Printing float array using for loop and pointers

Question

I'm starting out in CPP on arrays and pointers.

I'm trying to print a float array but I'm getting past the float array.

How can I stop printing the float array once i hit the end?

// all about arrays and pointers

#include <iostream>

using namespace std;

int main (int argc, char * argv []) {

    float inputs1 [] = { 12 };
    float inputs2 [] = { 12 , 4 };

    cout << "sizeof(inputs1)" << sizeof(inputs1) << endl;
    cout << "sizeof(inputs2)" << sizeof(inputs2) << endl;

    float *it = inputs2;
    cout << *it << endl;
    it++;
    cout << *it << endl;
    cout << endl;

    cout << "sizeof(inputs1): " << sizeof(inputs1) << endl;
    cout << "sizeof(inputs2): " << sizeof(inputs2) << endl;
    cout << "start address: " << &inputs2 << endl;
    cout << "end address: " << &inputs2[sizeof(inputs2)] << endl;

    for (float *it2 = inputs2; it2!=&inputs2[sizeof(inputs2)]; it2++) {
        cout << "it2 " << it2 << endl;
        cout << "*it2 " << *it2 << endl;
    }
}

inputs2[sizeof(inputs2)] - note that sizeof returns the size in terms of chars, not floats... — Oliver Charlesworth
– Oliver Charlesworth, Commented Jul 26, 2015 at 2:47

flogram_dev · Accepted Answer · 2015-07-26 02:53:38Z

4

sizeof returns the size of its argument in bytes — not the number of elements in it¹.

To get the number of elements, you have to divide sizeof(inputs1) by sizeof(inputs1[0]).

_{¹ unless the elements are one byte each, such as char}

edited Jul 26, 2015 at 2:53

answered Jul 26, 2015 at 2:48

flogram_dev

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