Thread: foo

i am trying to run the following program and i keep getting a compiling error that says that foo is not a function. What am I doing wrong?

Code:

#include <stdio.h>
#include <stdlib.h>

double* makeAnArray(double size);
	
int main (void)
	{
	double *anArray;
	int size=0;
	int i;
	double *foo;

	printf("Please enter the size of the array  ");
	scanf("%d", &size);

	anArray=makeAnArray(size);

for (i = 0; i < size; i++)
	{
		printf("%.1f  ", anArray[i]);
	}
	printf("\n");

foo(anArray,size);

	free(anArray);

	getchar();
return 0;	
}
double* makeAnArray(double size){
	double* array;
	array =calloc(size,sizeof(double));
	return array;
}

void foo (double* a, double size){
	int i;
	for (i=0; i<size; i++){
		*(a+i)=132;
	}
}

i got the prototype down and i indented, and I am trying to figure out why it still tells me that foo is not a function?

Code:

#include <stdio.h>
#include <stdlib.h>

double* makeAnArray(double size);
void foo(double*a, double size);
int main (void)
{
	double *anArray;
	int size=0;
	int i;
	double *foo;

	printf("Please enter the size of the array  ");
	scanf("&#37;d", &size);

	anArray=makeAnArray(size);

	for (i = 0; i < size; i++)
	{
		printf("%.1f  ", anArray[i]);
	}
	printf("\n");

	foo(anArray,size);

	free(anArray);

	getchar();
	return 0;	
}
double* makeAnArray(double size){
	double* array;
	array =calloc(size,sizeof(double));
	return array;
}

void foo (double* a, double size){
	int i;
	for (i=0; i<size; i++){
		*(a+i)=132;
	}
}

You're defining a local variable "double *foo"

i am sorry to keep bothering you but i am very new to this whole programing thing.
do i use void instead of double*

delete this line. u prototyped foo already. change the prototype to double instead of void if u want a return type double.

Code:

	double *foo;