Thread: Creating an scanf function inside an function

Hello,

I've an problem whit understanding why i can't make an scanf() work inside an custom function.

This is the programing:

PHP Code:

int main(int argc, char** argv) {

    int scan();
    return (EXIT_SUCCESS);
}

int scan(){
    int number;
    scanf("%d", &number);
    printf("%d\n", number);
} 


As you can see i've made main call the scan function which should promote the user to type in but it just skips it. The program works if i just make the scanf inside of main but not when it's called via an function.

How should i do in order to get this to work?

Thank you for your time.

Code:

    int scan();

is not a function call. It's called function prototype.
You call a function by its name and arguments if there is any.

like

Code:

  scan();
  x = sin( 3.0 );
  printf(" sin(%f) = %f\n",3.0, x);

You are already calling scanf() and printf().....

Originally Posted by anserudd

Hello,

I've an problem whit understanding why i can't make an scanf() work inside an custom function.

A few problems:

1) Since you are not using function prototypes you need to place your scan() function above main. This is so that by the time the compiler reads down to main() it already knows about scan().

2) when calling a function just use it's name. That is if you've defined int scan() then you call it simply by typing scan(); ... don't repeat the entire declaration. (hint: you are already doing this with scanf() and printf() just fine.)

3) the correct declaration for a function that does not take any input values is ... function(void) ... not function() ... the latter says "accepts an unknown number of parameters" and that's not what you are doing.

4) the correct declaration for a function that does not return a value is ... void function() ... if you declare it as int function(void), the compiler expects a returned value, which you aren't providing.

5) Since you aren't (yet) making any use of command line arguments the correct form for your main function is int main (void) ... that may change if you add stuff to your program.

Soooo...

Code:

void scan( void ){
    int number;
    scanf("%d", &number);
    printf("%d\n", number);
}

int main( void ) {

    scan();
    return (EXIT_SUCCESS);
}

When you call a function you have to supply a number of things (taking sin as an example):

Code:

double sin ( double arg ); /* sin's prototype */

/* function call a.k.a. function invocation */
double x;
x = sin( 3.0 );

Color coordinated for your convenience. Your scan() function would be similar, it is only the type and the number of arguments that are different. Once you understand how to call a function, you'll see problems with how you invoke scan. Currently, you specify it will return an int, but there is no place for an int to be stored where you call scan.