C++ 当我更改函数内的参数时，调用者的参数是否也会更改？
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When I change a parameter inside a function, does it change for the caller, too?
提问by Lisa
I have written a function below:
我在下面写了一个函数：
void trans(double x,double y,double theta,double m,double n)
{
m=cos(theta)*x+sin(theta)*y;
n=-sin(theta)*x+cos(theta)*y;
}
If I call them in the same file by
如果我在同一个文件中调用它们
trans(center_x,center_y,angle,xc,yc);
will the value of xcand ycchange? If not, what should I do?
会的价值xc和yc变化？如果没有，我该怎么办？
回答by Mark Rushakoff
Since you're using C++, if you want xcand ycto change, you can use references:
由于您使用C ++，如果你想xc和yc变化，您可以使用引用：
void trans(double x, double y, double theta, double& m, double& n)
{
m=cos(theta)*x+sin(theta)*y;
n=-sin(theta)*x+cos(theta)*y;
}
int main()
{
// ...
// no special decoration required for xc and yc when using references
trans(center_x, center_y, angle, xc, yc);
// ...
}
Whereas if you were using C, you would have to pass explicit pointers or addresses, such as:
而如果您使用 C，则必须传递显式指针或地址，例如：
void trans(double x, double y, double theta, double* m, double* n)
{
*m=cos(theta)*x+sin(theta)*y;
*n=-sin(theta)*x+cos(theta)*y;
}
int main()
{
/* ... */
/* have to use an ampersand to explicitly pass address */
trans(center_x, center_y, angle, &xc, &yc);
/* ... */
}
I would recommend checking out the C++ FAQ Lite's entry on referencesfor some more information on how to use references properly.
我建议查看C++ FAQ Lite 关于参考的条目，以获取有关如何正确使用参考的更多信息。
回答by Chris Jester-Young
Passing by reference is indeed a correct answer, however, C++ sort-of allows multi-values returns using std::tupleand (for two values) std::pair:
通过引用传递确实是一个正确的答案，但是，C++ 排序允许使用std::tuple和（对于两个值）返回多值std::pair：
#include <cmath>
#include <tuple>
using std::cos; using std::sin;
using std::make_tuple; using std::tuple;
tuple<double, double> trans(double x, double y, double theta)
{
double m = cos(theta)*x + sin(theta)*y;
double n = -sin(theta)*x + cos(theta)*y;
return make_tuple(m, n);
}
This way, you don't have to use out-parameters at all.
这样，您根本不必使用输出参数。
On the caller side, you can use std::tieto unpack the tuple into other variables:
在调用方，您可以使用std::tie将元组解包为其他变量：
using std::tie;
double xc, yc;
tie(xc, yc) = trans(1, 1, M_PI);
// Use xc and yc from here on
Hope this helps!
希望这可以帮助！
回答by Soufiane Hassou
You need to pass your variables by reference which means
您需要通过引用传递变量，这意味着
void trans(double x,double y,double theta,double &m,double &n) { ... }
回答by timB33
as said above, you need to pass by reference to return altered values of 'm' and 'n', but... consider passing everything by reference and using const for the params you don't want to be altered inside your function i.e.
如上所述，您需要通过引用来返回“m”和“n”的更改值，但是...考虑通过引用传递所有内容并使用 const 作为您不想在函数中更改的参数，即
void trans(const double& x, const double& y,const double& theta, double& m,double& n)
