Can't bring array values to function

Question

I have prompted the user for 5 inputs in the main and saved them into an array called array. I've checked the array in main and it prints out the values as they were entered. However, when I pass it to a function which I've included below, I get the output that the array contains all 0 values. What am I doing wrong?

conversion(float array[], int size)
{
float add = 0.0;
float change, num;

printf("\nThe array is: \n");
for (i=0;i < size;i++)
{
    printf("%.2f\n",&array[i]);
}


/*calculate and store the conversion values in a new array*/
for(i=0; i<s; i++)
{
    num = array[i];
    change = (num*100.50);
    for(j=0; j<1; j++)
    {
        printf("\n %.2f your number is %.2f in float percent\n", &num, &change);
    }
}
}

printf("%.2f\n",&cArray[i]); Remove the &. With it you are passing the address of the variable, when you want to pass the value. — 001
– 001, Commented Feb 21, 2018 at 22:07
Thank you, that did let me view the right print so at least now I know that the array is being passed. However, I'm still getting multiple "0.00 in C is 0.00 in F", do you have any advice for that? — Amanda6276
– Amanda6276, Commented Feb 21, 2018 at 22:09
Thank you so much, that's fixed it. Sorry for the seemingly simple question. I'm only starting C after using Java and I'm having some issues with the print functions. I'm used to printing anything in some ++. And all this %d/%f/%c and & etc is just confusing still. — Amanda6276
– Amanda6276, Commented Feb 21, 2018 at 22:13

Stephen Docy · Accepted Answer · 2018-02-21 22:16:27Z

2

&cArray[i]

you don't need to address of the ith element in order to print it, you just need the ith element

Change

printf("%.2f\n",&cArray[i]);

printf("\n %.2f in Celsius is %.2f in Fahrenheit\n", &temp, &con);

to

printf("%.2f\n",cArray[i]);

printf("\n %.2f in Celsius is %.2f in Fahrenheit\n", temp, con);

The same with con, printf() doesn't need the address of a scalar variable in order to print it.

edited Feb 21, 2018 at 22:16

answered Feb 21, 2018 at 22:11

Stephen Docy

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RyanArr · Accepted Answer · 2018-02-21 22:37:21Z

0

Like Johnny Mopp said you have an extraneous &,

But depending on the size of the array you're dealing with you may just want to pass a pointer to the function instead.

edited Feb 21, 2018 at 22:37

answered Feb 21, 2018 at 22:11

RyanArr

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1 Comment

Stephen Docy Over a year ago

In C, arrays are always passed as pointers, they are not copied.

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