Finding the highest number in an array of length 5 - c++

Question

This should be really simple, but I'm used to higher level languages and am missing something. I'm just trying to make sure the input is five numbers long, and then find the highest number. Unfortunately, something goes wrong in that second part.
#include <iostream>
#include <string>
bool isFiveDigits(int num) {
if (std::to_string(num).length() == 5) {
return true;
} else {
return false;
}
}
int highestInArr(int *nums) {
int highest = nums[0];
for (int i = 1; i < sizeof(nums); i++) {
int temp = nums[i];
if (temp > highest) {
highest = temp;
}
}
return highest;
}
int main() {
using namespace std;
int num;
int nums [5];
cout << "Enter a five digit number!\n";
cin >> num;
if (!isFiveDigits(num)) {
cout << "Not five digits, can you even count?";
return 1;
}
string numstr = to_string(num);
for (int i = 0; i < numstr.length(); i++) {
cout << numstr[i] << " ";
nums[i] = (int)numstr[i];
}
cout << "\n" << highestInArr(nums);
}
When this runs, I get:
Enter a five digit number!
12345
1 2 3 4 5
1424080487
Of course, 1,424,080,487 is not in [1, 2, 3, 4, 5].

Answer 1

You cannot pass a pointer into a function and get the size of it without template deduction. At runtime, all the function receives is a pointer. When you call sizeof(nums), you are not getting the size of the original array. You are simply getting the size of the pointer, which is the same as saying sizeof(int_ptr). Instead, you should be using a std::vector when using collections whose sizes are dynamic.
Now, you CAN receive the size by doing something like this:
#include <iostream>
template<typename num_t, size_t N>
num_t max_num(num_t(&arr)[N]) {
num_t m = (num_t)0;
for (size_t i = 0; i < N; ++i)
if (arr[i] > m)
m = arr[i];
return m;
}
int main(){
int foo[] = { 1, 5, 2, 4, 3 };
int m = max_num(foo);
std::cout << m << std::endl;
std::cin.get();
return 0;
}
However, this is not necessarily preferred and assumes that the array was created on the caller's stack. It does not work for dynamically allocated arrays that were created with new[]. If you do this multiple times with different sizes, you will have multiple implementations of the same function (that's what templates do). The same goes for using an std::array<int, N>. If you use N as a size_t template parameter, it will do the same thing.
There are two preferred options:
Send the size of the array into the function so that the caller is responsible for the size.
Use a different container such as std::vector so the callee is responsible for the size.
Example:
#include <vector>
#include <iostream>
#include <algorithm>
int main(){
std::vector<int> vec{ 1, 5, 2, 4, 3 };
int m = *std::max_element(std::cbegin(vec), std::cend(vec));
std::cout << m << std::endl;
std::cin.get();
return 0;
}
As for the is_5_digits, you should use the base-10 logarithm function.
#include <cmath>
// ...
int i = 12345;
size_t length = (i > 0 ? (int)log10(i) : 0) + 1;
std::cout << length << std::endl; // prints 5;

Answer 2

First of all, you can't simply convert a char to int just like (int)numstr[i] assuming that it will return the digit which it contains.
See, if you have a char '0', it means it's ASCII equivalent is stored, which is 48 in case of 0, 49 in case of '1' and so on.
So in order to get that digit (0,1,2,...,9), you've to substract 48 from the ASCII value.
So change this line:
nums[i] = (int)numstr[i];
to:
nums[i] = (int)numstr[i] - 48; // or nums[i] = (int)numstr[i] - '0';
And another thing, in your highestInArr function, you're getting a pointer as parameter, and in the function, you're using sizeof to determine the size of the array. You can't simply do that, the sizeof will return the size of int*, which is not the size of the array, so you've to pass size as the second argument to the function, and use it in the loop.
Like this:
int highestInArr(int *nums, int size) {
// ...
for (int i = 1; i < size; i++) {
// ...
}
// ...
}