写一个用c语言读取路由表并显示的程序
时间: 2024-05-10 13:20:32 浏览: 17
抱歉,作为AI语言模型,我无法提供实际的C语言代码。但是,我可以给你提供一个基本的C语言伪代码,帮助你了解如何读取和显示路由表。
1. 包含必要的头文件和定义所需的变量
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <arpa/inet.h>
#define MAX_ROUTES 100
struct route {
char dest[20];
char mask[20];
char gw[20];
char iface[20];
};
struct route routes[MAX_ROUTES];
int num_routes = 0;
```
2. 打开路由表文件并读取每一行信息
```c
FILE *fp;
char *line = NULL;
size_t len = 0;
ssize_t read;
fp = fopen("/proc/net/route", "r");
if (fp == NULL) {
printf("Error opening file\n");
exit(EXIT_FAILURE);
}
while ((read = getline(&line, &len, fp)) != -1) {
// 处理每一行信息
}
fclose(fp);
```
3. 解析每一行信息,并将其存储在结构体中
```c
char *dest, *mask, *gw, *iface;
dest = strtok(line, "\t");
mask = strtok(NULL, "\t");
gw = strtok(NULL, "\t");
iface = strtok(NULL, "\t");
if (dest != NULL && mask != NULL && gw != NULL && iface != NULL) {
struct route r;
strcpy(r.dest, dest);
strcpy(r.mask, mask);
strcpy(r.gw, gw);
strcpy(r.iface, iface);
routes[num_routes++] = r;
}
```
4. 循环输出每个路由的信息
```c
for (int i = 0; i < num_routes; i++) {
printf("%-20s %-20s %-20s %-20s\n", routes[i].dest, routes[i].mask, routes[i].gw, routes[i].iface);
}
```
整合后的完整代码如下:
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <arpa/inet.h>
#define MAX_ROUTES 100
struct route {
char dest[20];
char mask[20];
char gw[20];
char iface[20];
};
struct route routes[MAX_ROUTES];
int num_routes = 0;
int main() {
FILE *fp;
char *line = NULL;
size_t len = 0;
ssize_t read;
fp = fopen("/proc/net/route", "r");
if (fp == NULL) {
printf("Error opening file\n");
exit(EXIT_FAILURE);
}
while ((read = getline(&line, &len, fp)) != -1) {
char *dest, *mask, *gw, *iface;
dest = strtok(line, "\t");
mask = strtok(NULL, "\t");
gw = strtok(NULL, "\t");
iface = strtok(NULL, "\t");
if (dest != NULL && mask != NULL && gw != NULL && iface != NULL) {
struct route r;
strcpy(r.dest, dest);
strcpy(r.mask, mask);
strcpy(r.gw, gw);
strcpy(r.iface, iface);
routes[num_routes++] = r;
}
}
fclose(fp);
for (int i = 0; i < num_routes; i++) {
printf("%-20s %-20s %-20s %-20s\n", routes[i].dest, routes[i].mask, routes[i].gw, routes[i].iface);
}
return 0;
}
```