I am trying to call a function from the main function, but it gave me a random number when I compiled and ran it. I want to ask from the user to enter a number in the yrBorn function, then return it to the main function and finally display it in the terminal.
int yrBorn(int);
#include<stdio.h>
int yrBorn (int yearBorn)
{
printf("Enter the year you were born in: ");
scanf("%d", &yearBorn);
return yearBorn;
}
int main ()
{
printf("%d",yrBorn);
}
You have to understand your code like this :
int yrBorn(int); // function prototype say it must accept a int argument
#include<stdio.h>
int yrBorn (int yearBorn) //it must accept a argument of int type because of this definition
{
printf("Enter the year you were born in: ");
scanf("%d", &yearBorn);
return yearBorn;
}
int main ()
{
printf("%d",yrBorn); // You are passing function pointer here not a function call use like yrBorn(1994)
printf("%d",yrBorn(1994)); //it will work :)
}
#include<stdio.h>
int yrBorn();
int yrBorn()
{
int yearBorn = 0;
printf("Enter the year you were born in: ");
scanf_s("%d", &yearBorn);
return yearBorn;
}
int main()
{
printf("%d", yrBorn());
}
Modified your program to work as you expect!
Your Syntax is not correct.
Suppose For Example , I want to make a Function That takes a number and returns it square , I would go like ,
#include<stdio.h>
int square(int); //Function Prototype Declaration Telling Compiler
//That there would be function named square whose
//takes an integer and whose return type is integer.
int main()
{
int x = 4;
int y ;
y = square(x); //Calling Square Function passing Value `x`
//And Storing it in `y`.
printf("%d\n" , y); //Printing `y`
//Or You Could As Well Do This
printf("%d" , square(x)); //Without Storing Printing return
//value of square(x)
return 0;
}
int square(int k){ //Function Definition Contains The Code What The
int r = k*k; //Function Does , Here it take the value passed
return r; //to the function in this case `x` and store it
//in `k` , and then initialise `r` to be `k*k`
} //and then returns `
So in your code
int yrBorn(int);
#include<stdio.h>
int yrBorn (int yearBorn)
{
printf("Enter the year you were born in: ");
scanf("%d", &yearBorn);
return yearBorn;
}
int main ()
{
printf("%d",yrBorn); //Here You are not calling function.
printf("%d" , yrBorn(0)); //This will work as in your prototype
//declaration you have declared
//function to take a int parameter and
//return int
}
What You Would Wanted I Suppose Was this
#include<stdio.h>
int yrBorn(void);
int yrBorn (void) //Here Function is Expected to take no arguments
//and return int value.
{
int yearBorn;
printf("Enter Your Birth Year : ");
scanf("%d" , &yearBorn);
return yearBorn;
}
int main(){
printf("%d" , yrBorn()); //You Would Call it like this
}
Notice there is nothing between parenthesis it is because function does not accept any arguments.
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printf("%d",yrBorn);
This is not how a function is called. What you're doing instead of calling yrBorn is taking its address. That address is then being interpreted as an int and printed as such.
To call a function, you need to use the () operator. Any parameters to the function would go inside the parenthesis.
Also, your function is taking a parameter, but isn't doing anything with its value. Rather than take a parameter, this variable should be local to the function.
So with these fixes, your function would look like this:
int yrBorn ()
{
int yearBorn; // local instead of a parameter
printf("Enter the year you were born in: ");
scanf("%d", &yearBorn);
return yearBorn;
}
And you call it like this:
printf("%d",yrBorn());
