Code with pointers gives out 1 no matter what I tried c++

Question

So I'm doing a code which sums up all elements of field but I have to use pointers :( No matter what I tried I get 1 as output. Tried same code with multiplication but still nothing... Code:

#include <iostream>
#include <cmath>

using namespace std;

float vrat=0;
int suma (int pok2, int vel22){
    for(int z=0; z<vel22; z++){
            vrat+=pok2;
    }
    return vrat;
}

int main()
{
        /// 2. zadatak

    int vel2;
    int *vel22=&vel2;
    cout<<"Unesi broj elemenata koje hoces upisati"<<endl;
    cin>>vel2;

    int polje2[vel2];

    cout<<"Kreni unosit elemente "<<endl;
    for(int z=0; z<vel2; z++){
            cin>>polje2[z];
    }

    int *pok2=&polje2[vel2];
    suma(*pok2, *vel22);

    cout<<"Suma elemenata je "<<suma<<endl;

    return 0;
}

Thank you!

`int polje2[vel2];` -- This is not valid C++. Arrays in C++ must have their size denoted by a compile-time constant, not a runtime value such as `vel2`. In C++, this should be `std::vector polje2(vel2);` Everything else stays the same. — PaulMcKenzie, Jan 27 '21 at 20:19
@PaulMcKenzie I'm pretty sure the array he declared isn't static, so this isn't true. The array is allocated at runtime after receiving user input, otherwise the code I posted would cause a memory error. Please correct me if I'm wrong, though. — Kyle Burns, Jan 27 '21 at 20:29
@KyleBurns -- No. Declaring arrays with a runtime expression is not valid C++. [See this](https://rextester.com/PPWAK85180), and [this](https://stackoverflow.com/questions/1887097/why-arent-variable-length-arrays-part-of-the-c-standard) — PaulMcKenzie, Jan 27 '21 at 20:39
@KyleBurns Variable length arrays are a compiler extension that g++ and other compilers (unfortunately) have turned on by default. This confuses new programmers into believing that the code is valid, when it really isn't. By setting the proper compiler switches, g++ will then not recognize that as valid code (which to me should be the default setting). — PaulMcKenzie, Jan 27 '21 at 20:47
@PaulMcKenzie Wow. I had no idea about that! Thank you for responding:) I'll be changing the settings on my compiler — Kyle Burns, Jan 27 '21 at 20:54

score 0 · Answer 1 · answered Jan 27 '21 at 20:18

0

Ah! It looks like at the very end you're printing the function suma rather than the result. Try this instead:

    int result = suma(*pok2, *vel22);

    cout<<"Suma elemenata je "<< result <<endl;

I'm not sure why printing a function results in "1", but printing the actual result should work here.

answered Jan 27 '21 at 20:18

mattlangford

1,260
1
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12

Tried this and then it just spits out some big numbers, don't know why – Marko Šokčević Jan 27 '21 at 20:27

score 0 · Accepted Answer · answered Jan 27 '21 at 20:25

I think you meant to do something like this

#include <iostream>
#include <cmath>

using namespace std;

float vrat=0;
int suma (int *pok2, int vel22){
    for(int z=0; z<vel22; z++, pok2++){
            vrat+=*pok2;
    }
    return vrat;
}

int main()
{
        /// 2. zadatak

    int vel2;
    int *vel22=&vel2;
    cout<<"Unesi broj elemenata koje hoces upisati"<<endl;
    cin>>vel2;

    int polje2[vel2];

    cout<<"Kreni unosit elemente "<<endl;
    for(int z=0; z<vel2; z++){
            cin>>polje2[z];
    }

    int *pok2= polje2;
    suma(pok2, *vel22);

    cout<<"Suma elemenata je "<< vrat << endl;

    return 0;
}

this line

int *pok2=&polje2[vel2];

attempts to access memory outside of the bounds of polje2. Remember arrays are zero-indexed. Let me know if this isn't what you were intending to do.

Code with pointers gives out 1 no matter what I tried c++

2 Answers2