CODEWITHSUNDEEP
cfunction
Devin
Devin

Reputation: 23

How can I solve this simple C function?

I am taking a hybrid class of "Algorithms, architecture, and assembly language" at my local community college. Although it's an intro class and mainly focuses on how computers turn code into binary, we have some assignments for C code. Some of the stuff we've never even gone over in class, and I'm lost.

The instructions read:

Write a function named DoSomething, with no return value, it should do just one (1) thing: multiply the global variable my_result by 5 and store the result back into my_result

You do not have to call DoSomething, or initialize my_result, I will do that.

I have tried

int my_result;

dosomething (my_result) { 

    my_result = my_result * 5;

}

but this is incorrect. I have almost zero experience with C language and am stuck. I'm not asking anyone to do my homework, I just want to understand. This has not been covered in class.

Upvotes: 1

Views: 128

Answers (4)

Pablo
Pablo

Reputation: 13590

A correct function declaration must have a type, the name of the function and its arguments.

type function_name(type1 arg1, type2 arg2, type3 arg3, ...);

If a function does not return anything, then type must be void

void function_name(type1 arg1, type2 arg2, type3 arg3, ...);

If a function does not take any parameter, then the you can use void instead of the list of arguements:

type function_name(void);

Your dosomething function is missing the return type, which should be void (assignment says Write a function named DoSomething, with no return value) and it takes no arguments (at least the assignment does not specify any), so the correct prototype of the function must be:

void DoSomething(void);

So the correct program

#include <stdio.h>

int my_result;

void DoSomething(void)
{
    my_result = my_result * 5;
}

int main(void)
{
    my_result = 6; // initializing global variable

    DoSomething();

    printf("my_result: %d\n", my_result);

    return 0;
}

which will print my_result: 30.

Upvotes: 3

Eric Postpischil
Eric Postpischil

Reputation: 223632

The instructions read:

  • Write a function named DoSomething, with no return value,…

So code for that is:

void DoSomething(void)
  • … it should do just one (1) thing: Multiply the global variable my_result by 5 and store the result back into my_result.

And code for that is:

{
    my_result = my_result * 5;
}
  • You do not have to call DoSomething, or initialize my_result, I will do that.

Done. The total code requested is:

void DoSomething(void)
{
    my_result = my_result * 5;
}

You have indicated in your comments that you are submitting this code into some sort of automatic grading/checking software. So that software is designed to accept code that matches the assignment, no more and no less. Quite likely, it puts your code into a file and that compiles a source file that includes the former file with #include. That source file defines my_result and main, so your code should not, or it will cause compilation and/or link errors. You need to submit just the code requested in the instructions.

Notes about your code:

  • The instructions say to write a routine named DoSomething. You used dosomething. These are different in C; the case matters.
  • You declared the routine without specifying a return type. The instructions say the instruction has no return value, but that does not mean you should just omit any return type. You should explicitly say there is no return value by using void for the return type of the function, as in void DoSomething(…). (For historic reasons, if you omit the return type in a function declaration, it defaults to int. Letting the type default like that is old syntax and should be avoided.)
  • The instructions did not say whether the routine should take parameters. This is a shortcoming in the instructions. However, dosomething (my_result) is incorrect for two reasons. One, my_result is described as a global variable, not a parameter. Two, it is the wrong syntax for a parameter declaration. A parameter declaration must have a type, as in dosomething(int x). Since the routine needs no parameters, a proper declaration is void DoSomething(void). (Although there is some possibility the instructor intended void DoSomething(), but that would not generally be preferred.)

Upvotes: 1

Jorvis
Jorvis

Reputation: 386

There is nothing to get worried about. Relax, it's just part of the process in becoming a good developer.

To solve such problems, first note what the function expects and we want from the function to return.

  • Now, in your question, it is given that function would return nothing. So the return type of the function would be void.

  • Now, since we have to use a global variable, it means function expects no argument.

Hence, our code is :

#include <stdio.h>

int my_result;    // Our Global Variable

void doSomething (void)    // Our Function
{
    my_result = my_result * 5;
}

int main()
{
        /* Asking the value of my_result */
    printf("Please enter a value : ");
    scanf("%d", my_result);

    doSomething();

    printf("\nNew value of my_result is : %d\n", my_result);

    return 0;
}
// End of main.

Upvotes: 1

Stephen Docy
Stephen Docy

Reputation: 4788

You almost have it. Since my_result is global, you do not need to pass it into the function, it is accessible everywhere. Oh, and every function needs its return value specified. Use void to specify that there is no return value and no parameters.

int my_result;

void dosomething (void) { 

    my_result = my_result * 5;
}

Upvotes: 3

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