Thread: Error - function returns address of local variable

Code:

int main(void)
{
    char brand[20];

    printf("Enter brand: ");
   brand = validatebrand();

}

char * validatebrand()
{
   char brandName[20];	
.....
.......
......
	if( (strlen (brandName)) >= 1 && (strlen (brandName)) < 20)
	{	
	  return brandName; <-- this is the warning
	}		
	else
       {	
	return NULL;
       }
}

Ask user to key in brand, and use a validatebrand() to check whether it's valid or not, if it is valid, return the brandName. But when it is return, and i tried a printf to check what's returned in the main(), i found out tat it's not returned the exact data but the address. How i am going to make the validatebrand() to return the entire valid data after validated? Can anyone point it out?
Thanks

C doesn't pass arrays. It insteads reduces it to a pointer.

Now since the function is done by the time you would do a printf in main, all the local variables are no longer valid to read from.

Also look at your else condition, you have a logic error there. If the else condition is done then you are trying to assign NULL to an array, which if you then try to printf it off would (hopefully) seg fault.

Instead of returning the data try passing it the array.

Code:

int main(void)
{
    char brand[20];

    printf("Enter brand: ");
   validatebrand(brand);

}

void validatebrand(char *brandName)
{
	fgets(brandName, 20, STDIN);  // fgets won't let you overflow the buffer.
}

problem solved
thankss

Your problem might have been solved, but if you really want to write good robust code, dont ever return pointers to local variables.
That is suicide!

What if your program stack gets overwritten at the place where your local variable is defined?
You may get the correct output now, but never ever do it in a large program.