C：不兼容的类型赋值 [英] C: incompatible types in assignment

问题描述

我正在写一个程序来检查，看是否有端口处于C.特别拷贝参数char数组一一线路畅通。然而，当我尝试编译，它说：

I'm writing a program to check to see if a port is open in C. One line in particular copies one of the arguments to a char array. However, when I try to compile, it says:

错误：不兼容的类型
  分配

error: incompatible types in assignment

继承人的code。该错误是地址

Heres the code. The error is on the assignment of addr

#include <sys/socket.h>
#include <sys/time.h>
#include <sys/types.h>
#include <arpa/inet.h>
#include <netinet/in.h>
#include <errno.h>
#include <fcntl.h>
#include <stdio.h>
#include <netdb.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>

int main(int argc, char **argv) {
  u_short port;                /* user specified port number */
  char addr[1023];             /* will be a copy of the address entered by u */
  struct sockaddr_in address;  /* the libc network address data structure */
  short int sock = -1;         /* file descriptor for the network socket */

  port = atoi(argv[1]);
  addr = strncpy(addr, argv[2], 1023);
  bzero((char *)&address, sizeof(address));  /* init addr struct */
  address.sin_addr.s_addr = inet_addr(addr); /* assign the address */
  address.sin_port = htons(port);            /* translate int2port num */

  sock = socket(AF_INET, SOCK_STREAM, 0);
  if (connect(sock,(struct sockaddr *)&address,sizeof(address)) == 0) {
    printf("%i is open\n", port);
  }
  if (errno == 113) {
    fprintf(stderr, "Port not open!\n");
  }
  close(sock);
  return 0;
}

我是新的C，所以我不知道为什么会这么做。

I'm new to C, so I'm not sure why it would do this.

推荐答案

地址是一个数组，所以你不能直接分配给它。

addr is an array so you can't assign to it directly.

修改地址=函数strncpy（地址，ARGV [2]，1023）; 到函数strncpy（地址，ARGV [2]，1023）;

一个指针，你传给什么返回，但并不需要此值。到函数strncpy 通话单独将复制从字符串的argv [2] 到地址。

A pointer to what you passed in is returned, but this value isn't needed. The call to strncpy alone will copy the string from argv[2] to addr.

请注意：我注意到，有时你在数组的地址传递，有时你数组本身在传递，无需操作员的地址。

Note: I notice sometimes you pass in the address of your array and sometimes you pass in the array itself without the address of operator.

在参数只要求的char * ...

When the parameter only asks for char*...

虽然两者将工作在地址传递，而不是＆放大器;地址是比较正确的。 ＆放大器;地址给出了一个指向一个字符数组字符（*）[1023] ，而地址为您提供了一个的char * 这是第一个元素的地址。它通常并不重要，但如果你做指针算法，然后它会带来很大的区别。

Although both will work passing in addr instead of &addr is more correct. &addr gives a pointer to a char array char (*)[1023] whereas addr gives you a char* which is the address of the first element. It usually doesn't matter but if you do pointer arithmetic then it will make a big difference.

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