Why do my variables all have address 0 when I use "&"?
 

Ok, I am having a problem where I am attempting to obtain the address of some variables so I can pass them to a function using pointers. Here is part of my code:

float point_a[2], point_b[2], point_c[2], point_d[2];

point_a[0] = 1.5;
point_a[1] = 1.63;
point_b[0] = 0.5;
point_b[1] = 2.63;
point_c[0] = 7.0;
point_c[1] = 9.13;
point_d[0] = 8.0;
point_d[1] = 8.13;

printf("Original Coordinates: %f %f \n", &point_a[0], &point_a[1]);
printf("Original Coordinates: %f %f \n", &point_b[0], &point_b[1]);
printf("Original Coordinates: %f %f \n", &point_c[0], &point_c[1]);
printf("Original Coordinates: %f %f \n", &point_d[0], &point_d[1]);

When I run this code, printf will print out all zeros for the addresses corresponding to each point.

Does C++ not automatically assign addresses to the variables? And if so, how do I get the program to properly assign addresses.

Thanks.

I don't exactly know what you are trying to do. Why do you want to use the '&' operator? You can create a reference to another variable with it. So:


Look into how references work. However, if you're just trying to just print the value of the variable, then don't use the '&' operator.

The reason why I need the addresses of the variables to work is I have another, already working function, that takes the address of the first element of an array as an input, as opposed to simply taking the array as the input.

Specifically: pt_shift(&regressor[0],T,o,xy);

Where regressor is a two element array of float variables.

However, when I used my own arrays as opposed to the arrays the program was originally using(by using &point_a[0] in place of &regressor[0]), the function in question was returning nonsense. I nested printf statements within the function and found dereferenced values of the input within the function were completely different from those I passed to the function, even though this was before the function did anything to the input. So I checked to the addresses and found that before I passed the variable to the function, all the addresses were 0.

Also, I tried your code example:

int test;
int &testRef = test;

But it didn't work. The error is "missing ';' before '&'. Commenting out "int &testRef=test;" eliminates this error, so I know its not a ";" missing somewhere else in the code. I also get a "'Test Ref' undeclared identifier error. And several others.

Thanks for the help.

Maybe this little tutorial can some what help you with your problem...link

Drop the & in the printf statement. Your snippet doesn't call for the use of references (though other code might), so ignore that part for this snippet.

Familiarize yourself with pointers. Explaining errors in a couple of lines of code is pretty much wasted until you do. See the link regarding pointers in my signature. Pay some attention to "What is NOT a pointer". The "equivalence" of pointers and arrays is required by the standard and is not all pervasive.

Read the forum's rules/FAQ. The appropriate posting, including the use of code tags, is covered there, and is easier to understand than programming. It's also the polite thing to do.

See what DaWei says, above. It doesn't look like you're actually wanting to print addresses, so drop the address-of operator (&).
However, should you want to print addresses, you're still doing it wrong. Use the format specifier "%p" for pointers instead of "%f" (which is for floating-point numbers).

%p fixed the problem. Thanks for the help.