c - square roots with arrays

[Trezkez]

I know it's dumb of me to ask, but how do I make a program that calculates the square of the elements of an array of 10 numbers?

[helios]

1. Do you know how to get the square of a number?
2. Do you know how to apply a function to each element of an array?

[Trezkez]

This is what I did.

#include <stdio.h>  
#include <math.h>    
 
int main()
{
	int i;
	float num[10], square;
	
	printf("Enter 10 numbers to start calculating the square root: \n");
	
	for(i=0; i<10 ;i++)
	{
		scanf("%f", &num[i]);
		
		square=sqrt(num[i]);
		
		printf("%4.2f \n", square);
		printf("\n");
	}	
}

[tpb]

Those are square roots, not squares.
And if that's all you need to do then you don't need an array.

[Trezkez]

How do I do it with squares then?

[jonnin]

c++ has a pow() function but it is very heavy because it supports floating point powers, so it has to do a bunch of things that are totally unnecessary for simple squares.

if you wanted x to the pi power, use pow().
if yuo want x squared, a good old fashioned x*x is all you need.

square = num[i] * num[i]; in your code, then.

There seems to be some confusion.
your code's text says:
printf("Enter 10 numbers to start calculating the square root: \n");

but you ask about squares.
Be sure you know what it is you actually want to do here.

[helios]

The real question is, do you need the square, or the square root? Because it feels like you're using both terms interchangeably.

[Trezkez]

Well originally this is translated straight from spanish. Though, we asked our prof. if it was square or square root. He said square root. I'm not sure anymore.

[Ganado]

Then just do square root (raíz cuadrada), sqrt(). Your variable name (square/cuadrada) is just misleading, in that case.