Hello, I am getting up to speed with C during the week and I have written one of my first programs and I do not understand why the return of the function. If I input X as 3, Y as 4 and N as 6 I get the return as several million. I should be getting the return of check funtion as 1 with those inputs.

#include <stdio.h>

int check(int x,int y,int n);

int main(void)
{
int x,y,n;
printf("\nValue of X: ");
scanf("%d", &x);

printf("\nValue of Y: ");
scanf("%d", &y);

printf("\nValue of N: ");
scanf("%d", &n);

check(x,y,n);
printf("check = %d", check);
if (check == 1){
printf("\vx and y checked out.");
}
else if (check == 0){
printf("Digits failed the check");
}
}

int check(int x,int y,int n)
{
int i=0;
if (x>0 && x<(n-1)){
if (y>0 && y<(n-1)){
return 1;
}
}

}

Because if your IFs fail, you don't return anything at all.

Because if your IFs fail, you don't return anything at all.

Hi Kotik..

I think it's because you call check function directly without put it into variabel.
Please try this code , maybe it help you.

#include <stdio.h>

int main(void)
{
int x,y,n;
int z;
printf("\nValue of X: ");
scanf("%d", &x);

printf("\nValue of Y: ");
scanf("%d", &y);

printf("\nValue of N: ");
scanf("%d", &n);

z = check(x,y,n);
printf("check = %d", z);
if (z == 1){
printf("\vx and y checked out.");
}
else if (check == 0){
printf("Digits failed the check");
}
}

int check(int x,int y,int n)
{
int i=0;
if (x>0 && x<(n-1)){
if (y>0 && y<(n-1)){
return 1;
}
}

}

: Hello, I am getting up to speed with C during the week and I have
: written one of my first programs and I do not understand why the
: return of the function. If I input X as 3, Y as 4 and N as 6 I get
: the return as several million. I should be getting the return of
: check funtion as 1 with those inputs.
:
: #include <stdio.h>
:
: int check(int x,int y,int n);
:
: int main(void)
: {
: int x,y,n;
: printf("\nValue of X: ");
: scanf("%d", &x);
:
: printf("\nValue of Y: ");
: scanf("%d", &y);
:
: printf("\nValue of N: ");
: scanf("%d", &n);
:
: check(x,y,n);
: printf("check = %d", check);
: if (check == 1){
: printf("\vx and y checked out.");
: }
: else if (check == 0){
: printf("Digits failed the check");
: }
: }
:
: int check(int x,int y,int n)
: {
: int i=0;
: if (x>0 && x<(n-1)){
: if (y>0 && y<(n-1)){
: return 1;
: }
: }
:
: }
:
:
Hi Kotik..

I think it's because you call check function directly without put it into variabel.
Please try this code , maybe it help you.

#include <stdio.h>

int main(void)
{
int x,y,n;
int z;
printf("\nValue of X: ");
scanf("%d", &x);

printf("\nValue of Y: ");
scanf("%d", &y);

printf("\nValue of N: ");
scanf("%d", &n);

z = check(x,y,n);
printf("check = %d", z);
if (z == 1){
printf("\vx and y checked out.");
}
else if (check == 0){
printf("Digits failed the check");
}
}

int check(int x,int y,int n)
{
int i=0;
if (x>0 && x<(n-1)){
if (y>0 && y<(n-1)){
return 1;
}
}

}

you are getting such results in printf statement because you have written your statement as

printf("check = %d", check);

What this will do is print the address where your function is present in the memory. But if your statement as

printf("check = %d", check());

which I thing you intended for.

One more thing you should rewrite your check method/function, as if conditions are not satisfied you will end up in returning a garbage value to your return statement. For beginners it is prefered to code in the way our other fellow programmers have modified your code.