CODEWITHSUNDEEP
c
pdenlinger
pdenlinger

Reputation: 3917

Returns wrong answer

When I run this code, it iterates, but then returns "The answer is 0", when it should be "The answer is 10."

Why is this?

#import <Foundation/Foundation.h>

int main (int argc, const char * argv[])
{
     int i;
     for (int i = 0; i < 12; i++){ 
        if (i % 3 == 0) {
            continue;
         }
         printf("Checking i = %d\n", i);
         if (i + 90 == i * i) {
             break;
         }
     }
    printf("The answer is %d.\n", i);
    return 0;


}

Upvotes: 1

Views: 105

Answers (5)

abe312
abe312

Reputation: 2635

If you want to use i outside the for loop, do not declare it inside for loop

So your for loop should be like

for (i = 0; i < 12; i++)

Upvotes: 0

Fatih Donmez
Fatih Donmez

Reputation: 4347

There are two seperate declared int i's in your code.

Upvotes: 0

Zach Rattner
Zach Rattner

Reputation: 21353

You're actually declaring i again inside the for loop. Make your for loop like so:

for (i = 0; i < 12; i++)

Then, it will retain its value once you exit the loop.

Upvotes: 1

Kerrek SB
Kerrek SB

Reputation: 476990

You have two separate is in your code, the inner one hiding the outer one. Moreover, the print statement causes undefined behaviour, because the outer i is read uninitialized.

Say for (i = 0; i < 12; i++) to use the outer variable instead of declaring a new variable.

Upvotes: 5

Mat
Mat

Reputation: 206689

 int i;
 for (int i = 0; i < 12; i++){ 
      ^^^^^

The i inside the loop is not the same as the i outside the loop.

Repace that with:

 int i;
 for (i = 0; i < 12; i++){

Upvotes: 8

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