How to understand the declaration?

Xiangbuilder · 09-13-2003, 01:55 AM

How to understand the declaration, int c1=i%3==0; ?

Why

Code:

int i;
cin>>i;
cout<<c1=i%3==0;

is right while

Code:

cout<<1%3==0;

is wrong?

Thank you.

SaTaN · 09-13-2003, 02:18 AM

Code:

#include<iostream>
using namespace std;
main()
{
        int i;
        cout<<(15%3==0);
}

The output is 1 as required.

If we don't use the brackets then the value of 15%3 will be outputted.

If I haven't understood ur question plz be more specific

Xiangbuilder · 09-13-2003, 02:46 AM

Thank you.
I guess c1=i%3 means give the value of i%3 to the variable c1, if add "==0" at the end, what is the meaning of the clause, c1=i%3==0 ?
Another question:

Code:

 #include<iostream>
using namespace std;
main()
{
        int i;
        cout<<1%3==0;
}

why the output is 1 instead of 0?

SaTaN · 09-13-2003, 02:57 AM

The reason the output given is the ans for (1%3) .
The cout takes it that way due to I suppose operator precedence.
If you want entire thing i.e , 1%3==0 , you must enclose them in
brackets i.e , (1%3==0)

The meaning of this clause is (1%3==0) is it checks if the value
of 1%3 is equal to 0 and returns (0 for false, 1 for true)

Ex:- Suppose we say (14%3==0)
14%3 is 2 and it is not equal to 0
So , output will be "0" [False]

Xiangbuilder · 09-13-2003, 03:11 AM

Thank you.
I am poor in both c++ and English.

Now I want to know why the output of cout<<1%3==0; is 1 instead of some error information?
What is the computer's thinking about the clause, cout<<1%3==0;?
Does the computer consider cout<<1%3==0; as cout<<1;, and ignore %3==0?

SaTaN · 09-13-2003, 03:21 AM

No , the computer will think that
cout<<14%3==0 is same as

((cout<<14%3)==0);
So, the output will be 2.

Xiangbuilder · 09-13-2003, 04:08 AM

I feel 14%3 is equal to the integer 2,
does (cout<<14%3) is equal to the integer 2?
does the type of (cout<<i) is same as the type of i?

Why the output of ((cout<<14%3)==0); is 2?
Does the computer only compute (cout<<14%3), while ignore (2==0)?
Why thc computer don't think ((cout<<14%3)==0) in this process:
check the value of (cout<<14%3), if the value of (cout<<14%3) ia equal to 0, then the result of ((cout<<14%3)==0) is 1 (right), otherwise, is 0 (false)?

Another question:
Why cout<<1%3==0; can output 1, while I can not successfully compile cout<<1%3==0<<endl;?

SaTaN · 09-13-2003, 04:36 AM

cout <<14%3 will print value 2 .
The compiler isn't ignoring (2==0)

cout<<((cout<<14%3)==0);

This will first give output 2 and then 0 [since 2==0 is false]

Xiangbuilder · 09-13-2003, 04:49 AM

Thank you for your patience.