C++ code printing strange values

Question

i am having a very silly but strange problem. When i am trying to compile and run the following code my compiler is printing "ggl" but i think it shouldn't. It is strange that after doing so much programming i am stuck here. What exactly is the problem? Can someone please help me out? Thanks in advance !!

#include <iostream>
using namespace std;

int main() {
    int t=8;
    if(1<t<5){
        cout<<"ggl";
    }
    //cout<<aa;
    return 0;
}

score 5 · Accepted Answer · answered Jan 11 '15 at 15:46

5

This line doesn't do what you think it does

 if(1<t<5)

You would have to say

 if (1 < t && t < 5)

The first version says

if ((1 < t) < 5)

Which evaluates to

if (true < 5)
if (1 < 5)

Which is always true.

answered Jan 11 '15 at 15:46

Cory Kramer

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We appear to be typing the same things at the same time. I'll +1 yours anyway. – abligh Jan 11 '15 at 15:47
Thanks Cyber for the quick response !! Is this (1 – varunkr Jan 11 '15 at 15:54
@varunkr In general, most languages have the behavior that you just saw for the reason I described. One exception that comes to mind is `Python` that allows the syntax you originally wrote. – Cory Kramer Jan 11 '15 at 15:55

vaultah · Answer 2 · 2015-01-11T18:47:32.873

4

Your if condition effectively says if ((1 < t) < 5), which is always true, because (1 < t) is either 1 or 0 (1 < 8 evaluates to 1).

Since chained comparisons do not (usually) work in C++, you'll need to check the condition differently:

if (1 < t && t < 5) {
    cout << "ggl";
}

edited Jan 11 '15 at 18:47

answered Jan 11 '15 at 15:46

vaultah

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score 2 · Answer 3 · answered Jan 11 '15 at 15:46

This

if(1<t<5)

does not do what you think it does. It does not determine whether t is between 1 and 5. You want

if ((1<t) && (t<5))

What it actually does is take the value (1<t) (which will 1 if 1<t and 0 otherwise), then see if that value is less than 5, which it always will be.

C++ code printing strange values

3 Answers3