小型C ++代码激怒了我

Question

please consider this short C++ code

#include<iostream.h>
#include<conio.h>
#include<stdio.h>

void main()
{
    clrscr();
    char arr[]="MQMHSJKLSUGDUGIGIUDKLKO";
    for(int i=0;i<5;i++)
    {
       if(i%2==0)
       {
           cout<<arr[i];
       }
    }

    char a[]={'78','45','21','5'};
    cout<<(int)a;
    getch();
}

the output is 'MMS 18'

MMS is clear from the first loop, but how '18' is the output of second one ? even if you change the array elements, the answer remains 18. Please explain !

Answer 1

a converted to a pointer to its first element. You are trying to print the address of first element by casting it to int . Result may vary compiler to compiler.

Answer 2

'a' represent the address of the first element in the array.
As if in your code the address of the first element of the array 'a' is 18.If you try *a instead of a it would give you the answer as 78.The code you expect could be

#include<iostream.h>
#include<conio.h>
#include<stdio.h>
void main()
{
    clrscr();
    char arr[]="MQMHSJKLSUGDUGIGIUDKLKO";
    for(int i=0;i<5;i++)
    {
        if(i%2==0)
        {
           cout<<arr[i];
        }
    }

    char a[]={'78','45','21','5'};
    cout<<(*a);
    getch();
}