greatj0b greatj0b - 1 year ago 85
C++ Question

Function changing a variable, even when passed by value?

This is what I have, and it's causing some extremely weird discrepancies.

using namespace std;

void input_data(int data[], short size);

int main(){
short size;
int data[1000];

cout << "How many values would you like to store? ";
cin >> size;

while (size > 0){
cout << "Enter your " << size << " values:" << endl;
input_data(&data[1000], size);
cout << size << " values stored in array." << endl;

void input_data(int data[], short size)
int i;
for (i=0;i<size;i++){
cin >> data[i];

So this code should be pretty straight forward, but any time I enter a value other than 3 for
, the line
cout << size << " values stored in array." << endl;
ends up saying "0 values stored in array."

I don't understand how
could be changing the value of
, since I am passing by value, not by reference.

Why is this happening, and better yet, why is it working with the number 3?

Answer Source

You are overwriting the stack.

First you declare a function that takes an array:

void input_data(int data[], short size);

And then you call it. Instead of passing a pointer to an actual array like this:

input_data(data, size);

You are passing a pointer to the space immediately AFTER the array (which includes many things including size):

input_data(&data[1000], size);

While a parameter of data and &data[0] are essentially the same, &data[1000] points to a completely different thing - it points to the address of data[1000] which doesn't exist in the array because the array only has 1000 elements (from 0 to 999)

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