func return 2 values

[titosd]

Hello,

I have a question, How I can write a function that return 2 values.
the function should take 3 numbers and do between the first and the second number an arethmetic operation and check if the result will be the third number, if yes the function return 1 and return the character of the arethmetic operation.
for example: the function take the numbers 3 5 15, the function will check the three numbers, and will fint that 3*5= 15, so it will return 1 and return *.

I hope that you understand me, and I hope that you can help me with the first thread :)

[grumpier]

A common way of return multiple values is by returning a struct which has multiple members, or for the function to accept a pointer that receives additional data from the function.

You don't actually need to return multiple values though: return 0 if there is no operation, '*' if multiplication is the operation, '+' if addition, etc etc. '*', '+', '-', and '/' are non-zero characters with all character sets that I know of.

[titosd]

thank you grampier
but by the question that i have I should reurn 2 values if there is operation return 1 and the character, I asked somebody the all tell me that I should use the pointers but anyone of them know how should write the function.

[minas1]

You can't return two values from a functio using the return statement. Use a struct as you've been told to above:

typedef struct
{
int x, y;
} P;

P getPoint()
{
int a, b;
//a = .... b = .....
P temp;
temp.x = a;
temp.y = b;
return temp;
}

[titosd]

thanks minas
but I want use the poiner, I sure that I can use the pointer, but I didn't know how to use it !! :(
is anyone know ?

[grumpier]

You need to put some effort into understanding the answers you receive, rather than giving up if you're not given code examples.

The approach with using pointers is;

SomeTypeA Function( <arguments>,   SomeTypeB *x)
{
       *x = whatever_data_needed_other_than_return_value();
       return return_value();
}

The "SomeTypeB *x" argument is what I was referring to when I said "or the function to accept a pointer that receives additional data from the function".

[minas1]

So 1 has to be returned for you to know that the result is the third number. You can do it with only one return value:

char arithmeticOperation(int a, int b, int c)
{
	// +
	if(a + b == c) return '+';

	// -
	if(a - b == c) return '-';

	//you also do this if the order you pass the numbers isn't critical
	if(b - a == c) return '-';

	// *
	if(a * b == c) return '*';

	// /
	if(a / b == c) return '/';
	if(b / a == c) return '/';

	// %
	if(a % b == c) return '%';
	if(b % a == c) return '%';

	//if nothing matches
	return ' '; // return a space
}

int main(void)
{
	char c = arithmeticOperation(4, 2, 0);

	if(c == ' ') // check if it's a space which means there isn't an operation
		printf("No arithmetic operation found.\n");
	else
		printf(%c, "Operation ", c, " found.\n");
}

[titosd]

thank you grumpier
but I didn't understand the example, if you can give me actual example plz.

minas thank you, but I said that we should write the function with pointers, and te function should return 2 values (the character and the number 1).

I can write it by the structer but we should write it just with pointers !!!

[minas1]

Do you want the function to take pointers as parameters or return a pointer or both?
And why do you want to return 1?

[grumpier]

but I didn't understand the example, if you can give me actual example plz.

I can, but I won't. As horrifying as the concept clearly is, you need to apply some effort to understand the answers you've been given.

[minas1]

Ok, let's go step by step.
We will create a struct that has two members, a char and an int.
Go on and declare the struct.

[titosd]

grumpier
I didn't ask you to write the func that i want, but I want another example but with parameters to understand the way.
minas I have write the func with structer

#include<stdio.h>
struct mytype
{
int ok;
char ch;
};
 
struct mytype fundCalculation(int a,int b,int c)
{
struct mytype tmp;
if(c==a+b)  {tmp.ok=1;tmp.ch='+';}
else if(c==a-b) {tmp.ok=1;tmp.ch='-';}
else if(c==a*b) {tmp.ok=1;tmp.ch='*';}
else if(c==a/b) {tmp.ok=1;tmp.ch='/';}
else {tmp.ok=0;tmp.ch=' ';}
 
return tmp;
 
}
main()
{
 
 
printf("%i %c ",fundCalculation(3,4,12).ok,fundCalculation(3,4,12).ch);
 
getchar();      
}

[minas1]

Well it works, good job!

Some tips:
Use this form when declaring a struct.

typedef struct
{
int ok;
char ch;
} mytype;

So you don't have to prefix mytype with struct all the time. Now you can do this:

mytype fundCalculation(int a,int b,int c)
{
  /* ....... */
}

[titosd]

yes I know that it works, but in the question we should write the func just with pointers !!!

[minas1]

Then instead of ints in the argument list, use pointers to ints.

a pointer to int: int *a = 0;

when you call the function, you need to pass addresses.

so

int x = 4;
callSomeFuncTakingPointers(&x);

when you need to access a value stored in a pointer, you the * operator.
int x = 3;
int *p = &x;
printf(%d, *p);

If you want to learn more about pointers,
http://www.cplusplus.com/doc/tutorial/pointers.html

[titosd]

like this ?

int findCalculation(int *a, int *b, int *c)
{
	char ch;
	if (*a - *b == *c)
		return 1;
	else return 0;

	if (*a + *b == *c)
		return 1;
	else return 0;

	if (*a * *b == *c)
		return 1;
	else return 0;

	if (*a / *b == *c)
		return 1;
	else return 0;

}

[minas1]

Yes, but use our function that returns the struct.

[titosd]

How can I do this :( ?!!

[minas1]

#include<stdio.h>
struct mytype
{
int ok;
char ch;
};
 
struct mytype fundCalculation(int a,int b,int c)
{
struct mytype tmp;
if(c==a+b)  {tmp.ok=1;tmp.ch='+';}
else if(c==a-b) {tmp.ok=1;tmp.ch='-';}
else if(c==a*b) {tmp.ok=1;tmp.ch='*';}
else if(c==a/b) {tmp.ok=1;tmp.ch='/';}
else {tmp.ok=0;tmp.ch=' ';}
 
return tmp;
 
}
main()
{
 
 
printf("%i %c ",fundCalculation(3,4,12).ok,fundCalculation(3,4,12).ch);
 
getchar();      
}

Instead of passing ints, pass pointers to ints. you did it above; try it and show us, even if it doesn't compiles.

[titosd]

like this ??

#include <stdio.h>
struct mytype
{
int ok;
char ch;
} mytype;
 
struct mytype fundCalculation(int *a,int *b,int *c)
{
struct mytype tmp;
if(*c==*a+*b)  {tmp.ok=1;tmp.ch='+';}
else if(*c==*a-*b) {tmp.ok=1;tmp.ch='-';}
else if(*c==*a * *b) {tmp.ok=1;tmp.ch='*';}
else if(*c==*a / *b) {tmp.ok=1;tmp.ch='/';}
else {tmp.ok=0;tmp.ch=' ';}
 
return tmp;
 
}
main()
{
 
 
printf("%i %c ",fundCalculation(3,4,12).ok,fundCalculation(3,4,12).ch);
 
getchar();      
}

[minas1]

The function is correct, but you don't call it correctly.
Create 3 variables in main x, y, z, give them values and pass their address to the function.

[titosd]

like this ??

if yes, but we use the structer, and we should just use in the function the pointers without anything else ....

this is the code:

#include <stdio.h>
struct mytype
{
int ok;
char ch;
} mytype;
 
struct mytype fundCalculation(int *a,int *b,int *c)
{
struct mytype tmp;
if(*c==*a+*b)  {tmp.ok=1;tmp.ch='+';}
else if(*c==*a-*b) {tmp.ok=1;tmp.ch='-';}
else if(*c==*a * *b) {tmp.ok=1;tmp.ch='*';}
else if(*c==*a / *b) {tmp.ok=1;tmp.ch='/';}
else {tmp.ok=0;tmp.ch=' ';}
 
return tmp;
 
}
main()
{
 
int x=3;
int y=4;
int z=12;

printf("%i %c ",fundCalculation(&x,&y,&z).ok,fundCalculation(&x,&y,&z).ch);
 
getchar();      
}

[minas1]

Yes.

In main, it's a better idea to actually create an oject
int main()
{
struct mytype mt = fundCalculation(....)

[titosd]

thank you very much minas1 :)

thanks again for helping and tips :icon_wink:

[minas1]

Nothing. Now show us the complete program. also use the typedef for the struct like I told you.

[titosd]

Ok, I will.
but the program is very long. contain some function, when I finish it I will put it here :)

[death_oclock]

I think what you want to do is just return the one or zero rather than a struct. The operation can be passed as a pointer to a char (in the parameters). a, b, and c don't need to be pointers.