64,647
社区成员
发帖
与我相关
我的任务
分享初级菜鸟问一个弱智的类内函数指针数组定义问题，求指点
class Widget
{
Widget()
{
pa[0] = &f;
pa[1] = &g;
pa[2] = &i;
pa[3] = &j;
}
void select(int idx)
{
(this->*pa[idx])();
}
private:
void f(){cout<<"void f()"<<endl;}
void g(){cout<<"void g()"<<endl;}
void i(){cout<<"void i()"<<endl;}
void j(){cout<<"void j()"<<endl;}
enum {cnt = 4};
void (Widget::*pa[cnt])() ={f,g,i,j};
};void (Widget::*pa[cnt])() ={f,g,i,j}; //这一行一直报错，Widget::Widget::pa”: 无法指定数组的显式初始值设定项，编译器用的QT
...全文
请发表友善的回复…
发表回复
做或不做 2017-08-11
- 打赏
- 举报
以下观点仅是个人对于问题解决方法的倾向 并不是想误导别人。仅供参考。
首先纠正下你一个问题 编译器用的qt这句话 总感觉怪怪的。 不知道上面的人 为什么没有纠正。
不讨论问题本质。如果你使用的是VS2015的编译器的话 我猜测大概率你是可以编译过去的。
如果你是在widows下编程 那么你编译器用的只能是 VS2010 VS2012 VS2013 VS2015等其中的一个吧。
msvc是编译器 ，QT只是一个界面框架。 特别的就是实现中采用了有别于C++模板的元对象编程技术。
给你讲一个故事 就是说美国人为了在太空设计一种可以使用的墨水钢笔 花费了100万研究出来
苏联人不使用这种带墨水的钢笔 直接使用电子笔。
其实我想说的是就算是C++11的标准 也可以通过非C++11标准的C++重写成别的样子。不要纠结怎么可以编译过去。
假设就算打死你都不可能编译的过去 你就算怎么改 感觉也是无济于事的。因为前面根本就没有路。
Really_want 2017-08-10
- 打赏
- 举报
#include <iostream>
using namespace std;
//g++ -g -Wall -std=c++11 -o a.exe main.cpp
struct A
{
int a[2] = {1,2}; //C++11支持这种语法
};
int main(int argc,char* agrv[])
{
A a;
cout<< a.a[0] << "," << a.a[1] <<endl;
return 0;
}
constCpp 2017-08-08
- 打赏
- 举报
成员变量能直接初始化吗？
jena_wy 2017-08-08
- 打赏
- 举报
只有静态常量整型数据成员才可以在类中初始化
呵呵程序喵 2017-08-07
- 打赏
- 举报
您说的我还不太明白，我改用了public试了，依然不行，这跟C++11关系大吗？我用的是比较新的QT版本
呵呵程序喵 2017-08-07
- 打赏
- 举报
多谢！我再自己分析一下
Really_want 2017-08-07
- 打赏
- 举报
好吧，其实跟是否private没关系，重点是c++11
我排错时，在还没有用C++11前有private的提示，误以为有区别了。
Really_want 2017-08-07
- 打赏
- 举报
1. 我说了用private不行！！！我代码中使用的是public。
2. 你可能没注意 我图片中的标识，需要C++11标准支持
赵4老师 2017-08-07
- 打赏
- 举报
再供参考：
//C++ Operators
// Operators specify an evaluation to be performed on one of the following:
// One operand (unary operator)
// Two operands (binary operator)
// Three operands (ternary operator)
// The C++ language includes all C operators and adds several new operators.
// Table 1.1 lists the operators available in Microsoft C++.
// Operators follow a strict precedence which defines the evaluation order of
//expressions containing these operators. Operators associate with either the
//expression on their left or the expression on their right; this is called
//“associativity.” Operators in the same group have equal precedence and are
//evaluated left to right in an expression unless explicitly forced by a pair of
//parentheses, ( ).
// Table 1.1 shows the precedence and associativity of C++ operators
// (from highest to lowest precedence).
//
//Table 1.1 C++ Operator Precedence and Associativity
// The highest precedence level is at the top of the table.
//+------------------+-----------------------------------------+---------------+
//| Operator | Name or Meaning | Associativity |
//+------------------+-----------------------------------------+---------------+
//| :: | Scope resolution | None |
//| :: | Global | None |
//| [ ] | Array subscript | Left to right |
//| ( ) | Function call | Left to right |
//| ( ) | Conversion | None |
//| . | Member selection (object) | Left to right |
//| -> | Member selection (pointer) | Left to right |
//| ++ | Postfix increment | None |
//| -- | Postfix decrement | None |
//| new | Allocate object | None |
//| delete | Deallocate object | None |
//| delete[ ] | Deallocate object | None |
//| ++ | Prefix increment | None |
//| -- | Prefix decrement | None |
//| * | Dereference | None |
//| & | Address-of | None |
//| + | Unary plus | None |
//| - | Arithmetic negation (unary) | None |
//| ! | Logical NOT | None |
//| ~ | Bitwise complement | None |
//| sizeof | Size of object | None |
//| sizeof ( ) | Size of type | None |
//| typeid( ) | type name | None |
//| (type) | Type cast (conversion) | Right to left |
//| const_cast | Type cast (conversion) | None |
//| dynamic_cast | Type cast (conversion) | None |
//| reinterpret_cast | Type cast (conversion) | None |
//| static_cast | Type cast (conversion) | None |
//| .* | Apply pointer to class member (objects) | Left to right |
//| ->* | Dereference pointer to class member | Left to right |
//| * | Multiplication | Left to right |
//| / | Division | Left to right |
//| % | Remainder (modulus) | Left to right |
//| + | Addition | Left to right |
//| - | Subtraction | Left to right |
//| << | Left shift | Left to right |
//| >> | Right shift | Left to right |
//| < | Less than | Left to right |
//| > | Greater than | Left to right |
//| <= | Less than or equal to | Left to right |
//| >= | Greater than or equal to | Left to right |
//| == | Equality | Left to right |
//| != | Inequality | Left to right |
//| & | Bitwise AND | Left to right |
//| ^ | Bitwise exclusive OR | Left to right |
//| | | Bitwise OR | Left to right |
//| && | Logical AND | Left to right |
//| || | Logical OR | Left to right |
//| e1?e2:e3 | Conditional | Right to left |
//| = | Assignment | Right to left |
//| *= | Multiplication assignment | Right to left |
//| /= | Division assignment | Right to left |
//| %= | Modulus assignment | Right to left |
//| += | Addition assignment | Right to left |
//| -= | Subtraction assignment | Right to left |
//| <<= | Left-shift assignment | Right to left |
//| >>= | Right-shift assignment | Right to left |
//| &= | Bitwise AND assignment | Right to left |
//| |= | Bitwise inclusive OR assignment | Right to left |
//| ^= | Bitwise exclusive OR assignment | Right to left |
//| , | Comma | Left to right |
//+------------------+-----------------------------------------+---------------+
赵4老师 2017-08-07
- 打赏
- 举报
仅供参考：
#include <stdio.h>
double A[300][100];
double valuex[300];
double valuey[300];
int i;
double fun00(double x,double y) {return x + y ;};
double fun01(double x,double y) {return x*x+ 3*y ;};
double fun02(double x,double y) {return x + 2*y*x;};
//... fun03(double x,double y) {return ...+... ;};
//...
//... fun98(double x,double y) {return ...+... ;};
double fun99(double x,double y) {return x/2+20*y ;};
double (*funNN[100])(double,double)={
fun00,
fun01,
fun02,
// fun03,
// ...
// fun98,
fun99,
};
int main() {
double x,y;
int f,FN;
for (i=0;i<300;i++) {
valuex[i]=(double)i;
valuey[i]=(double)i;
}
FN=3;
for (i=0;i<300;i++) {
x=valuex[i];
y=valuey[i];
for (f=0;f<FN;f++) A[i][f]=funNN[f](x,y);
}
for (i=0;i<3;i++) {
for (f=0;f<FN;f++) printf("%lg ",A[i][f]);
printf("\n");
}
return 0;
}
//0 0 0
//2 4 3
//4 10 10
sdghchj 2017-08-07
- 打赏
- 举报
void (Widget::*pa[cnt])() ={&f,&g,&i,&j};
赵4老师 2017-08-07
- 打赏
- 举报
《深度探索C++对象模型》
《C++反汇编与逆向分析技术揭秘》
呵呵程序喵 2017-08-07
- 打赏
- 举报
多谢大家回复！
@赵4老师 加static依然不行
@ccssddnn218 我改用了private 依然不行，
完整程序如下，用的QT，我看的王桂林老师的视频，他在讲课时这样做就行
#include <iostream>
using namespace std;
class Widget
{
public:
Widget()
{
pa[0] = &f;
pa[1] = &g;
pa[2] = &i;
pa[3] = &j;
}
void select(int idx)
{
(this->*pa[idx])();
}
private:
void f(){cout<<"void f()"<<endl;}
void g(){cout<<"void g()"<<endl;}
void i(){cout<<"void i()"<<endl;}
void j(){cout<<"void j()"<<endl;}
enum {cnt = 4};
void (Widget::*pa[cnt])() ={f,g,i,j};
};
int main()
{
Widget w;
w.select(2);
return 0;
}Really_want 2017-08-07
- 打赏
- 举报
#include <iostream>
using namespace std;
class Widget
{
public: //构造函数要public
Widget()
{
pa[0] = &f;
pa[1] = &g;
pa[2] = &i;
pa[3] = &j;
}
void select(int idx)
{
(this->*pa[idx])();
}
public: //private不行
void f(){cout<<"void f()"<<endl;}
void g(){cout<<"void g()"<<endl;}
void i(){cout<<"void i()"<<endl;}
void j(){cout<<"void j()"<<endl;}
enum {cnt = 4};
void (Widget::*pa[cnt])() ={f,g,i,j};
};
int main()
{
Widget w;
w.select(2);
return 0;
} 
赵4老师 2017-08-07
- 打赏
- 举报
加static ？
在工程中我们少不了要定义类或者结构去储存数据，这些数据将被临时地储存在内存中，现在我们想要对其完成一些类似于查找、过滤等等常见的任务的时候，我们该如何去做呢？ 我们可以自己写代码去对集合中的每个...
