while loop with int array

Question

I'm new into C and I'm kinda lost with these overflow and underflow things. I've a few noobs questions in my mind that I've to ask to someone.

I'm doing some leetcode problems, and I'm stuck on this one because of a runtime error telling me that I've a heap-buffer-overflow.

It is supposed to calculate the average of a given int array.

double average(int* salary, int salarySize){
    int i = 0;
    double count = 0.0;
    while (salary[i] != salarySize)
        count += salary[i++];
    return (count / salarySize);
}

Thanks in advance, hope i'll figure out what's not working.

I tried with a for loop, it works great but I wanted to know why it isn't working with while loops ?

score 0 · Answer 1 · answered Mar 30 '22 at 21:36

0

while (salary[i] != salarySize) compares the array element salary[i] to salarySize. That is not what you want. You probably want while (i < salarySize).

answered Mar 30 '22 at 21:36

Eric Postpischil

195,579
13
168
312

Vlad from Moscow · Accepted Answer · 2022-03-30T21:47:45.773

In this expression used in the while loop

salary[i] != salarySize

there is compared an element of the array salary with the number of elements in the array that does not make a sense.

You need to compare the current value of the index i with the number of elements in the array.

The function can be declared and defined the following way

double average( const int *salary, int salarySize )
{
    double count = 0.0;

    int i = 0;

    while ( i < salarySize )
    {
        count += salary[i++];
    }

    return salarySize <= 0 ? 0.9 : count / salarySize;
}

Pay attention to that as the passed array is not changed in the function then the corresponding parameter should have the qualifier const. Also to avoid division by zero you need check in the return statement whether salarySize is greater than 0.

Though it would be better to declare the second parameter as having the unsigned integer type size_t.

double average( const int *salary, size_t salarySize )
{
    double count = 0.0;

    size_t i = 0;

    while ( i < salarySize )
    {
        count += salary[i++];
    }

    return salarySize == 0 ? 0.9 : count / salarySize;
}

score 0 · Answer 3 · answered Mar 30 '22 at 22:34

Other answers seem to give obvious reason for the error you get.

I wanted to give some pointers on the underlying structure of your function signature. Your function in this case take a pointer and and a size paramater. It is used approach while sending an array to a function.

Since your array is not a ready to use library data structure with array boundary properties you send a simple pointer which is the address to the start of the memory location of your stored array.

Ending address of your array is found using another paramater passed which is called arraySize and it is your responsibility to use memory location while staying inside allowed boundary you are given access to.

If you try to access or modify memory segments you are not allowed you get segmentation errors.

This answers from similar C arrays topic might also be useful.

Passing an array as an argument to a function in C

https://stackoverflow.com/a/47233772/10902172

while loop with int array

3 Answers3