Question

c++,类成员,string类型数组的初始化问题

代码如下
class supplier: public people //people只是一个抽象类
{
private:
string med[10];
string name;
string add;

public:
supplier();
supplier(string,string);
virtual void show() ;
virtual void set() ;
};

supplier::supplier()
{
med[10]={"x","x","x","x","x","x","x","x","x","x"}; //出问题的语句

name='x';
add='x';
}

error C2059: 语法错误 : “{”
error C2143: 语法错误 : 缺少“;”(在“{”的前面)
error C2143: 语法错误 : 缺少“;”(在“}”的前面)

尝试把错误语句换成这个：
for (int i=0;i<10;i++)
{
med[i]="x";
}
然后编译通过了，但是调试的时候没有赋值成功
求前辈解答疑惑。。。

Answer 1

med[10]={"x","x","x","x","x","x","x","x","x","x"};这样的赋值只有在定义数组的时候同时初始化才可以
比如 string med[10]={"x","x","x","x","x","x","x","x","x","x"};
对于类成员，早被定义了，所以这样赋值是非法的
不过
用
for (int i=0;i<10;i++)
{
med[i]="x";
}
这种方法应该是可行的
验证程序
#include <iostream>
#include <string>
using namespace std;
int main()
{
string a[3] ;
for(int i = 0; i < 3; i ++)
a[i] = "x";
for(int i = 0; i < 3; i ++)
cout << i << ": " << a[i] << endl;
return 0;
}

输出
0: x
1: x
2: x
你说的没赋值成功是怎么知道的？应该是其他地方有问题
你可以在赋值后，把med[i]的值打出来，应该是可以的

Answer 2

　　在构造函数体里面已经不是初始化了，是赋值 ”med[10]={"x","x","x","x","x","x","x","x","x","x"};
“ 整体赋值好像不符合c++语言规则