Using memset with multidimensional arrays in C++

Question

I am trying to use memset to set a dynamic array of size rownum x rownmum. However, when I call the showarr function as shown below, instead of getting the output of all zeros (i.e. ASCII 48), I am getting random/garbage values.

void showarr(int **a, int rownum)
{
        for ( int i = 0; i < rownum; i++) {
                for ( int j = 0; j < rownum; j++) {
                        cout<<a[i][j]<<endl;
                }
        }
}

int main(int argc, char *argv[])
{
        int n;
        cin>>n;

        int rownum;
        cin>>rownum;
        int **a = new int* [rownum] ;
        for ( int i = 0; i <rownum; i++) {
             a[i] = new int[rownum];
        }
        for (int i = 0; i < rownum; i++) {
              memset(a[i],48,sizeof(a[i]));
        }
        showarr(a,rownum);

        return 0;
}

sizeof(a) == size(int **), meaning either 4 or 8 bytes, depending on your processor. — Fred Foo
– Fred Foo, Commented Jan 25, 2012 at 12:51
What are you trying to do? Generally speaking a smart pointer is better than a raw pointer, a vector is better than an array, a reference to a pointer is better than a pointer to a pointer, and std::fill() is better than memset(). — wilhelmtell
– wilhelmtell, Commented Jan 25, 2012 at 13:02
Using sizeof like you try to do only works on explicit arrays (i.e. variables declared like int a[10]). — Some programmer dude
– Some programmer dude, Commented Jan 25, 2012 at 13:02

Fred Foo · Accepted Answer · 2012-01-25 13:02:38Z

5

sizeof(a[i]) == sizeof(int *); the sizeof operator cannot be used to get the length of a dynamically allocated array. Do

memset(a[i], '0', sizeof(a[i][0]) * rownum);

instead.

(And please refrain from using ASCII values as magic numbers in code.)

answered Jan 25, 2012 at 13:02

Fred Foo

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Comments

stefaanv · Accepted Answer · 2012-01-25 13:17:17Z

0

See larsmans answer as to what the issue is. This is easily replaced with vectors (with no chance of memory leaking) (not compiled/tested):

#include <vector>
#include <iostream>

typedef std::vector<int> vector1D;
typedef std::vector<vector1D> vector2D;

void showarr(const vector2D& v)
{
        for (vector2D::iterator it1 = v.begin(); it1 != v.end(); ++it1) {
                for (vector1D::iterator it2 = it1.begin(); it2 != it1.end(); ++it2) {
                        cout<<*it2<<endl;
                }
        }
}

int main()
{
        int rownum;
        cin>>rownum;
        vector2D a();
        for (int i = 0; i <rownum; i++) {
             a.push_back(vector1D(rownum, '0'));
        }
        showarr(a);
}

answered Jan 25, 2012 at 13:17

stefaanv

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2 Comments

Fred Foo Over a year ago

The behavior of this piece of code is different, though: it will set each int to '0' rather than each byte within each int.

stefaanv Over a year ago

@larsmans: correct, although in this case this is more in line with what is expected. (and part of the not compiled/tested).

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