Variable not defined in loop?

Question

It's saying on line 13: name lookup of 'i' changed for ISO 'for' scoping

#include <iostream>
using namespace std;

int main()
{
    int Fib[40];
    Fib[0] = 1;
    Fib[1] = 1;
    cout << Fib[0] << Fib[1] << endl;

    for (int i=2; i<40; i++)
    Fib[i]=Fib[i-1]+Fib[i-2];
        cout << Fib[i] << endl;
}

You're aware that because you didn't use curly braces only the first statement is inside the loop yes? the cout << Fib[i] << endl; line is outside the for loop. Also fix your indentation. — Borgleader
– Borgleader, Commented Dec 8, 2012 at 4:19

Nikos C. · Accepted Answer · 2012-12-08 04:19:31Z

2

You forgot to scope the instructions you want to be executed in the loop:

for (int i=2; i<40; i++) {
    Fib[i]=Fib[i-1]+Fib[i-2];
    cout << Fib[i] << endl;
}

Loops will only execute a single instruction or a single scope of instructions. In your code, only the first line was part of the loop. The second was not, and therefore i was out of scope. Whether you indent the lines or not doesn't matter at all in C++. The compiler doesn't care about indentation; it is used only to aid the programmer, not the compiler.

answered Dec 8, 2012 at 4:19

Nikos C.

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Comments

kaushik gandhi · Accepted Answer · 2012-12-08 04:23:28Z

1

The last line has [i] it is not inside the loop cout << Fib[i] << endl; the scope of i is only within the for loop as i is declared within the loop according to new ISO standard so the correct line would be

for (int i=2; i<40; i++) {
    Fib[i]=Fib[i-1]+Fib[i-2];
    cout << Fib[i] << endl;
}

answered Dec 8, 2012 at 4:23

kaushik gandhi

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