段'initialize_graph'声明为void | C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 22 |错误：在'，'令牌之前预期的primary-expression | C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 22 | error：在'int'|之前预期的primary-expression C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 23 | error：变量或字段'read_graph'声明为void | C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 23 |错误：在'，'标记之前预期的primary-expression | C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 23 | error：在'int'|之前预期的primary-expression C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 24 | error：变量或字段'insert_edge'声明为void | C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 24 |错误：在'，'令牌之前预期的primary-expression | C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 24 |错误：在'int'|之前预期的primary-expression C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 24 |错误：在'int'|之前预期的primary-expression C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 24 |错误：在'int'|之前预期的primary-expression C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 25 | error：变量或字段'print_graph'声明为void | C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 25 |错误：期望' - '之前的primary-expression'令牌| C：\ Users \ Garg \ Desktop \ try \ Stack.cpp ||在函数'int main（）'中：| C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 32 |错误：'read_graph'未在此范围内声明 C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 33 |错误：'print_graph'未在此范围内声明 C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 36 | error：变量或字段'initialize_graph'声明为void | C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 36 |错误：'g'未在此范围内声明 C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 36 | error：在'int'|之前预期的primary-expression C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 46 | error：变量或字段'read_graph'声明为void | C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 46 |错误：'g'未在此范围内声明 C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 46 | error：在'int'|之前预期的primary-expression || ===构建完成：21个错误，0个警告=== |

这是我的计划：

#include<iostream>
#define MAXV 1000       /* maximum number of vertices */
using namespace std;
struct node
{ 
int y;              /*adjacency info*/
int weight;             /* edge weight, if any */
struct node *next;      /* next edge in list */
} edgenode;

struct graph{
node *edges[MAXV+1];     /* adjacency info */
int degree[MAXV+1];             /* outdegree of each vertex */
int nvertices;          /* number of vertices in graph */
int nedges;             /* number of edges in graph */
int directed;           /* is the graph directed? */
} graph;
void initialize_graph (graph *, int);
void read_graph (graph *, int);
void insert_edge (graph *, int, int, int);
void print_graph (graph *);

int main()
{
struct graph *g = NULL;
cout << "Now reading graph";
read_graph(g, 1);
print_graph(g);
return 0;
}
void initialize_graph(graph *g, int directed)
{
int i;
g -> nvertices = 0;
g -> nedges = 0;
g -> directed = directed;
for (i=1; i<=MAXV; i++) g->degree[i] = 0;
for (i=1; i<=MAXV; i++) g->edges[i] = NULL;
}

void read_graph(graph *g, int directed)
{
int i;
int m;
int x, y;           /* vertices in edge (x,y) */
initialize_graph(g, directed);
cout << "Enter the number of vertivces and edges";
cin >> g->nvertices;
cin >> m;
cout << "Enter the vertices for the edge and the weight of the edge";
for (i=1; i<=m; i++) {
cin >> x;
cin >> y;
insert_edge(g,x,y,directed);
}
}

void insert_edge(graph *g, int x, int y, int directed)
{
struct node *p;
p = malloc(sizeof(struct node));
p->weight = NULL;
p->y = y;
p->next = g->edges[x];
g->edges[x] = p; /* insert at head of list */
g->degree[x] ++;
if (directed == 0)
insert_edge(g,y,x,1);
else
g->nedges ++;
}



void print_graph(graph *g)
{
int i; /* counter */
edgenode *p; /* temporary pointer */
for (i=1; i<=g->nvertices; i++) {
cout << i;
p = g->edges[i];
while (p != NULL) {
cout << p->y;
p = p->next; 
}
cout << "\n";
}
}

此外，当我在main之前定义函数时，我得到以下结果：

C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 23 | error：在'（'token |'之前的预期构造函数，析构函数或类型转换 C：\ Users \ Garg \ Desktop \ try \ Stack.cpp | 33 |错误：在'（'token |'之前的预期构造函数，析构函数或类型转换 || ===构建完成：2个错误，0个警告=== |

#include<iostream>
#define MAXV 1000       /* maximum number of vertices */

using namespace std;

struct node
{
int y;              /*adjacency info*/
int weight;             /* edge weight, if any */
struct node *next;      /* next edge in list */
} edgenode;


struct graph{
node *edges[MAXV+1];     /* adjacency info */
int degree[MAXV+1];             /* outdegree of each vertex */
int nvertices;          /* number of vertices in graph */
int nedges;             /* number of edges in graph */
int directed;           /* is the graph directed? */
} graph;


initialize_graph(graph *g, int directed)
{
int i;
g -> nvertices = 0;
g -> nedges = 0;
g -> directed = directed;
for (i=1; i<=MAXV; i++) g->degree[i] = 0;
for (i=1; i<=MAXV; i++) g->edges[i] = NULL;
}

read_graph(graph *g, int directed)
{
int i;
int m;
int x, y;           /* vertices in edge (x,y) */
initialize_graph(g, directed);
cout << "Enter the number of vertivces and edges";
cin >> g->nvertices;
cin >> m;
cout << "Enter the vertices for the edge and the weight of the edge";
for (i=1; i<=m; i++) {
cin >> x;
cin >> y;
insert_edge(g,x,y,directed);
}
}

insert_edge(graph *g, int x, int y, int directed)
{
struct node *p;
p = malloc(sizeof(struct node));
p->weight = NULL;
p->y = y;
p->next = g->edges[x];
g->edges[x] = p; /* insert at head of list */
g->degree[x] ++;
if (directed == 0)
insert_edge(g,y,x,1);
else
g->nedges ++; 
}



print_graph(graph *g)
{
int i; /* counter */
edgenode *p; /* temporary pointer */
for (i=1; i<=g->nvertices; i++) {
cout << i;
p = g->edges[i];
while (p != NULL) {
cout << p->y;
p = p->next;
}
cout << "\n";
}
}

int main()
{
struct graph *g = NULL;
cout << "Now reading graph";
read_graph(g, 1);
print_graph(g);
return 0;
}

指出我做错了什么？

Answer 1

你有一个结构和一个名为graph的变量。不要那样做。

当您声明这些函数时，编译器会尝试使用变量...

Answer 2

我认为代码

struct node
{ 
    int y;              /*adjacency info*/
    int weight;             /* edge weight, if any */
    struct node *next;      /* next edge in list */
} edgenode;

在typedef之前缺少struct。尝试将其添加到两个结构定义中。

没有typedef，它说：这是一个结构，顺便创建一个这种类型的变量。使用typedef，您实际上是在创建类型edgenode和graph。

Answer 3

忘记声明函数的返回类型？例如，在您的最后一串代码中initialize_graph应为

void initialize_graph(graph *g, int directed);

<强>更新

您对变量graph的声明会使struct同名。因此，当您声明函数以graph*作为参数时，您指的不是类型而是变量。

另一个更新：

您声明指向graph的指针并将其初始化为NULL。然后尝试从函数中取消引用此类空指针。您必须使指针指向有效图形，例如通过指定它new graph();

预期构造函数，析构函数或类型转换之前的＆＃39;（＆＃39;令牌

3 个答案: