注释掉未使用的行后,开关案例将无法编译

Question

这是我的代码:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <netdb.h>
#include <sys/types.h>
#include <sys/socket.h>
#include <arpa/inet.h>

int main (void) {

  struct addrinfo hints; 
  memset (&hints, 0, sizeof hints);

  hints.ai_family = AF_UNSPEC; 
  hints.ai_socktype = SOCK_DGRAM;  
  hints.ai_flags = AI_CANONNAME;   

  struct addrinfo *res;

  getaddrinfo ("example.com", "http", &hints, &res);
  printf ("Host: %s\n", "example.com");

  void *ptr;

  while (res != NULL) {
    printf("AI Family for current addrinfo: %i\n", res->ai_family);
    switch (res->ai_family) {
      case AF_INET:
        ptr = (struct sockaddr_in *) res->ai_addr;
        struct sockaddr_in *sockAddrIn = (struct sockaddr_in *) res->ai_addr;
        break;
    }
    res = res->ai_next;
  }
  return 0;
}

编译好.

但是,当我注释掉这一行时:

//ptr = (struct sockaddr_in *) res->ai_addr;

我会得到:

$ gcc ex4.c
ex4.c:30:9: error: expected expression
        struct sockaddr_in *sockAddrIn = (struct sockaddr_in *) res->ai_addr;
        ^
1 error generated.

我错过了什么？

Answer 1

从技术上讲,switch语句中的每个case都是一个标签.出于某些模糊和旧的原因,您不能将变量声明作为标签后的第一行.通过评论分配

ptr = (struct sockaddr_in *) res->ai_addr;

这条线

struct sockaddr_in *sockAddrIn = (struct sockaddr_in *) res->ai_addr;

成为标签之后的第一行AF_INET:,正如我所说,在C中是非法的.

解决方案是将所有case语句包装在大括号中,如下所示:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <netdb.h>
#include <sys/types.h>
#include <sys/socket.h>
#include <arpa/inet.h>

int main (void) {

  struct addrinfo hints; 
  memset (&hints, 0, sizeof hints);

  hints.ai_family = AF_UNSPEC; 
  hints.ai_socktype = SOCK_DGRAM;  
  hints.ai_flags = AI_CANONNAME;   

  struct addrinfo *res;

  getaddrinfo ("example.com", "http", &hints, &res);
  printf ("Host: %s\n", "example.com");

  void *ptr;

  while (res != NULL) {
    printf("AI Family for current addrinfo: %i\n", res->ai_family);
    switch (res->ai_family) {
      case AF_INET:
      {
        ptr = (struct sockaddr_in *) res->ai_addr;
        struct sockaddr_in *sockAddrIn = (struct sockaddr_in *) res->ai_addr;
        break;
      }
    }
    res = res->ai_next;
  }
  return 0;
}

无论如何,我认为这是更好的编码实践.

Answer 2

作为已接受答案的补充,您可以在案例标签之前声明变量.

switch(a) {
    int b; //can't initialize variable here
    case 0:
    ...
}

或者只使用一个空语句.