plzz help

int ages[10];
write a function called set_ages, that will take the ages array, an age, and a parameter of the size of the array and set all the elements in the array to be the same age.
such as to set all the ages to 21.

plz help me how to do this i don't understand if sum1 can explain it 2 me plz i have midterm coming up please help

It's funny, there isn't a question or even a question mark in that entire post. Haven't read the forum guidelines, now have you ?

Here's a hint, use a loop.

Surely you were taught this anyway? Its quite a simple thing to do and one-dimensional arrays are quite easy to work with

only if sum1 can show me

plz read forum guidelines & homewrok policy from teh top of teh page 1st

Originally Posted by aj11

only if sum1 can show me

I think that is partly what the professor is getting paid for.

well im new to c++ as well, and i thought this might be good practice for me,
i tried to do this and for some reason it only works halfway

i spent the time on it, and i would like to know what im doing wrong, if possible help me

here is my output (it is supposed to display TEN 21s)

Code:

 21 21 -1 21 2009315348 4007008 21 8 2009312941 -1724333122

the program the guy described in the first post says to pass an array to a function and set it full of 21s

Code:

#include <iostream>
using namespace std;

void age_set(int* y,int age,int size)
{
     for(int x=0;x<size;x++)
     {
             y+=x;
             (*y)=age;
     }
}

int main()
{
    int ageset[10];
    int asize = (sizeof(ageset))/4;  //divide by 4 since ints are held in packs of 4 bits
    int* y;
    y = &ageset[0]; //y pointer now holds the address of start of array in memory
    age_set(y,21,asize);
    
    for(int x=0;x<asize;x++)
    {
            cout << " "<<ageset[x];
    }
}

I would change:

Code:

int asize = (sizeof(ageset))/4;  //divide by 4 since ints are held in packs of 4 bits

To this:

[code]

Code:

int asize = (sizeof(ageset))/sizeof(asize);

That way you maintain portability and you don't have to know or rely on an int being 32-bits (although it generally will be nowadays).

Inside your function, you have the right idea except you're adding to y by way too much.

Pretend this is an array of size 5 that contain garbage:

Code:

[1321], [2356], [9234], [2332423], [-7312]

In your function, it's like this:

Code:

y -> [1321], [2356], [9234], [2332423], [-7312]

If you add one to y, since y is a pointer, it's taken to mean that you want to point to the next element, like this:

Code:

[b][1321], y -> [2356], [9234], [2332423], [-7312]

Now take a look at what you're adding to y in your function and see how this differs.

ah i didnt catch that, thanks

Code:

     for(int x=0;x<size;x++)
     {
             
             (*y)=age;
             y+=1;
     }

thanks guyz u rock