Compute expression

Question

I have:

#include <iostream>
int main()
{
    static int i, arr[10];
    cout<<(i==0) && (arr[i]<0);
}

Which means that both i and all of the elements oft are automatically initialized with 0. Why does this expression (i==0) && (t[i]<0) returns true? Even this returns true:

#include <iostream>
int main()
{
    static int i;
    cout<<(i==0) && (i==1);
}

I got confused when I red this question which supposedly has the correct answer a:

Given the declarations:

static int i, t[10];

and assuming that neither i nor t are explicitly initialized, the value of the expression (i==0) && (t[i]<0)

(a) is 1

(b) is 0

(c) depends on the context

I edited the question so you can understand why I did not used `cout<< (i==0 && i==1);`. — Alexandru Cimpanu, Jun 17 '14 at 12:20

score 8 · Accepted Answer · answered Jun 17 '14 at 12:08

8

Your problem is with operator precedence. The && operator is evaluated after <<. Thus your print expression becomes: (cout<<(i==0)) && (i==1);. Correct the precedence and it prints 0 as expected: cout<<((i==0) && (i==1));

answered Jun 17 '14 at 12:08

Dark Falcon

43,592
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2

If in doubt: More parenthesis! – Baum mit Augen Jun 17 '14 at 12:12
Thank you for the clarification. I know I should surround the whole expression with parenthesis to be sure. I edited the question to explain why I tried that way. – Alexandru Cimpanu Jun 17 '14 at 12:23
The C++ standard required that [both file-scoped and local static variables are zero-initialized](http://stackoverflow.com/a/17803616/2101267). If they gave the answer of (a) as you said, they are wrong. – Dark Falcon Jun 17 '14 at 12:25
That's what I was thinking but trying `cout<<(i==0) && (i==1);` got me confused. – Alexandru Cimpanu Jun 17 '14 at 12:29

Compute expression

1 Answers1