I am still learning C programming and had a question regarding pointers. If I create a pointer in main and assign a value, then pass the pointer to another function, should it contain the same address? That is, if I print the address in main, then print the address in the second function, should the addresses be the same?
When you have a variable such as
and you pass it to a functionCode:int var = 1;
A copy of the value stored inside var is passed to function. In this case the integer 1.Code:function(var);
If instead you were to pass a pointer to var to function
A copy of the value stored inside ptr is passed to function. In this case the address of var.Code:int var = 1; int *ptr = &var; function(ptr);
So if inside main you had
and inside function you hadCode:printf("%p\n", &var); // could also use ptr
You would see they would print the same addresses which is to be expected.Code:printf("%p\n", ptr);
Hope this helps.
Last edited by wbpribs; 05-30-2021 at 06:58 PM.
What wbpribs said. But also, the address passed to your function is a copy of the address held in the pointer variable you created in main(). So if your function updates the pointer copy, the original pointer will not be changed.
Pointers are the core of C, where all the power lies, but they are also where most newbies come unstuck. Use pointers, don't steer clear of them. But learn what they are and how they're used, or they will bite you, hard. Unit-testing/TDD/TFD will probably help in trying things out, and confirming they work as you think they do/should.
Good luck with C - the old ones are still the best!
Be careful when printing with printf(), using the wrong type of variable for the format specifier can invoke undefined behavior. The "%p" specifier expects a void * so you need to cast your pointers to this type:
Code:int var = 1; int *ptr = &var; printf("%p\n", (void*)&var); printf("%p\n", (void*)ptr);
Thanks, I forgot about the cast.Originally Posted by jimblumberg
