how to return an array from the function

**johny10151981** · Dec 16 '10, 10:40 AM

with your current development can only print the first data in INT array.

to print all you will have to follow loop

**Siddarth777** · Dec 16 '10, 10:49 AM

thanks a lot for ur reply.......... ..........
but the problem is,it is not even printing the first value

**johny10151981** · Dec 16 '10, 12:54 PM

i tested your program in my machine. it print the first value.

well try this way. return ->

Code:

return &a[0];

**manontheedge** · Dec 16 '10, 01:02 PM

this code below will do what you're trying to do. I ran it in a C++ compiler just to make sure it was working. I tried to comment to help you out. Take a look, hope this helps ...

Code:

#include<iostream>
#include<stdio.h>

//int arr(int *n)
void arr(int a[], int size) // need to pass an array AND the size of it
{
	//int a[20]; don't need this
	printf("ENTER THE ELEMENTS OF ARRAY:");
	for( int i=0; i<size; i++ )
	//scanf("%d",a[i]);
	std::cin >> a[i];
	//return a; // you can't return an array, must use passing by value
}
void main()
{
//	int a[50]; // it's a good idea not to hard code the size of an array if you can avoid it
//	int i;	// not being used
	int size = 0;
//	int *val; // don't need this
	printf("ENTER THE SIZE OF THE ARRAY:");
	//scanf("%d",size);
	std::cin >> size; 

	int* a = new int[size]; // here, a new, DYNAMICALLY allocated array is created

	//val=arr(&s); 
	// instead of returning, you pass the array BY VALUE, along with the array's SIZE
	//	this allows the array 'a' to be changed in another function and everyone can
	//	see the changes
	arr(a, size);

	fflush(stdout);
	
	for( int i = 0; i < size; i++ ) // added this to print ALL of the values
	{
		printf("THE RETURNED ARRAY IS:%d",a[i]);
	}

	int x;
	std::cin >> x;
}

**donbock** · Dec 16 '10, 03:31 PM

Referring to your original source code.

Use square brackets (not angle brackets) around your CODE tags.
Function arr fills local array a[20]. It would very bad if arr were called with a parameter value greater than 20. This function should protect itself.
By the way, what's the benefit of using call-by-reference instead of call-by-value in function arr? There's no harm in doing it your way, although you might want to make the argument a pointer-to-const to guarantee arr can't change the caller's parameter value.
The definition of function arr says that it returns an integer, but you return an array of int, which actually returns the address of the first element of the array (that is, a pointer-to-int). This is the source of the first warning.
Function arr returns the address of local array a. Local variables cease to exist once their enclosing function returns. This means that main is referencing ghost variables if it tries to dereference the returned pointer. That is undefined behavior -- there is no way to predict what will happen. This is the source of the second warning.
You defined the main function without specifying its return type. Without an explicit return type, the compiler assumes the function returns an int. This is the source of the third warning. By the way, the return type mandated for main by the Standard is int.
Function main assigns the return value of function arr to variable val. Function arr is defined to return an int, but variable val is a pointer-to-int. This discrepancy is the source of the fourth warning.
Function main defines local variables a[50] and i, but never uses them. This is the source of the fifth and sixth warnings.
Function main has [implicit] return type int, but execution falls through to the bottom of the function without encountering a return statement. What int value should it return? This is the source of the seventh warning.

**Siddarth777** · Dec 18 '10, 07:04 AM

thanks a lot everyone for your valuable replies
especially DONBOCK thanks a lot for your explanation of the warnings

but it did not help
am unable to get the output, i.e to print the complete array
am not getting any errors or warnings

this is the code

Code:

#include<stdio.h>
 int* arr(int *n)
{
int a[20],i,*add;	//'add' is for storing the address of the first location in the array
printf("ENTER THE ELEMENTS OF ARRAY:");
for(i=0;i<*n;i++)
scanf("%d",&a[i]);
add= &a[0];
return add;	//returning the address of the 'a[0]'
}
int main()
{
int s,i;
int *val;
printf("ENTER THE SIZE OF THE ARRAY:");
scanf("%d",&s);
val=arr(&s);
fflush(stdout);
for(i=0;i<s;i++)
printf("THE RETURNED ARRAY IS:%d",*val);	//printing the  complete array
return 0;
}

please help
thanks a lot in advance

**donbock** · Dec 19 '10, 01:28 AM

Refer to item #5 in my earlier post. You haven't fixed this yet. Note what I said: "there is no way to predict what will happen". You must fix this.
Refer to the loop in lines 19 and 20 of your new source code. This loop prints the first entry in the array over and over again.

**Siddarth777** · Dec 21 '10, 02:26 PM

Thanks a lot for all,who helped me in my code development
i have figured a way to return the complete array it is working

heres the code i am posting

any enhancements needed?
please let me know

here is my code:

Code:

#include<stdio.h>

//#include<conio.h>

int siddu(int *x)

{

int i;

int a[20];

//printf("enter the size of array:");

//scanf("%d",x);

fflush(stdin);

printf("ENTER THE ARRAY ELEMENTS:");

for(i=0;i<*x;i++){

scanf("%d",&a[i]);

//for(i=0;i<*x;i++)

return a[i];

}

}

int siddu2(int *x)

{

int a[20],i;

for(i=0;i<*x;i++){

scanf("%d",&a[i]);

//for(i=0;i<*x;i++)

return a[i];

}

}

void main()

{

int n,j,ret[20];

//clrscr();

printf("enter the no of sides:");

scanf("%d",&n);

for(j=0;j<n;j++){

if(j>0){

ret[j]=siddu2(&n);

continue;

}

ret[j]=siddu(&n);

}

fflush(stdout);

for(j=0;j<n;j++)

printf("%d",ret[j]);

//getch();

}

**donbock** · Dec 21 '10, 03:45 PM

What do you intend to be the difference between siddu and siddu2? I don't see any practical difference between them. siddu calls fflush on stdin, but that is undefined behavior. The other difference is that siddu prints an annunciator string before entering the data entry loop.

Both siddu and siddu2 enter a data entry loop but return to the caller after the first pass through the loop. Why have code for a loop that you don't intend to execute? Similarly, why allocate a 20-entry array when you don't intend to use more than its first entry?

Line 51: it is illegal for main to return void. It must return int. If you change the definition on line 51 then you need to terminate the function with a return statement.

The loop between lines 63 and 75 has complicated logic that has the effect of calling siddu the first time through the loop and siddu2 for all other iterations. Presumably the reason for doing this is so the siddu annunciator string (line 19) is only printed once. It would be much easier to do that by moving line 19 to somewhere between lines 61 and 63 and simplifying the loop.

The essence of siddu and siddu2 is their calls to scanf (lines 23 and 41). What do you gain by having main call siddu/siddu2 rather than simply calling scanf directly?

Some C platforms have buffered output on stdout. If so, then all printed characters go to an internal buffer. They don't go to the output device until either fflush is called or a newline is printed. You should insert a line between lines 81 and 85 to print a newline (or move line 77 there).

The user does not have to enter what you expect. If the user types in something crazy then scanf will fail and the variable you are trying to load won't change. This would be especially bad if it happened at line 61. You should be checking the return code from scanf.

how to return an array from the function

how to return an array from the function

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