int id1; //variables used in the isr should be volatile
int chid;
void *ConfigureISR(void) //void *ISR (void *arg)
{
/* the software must tell the OS that it wishes to associate the ISR with a particular source of interrupts
* / * On x86 platforms, there are generally 16 hardware Interrupt Request lines (IRQs) */
StartInterruptTime = GetTimeStamp(); //startTime of the interrupt
// volatile int irq = 11; //0 : A clock that runs at the resolution set by ClockPeriod()
// InterruptEnable();
ThreadCtl (_NTO_TCTL_IO, NULL); // enables the hardware interrupt
// Initialize event structure
//
// Setup COID and event
//
/* Tell the kernel to attach an interrupt signal event to this thread */
// chid = ChannelCreate( 0 );
SIGEV_INTR_INIT( &event);
//id1 = InterruptAttach(0, ISR, NULL, 0, 0); // ISR is the interrupt service routine
id1 = InterruptAttach(_NTO_INTR_CLASS_SYNTHETIC, ISR, NULL, 0, 0);
if (id1 == -1)
{
fprintf(stderr, "can't attach to IRQ\n");
perror (NULL);
exit (EXIT_FAILURE);
}
while(1)
{
InterruptWait(NULL, NULL);
}
EndInterruptTime = GetTimeStamp();
InterruptLatency = (EndInterruptTime - StartInterruptTime);
printf("Inerrupt latency is %llu",InterruptLatency); //I am getting warning
measurements[17] = InterruptLatency;
}
我在另一个 .c 程序中获取 StartInterruptTime，并在上述程序中使用该时间戳来计算两者之间的差异(即 InterruptLatency)。我在调试程序时收到以下警告，并且 EndInterruptTime 将为零，但启动中断具有 double 值。
WARNING : Multiple markers at this line
- Unresolved breakpoint
有人可以帮我吗？
最佳答案
好的。如果 InterruptLatency 类型不是 int64_t，请使用此 printf("Inerrupt Latency is %lu",InterruptLatency);。
事实上，在 pintf("Inerrupt Latency is %Type_of_InterruptLatency",InterruptLatency); 中使用正确的 InterruptLatency 类型；
关于c++ - 如何解决这个 Unresolved 断点警告？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/23547538/