c ++默认函数参数不起作用

时间:2015-01-11 10:42:14

标签: c++ default-arguments

我有一个需要默认参数的函数:

LINE 13:
    void genRand(double offset_x = 0.0, double offset_y = 0.0);

这是功能:

LINE 84:
    void game::genRand(double offset_x = 0.0, double offset_y = 0.0) {
        perlin.SetFrequency(0.1);
        for(int _x=0; _x<dimensions.x/32; _x++){
            for(int _y=0; _y<dimensions.y/32; _y++){
                vec.push_back((perlin.GetValue(_x+offset_x, _y+offset_y, 0.2)+1.f)*2/2);
            }
        }
    }

错误:

make
g++ main.cpp -w -std=c++11 -lsfml-graphics -lsfml-window -lsfml-system -lnoise -o main
main.cpp:84:64: error: default argument given for parameter 1 of ‘void game::genRand(double, double)’ [-fpermissive]
 void game::genRand(double offset_x = 0.0, double offset_y = 0.0) {
                                                                ^
main.cpp:13:14: note: previous specification in ‘void game::genRand(double, double)’ here
         void genRand(double offset_x = 0.0, double offset_y = 0.0);
              ^
main.cpp:84:64: error: default argument given for parameter 2 of ‘void game::genRand(double, double)’ [-fpermissive]
 void game::genRand(double offset_x = 0.0, double offset_y = 0.0) {
                                                                ^
main.cpp:13:14: note: previous specification in ‘void game::genRand(double, double)’ here
         void genRand(double offset_x = 0.0, double offset_y = 0.0);
              ^

我不明白我做错了什么。

1 个答案:

答案 0 :(得分:0)

当您单独编写函数(正文)的定义时,不应再带default parameter

实际上,默认参数值必须出现在声明中,因为这是调用者看到的唯一内容。

最好在重复的函数参数列表中注释默认值:

void foo(int x = 42,
         int y = 21);

void foo(int x /* = 42 */,
         int y /* = 21 */)
{
   ...
}