CODEWITHSUNDEEP
c++pointers
alansiqueira27
alansiqueira27

Reputation: 8506

Pointer - Reference vector

I'm used to program in C# or Java, so I'm doing really bad in C++. I believe it's easy but I just can't make this work. Please help me.

I have this:


void swap(vector  * vet, int i, int j)
{
    int temp = vet[i];
    vet[i] = vet[j];
    vet[j] = temp;
}

I'm calling the method this way:


swap(&vet, j, j - 1);

What I want is to pass the vector using pointers instead of using value.

Obs: The code compiles well without the "*" and "&".

Please don't say that I have to at least try to study pointers, because I did. I just can't make this damn thing work!

Upvotes: 1

Views: 1355

Answers (4)

Ed Swangren
Ed Swangren

Reputation: 124632

Everyone has thus far responded by telling you to use references, which is correct, but they fail to explain why your code doesn't work. The problem here is that you do not understand pointer arithmetic.

Let's say we have a a pointer to 10 ints:

// best to use a vector for this, but for the sake of example...
int *p = new int[10];

Now, if we want to change the value of the second int in that chunk of memory we can write:

*(p + 1) = 20;

Or, the equivalent:

p[1] = 20;

See? Those two lines do the same thing. Adding n to a pointer increases the address of the pointer by n * sizeof *p bytes. Pointer arithmetic is convenient because it hides the sizeof bit from you and allows you to work with logical units (elements) instead of bytes.

So, knowing that, back to your broken code:

vet[i] = vet[j];

This indexes i * sizeof *vet bytes away from the pointer, i.e., i full vectors away from the base address. Obviously that is wrong, you wanted to invoke operator[] on the vector, i.e., treat it as an array. It is not an array however, so the correct syntax would be:

(*vec)[i]

Or

vec->operator[](i);

That said... just use a reference. Safer (object guaranteed to be valid) and idiomatic.

Upvotes: 4

sehe
sehe

Reputation: 392833

In fact, in C++ you'd just say

using std::swap;

swap(vet[i], vet[j]);

Upvotes: 2

Recker
Recker

Reputation: 1973

You can try something like....

void swap(vector<int> &vet, int i, int j)
{
    int temp = vet[i];
    vet[i] = vet[j];
    vet[j] = temp;
}

and call your swap function as

swap(vet,i,j);

Bottomline: Use reference variables.They are more like reference in Java.

Upvotes: 3

111111
111111

Reputation: 16148

You should take the vector by reference rather than "pass by pointer".

void swap(std::vector<int>& vet, std::size_t i, std::size_t j)
{
    using std::swap;
    swap(vet[i], vet[j]);
}

http://en.cppreference.com/w/cpp/algorithm/swap

Note the more idiomatic:

http://en.cppreference.com/w/cpp/algorithm/iter_swap

Upvotes: 5

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