how to initialize the “min” variable properly?

Question

I have a small problem in my code for finding the minimum value from a series of numbers. When I initialize min = 0, the minimum value turns out as 0. But when I do

Answer 1

When I initialize min = 0, the minimum value turns out as 0.

because in that case, if (min > a[i]) is always false. Remeber, a[i] is > = 100.

But when I don't initialize min, the answer is correct!

if you don't initalize the local variable, it's value in indeterminent and contains some garbage value (possibly a large one). So, seemingly, your logic works and you're apparently getting the right answer. However, using values of uninitalized local variable invokes undefined behaviour.

Solution: Initialize min with the largest possbile value, INT_MAX present in header file.

Answer 2

If you don't initialize a local variable, its value is indeterminate, and using that will lead to undefined behavior.

What you should do is initialize it to a large value, just like you initialize max to a small value. Use INT_MAX from the <limits.h> header file.

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Also, if your variables can't be negative, you should probably use unsigned int, and initialize min to UINT_MAX.

Answer 3

You must initialize min with a large number, otherwise 0 might remain as the smallest number.

You can initialize min as

int min = INT_MAX;

INT_MAX is the largest value that can be held by an int( Must add <limits.h> for INT_MAX ).

If you don't initialize min with a value, then it will have an indeterminate value, which can be anything, so always initialize your local variables before using their values..

Answer 4

The compiler warning is because accessing the value of uninitialised variables gives undefined behaviour.

A common technique to find a minimum value in an array is to initialise the min to be the first element in the array, then iterate over subsequent elements

 min = a[0];
 for (i = 1; i < 20; ++i)     /* assume 20 elements */
     if (a[i] < min) min = a[i];

This works for all numeric types that can be compared using the < operator, without having to muck around trying to find a value that exceeds all possible values in the array.

The other thing to watch is that array indexing starts at zero. So an array with 20 elements (as in int a[20]) has valid indices 0 to 19. Running a loop from 1 to 20 - and therefore accessing a[1] through to a[20] gives undefined behaviour, since a[20] does not exist.