The above code is overly complicated and less efficient than a simple solution. You absolutely do not need a complicated sorting algorithm to find the minimum and maximum numbers in an array (besides, heapsort is O(n log n) and finding minimum and maximum of an array is just O(n) efficiency).

The problem with the code is this section

for(i=1;i<n;i++)
{
if(list[0]<list[ i ])
{
maximum=list[0];
}
if(list[1]>list[ i ])
{
minimum=list[1];
}//if closed.
}//loop closed

The problem is that you are always comparing the current list item with the first or second (indexes 0 and 1), wihch means you won't end up with the accurate maximum or minimum. To get this to work, you need some base value for minimum and maximum to start out as, before the loop even begins. The first element in the array or list or whatever works (in this case set minimum and maximum to

list[0]

). Now you can iterate through the list. If the current element (

list[i]

) is less than minimum, simply set minimum to

list[i]

. Likewise if

list[i]

is greater than maximum, set maximum to

list[i]

. Also the for loop should start at i=0 not i=1.

@52, I did already say it was overcomplicated... but it is a good way of killing two birds with one stone. He can learn algorithms/data structures as well as this at the same time.

Give this a try:

#include <math.h>
// ...
if (val != (int)val) { // check for fractional part
float iptr;
float integral_part = modf(val, &iptr);
if (iptr > 0.5) {
val += 0.5;
} else if (iptr < -0.5) {
val -= 0.5;
}
}

It may not work on the else condition; the docs say:

Source: https://en.cppreference.com/w/c/numeric/math/modf

@pauliyobo:
Here's the code:

#include <stdio.h>
int main(void)
{
int i, n;
long result, dividend, divizer;
printf("Enter the numbers of pairs which you wish to divide:");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("Enter the numbers for divide:");
scanf("%d%d",&dividend,&divizer);
result=dividend/divizer;
if(result>0)
{
result=result+1;
}

else
{
result=result-1;
}
printf("the result is %d\n",result);
}
return 0;
}

The result is supposed to be a rounded intijer if there is a fractional part, (Like 1.2 3.9 with the intiger rounded to 1 or 3, with 0.5 added to it.)

hope that clears things.

Thanks for clearing that up -- I thought my solution was the one.

@61, appologies if I overwhelm you -- that is not my intent.
The expression

if(result != (int)result && result > 0.0)

determines if the floating point number has a fractional part. The fractional part is the number after the decimal point. So if we had a number like 2.5220, the fractional part would be 5220. (We could do that with strings, but that would be unnecessarily over complicated.) This expression, in particular, compares result to its integral version of itself. Naturally, a number with a fractioanl part and one without are entirely different. So, this translates to:

if result has a fractional part and result is greater than zero

.
HTH

Sorry. replace the printf statement with this one
printf("the result is %f\n", result);
@62 thanks for taking the time to explain this, I was in a hurry when I posted the code and didn't explain as detailed as I had to.

So, in a problem to count vowels within a string, I must test multiple strings, and count the vowels present within them.

The issue is, the code below does not tests the entire string, it exits after finding one vowel, and it only tests one string, doesn't matter whether I entered four or five strings which I wish to test.

#include <stdio.h>
int main()
{
int i, n, temp=0, vowels=0;
char string[100];
printf("How many strings do you wish to test?");
scanf("%d",&n);
for(i=0;i<n;i++)//This loop is for the number of strings which the user would like to test.
{
printf("Plese enter strings one by one:");
scanf("%s",string);
while(string[temp] !='\0')//this is where the condition to find the vowels is.
{
if(string[temp]=='a'||string[temp]=='e'||string[temp]=='i'||string[temp]=='o'||string[temp]=='u')
{
vowels++;
}
temp++;
}
printf("there are %d vowels within this string",vowels);
}
return 0;
}

Thanks Hijacker. strings is the topic which I have the least amount of knowledge, of course I am rectifying this. hope you guys would still be patient with me by the time I am done with strings.

Also, any tips for formatting my code? indentation and such are impossible, since  I have zero percent vision. but I am willing to learn other ways.

Ethin is right, even though you're blind, indentation should become normal for you. Some programming languages like Python require you to indent properly, because they don't have blocks and the indentation decides upon the code execution for you. Its simply a matter of training and getting used to it.
Another thing is to prevent too long lines of code (70 to 80 characters a line, including indentation, should be the maximum. Developers tend to align their screens not vertically, but horizontally, which means that they'd have to scroll to read longer lines, which takes up much time, so keep your lines shorter and break it if possible, just as I did in your code example above.
Best Regards.
Hijacker

Notepad is fine, In fact, I'm using it myself all the time, or at least a close alternative, Notepad2 right now. Some IDEs may be accessible and helpful, but I didn't find one I liked to use yet. The accessible ones like Eclipse tend to be rather slow compared to simple editors, and I don't like that. Most languages support CMD and thus you can stick with your way, but i'd recommend to format your code properly anyway .
Best Regards.
Hijacker

So I was trying to solve this problem about evaluating expressions on CodeWars, and I did actually managed to code the solution.

Except that the solution I coded is giving me an error with which I am unfamiliar with, (The error is described below,) and I am not sure how to look that up on the google.

As an experiment, I tested my solution with a full program, (CodeWars only gives you the functions to solve,) and my solution is working in my system without error, so I am not sure what is wrong.

Below I have given the entire program, with the line which is giving me error having a comment just after it describing what the error is.

#include <stdio.h>
int expression_matters(int, int, int);
int main(void)
{
int r, a, b, c;
printf("Enter three values:");
scanf("%d%d%d", &a, &b, &c);
r= expression_matters(a, b, c);
printf("%d is the largest value of this expression.", r);
return 0;
}
int expression_matters(int a, int b, int c)
{
int d, e, f, g;
d = a* (b+c);
// Above line gives the error called object type 'int' is not a function or function pointer.
e = a * b * c;
f = a * b +c;
g = (a + b) * c;
if(d > e &&
d > f &&
d > g)
{
return d;
}
if(e > d &&
e > f &&
e > g)
{
return e;
}
if(f > d&&
f > e &&
f > g)
{
return f;
}
if(g > d &&
g > e  &&
g > f)
{
return g;
}
}