printf giving me incorrect output in C

Question

This is probably a very elementary problem, but I cannot find the answer anywhere, and this is the first time I\'ve had the problem after several weeks of programming in C. In e

Answer 1

The calling conventions are different in printf, and scanf.

Printf's arguments are typically by value (no &), but scanf has to have pointers. (&)
(You can't put a new value to e.g. 1.0, but you can print it...)

Answer 2

You have to do:

printf("size is %d", size);

instead. This prints the value of the int object size.

But

 printf("size is %d", &size);

is undefined behavior.

Answer 3

You are printing the address of size, i.e., &size.

Just pass a plain size.

Answer 4

Try

printf("size is %d", size);

& gives you the memory location (address) of a variable.

printf("size is %d", &size);

So, the above will print the memory location(address) of size, not the value stored in size.

Answer 5

Remove the & in the printf statement as &size --> prints the address.

      printf("size is %d", size);