#include <iostream>
using namespace std;
class MyClass
{
private :
char str[848];
public :
MyClass()
{
}
MyClass(char a[])
{
str[848] = a[848];
}
MyClass operator () (char a[])
{
str[848] = a[848];
}
void myFunction(MyClass m)
{
}
void display()
{
cout << str[848];
}
};
int main()
{
MyClass m1; //MyClass has just one data member i.e. character array named str of size X
//where X is a constant integer and have value equal to your last 3 digit of arid number
MyClass m2("COVID-19") , m3("Mid2020");
m2.display(); //will display COVID-19
cout<<endl;
m2.myFunction(m3);
m2.display(); //now it will display Mid2020
cout<<endl;
m3.display(); //now it will display COVID-19
//if your array size is even then you will add myEvenFn() in class with empty body else add myOddFn()
return 0;
}
我不能使用string
,因为有人告诉我不要这样做,因此,我需要知道如何制作它才能显示所需的输出
如何在构造函数中初始化char
数组?
MyClass(char a[])
{
//make sure that sizeof(a) <= to sizeof(str);
// you can not do sizeof(a) here, because it is
// not an array, it has been decayed to a pointer
for (int i = 0; i < sizeof(str); ++i) {
str[i] = a[i];
}
}
std::copy
中的<algorithm>
const int size = 848;
std::copy(a, a + size, str);
std::copy
优先于strcpy
,如果您必须使用strcpy
,则优选strncpy
。您可以为其提供大小,以帮助防止错误和缓冲区溢出。
MyClass(char a[])
{
strncpy(str, a, sizeof(str));
}
std::array
。它具有多种优势,例如,您可以像普通变量一样直接分配它。示例:std::array<char, 848> str = {/*some data*/};
std::array<char, 848> str1;
str1 = str;
要复制字符串,您必须使用std::strcpy,而不是str[848] = a[848]
。
[str[848] = a[848]
仅复制一个元素,但由于您的数组具有索引[0,847]。在您的情况下,这是一个错误。
尝试
class MyClass
{
private :
char str[848];
public :
MyClass()
{
}
MyClass(char a[])
{
std::strcpy(src, a);
}
MyClass operator () (char a[])
{
std::strcpy(src, a);
}
void myFunction(MyClass m)
{
}
void display()
{
cout << str;
}
};