我想在结构中使用一个变量作为函数的默认参数
例如,像这样
struct A {
int b;
A () {
b = 0;
}
int func(int t = b) {
...
}
} 很抱歉我的代码不好,这是我第一次发布问题
但我一直收到错误,我也尝试使用静态int,但我得到运行时错误。
有没有办法使用b作为默认参数?!
当对b使用静态int时,我得到:(mx是b,wavelet_tree是A)
/tmp/cc3OBfq8.o: In function `main':
a.cpp:(.text+0x2dc): undefined reference to `wavelet_tree::mx'
/tmp/cc3OBfq8.o: In function `wavelet_tree::wavelet_tree()':
a.cpp:(.text._ZN12wavelet_treeC2Ev[_ZN12wavelet_treeC5Ev]+0x42): undefined reference to `wavelet_tree::mx'
/tmp/cc3OBfq8.o: In function `wavelet_tree::build(int*, int, int)':
a.cpp:(.text._ZN12wavelet_tree5buildEPiii[_ZN12wavelet_tree5buildEPiii]+0x66): undefined reference to `wavelet_tree::mx'
a.cpp:(.text._ZN12wavelet_tree5buildEPiii[_ZN12wavelet_tree5buildEPiii]+0x73): undefined reference to `wavelet_tree::mx'
a.cpp:(.text._ZN12wavelet_tree5buildEPiii[_ZN12wavelet_tree5buildEPiii]+0x7f): undefined reference to `wavelet_tree::mx'
collect2: error: ld returned 1 exit status发布于 2019-12-26 20:35:25
只需重载函数即可
struct A
{
int b;
A () : b(0) {};
int func(int t) { return whatever();};
int func() {return func(b);};
};这样,b可以是A的任何成员(无论是否为static ),也可以是定义func()时可访问的任何变量,等等。
https://stackoverflow.com/questions/59488414
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