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#2
lz10919149992012-03-10 15:29
方法有很多，举个例子：比如我写一个函数，它接收两个整形参数，返回这两个数的“和”和这两个数的“积”：
 程序代码：#include <iostream> using namespace std; class Result { public: Result(int s, int p) : sum(s), pdt(p) {} int getSum() const { return sum; } int getProduct() const { return product; } private: int sum; int pdt; }; const Result f(int a, int b) { return Result(a + b, a * b); } int main() { int a = 2, b = 3; Result r = f(a, b); cout << "The sum of a & b is: " << r.getSum() << endl; cout << "The product of a & b is: " << r.getProduct() << endl; } 或者： 程序代码：#include <iostream> using namespace std; void f(int a, int b, int& sum, int& pdt) { sum = a + b; pdt = a * b; } int main() { int a = 2, b = 3, sum, product; f(a, b, sum, product); cout << "The sum of a & b is: " << sum << endl; cout << "The product of a & b is: " << product << endl; } |
有个程序片段，其调用程序为：
Elas3DT4 eq(theElement);
eq.Deviator();
其中Elas3DT4是类，eq是对象，Deviator是类的成员，这个成语函数的定义为：
void Elas3DT4::Deviator(double coef)
{
double c1=_G*_det*coef, c2=2*c1;
for (size_t j=1; j<=4; j++) {
double db11 = c2*_dSh(j).x;
double db22 = c2*_dSh(j).y;
double db33 = c2*_dSh(j).z;
double db42 = c1*_dSh(j).z;
double db43 = c1*_dSh(j).y;
double db53 = c1*_dSh(j).x;
for (size_t i=1; i<=4; i++) {
eMat(3*i-2,3*j-2) += _dSh(i).x*db11 + _dSh(i).z*db42 + _dSh(i).y*db43;
eMat(3*i-2,3*j-1) += _dSh(i).y*db53;
eMat(3*i-2,3*j ) += _dSh(i).z*db53;
eMat(3*i-1,3*j-2) += _dSh(i).x*db43;
eMat(3*i-1,3*j-1) += _dSh(i).y*db22 + _dSh(i).z*db42 + _dSh(i).x*db53;
eMat(3*i-1,3*j ) += _dSh(i).z*db43;
eMat(3*i ,3*j-2) += _dSh(i).x*db42;
eMat(3*i ,3*j-1) += _dSh(i).y*db42;
eMat(3*i ,3*j ) += _dSh(i).z*db33 + _dSh(i).y*db43 + _dSh(i).x*db53;
}
}
}
现在问题来了，由于这个成语函数是void类型，也就是没有返回值的，哪在调用语句eq.Deviator();中它把什么值传递出来呢？可能有几种传值的方式呢？请高手提点。
