第三十八課：邏輯操作符的陷阱----------狄泰軟件學(xué)院

2019-11-14 12:25:14

字體：大中小

供稿：網(wǎng)友

一、邏輯運算符的原生語義

1.操作符只有兩種值（true和false） 2.邏輯表達(dá)式不用完全計算就能確定最終值 3.最終結(jié)果只能是true或者false

#include<iostream>using namespace std;int fun(int i){ cout<<"int fun(int i): i="<<i<<endl; return i;}int main(){ if(fun(0) && fun(1)) { cout<<"the result is true"<<endl; } else { cout<<"the result is false"<<endl; } cout<<endl; if(fun(0) || fun(1)) { cout<<"the result is true"<<endl; } else { cout<<"the result is false"<<endl; } return 0;}打印結(jié)果：int fun(int i): i=0the result is falseint fun(int i): i=0int fun(int i): i=1the result is true

二、重載邏輯操作符

#include<iostream>using namespace std;class Test{PRivate: int i;public: Test(int i) { this->i = i; } int getI() const { return i; }};bool Operator && (const Test& l, const Test& r){ return (l.getI() && r.getI());}bool operator || (const Test& l, const Test& r){ return (l.getI() || r.getI()); }Test fun(Test i){ cout<<"Test fun(Test i): i="<<i.getI()<<endl; return i;}int main(){ Test t0(0); Test t1(1); if(fun(t0) && fun(t1))//operator && (fun(t0), fun(t1)) 則先要算出參數(shù)fun(t0)和fun(t1)的值，且這兩個的計算順序不確定 { cout<<"the result is true"<<endl; } else { cout<<"the result is false"<<endl; } if(fun(t0) || fun(t1)) { cout<<"the result is true"<<endl; } else { cout<<"the result is false"<<endl; } return 0;}打印結(jié)果：Test fun(Test i): i=1Test fun(Test i): i=0the result is falseTest fun(Test i): i=1Test fun(Test i): i=0the result is true