CODEWITHSUNDEEP
carraysfunction
Amanda6276
Amanda6276

Reputation: 33

Can't bring array values to function

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);
    }
}
}

Upvotes: 2

Views: 42

Answers (2)

RyanArr
RyanArr

Reputation: 93

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.

Upvotes: 0

Stephen Docy
Stephen Docy

Reputation: 4788

&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.

Upvotes: 2

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