我写了一个小的ppoll测试,但我对信号处理感到困惑。该手册页说:
poll()和ppoll()之间的关系类似于select(2)和pselect(2)之间的关系:如pselect(2),ppoll()允许应用程序安全地等待,直到文件描述符变为准备好或直到信号被捕获。
在我的情况下,在超时后处理信号时立即处理fd更改。我在进程上下文中使用了ctr-c并从另一个shell中杀死但总是有相同的行为。
我做错了什么?我错过了什么吗?
#include <iostream>
#include <poll.h>
#include <signal.h>
#include <unistd.h>
void handleSignal(int signal)
{
switch(signal)
{
case SIGINT:
std::cerr << "sigint" << std::endl;
break;
case SIGTERM:
std::cerr << "sigterm" << std::endl;
break;
default:
std::cerr << "sig=" << signal << std::endl;
break;
}
}
int main(int argc, char* argv[])
{
int result;
FILE *fd = fopen("/tmp/tmpFile", "r");
result = fileno(fd);
if(-1 == result)
{
std::cerr << "could not open temp file" << std::endl;
return EXIT_FAILURE;
}
pollfd fds[2];
fds[0] = { 0, POLLIN, 0 };
fds[1] = {result, POLLIN, 0 };
sigset_t mySigset, oldSigset;
sigemptyset(&mySigset);
sigaddset(&mySigset, SIGINT);
sigaddset(&mySigset, SIGTERM);
struct sigaction mySigHandler;
mySigHandler.sa_handler = &handleSignal;
mySigHandler.sa_flags = 0;
sigemptyset(&mySigHandler.sa_mask);
sigaction(SIGINT, &mySigHandler, NULL);
sigaction(SIGTERM, &mySigHandler, NULL);
char buffer[1024];
timespec time;
while(true)
{
time = {20, 0};
result = ppoll(&fds[0], 2, &time, &mySigset);
if(0 == result)
{
// timeout
std::cout << "ppoll timeout" << std::endl;
}
else if(0 > result)
{
// error
std::cerr << "ppoll error" << std::endl;
}
else
{
// at least one fd changed
std::cout << "active fds: " << result << std::endl;
for(int i = 0; i < 2; i++)
{
if(fds[i].revents & POLLIN)
{
result = read(fds[i].fd, buffer, 1024);
if (-1 == result)
{
std::cerr << "error while reading fd " << fds[i].fd << std::endl;
}
else if(0 < result)
{
buffer[result] = '\0';
std::cout << "read fd " << fds[i].fd << ": " << buffer << std::endl;
}
}
}
}
}
return EXIT_SUCCESS;
}
答案 0 :(得分:0)
当您致电ppoll
时,您告诉它阻止SIGINT
和SIGTERM
将其传递到信号掩码中。