c++ cannot convert 'double' to 'double*' for argument 1 to 'void sort(double*,int)' error

Question

I'm a student listening to c programming lesson, and I'm using c++ to 'call-by-reference'. I don't know how to use c++ exactly, so I use c and save it into .cpp file. Anyway, I used a function to sort an array, and I now have an error. What should I do to solve this error?

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

double round(double value);
void sort(double a[],int cnt);
void swap(double& x,double& y);

int main()
{
    int i;
    double array[3];
    for(i=0;i<3;i++){
    scanf("%lf",&array[i]);
    }
    sort(array[3],3);
    printf("%d %d %d",ceil(array[0]),floor(array[2]),round(array[1]));
    return 0;
}

double round(double value)
{
    return floor(value+0.5);
}

void sort(double a[],int cnt)
{
    int i,j;
    for(i=0;i<cnt-1;i++){
        for(j=i+1;j<cnt;j++){
                if(a[i]<a[j]){
                    swap(a[i],a[j]);
                }
            }
        }
}

void swap(double& x,double& y)
{
    int imsi=x;
    x=y;
    y=imsi;
}

Answer 1

Your sort(double a[], int cnt) function takes as first argument double a[] , which is syntactic sugar for a pointer double* to the first element of an array. However in main() you invoke it as

sort(array[3], 3); // you pass a double here, not a double*

In the call above you pass the 4-th element of the array array , ie a double , not a pointer double* . To pass a pointer to the first element of the array, replace the above call with

sort(array, 3); // you now pass a double* (i.e. pointer to first element of the array)

The compiler is basically tell you what's wrong:

error: cannot convert 'double' to 'double*' for argument '1' to 'void sort(double*, int)' sort(array[3],3);

It expects a double* but you pass a double . It attempts to convert double to double* , but such conversion is impossible, hence the error.

Answer 2

In addition to your main concern (already answered by another user), you should probably change your swap() function so that you aren't using an int to store a double . This could result in unexpected behavior.

Answer 3

Next problem which you'll facing with:

printf("%d %d %d",ceil(array[0]),floor(array[2]),round(array[1]));

for print float you may use "%f" .

Simply to use cin and cout from library iostream instead of scanf and printf

for(i=0;i<3;i++) {
    std::cin>>array[i]>>' ';
}

for(i=0;i<3;i++) {
    std::cout>>array[i]>>' ';
}

btw http://www.cplusplus.com/reference/ could be useful for you on many reasons.