array parts

I i am just studying modulization in C. Now im a bit confused.
What i want to do is the user inputs 10 numbers and they are stored in an array. Then the numbers are each passed to a module (procedure) to determine if its the maximum value. How would i go about the modules?
Below i created a brief try, ignoring the like syntax errors, is this how i could pass the array to another module etc?
Thanks

Code:

int main(int argc, char ** argv)
{
int i;
float array[i];
	for(i=0;i<10;i++)
	{
		fprintf(output,"Please enter your values");
		fscanf(input,"%f",&array[i]);
		maximum_num(array[i]);
	}
	fprintf(output,"Maximum number is %f\n",&maximum_num);
}

int maximum_num(float array_size);
{
float current_max = 0;
if(array[i]>current_max)
	return current_max = array[i];
}

generally, you do something like this:

Code:

int maxnum(int *arr, int size);  // prototype. 
int main()
{
    int array[10];  // In C the size of arrays is always a numeric constant, not a variable. 
    int max;
    ...
    max = maxnum(arr, 10);
}

int maxnum(int *arr, int size)
{
   ... 
}

You should probably use a #define for the number 10 so that you don't have to repeat it in different places.

--
Mats

ok i get that but still a bit confused, i tried this now:

Code:

#include <stdio.h>
FILE * output =NULL;
FILE * input =NULL;
static int max(float array[i])
{

    float current_max = 0.0;
    if(array[i]>current_max)
	current_max=array[i]  

}
static int display(float array[i])
{
    for(i=1; i<10; i++)
    	fprintf(output,"%f ",array[i]);    
    return();
}


int main(int argc, char * argv[])
{
float array[i];

    output = stdout;
    input = stdin;

    for(i=0;i<10;i++)
    fprintf(output,"Please enter value: ");
    fscanf(input,"%f",&array[i]);

}

Now i am a bit confused how i can take a value from the array of values and the pass it to one fuction to determine if its the max value, then continue that until they have all been down, finally pass the numbers to another procedure to display all the numbers??

You can't use i as a "size" for arrays, as I explained in the previous post. For passing arrays, you either use "float *array" or "float array[]" - in either case, you need some way to determine that the array ends. Passing the number of elements as a parameter to your function is one way. Another is to mark the end with a special marker, such as -1 (assuming for example valid numbers are all positive).

--
Mats

otherwise it should work?

so how do i pass an array of elements to another module to then perform a task?

Code:

void blah(type *arr);
... 
int main()
{
    type array[10];
....
    blah(array);
...
}

void blah(type *arr) 
{
   ...  ... arr[x] ... ... 
}

--
Mats

alright i fixed it up, jus one little problem that i have no idea:

Code:

#include <stdio.h>
FILE * output =NULL;
FILE * input =NULL;

void display(float array[10])
{
int i;
	for(i=0; i<10; i++)
	{
		fprintf(output,"%.4f ",array[i]);
		fprintf(output,"\n");
	}
}

float max(float array[10])
{
int i;
	float max=0.0;
	//for(i=0; i<10; i++)
	//{
	if(array[i]>max)
	{
	max=array[i];
	}
	//}
	fprintf(output,"The maximum value is %f",&max);

}

int main(int argc, char * argv[])
{
int i;
float n;
float array[10];

    output = stdout;
    input = stdin;
	for(i=0; i<10; i++)
	{
    	fprintf(output,"Enter value ");
    	fscanf(input,"%f",&array[i]);
	}
	display(array);
	//max(array);
	fprintf(output,"%f\n",max(array));


}

Now the program is supposed to pass the array to the max module to find the biggest number, but doesnt work. if i put:

Code:

fprintf(output,"%f\n",max(array));

in the for loop in the main module it works, but once i put it outside like how it looks now it doesnt work, im not sure how to fix that?

What compiler are you using? Don't you have warnings enabled on it?

Your compiler should tell you that you are not returning a value from your max function (and you also have used the same name for a function and a variable, whilst this is valid as such, it's a bad thing to do, because it's very confusing to anyone reading the code).

Code:

fprintf(output,"The maximum value is %f",&max);

I bet this doesn't print the right value at all!

--
Mats

could you help me out with that, like how would i give the array to the max function and then find the max num?

(the code above might not work correctly because i was playing around and cant compile it because i dont have linux)

What do you mean "I haven't got Linux"? None of the code in your example is relying on Linux, as far as I can see.

And yes, I could fix the code up, but I don't see why I should do that - I have given you enough suggestions that you should be able to do it yourself.

--
Mats

Code:

float max(float * arr, size_t n)
{
    size_t i = 0;
    float m = 0.0f;

    /* maybe some loop here? */

    return m;
}

Something like that, you're close - just listen to what matsp says!
Guessing is no way to get something done, I also don't see what Linux has to do with anything

well i use linux to compile it

Originally Posted by zacs7

Code:

float max(float * arr, size_t n)
{
    size_t i = 0;
    float m = 0.0f;

    /* maybe some loop here? */

    return m;
}

Something like that, you're close - just listen to what matsp says!
Guessing is no way to get something done, I also don't see what Linux has to do with anything

what do u mean by the float m = 0.0f??

and

do i need the * arr etc? because i used it somewhere else without it and it works?

You can "receive" an array in several different ways in C:

Code:

func1(int *arr);
func2(int arr[]);
func3(int arr[10]);

All of which are essentially identical - the last one enforces the size of the array passed, so you can't for example pass a 20 element array to that function. The first two take "any size array".

This is also why zacs7 suggested you pass in a size_t n to indicate the size of the array. That way, your function works for all sorts of number of values, rather than ALWAYS having to have 10 values in the array.

m = 0.0f means set the value m to zero-point-zero, and it's a floating point number (rather than a double precision that 0.0 would be).

--
Mats