我学习如何模拟数字逻辑电路。 我在这里展示我第一次尝试的源代码。 它是用于模拟电路的小程序 AND,OR和NOT gate。
此代码适用于没有循环的电路。 当引入电路循环时,由于无休止的递归,它会导致堆栈溢出。 请帮我删除这个错误。
请注意,这是一个爱好项目,任何帮助将不胜感激。
源代码:
#include <cstdlib>
#include <iostream>
using namespace std;
class LogicGate
{
int type;//gate type: 0 for NOT, 1 for OR, 2 for AND
//pins
bool ina;//input a
bool inb;//input b::this pin is not used for NOT gate
bool outc;//output
//fan-in
LogicGate* ga;//gate connected to input a
LogicGate* gb;//gate connected to input b
//fan-out
LogicGate* gc;//gate connected to output
int gctarget;//target input to which the gate "gc" is connected, 0 for input a, 1 for input c
public:
char* name;
LogicGate()
{
ina = inb = outc = false;
ga = gb = gc = (LogicGate*)0;
type = 0;
}
LogicGate(bool a, bool b)
{
ina = a; inb = b; outc = false;
ga = gb = gc = (LogicGate*)0;
type = 0;
}
//set logic
void settype(int t){if(t>=0&&t<3)type=t;else type=0;}
//set input
void seta(bool a){ ina = a; }
void setb(bool b){ inb = b; }
void setab(bool a, bool b){ina = a; inb = b; }
//connect gate
void cona(LogicGate* cga){ ga = cga; }
void conb(LogicGate* cgb){ gb = cgb; }
void conab(LogicGate* cga, LogicGate* cgb){ ga = cga; gb = cgb; }
//connect the output of this gate to another gate's input
void chainc(LogicGate* cgc, int target)
{
gc = cgc;
gctarget = target;
if(target==0) cgc->cona(this); else cgc->conb(this);
}
//calculate output
bool calcout()
{
//if the input is not available make it available by forcing the connected gates to calculate
if(ga){ ina = ga->calcout(); } //BUG:this may cause Stack overflow for circuits with loops
if(gb){ inb = gb->calcout(); }//BUG:this may cause Stack overflow for circuits with loops
//do the logic when inputs are available
switch(type)
{
case 0:
outc = !ina; break;
case 1:
outc = ina || inb; break;
case 2:
outc = ina && inb; break;
}
//if a gate is connected to output pin transfer the output value to the target input pin of the gate
if(gc){
if(gctarget==0){
gc->seta(outc);
}else{
gc->setb(outc);
}
}
//for debugging only
cout<<name<<" outputs "<<outc<<endl;
return outc;
}
};
int main(int argc, char *argv[])
{
LogicGate x,z;
//AND gate
z.settype(2);
z.seta(false);
z.setb(true);
z.name = "ZZZ";
//OR gate
x.settype(1);
x.cona(&z); // take one input from AND gate's output
x.setb(true);
x.name = "XXX";
//z.cona(&x);// take one input from OR gate's output to make a loop:: results in stack overflow
x.chainc(&z,0);//connect the output to AND gate's input "a" to form loop:: results in stack overflow
cout<<"final output"<<x.calcout()<<endl;
return 0;
}
答案 0 :(得分:0)
这里的问题是你无限循环。程序的行为与真正的逻辑门不同。我在这里看到两种可能性:
1)实施周期
你可以像cpu一样实现它。对calcout的调用仅计算到一个门的输出并迭代到下一个门。您可以为您的门创建一个Container类:
class GateContainer
{
//Contains all gates of your "circuit"
std::vector<LogicalGate> allGates;
//Contains all current gates to be processed
std::queue<LogicalGate*> currentGates;
void nextStep();
}
nextStep函数可能如下所示:
void GateContainer::nextStep()
{
//Get first gate from queue
LogicalGate *current = currentGates.front();
currentGates.pop();
//Do all the calculations with th current gate
//Push the gate connected to the Output to the list
currentGates.push(current->gc);
}
请注意,此代码未经测试,可能还需要进行一些错误检查
2)尝试捕获循环
您还可以尝试在calcout
中捕获循环。你可以通过在LogicalGate中创建一个标志来实现这一点,并在每次调用calcout之前重置它:
class LogicalGate
{
...
bool calculated;
...
}
在致电calcout()
之前,您需要为每个门设置计算到false
。然后,calcout看起来像这样:
bool calcout()
{
calculated = true;
if(ga && !ga->calculated){ ina = ga->calcout(); }
if(gb && !ga->calculated){ inb = gb->calcout(); }
...
}