Is it possible to return a char datatype from main() in c?

Question

https://code.sololearn.com/c7UmZvFO4vNd/?ref=app
I tried returning a char datatype from main() to my function name() but it shows a lot of errors. Is returning char allowed in c or did I go wrong in this code?

Answer 1

It is recommended to use `int` for `main` function return type, details here https://en.cppreference.com/w/c/language/main_function

Your main function return type is `int`, not `char` ~ (code edited).

The errors came from incorrect parameter type for `name` function definition and forward declaration. Change type of parameter <t> from `char` to `char*` to get the code to run.

(Edited)

Answer 2

Ipang Oh thanks that worked!

But could you please explain why char* should be used instead of just char in the parameters?

And still I'm getting one warning saying main() return is not int but char, how can I rectify this?

Answer 3

Verstappen,
A string in C is `char` array, and variable <b> in `main` is a string (`char` array). When passing a string as argument for a function, `char*` is used.

But if the variable is just a `char` (not `char` array), then it is safe to define argument type as `char`.

Passing variable as argument to function (by using pointer) also means the changes to the argument will take direct effect, and reflects in function caller scope. So changes to argument <t> in `name` function will reflect in `main` function (caller).

BTW, last time I read your code the `main` function had `int` as return type, now it is changed to `char`. Just change the return type back to `int` to get rid of the warning.

Answer 4

Ipang thanks for this great explaination.

So as passing variable as argument (using pointer) to function will have direct effect on the caller, does that mean I don't have to return the value as the changes would have already occured in the main() ?

Answer 5

Verstappen,
I meant to say that any changes made to the argument <t> inside `name` function will reflect in actual variable <b> in `main` function, because they are one thing under different name (<b> in `main`, <t> in `name`).

As changes to arguments passed by pointer are directly effective, it's at your discretion whether or not to return the changed argument.

Answer 6

Ipang ohh okay got it, that was really helpful!! Thanks -))

Answer 7

Verstappen,
Look at this little snippet. Here in changes() <number> parameter is readable and writable, but in no_changes() <number> parameter is read-only, and will trigger error cause we try to change the data referred by the pointer.

#include <stdio.h>

void changes( int *number )
{
*number *= 2;
}

void no_changes( int const *number )
{
*number *= 2; // Error: can't change const argument
}

int main()
{
int n = 5;

changes( &n ); // works okay
printf( "after changes() n = %d\n", n );
no_changes( &n ); // compile error
printf( "after no_changes() n = %d\n", n ); // not displayed cause above error
return 0;
}

Answer 8

Ipang so we didn't use return here as we used pointer, as the pointer we used here directly changes the value of the address which is passed. Am I right on this?
As using return here is not even necessary .

And this pointer changes directly to the main(), is for all datatypes right? not only for int and char

Answer 9

Verstappen,
Pointers is a topic big enough to be written in a book, and I am also still not fully understanding it but the basic use cases. So here a link to an e-book that might explain it better than I could ever did with my limited knowledge.

http://www.iitk.ac.in/esc101/current/Lectures/ESc101Lec25_26pointers.pdf

Answer 10

Ipang ohh sure will check it 👍and yeah thanks a lot for helping me out !!