Thread: average problem

You should compile your code at a high warning level, then pay attention to the warnings. For example:

Code:

test.c: In function ‘main’:
test.c:16:9: warning: format ‘%f’ expects argument of type ‘double’, but argument 2 has type ‘float *’ [-Wformat=]
         printf("In %.2f tests your average is %.2f\n", &numberOftests, &average);
         ^

Basically, the compiler is saying that on this line:

Code:

printf("In %.2f tests your average is %.2f\n", &numberOftests, &average);

Instead of passing numberOftests and average as arguments, you passed &numberOftests and &average.

Originally Posted by Th0rh4memer

However i did not understand why it should have an"&".

The compile error indicates that it should not have a "&".

Originally Posted by Th0rh4memer

I learned that every time that its not an array i should put an "&", in this case it is a a float not an array thats why i wrote it. Can you explain when and why should i put "&"?

You need to understand a few things:

• Given an object x, &x results in a pointer to x, i.e., in this context, & is the address operator.
• When an array y is passed as an argument, it is converted to a pointer to y's first element.
• To print an object x with printf, you need to have a format specifier that matches the type of x, and then pass x as the corresponding argument.
• To read into an object x with scanf, you need to have a format specifier that matches the type of x, and then pass a pointer to x as the corresponding argument.

So, what you had was a printf call. Since you are not supposed to pass pointers to the objects as the corresponding arguments, passing &numberOftests and &average is wrong. If you had a scanf call instead, then they would be correct, except that then the format specifiers should have been %lf instead of %f.

However, if you had an array of char named str, and if you want to read a string into str with scanf, you would not pass &str, but rather you would still pass str since str would be converted to a pointer to its first element.

Originally Posted by Th0rh4memer

i did not understand it :/ can u try to say in other words please?

Do you understand pointers?

If not, then blindly follow these rules of thumb for now:

• When using printf, do not use &.
• When using scanf with %s to read a string, do not use &.
• Otherwise, when using scanf, use &.