the defined variable is different

Question

why is the variable in the average function is different from the same variable in the main function? I do not know how to solve it because I'm just learning.

#include <stdio.h>
#define  SIZE (sizeof(number)/sizeof(int))

int number[] = {5,65,84,25,644,2};

int average(int *number)
{
    printf("size = %ld\n",SIZE);
    int sum=0;
    for(int i=0;i<=SIZE ;i++)
    {
    sum += number[i];
    }
    return sum/SIZE;
}

int main()
{
printf("average : %d\n",average(number));
printf("number of elements: %ld\n", SIZE);


return 0;
}

If one of the answered given solved your problem, you should [accept it](https://stackoverflow.com/help/accepted-answer). — dbush, Nov 12 '18 at 19:10

score 1 · Accepted Answer · answered Nov 12 '18 at 19:02

1

You use a macro (i.e. #define ...) to specify SIZE. The macro is expanded "textually" wherever you make use of it, and it is interpreted in the respective context, i.e. it's meaning depends on where you place it.

So in main, number as part of your macro refers to the global variable number, whereas in the context of function average, it refers to the function argument number (which hides the equally named global variable).

Anyway, it's impossible to deduce any array size from a pointer variable. You'll have to pass the number of elements as an extra argument.

answered Nov 12 '18 at 19:02

Stephan Lechner

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WoW! I didn't see the macro mistake. +1 for that – scipsycho Nov 12 '18 at 19:08
you're welcome. If it helped, don't forget to mark one of the answers as "acceped" by pressing the green arrow. Thanks! – Stephan Lechner Nov 12 '18 at 19:15

score 0 · Answer 2 · answered Nov 12 '18 at 18:59

0

Within main, number is a global variable which is an array. So sizeof(number) gives you the size of the whole array in bytes.

In the average function however, number is a parameter to the function and has type int *. So sizeof(number) in that case gives you the size of the pointer, not the size of the array.

Either don't pass a parameter and just use the global, or pass the size of the array as a separate parameter to the function.

answered Nov 12 '18 at 18:59

dbush

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Is this an example of why using globals can be problematic? – Random Davis Nov 12 '18 at 19:00
@RandomDavis In a way yes, as it illustrates how a variable at an inner scope can mask a variable with the same name at a higher scope. – dbush Nov 12 '18 at 19:01

score 0 · Answer 3 · answered Nov 12 '18 at 19:06

You have defined a global variable number which is an array.

In the function average , you defined another variable with the same name/identifier number (although of different type i.e. a pointer).

Now, when you use number inside the main function you are actually using the pointer number and not the global variable number. This is also known as variable hiding.

If you want to use the global variable change the name of the argument of the average function to something else or you can go the hard way and do something like this.

Happy Coding;)

the defined variable is different

3 Answers3