Question

我该如何执行以下操作？ full_agentstring需要指向cdr-＆gt; channel或cdr-＆gt; dstchannel：

....
char *cdr_channel = cdr->channel;
char *cdr_dstchannel = cdr->dstchannel;
char *full_agentstring = NULL;
if(strstr(cdr_channel, agent_pattern) != NULL) {
      *full_agentstring = cdr->channel;
} else if(strstr(cdr_dstchannel, agent_pattern) != NULL) {
      *full_agentstring = cdr->dstchannel;

我也试过full_agentstring = &cdr_dstchannel但仍然收到警告：

从不兼容的指针类型[默认启用]

进行分配

Answer 1

if(strstr(cdr_channel, agent_pattern) != NULL) {
      full_agentstring = cdr->channel;
} else if(strstr(cdr_dstchannel, agent_pattern) != NULL) {
      full_agentstring = cdr->dstchannel;

或者，如果我不正确地理解你的问题，那么这是第二种方法

char *cdr_channel = cdr->channel;
char *cdr_dstchannel = cdr->dstchannel;
char **full_agentstring = NULL;
if(strstr(cdr_channel, agent_pattern) != NULL) {
      full_agentstring = &cdr->channel;
} else if(strstr(cdr_dstchannel, agent_pattern) != NULL) {
      full_agentstring = &cdr->dstchannel;

根据C中的条件更改指针（星号1.4 cdr struct）

1 个答案: