C 指针参数分配给具有不同值的参数

Question

I was trying to write a seemingly trivial function in C to get the maximum value in an array of integers. I'm trying to do so using pointers to get a better understanding. (I know how to do this passing the size of the array, the question is not how to do this another way)

#include <stdio.h>

int gimme_largest(int *, int *);

int main(void)
{
  int arr[] = {1, 22, 3, 44, 5};
  int largest;

  printf("In main, arr:\t\t\t %p \n", arr);
  printf("In main, arr + sizeof(arr):\t %p\n", arr + 5);
  putchar('\n');

  largest = gimme_largest(arr, arr + sizeof(arr));
  // printf("\n\nThe largest is %d", largest);

  return 0;
}

int gimme_largest(int *a, int *lastPlusOne)
{
  int largest = *a; // let's assume the first element is the largest one

  printf("In gimme_largest, a:\t\t %p\n", a);
  printf("In gimme_largest, a + 5:\t %p\n", a + 5);
  printf("In gimme_largest, lastPlusOne:\t %p\n", lastPlusOne);

  while (a < lastPlusOne) {
    if (*a > largest)
      largest = *a;
    a++;
  }

  return largest;
}

The problem is that lastPlusOne does not equal a + 5 in the invoked function, but a higher address; can't see why that's happening.

Answer 1

This call

largest = gimme_largest(arr, arr + sizeof(arr));

is invalid. sizeof( arr ) in the case of your program is equal to 5 * sizeof( int ) while for the correct pointer arithmetic you need to use only the number of elements in the array.

That is there shall be

largest = gimme_largest(arr, arr + sizeof(arr) / sizeof( *arr ) );

Thus the expression sizeof(arr) / sizeof( *arr ) is equal to the value 5 that you are using in this call of printf

  printf("In main, arr + sizeof(arr):\t %p\n", arr + 5);

Pay attention to that the user can call the function with an empty range when a is equal to lastPlusOne . In this case the function will have undefined behavior because for an empty range there is no largest elemenet.

The function should return a pointer to the largest element.

Also as in C there is no function overloading then the function should be called for constant and non-constant arrays.

Here is a demonstrative program that shows how the functipon can be defined.

#include <stdio.h>

int * gimme_largest( const int *first, const int *last )
{
    const int *largest = first;
    
    if ( first != last )
    {
        while ( ++first != last )
        {
            if ( *largest < *first ) largest = first;
        }
    }
    
    return ( int * )largest;
}

int main(void) 
{
    int arr[] = {1, 22, 3, 44, 5};
    const size_t N = sizeof( arr ) / sizeof( *arr );
    
    int *largest = gimme_largest( arr, arr + N );
    
    printf( "The largest number is %d\n", *largest );
    
    return 0;
}

The program output is

The largest number is 44

Answer 2

To get the end address of the array, you can do this by arr + sizeof(arr) / sizeof(arr[0]) :

#include <stdio.h>

int gimme_largest(int *, int *);

 int main(void)
{
 int arr[] = {1, 22, 3, 44, 5};
  int largest;

  printf("In main, arr:\t\t\t %p \n", arr);
  printf("In main, arr + sizeof(arr):\t %p\n", arr + 5);
  putchar('\n');

  largest = gimme_largest(arr, arr + sizeof(arr) / sizeof(arr[0]));
  printf("\n\nThe largest is %d", largest);

  return 0;
  }

int gimme_largest(int *a, int *lastPlusOne)
{
 int largest = *a; // let's assume the first element is the largest one

  printf("In gimme_largest, a:\t\t %p\n", a);
  printf("In gimme_largest, a + 5:\t %p\n", a + 5);
  printf("In gimme_largest, lastPlusOne:\t %p\n", lastPlusOne);

  while (a < lastPlusOne) {
    if (*a > largest)
       largest = *a;
    a++;
  }

 return largest;
}

output:

In main, arr:                    000000000062FE00
In main, arr + sizeof(arr):      000000000062FE14

In gimme_largest, a:             000000000062FE00
In gimme_largest, a + 5:         000000000062FE14
In gimme_largest, lastPlusOne:   000000000062FE14


The largest is 44