CODEWITHSUNDEEP
c++pointerspointer-arithmetic
jameswoo
jameswoo

Reputation: 396

Why am I getting different addresses when using similar logic to increment two different pointer types?

I don't understand why the addresses of my floats go up by 16, when the size of my floats are 4. Could someone please explain?

Code:

char* mychar   = new char[SIZE];
float* myfloat = new float[SIZE];

for(int i = 0; i < SIZE; i++)
{
    mychar[i] = 'A' + i;
    myfloat[i] = 101 + i; 
}

for(int i = 0; i < SIZE; i++)
    {
        cout << setw(12) << "new char @ <"  << static_cast<void*>(mychar) + sizeof(char)*i  << ">=<" << mychar[i]  << ">" 
             << setw(14) << "   new float @ <" << myfloat + sizeof(float)*i                    << ">=<" << myfloat[i] << ">\n"; 
    }

cout<< "Size of float: " << sizeof(float) << endl << "Size of char: " << sizeof(char);

Output:

new char @ <0x800102e8>=<A>   new float @ <0x800102f8>=<101>
new char @ <0x800102e9>=<B>   new float @ <0x80010308>=<102>
new char @ <0x800102ea>=<C>   new float @ <0x80010318>=<103>
new char @ <0x800102eb>=<D>   new float @ <0x80010328>=<104>
Size of float: 4
Size of char: 1

Upvotes: 2

Views: 85

Answers (1)

templatetypedef
templatetypedef

Reputation: 372664

Notice that myfloat is a float*. This means that if you write an expression like

myfloat + n

the result will be a pointer to the float that is n positions down in the array of floats whose address starts at myfloat. In other words, adding n here will automatically increment the pointer by n * sizeof(float) bytes, since pointer addition is aware of the size of the object.

Consequently, if you write

myfloat + n * sizeof(float)

you are not jumping forward by n * sizeof(float) bytes, but rather than n * sizeof(float) * sizeof(float) bytes, which happens to be 16n bytes.

Hope this helps!

Upvotes: 6

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