C language problem?

[flyk2luna]

Hiii i am working on simple functions. and I keep getting weird syntex errors when i complie. basically i can divided them into two categories. one bunch of errors (about five or six) comes from one function body 0_o. and the second one is about using math.h.

So here is my code( its not hard to read), there is my introduction on C and that might help you going through. All the first errors come at the int input_data(double arr[], double value) block, where i try to input values and store them in array (-1 as sentiel value). Second type of error is that i get messeges that saying too many parameters in sqrt. so i made it simple enough, but it still doesnt work. Below the code are my errors.


#include <stdio.h>
#include <math.h>

int input_data(double arr[], double value);
void print_array(double arr[], int k);
void sort(double arr[], int k);
double average_function(double arr[], int k);
double sum_of_xi_x_squared(double arr[], double average, int k);

main ()
{
double arr[31], average, median1, median2, SD, mid, value;
int k, s;

printf ("Don't enter more than 30 values.\n");
printf ("Put 3 values at least, or an error messege appears.\n");
printf ("This program does following: original array, sorted array(in ascending order), average, median(s), and standard deviation.\n");
printf ("The standard deviation used in this program is SD = ( (1/n) * sigma (xi - x)^2 )^1/2, where n is the number of values, xi is each array element, and x is the average value.\n");


input_data(arr, value);
s = k;
if (s >= 3)
{
print_array(arr, s);
sort(arr, s);
average = average_function(arr, s); /* not a function?*/
printf ("the average is %.1lf.\n", average);
if (s/2 == 0)
{
median1 = arr[s/2];
median2 = arr[(s/2)-1];
printf ("The two medians are %.1lf and %.1lf.\n", median1, median2);
}
else
{
median1 = arr[s/2];
printf ("the median is %.1lf.\n", median1);
}
mid = (sum_of_xi_x_squared(arr, average, s));
SD = sqrt(mid, 1/2);
printf ("The standard deviation is %.1lf.\n", SD); /*error here*/
}
else
{
printf ("Sorry error. Less than 3 values.\n");
printf ("End of program\n");
}
printf ("The last value is %.1lf.\n", value);
return 0;
}

int input_data(double arr[], double value)
{

int k, i;
k = 0;
i = 0;

printf ("Now please enter values: \n");
scanf ("%lf," &value); /*error here*/
while (value != -1)
{
arr[i] = value;
k = k + 1;
i = i + 1;
scanf ("%lf," &value); /*error here*/
}
return k;
}

void print_array(double arr[], int s)
{
int i, n;
i = 0;
n = 1;
while ((i < 6*n) && (i < s))
{
if (i == 6*n - 1)
{
n = n + 1;
printf ("\n");
}
printf ("%.1lf", arr[i]);
i = i + 1;
}
}

void sort(double arr[], int s)
{
int i, j;
double temp;

for (i = 0; i <s - 1 ; i = i + 1)
for (j = 0; j < s - 1 - i; j = j + 1)
if (arr[j] > arr[j+1])
{
temp = arr[j];
arr[j] = arr[j+1];
arr[j+1] = temp;
}
}

double average_function(double arr[], int s)
{
int i;
double sum, ave;
i = 0;
while (i < s)
{
sum = sum + arr[i];
i = i + 1;
}
ave = sum/s;
return ave;
}

double sum_of_xi_x_squared(double arr[], double average, int s)
{
int i;
double difference, sum;
i = 0;
while (i < s)
{
difference = arr[i] - average;
sum = (difference*difference) + sum;
i = i + 1;

}
return sum;
}


errors i see(please refer to /*error here*/ above :


warning C4020: 'sqrt' : too many actual parameters


'&' : illegal, left operand has type 'char [5]'
&' : illegal, right operand has type 'double '
'scanf' : too few actual parameters
( these three errors are one of the scanf error. and it repeats again for the second scanf, so i have 7 errors total)

how can i fix this?

==================================================
hold on! so i tried scanf ("%lf", value) without & in front of value,
but it still gives me errors !

syntax error : missing ')' before identifier 'value'
syntax error : ')'

but i dont see any ) missing anywhere

[weaknessforcats]

First, the sqrt() function has only one argument which is the number to take the square root of.

Second, when you scanf() into a double you do not need the &.

Third, your main() needs to return an int.

[JosAH]

Second, when you scanf() into a double you do not need the &.

Yes you do; the scanf function needs to know where it needs to store the number.

kind regards,

Jos

[weaknessforcats]

the scanf function needs to know where it needs to store the number.

So how come Visual Studio.NET 2008 compiles both without error?

double data = 0.0;
scanf("%lf", &data);
scanf("%lf", data);
 

[boxfish]

Both ways compile fine, but without the ampersand, I get an access violation.

[JosAH]

So how come Visual Studio.NET 2008 compiles both without error?

double data = 0.0;
scanf("%lf", &data);
scanf("%lf", data);
 

scanf() and (and its friends) is a varargs function; the compiler can't say anything
about the arguments following the format char* argument.

kind regards,

Jos

[donbock]

The other respondents are more intrepid than I. I didn't have the patience to try and figure out what is wrong with the OP's code. I humbly request that he put Code tags around his code so that, among other things, we see line numbers. I further request that the OP add line numbers to the provided compiler warnings (being careful to translate the line numbers so they correspond to those in the Code block). Those simple actions dramatically decrease the effort needed to find the problems.

[newb16]

You have comma inside string in scanf, like
..( "%lf, " value );
instead of
...("%lf", value );