【发布时间】:2015-08-27 12:11:57
【问题描述】:
ZeroMQ 官方网站 here 提供了一个基本示例(向下滚动到“询问,您将收到”部分):
// Hello World server
#include <zmq.h>
#include <stdio.h>
#include <unistd.h>
#include <string.h>
#include <assert.h>
int main (void)
{
// Socket to talk to clients
void *context = zmq_ctx_new (); // LINE NUMBER 9
void *responder = zmq_socket (context, ZMQ_REP);
int rc = zmq_bind (responder, "tcp://*:5555");
assert (rc == 0);
while (1) {
char buffer [10];
zmq_recv (responder, buffer, 10, 0); // LINE NUMBER 15
printf ("Received Hello\n");
zmq_send (responder, "World", 5, 0); // LINE NUMBER 17
sleep (1); // Do some 'work'
}
return 0;
}
我使用的是 Ubuntu 12.04 LTS,并从标准存储库安装了 ZeroMQ。但是当我尝试编译上面的例子时,我得到了以下错误:
./test.c: In function ‘main’:
./test.c:9:21: warning: initialization makes pointer from integer without a cast [enabled by default]
./test.c:15:9: warning: passing argument 2 of ‘zmq_recv’ from incompatible pointer type [enabled by default]
/usr/include/zmq.h:228:16: note: expected ‘struct zmq_msg_t *’ but argument is of type ‘char *’
./test.c:15:9: error: too many arguments to function ‘zmq_recv’
/usr/include/zmq.h:228:16: note: declared here
./test.c:17:9: warning: passing argument 2 of ‘zmq_send’ from incompatible pointer type [enabled by default]
/usr/include/zmq.h:227:16: note: expected ‘struct zmq_msg_t *’ but argument is of type ‘char *’
./test.c:17:9: error: too many arguments to function ‘zmq_send’
/usr/include/zmq.h:227:16: note: declared here
于是我在/usr/include/zmq.h中查找了zmq_recv和zmq_send的定义,发现编译器确实说的是实话。以下是签名:
int zmq_bind (void *s, const char *addr);
int zmq_send (void *s, zmq_msg_t *msg, int flags);
int zmq_recv (void *s, zmq_msg_t *msg, int flags);
那么,我认为该站点上的文档不正确(或者可能已过时)是否正确?有没有其他人遇到过这个问题?
【问题讨论】:
-
所以首先确保您拥有正确的版本 - 默认存储库中的版本可能不适合您正在查看的文档。其次,您是否使用了正确的标志?
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@EiyrioüvonKauyf,确实如此。我查看了源文件,似乎
Ubuntu 12.04的存储库只有version 2.1.11。本教程假定版本号至少为3.2。我真傻;我只是假设像 Ubuntu 这样的主要发行版会有一个相当最新的版本。