Simple IF statement scanf

Question

For an assignment I was asked to create a small program which asks for the users input which determines the converter they wish to operate. My question is why doesn't the program ask for the users input, AFTER they have entered which convertor they wish to use (1 or 2). Instead of calling scanf, it just runs the entire statement in one go.

#include <stdio.h>

int main()
{
float cm;
float inches;
int operation;

printf("Hello and welcome to the Inches to Centimetres converter. \n");
printf("Choose from the options below by entering the corresponding number:\n");
printf("Inches to CM converter (1)\n");
printf("CM to Inches converter (2)\n");
scanf("&d", &operation);

if (operation == 1)
{
    printf("Please enter the amount of Inches you wish to convert : \n");
    scanf("%f", &inches);

    cm = inches * 2.54;


    if (inches <= 0)
        {
        printf("Invalid number");
        }
    else
        printf("%f inches is equal to %f centimetres.", inches, cm);
}
else if (operation == 2);
{
    printf("Please enter the amount of Centimetres you wish to convert : ");
    scanf("%f", &cm);

    inches = cm / 2.54;


    if (cm <= 0)
        {
        printf("Invalid number");
        }
    else
        printf("%f centimetres is equal to %f inches.", cm, inches);
}


}

Output

Answer 1

Two problems here. First:

scanf("&d", &operation);

There's a typo, "&d" should be "%d", and it's why you get prompted twice right away. You want:

scanf("%d", &operation);

Second is this:

}
else if (operation == 2);
{

The ; immediately ends the else block. So the block in braces will always run. Get rid of the ;

}
else if (operation == 2)
{

Better yet:

} else if (operation == 2) {

Formatting your braces this way will practically eliminate this type of error.