C++ programming -- dynamic arrays?

Question

Okay, so here's the link for the problem http://www.pic.ucla.edu/~nathan/cgi-bin/moin.cgi/la1
I'm having trouble on the first part.
For the first function

intialize_array: I'm confused on how I can get the print_array function to print out NA's.. If I just initialize all values in the array to a default value such as -1, then ask the print_array function to print out the values, the values will read 'NA'?

for example:
(initialize_array(int a[],int n1) {
for (int i =0; i < a[n1]; i++)
a[i] = -1;
a[n1] = n1;


Also, for the copy_array function.. i wasn't sure if i was approaching this right

copy_array(int a[], int n1, int b[], int n2);
for(int i =0; i < n1+1; i++)
b[i] = a[i];
b[n2] = n2;

thanks so much

thanks for your help..
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good luck!

Answer 1

First, you need to implement your array as vector. Second, you have to use your array.size() to know *where* to put some value.
Read this, it will help:
http://www.sgi.com/tech/stl/Vector.html

Answer 2

I am 15 so here its goes. I am a beginner.

first create a for loop that will deposit the numbers like

for (int i =0; i < a[n1];){
cout << "ENTER A num:";
cin >> array[i]; // I did a array[ i] because the for loop will try all the possible nums it will be less than a[n1] so if it start from 1 then the first num be story in arry[1]
cout << "WANNA ENTER ANOTHER NUM[Y/N]";
cin >> more; // create a char called cin so u can desposit more num if the user wants.
if(more == 'n')
break;
}else{
i++;
}

then later if u want to recall the array back use this loop

for (int i =0; i < a[n1];){
cout << "array" << i << " is " << array[i] << endl; // will say Array 1 = etc and so on.
}

HOPE THIS HELPS.

~~HELP ME~~
I am 15 and I am learning programming. I have a 4.0 GPA but I think I won't get higher than 700 on the SAT so How do i get into a good college.

Answer 3

i'm no longer one hundred%. Why no longer loop from the starting up handle (i.e. ptr) till eventually you hit a null (which shows the end of the dispensed area) and count number the type of ones and zeros. at the same time as(*ptr != null){ if(*ptr == a million){ improve the a million's counter; } else if (*ptr == 0){ improve the 0's counter; } ptr++; }