Why I am getting garbage(unwanted) output here?

Question

Whenever I am writing this following code, I am getting garbage(unexpected) output in some online compiler, but if I use code block then getting satisfied output. So my question is why I am getting this type of output?

for example, if I input

then I am getting

22 1

in the code block. But in the online compiler, it's something else.

#include<iostream>
#include<cstdlib>
using namespace std;
int main()
{
    int have, n, i;
    int kid=0;

    cin>>n>>have;

    int line[n];

    for(i=0;i<n;i++)
    {
        cin>>line[i];

        if(line[i]>=0)
            have+=line[i];
        else
        {
            if(have>=abs(line[i]))
                have+=line[i];
            else
                kid++;
        }
    }

    cout<<have<<" "<<kid<<endl;

}

we dont know what you expect or what satisfies you unless you tell us — 463035818_is_not_an_ai, Feb 13 '19 at 14:59
There is no point in storing the inputs in an array since you're only using the last value you read. — molbdnilo, Feb 13 '19 at 15:00
I am getting a different type of output with the same input. For example, once I am getting 7 and then 1855857143 — mahin, Feb 13 '19 at 15:01
@mahin And what are the inputs? (And what is the online compiler?) — molbdnilo, Feb 13 '19 at 15:01
Please include: input, expected output and actual output in the question — 463035818_is_not_an_ai, Feb 13 '19 at 15:04
5 7 + 5 - 10 - 20 + 40 - 20 these are input, and I am getting 22 1 as output using code block. But if I use online compiler then it doesn't work@molbdnilo — mahin, Feb 13 '19 at 15:06
@mahin You can't read non-integers into an `int`. I'm guessing that you had whitespace between the signs and the numbers in the failing cases but not when it worked. — molbdnilo, Feb 13 '19 at 15:07
Please [edit](https://stackoverflow.com/posts/54673114/edit) your question to include: input, expected output and actual output — 463035818_is_not_an_ai, Feb 13 '19 at 15:08
_"it's something else"_, you should tell us __what__ exactly. — Jabberwocky, Feb 13 '19 at 15:14
once I am getting 7 and then 1855857143 and if I run again then getting another garbage value. this value changes once I rerun the code@Jabberwocky — mahin, Feb 13 '19 at 15:16
Maybe you should simply not put a space between the sign and the numbers during input: try to input `+5` instead of `+5` — Jabberwocky, Feb 13 '19 at 15:17
Unfortunately, this is for an online judge, So I can't do that@Jabberwocky — mahin, Feb 13 '19 at 15:18
If the input *has to be* in the form "", your code is unable to read it. `cin>>line[i];` will fail when trying to read a non integer input like `'+'`. So maybe what you need is something like `std::string op; cin >> op >> number; if (op == "+") {... `. — Bob__, Feb 13 '19 at 16:06

Constantinos Glynos · Accepted Answer · 2019-02-14T12:56:55.980

1

The main problem I can see in your code is this:

int line[n];

This is known as a VLA (Variable Length Array) and it is not supported in C++. It is valid in C. Most compilers still allow this behaviour due to the fact that C++ is based on C, but it is not valid C++ code. In a previous question, I found out that clang supports designated initializers, when gcc and vc++ did not. The reason is because some compilers like clang, support c99-extensions by default. My point is that just because the code compiles, it doesn't mean it's always right.

If you compile with the -pedantic argument, you will see that the compiler is warning you about this being a C99 feature. Have a look at the rextester example here. From the comments below, using -pedantic-errors in the compiler flags, will prompt an error.

If you know the size of the array before run-time, then you should use a static array int line[4];, but if you don't then you need to use a dynamic array. std::vector is essentially a dynamic array that also handles memory for you. It's easy to use and very efficient. std::vector<int> line;

You can read more about the vector container here: http://www.cplusplus.com/reference/vector/vector/

Btw, I tried your code in rextester, ideone and repl.it and I got the same results: 22 1. I think what you are witnessing it undefined behaviour.

Also, you can qualify int n with constexpr and it'll be fine.

constexr int n = 200;
int line[n]; //now it's ok.

But this again means that you know the size of the array at compile time.

edited Feb 14 '19 at 12:56

answered Feb 13 '19 at 16:11

Constantinos Glynos

2,952
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`int[n];` for non compile time constant `n` is either forbidden or allowed as compiler extension, but it does not invoke undefined behviour – 463035818_is_not_an_ai Feb 13 '19 at 18:04
@user463035818: The C++ standard does not accept VLAs which makes it forbidden, but compilers support them and not always with the need for an extension. So what happens when `int [n]` is both forbidden and supported? I guess in that case, C++ would say: "I do not accept this code, but if you still want to go for it, then you're on your own!". That's what I meant by undefined behavior. And again, it's only a guess... – Constantinos Glynos Feb 13 '19 at 23:20
when the compiler defines what happens for VLAs then it is not undefined. Its just not standard c++ – 463035818_is_not_an_ai Feb 14 '19 at 09:18
@user463035818: I see your point. One can argue however that ok, it is defined by the compiler (if you're lucky because not all compilers support them - and what if it's a custom compiler), but it is not defined in the standard. So, defined by the compiler, undefined by the standard. Overthinking it.... – Constantinos Glynos Feb 14 '19 at 09:49
well I have to admit, i was not completely right, `int[n];` only produces an error with `-pedantic-errors` but not with `-pedantic` (only a warning), while `-pedantic` should be enough to make the compiler behave standard compliant, I am still a bit confused why `int[n]` is not required to result in an error, will have to do some more reading... – 463035818_is_not_an_ai Feb 14 '19 at 10:51
@user463035818: Good point about the `-pedantic-errors`! I'll need to put that in the answer too. I think it may have something to do with it being valid in C and most compilers tend to allow C code. Not sure though. Please post a comment if you find something on this. – Constantinos Glynos Feb 14 '19 at 11:35
@user463035818 `-pedantic` *does* make it standard compliant. The standard requires a diagnostic message if the code is ill-formed, but it doesn't have to be a error; a simple warning is enough. – HolyBlackCat Feb 14 '19 at 11:40

Why I am getting garbage(unwanted) output here?

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