Weird pointer syntax

Question

I was looking at some code from google and something caught my eye.

#include <iostream>

using namespace std;

const int kStudents = 25;
const int kProblemSets = 10;

// This function returns the highest grade in the Problem Set array.
int get_high_grade(int *a, int cols, int row, int col) {

  int i, j;
  int highgrade = *a;

  for (i = 0; i < row; i++)
    for (j = 0; j < col; j++)
      if (*(a + i * cols + j) > highgrade)  // How does this line work?
        highgrade = *(a + i*cols + j);

  return highgrade;
}

main() {

 int grades[kStudents][kProblemSets] = {

   {750, 700, 85, 720, 84},
   {85, 92, 93, 96, 86},
   {95, 90, 103, 76, 97},
   {65, 62, 73, 84, 73}

 };

 int std_num = 4;
 int ps_num = 5;
 int highest;

 cout << *(int *)grades << endl;

 highest = get_high_grade((int *)grades, kProblemSets, std_num, ps_num);
 cout << "The highest problem set score in the class is " << highest << endl;
}

It was (int *)grades. I've never seen that before in any tutorial before so it caught me off guard. I used cout << *(int *)grades << endl; to determine that it was a pointer but I've never seen a pointer with that syntax.

Can anyone explain how it works? Thanks.

Keith Thompson · Accepted Answer · 2016-10-10 19:59:56Z

2

grades is an array of array of int objects.

The pointer expression grades is implicitly converted in most contexts to a pointer to the array's initial (0th) element, so the expression grades by itself is of type int(*)[kProblemSets], a pointer to an array of int.

The (int*) cast converts this pointer value from one pointer type to another, and the * dereferences the resulting converted pointer, yielding an int result.

This is questionable; I don't think there's any guarantee that the code will do what you expect. If the intent is to print the value of the first element of the first row of the array, just use:

std::cout << grades[0][0] << std::endl;

(Incidentally, main() should beint main(). C++ has no "implicit int rule; you have to specify the return type explicitly.)

answered Oct 10, 2016 at 19:59

Keith Thompson

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4 Comments

TCG Over a year ago

If you change 750 to 70 then cout << *(int *)grades << endl; will output 720, a.k.a the highest value.

Keith Thompson Over a year ago

I wasn't looking at the code that computes the highest value, just the cout << *(int *)grades << endl; that you asked about. When I change 750 to 70, that statement prints 70.

TCG Over a year ago

Oh, my misunderstanding. Sorry. You're right. Thanks for clarifying.

M.M Over a year ago

*(int *)grades is guaranteed to work , there could be debate over whether the function performs an out-of-bounds array access (reading off the end of the first row)

UKMonkey · Accepted Answer · 2016-10-10 19:51:49Z

2

As grades is a 2d array, when cast to a pointer, you get a pointer to the first item of the array. This is what the (int*) does. The * at the front gets the value of the pointer - which as said, is the first item, which in this case has the value 750.

It's the equivalent to grades[0][0].

answered Oct 10, 2016 at 19:51

UKMonkey

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3 Comments

TCG Over a year ago

When you get rid of all the code an simply have the 2d array and cout << *(int *)grades << endl; then it will give a compilation error. And if you change the position of the highest value in the program it will output that instead of the first.

Keith Thompson Over a year ago

@TCG: Are you sure about that? grades is an array expression, which is converted to a pointer. The (int*) cast converts it to a different pointer type which is questionable but legal. The * dereferences the resulting pointer, yielding a result of type int. I don't think the behavior is well defined, but it should compile. (A compiler could warn about it, but is not required to.)

TCG Over a year ago

I checked it without the function get_high_grade and the last two lines of code in the main function. It did not compile as it gave an error, that's why I was so confused.

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