Why does my code need cout for printf() to work? - c++

Question

I'm pretty new to coding and stumbled upon this issue when trying to code a program that prints the biggest number in an array.
Basically, if I don't include cout<<i;, the printf() will print the array location instead of the number 20. Any ideas why? (I'm guessing it is something stupid I overlooked, so sorry in advance).
#include <iostream>
using namespace std;
int maxinlst(int lst[], int size) {
int maxNum;
for (int i = 0; i < size; i++) {
cout << i;
if (lst[i] == lst[0])
int maxNum = lst[i];
else if (maxNum < lst[i]) {
maxNum = lst[i];
}
}
return maxNum;
}
int main() {
int lst[] = {-19, -3, 20, -1, 5, -25};
printf("%i", maxinlst(lst, 6));
}

Answer 1

You shadow maxInt by declaring another variable with the same name. See my comments here:
int maxinlst(int lst[], int size) {
// First declaration
int maxNum;
for (int i = 0; i < size; i++) {
cout << i;
if (lst[i] == lst[0])
// Second declaration
int maxNum = lst[i];
else if (maxNum < lst[i]) {
maxNum = lst[i];
}
}
return maxNum;
}
This is legal in C++, and in most languages with block scope. The second declaration creates a new variable, but it's never used because it goes out of scope immediately, so the whole assignment, along with the conditional, can be eliminated by the compiler. If you enable compiler warnings, you should get a warning about the second declaration because that variable is never used again:
test.cpp: In function ‘int maxinlst(int*, int)’:
test.cpp:8:17: error: unused variable ‘maxNum’ [-Werror=unused-variable]
int maxNum = lst[i];
^~~~~~
This also means that the value of the outer maxNum after the first iteration of the loop is indeterminate and reading it could be undefined behavior so the second loop iteration is either going to (a) use an indeterminate value since the outer maxNum was never assigned or (b) something else entirely because of UB.
If the second conditional is never true (assuming an indeterminate value and not UB) then the value returned by this function will also be indeterminate -- whatever unpredictable number maxNum happened to be.
The correction here would be to remove int in the second declaration.
You could also rewrite this to avoid the first in-loop conditional:
int maxinlst(int lst[], int size) {
// First declaration
int maxNum = lst[0];
for (int i = 1; i < size; i++) {
if (maxNum < lst[i]) {
maxNum = lst[i];
}
}
return maxNum;
}
As far as why cout << i changes the value you see, that's exactly the nature of using indeterminate values / undefined behavior. You can't reason about what's happening. Adding or removing other code could also change the value returned by the function. You may even see different values if you run the program multiple times without making any changes to it.

Answer 2

You actually have two variables called maxNum. Inside your if statement, int maxNum = lst[i] defines a new variable called maxNum which hides the outer maxNum variable. The inner, newly defined maxNum is destroyed as soon as the program exits the if statement.
What makes you think that the array location is being print?
maxNum is not set to a particular value when it is defined. Because of that it's just going to have whatever was stored in the memory before.
If that value in memory is large, then maxNum will be larger than all the values in your array, then maxNum will never change and it will keep the garbage value.
So, when you get to your printf statement, the garbage value will be output. I doubt that the location of the array is being output. I bet that it's just whatever garbage was in memory.