[英]cannot convert ‘char**’ to ‘char*’
I have an index for the network interface I got a packet from (ie 2), and need to find the name of interface, which should return "eth0"
. 我有一个网络接口的索引,该网络接口是我从中获得数据包的(即2),并且需要找到接口名称,该名称应返回"eth0"
。 I'm using if_indextoname()
. 我正在使用if_indextoname()
。
I'm not much familiar with C++ on Ubuntu, but my code drops an error: 我对Ubuntu上的C ++不太熟悉,但是我的代码删除了一个错误:
cannot convert
char**
tochar*
for argument 2 tochar* if_indextoname(unsigned int, char*)
无法将参数2的char**
转换为char*
到char* if_indextoname(unsigned int, char*)
Can someone help me to fix it? 有人可以帮我解决吗?
#include <net/if.h>
#include <iostream>
int main()
{
unsigned int ifindex = 2;
char *ifname[10];
std::cout << if_indextoname(ifindex, ifname);
std::cout << ifname << std::endl;
}
char *ifname[10];
declares 10 char pointers. 声明10个char指针。
I guess what you need is a char pointer. 我猜您需要的是一个char指针。
char* ifname = new char[IF_NAMESIZE+1]
should solve your problem. char* ifname = new char[IF_NAMESIZE+1]
应该可以解决您的问题。
Alternatively, you could just allocate an auto char buffer, if you do not want to pass it to other functions. 另外,如果您不想将其传递给其他函数,则可以分配一个自动char缓冲区。
char ifname[IF_NAMESIZE+1]
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