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return statement in c ??? why this function always returning 20?

Solutons:


You have a problem in your code. Change

scanf("%d",arr[i]);

To

scanf("%d",&arr[i]);

This is done because scanf expects an argument of type int* but you provide argument arr[i] which is of type int. Also add a check that ends the program if user inputs a number which is greater than 10 for the first scanf.

There can be two reasons.

Case 1 [Much likely for _always_]

Simple. Because your if(arr[i]==n) condition is not met, and i<m became false. It came out of for() loop and hence, return 20.

case 2 [Less likely for _always_]

By chance, the value of n is present at the 21st location [index 20] in the input array.

Apart from the coding aspect, did you understand what’s the logical purpose of this function? If not, begin with that. It searches for a specific value in an array of given length, and if no element of the array matches that value, it returns 20.

Now analyze your case, based on your input.

EDIT:

After seeing the complete code, as Mr. CoolGuy has pointed out, use

scanf("%d",&arr[i]);

Just for more reference, you can look at Chapter 7.19.6.2, paragraph 12 , %d format specifier, which goes like

… The corresponding argument shall be a pointer to signed integer.

In your code, arr[i] is of type int. What you need is a int *, i.e., &arr[i].

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