Programming Forums > Application Development > C++

ViZioN · Feb 21st, 2005, 3:19 PM

Hey all,

I'm currently writing a calculator program for Uni which takes in 2 Roman Numeral expressions and displays the answer Roman Numerals. The answer is then displayed in Roman Numerals

i.e IX+XV = XXIV

The problem i'm having is trying to return an array from a function. The function takes in the integer answer worked out earlier in the program, and then converts this to roman numerals, saving each character in an array, which is created by the function. I wish to return the array back into the main program but i'm having problems

Sorry if this has been asked before, I tried searching the forums but couldn't find anything that similar .
The code for my sub-routine is as follows:

(Toggle Plain Text)

char convert(int answer)
{
int j=0;
char numeral[20];

	// Calculate M's
   while((answer-1000)>=0)
   {
	answer=answer-1000;
	numeral[j++]='M';
   }

   if(answer>=900)
   {
   numeral[j++]='C';
   numeral[j++]='M';
   answer=answer-900;
   }

   // Calculate D's

	while((answer-500)>=0)
	{
   answer=answer-500;
	numeral[j]='D';
   j++;
   }

   if(answer>=400)
   	{
   	numeral[j++]='C';
   	numeral[j++]='D';
   	answer=answer-400;
   	}

   // Calculate C's

	while((answer-100)>=0)
   {
	answer=answer-100;
	numeral[j++]='C';
   }

   if(answer>=90)
   {
   numeral[j++]='X';
   numeral[j++]='C';
   answer=answer-90;
   }

   //Calcalute L's

while((answer-50)>=0)
   	{
	answer=answer-50;
	numeral[j++]='L';
   }

   if(answer>=40)
   {
   numeral[j++]='X';
   numeral[j++]='L';
   answer=answer-40;
   }

   // Calculate X's

   while((answer-10)>=0)
	{
	answer=answer-10;
	numeral[j++]='X';
   }

   if (answer==9)
   {
   numeral[j++]='I';
   numeral[j++]='X';
   answer=answer-9;
   }

   // Calculate V's

   while((answer-5)>=0)
	{
	answer=answer-5;
	numeral[j++]='V';
   }

   if(answer==4)
   {
   numeral[j++]='I';
   numeral[j++]='V';
   answer=answer-4;
   }

   // Calculate j's

while((answer-1)>=0)
   {
	answer=answer-1;
	numeral[j++]='I';
   }

   numeral[j]  = '\0';
   printf("\n");
   return numeral;
}

I want to return the array numeral so that it can be displayed on screen and the code can be used to convert other numbers.

These are the errors that the compiler is displaying:

Warn : roman.c(207,19)

uspicious pointer conversion
Error: roman.c(207,19):Nonportable pointer conversion

Can someone help me out so that i can get my program working properly?

Thanks in advance,

Dave

tempest · Feb 21st, 2005, 3:37 PM

(Toggle Plain Text)

char *convert(int answer)
{
int j=0;
char numeral[20];

	// Calculate M's
   while((answer-1000)>=0)
   {
	answer=answer-1000;
	numeral[j++]='M';
   }

   if(answer>=900)
   {
   numeral[j++]='C';
   numeral[j++]='M';
   answer=answer-900;
   }

   // Calculate D's

	while((answer-500)>=0)
	{
   answer=answer-500;
	numeral[j]='D';
   j++;
   }

   if(answer>=400)
   	{
   	numeral[j++]='C';
   	numeral[j++]='D';
   	answer=answer-400;
   	}

   // Calculate C's

	while((answer-100)>=0)
   {
	answer=answer-100;
	numeral[j++]='C';
   }

   if(answer>=90)
   {
   numeral[j++]='X';
   numeral[j++]='C';
   answer=answer-90;
   }

   //Calcalute L's

while((answer-50)>=0)
   	{
	answer=answer-50;
	numeral[j++]='L';
   }

   if(answer>=40)
   {
   numeral[j++]='X';
   numeral[j++]='L';
   answer=answer-40;
   }

   // Calculate X's

   while((answer-10)>=0)
	{
	answer=answer-10;
	numeral[j++]='X';
   }

   if (answer==9)
   {
   numeral[j++]='I';
   numeral[j++]='X';
   answer=answer-9;
   }

   // Calculate V's

   while((answer-5)>=0)
	{
	answer=answer-5;
	numeral[j++]='V';
   }

   if(answer==4)
   {
   numeral[j++]='I';
   numeral[j++]='V';
   answer=answer-4;
   }

   // Calculate j's

while((answer-1)>=0)
   {
	answer=answer-1;
	numeral[j++]='I';
   }

   numeral[j]  = '\0';
   printf("\n");
   return numeral;
}

Example:

(Toggle Plain Text)

char retFromFunc[20];
strcpy(retFromFunc, convert(3));

ViZioN · Feb 21st, 2005, 3:53 PM

I'm not completly sure where to add in youe example code

The code that calls the function at the moment is,

(Toggle Plain Text)

answer2=convert(answer);
  	printf("\n");
   printf("%s", answer2);

and my function is declared at the top of my program as

(Toggle Plain Text)

char convert(int)

I'm new to C and pointers (only been learning C 4 weeks). Can you explain how to alter the other lines of my code so that the program works correctlly.

Thanks in advance,

Dave

Dizzutch · Feb 21st, 2005, 5:17 PM

you can't return an array in C you'll have to return a pointer to the array.

ViZioN · Feb 21st, 2005, 6:01 PM

Yeah Dizzutch i've been told that already today but i'm not too sure how to do that

I've only been learning C for 4 weeks

Are you able to help me out?

Xero · Feb 21st, 2005, 6:45 PM

Edit: Actually Im not sure if this applies for C. Works for C++.

Umm Arrays automatically pass by reference (theres another term used for arrays, cant remember). Theres no need for a pointer.

Feb 21st, 2005, 3:19 PM	#1
ViZioN Newbie Join Date: Feb 2005 Posts: 3 Rep Power: 0	Returning An Array From a Function Hey all, I'm currently writing a calculator program for Uni which takes in 2 Roman Numeral expressions and displays the answer Roman Numerals. The answer is then displayed in Roman Numerals i.e IX+XV = XXIV The problem i'm having is trying to return an array from a function. The function takes in the integer answer worked out earlier in the program, and then converts this to roman numerals, saving each character in an array, which is created by the function. I wish to return the array back into the main program but i'm having problems Sorry if this has been asked before, I tried searching the forums but couldn't find anything that similar . The code for my sub-routine is as follows: (Toggle Plain Text) char convert(int answer) { int j=0; char numeral[20]; // Calculate M's while((answer-1000)>=0) { answer=answer-1000; numeral[j++]='M'; } if(answer>=900) { numeral[j++]='C'; numeral[j++]='M'; answer=answer-900; } // Calculate D's while((answer-500)>=0) { answer=answer-500; numeral[j]='D'; j++; } if(answer>=400) { numeral[j++]='C'; numeral[j++]='D'; answer=answer-400; } // Calculate C's while((answer-100)>=0) { answer=answer-100; numeral[j++]='C'; } if(answer>=90) { numeral[j++]='X'; numeral[j++]='C'; answer=answer-90; } //Calcalute L's while((answer-50)>=0) { answer=answer-50; numeral[j++]='L'; } if(answer>=40) { numeral[j++]='X'; numeral[j++]='L'; answer=answer-40; } // Calculate X's while((answer-10)>=0) { answer=answer-10; numeral[j++]='X'; } if (answer==9) { numeral[j++]='I'; numeral[j++]='X'; answer=answer-9; } // Calculate V's while((answer-5)>=0) { answer=answer-5; numeral[j++]='V'; } if(answer==4) { numeral[j++]='I'; numeral[j++]='V'; answer=answer-4; } // Calculate j's while((answer-1)>=0) { answer=answer-1; numeral[j++]='I'; } numeral[j] = '\0'; printf("\n"); return numeral; } char convert(int answer) { int j=0; char numeral[20]; // Calculate M's while((answer-1000)>=0) { answer=answer-1000; numeral[j++]='M'; } if(answer>=900) { numeral[j++]='C'; numeral[j++]='M'; answer=answer-900; } // Calculate D's while((answer-500)>=0) { answer=answer-500; numeral[j]='D'; j++; } if(answer>=400) { numeral[j++]='C'; numeral[j++]='D'; answer=answer-400; } // Calculate C's while((answer-100)>=0) { answer=answer-100; numeral[j++]='C'; } if(answer>=90) { numeral[j++]='X'; numeral[j++]='C'; answer=answer-90; } //Calcalute L's while((answer-50)>=0) { answer=answer-50; numeral[j++]='L'; } if(answer>=40) { numeral[j++]='X'; numeral[j++]='L'; answer=answer-40; } // Calculate X's while((answer-10)>=0) { answer=answer-10; numeral[j++]='X'; } if (answer==9) { numeral[j++]='I'; numeral[j++]='X'; answer=answer-9; } // Calculate V's while((answer-5)>=0) { answer=answer-5; numeral[j++]='V'; } if(answer==4) { numeral[j++]='I'; numeral[j++]='V'; answer=answer-4; } // Calculate j's while((answer-1)>=0) { answer=answer-1; numeral[j++]='I'; } numeral[j] = '\0'; printf("\n"); return numeral; } I want to return the array numeral so that it can be displayed on screen and the code can be used to convert other numbers. These are the errors that the compiler is displaying: Warn : roman.c(207,19)uspicious pointer conversion Error: roman.c(207,19):Nonportable pointer conversion Can someone help me out so that i can get my program working properly? Thanks in advance, Dave

Feb 21st, 2005, 3:37 PM	#2
tempest Programming Guru Join Date: Oct 2004 Posts: 1,041 Rep Power: 5	(Toggle Plain Text) char convert(int answer) { int j=0; char numeral[20]; // Calculate M's while((answer-1000)>=0) { answer=answer-1000; numeral[j++]='M'; } if(answer>=900) { numeral[j++]='C'; numeral[j++]='M'; answer=answer-900; } // Calculate D's while((answer-500)>=0) { answer=answer-500; numeral[j]='D'; j++; } if(answer>=400) { numeral[j++]='C'; numeral[j++]='D'; answer=answer-400; } // Calculate C's while((answer-100)>=0) { answer=answer-100; numeral[j++]='C'; } if(answer>=90) { numeral[j++]='X'; numeral[j++]='C'; answer=answer-90; } //Calcalute L's while((answer-50)>=0) { answer=answer-50; numeral[j++]='L'; } if(answer>=40) { numeral[j++]='X'; numeral[j++]='L'; answer=answer-40; } // Calculate X's while((answer-10)>=0) { answer=answer-10; numeral[j++]='X'; } if (answer==9) { numeral[j++]='I'; numeral[j++]='X'; answer=answer-9; } // Calculate V's while((answer-5)>=0) { answer=answer-5; numeral[j++]='V'; } if(answer==4) { numeral[j++]='I'; numeral[j++]='V'; answer=answer-4; } // Calculate j's while((answer-1)>=0) { answer=answer-1; numeral[j++]='I'; } numeral[j] = '\0'; printf("\n"); return numeral; } char convert(int answer) { int j=0; char numeral[20]; // Calculate M's while((answer-1000)>=0) { answer=answer-1000; numeral[j++]='M'; } if(answer>=900) { numeral[j++]='C'; numeral[j++]='M'; answer=answer-900; } // Calculate D's while((answer-500)>=0) { answer=answer-500; numeral[j]='D'; j++; } if(answer>=400) { numeral[j++]='C'; numeral[j++]='D'; answer=answer-400; } // Calculate C's while((answer-100)>=0) { answer=answer-100; numeral[j++]='C'; } if(answer>=90) { numeral[j++]='X'; numeral[j++]='C'; answer=answer-90; } //Calcalute L's while((answer-50)>=0) { answer=answer-50; numeral[j++]='L'; } if(answer>=40) { numeral[j++]='X'; numeral[j++]='L'; answer=answer-40; } // Calculate X's while((answer-10)>=0) { answer=answer-10; numeral[j++]='X'; } if (answer==9) { numeral[j++]='I'; numeral[j++]='X'; answer=answer-9; } // Calculate V's while((answer-5)>=0) { answer=answer-5; numeral[j++]='V'; } if(answer==4) { numeral[j++]='I'; numeral[j++]='V'; answer=answer-4; } // Calculate j's while((answer-1)>=0) { answer=answer-1; numeral[j++]='I'; } numeral[j] = '\0'; printf("\n"); return numeral; } Example: (Toggle Plain Text) char retFromFunc[20]; strcpy(retFromFunc, convert(3)); char retFromFunc[20]; strcpy(retFromFunc, convert(3)); __________________

Feb 21st, 2005, 3:53 PM	#3
ViZioN Newbie Join Date: Feb 2005 Posts: 3 Rep Power: 0	I'm not completly sure where to add in youe example code The code that calls the function at the moment is, (Toggle Plain Text) answer2=convert(answer); printf("\n"); printf("%s", answer2); answer2=convert(answer); printf("\n"); printf("%s", answer2); and my function is declared at the top of my program as (Toggle Plain Text) char convert(int) char convert(int) I'm new to C and pointers (only been learning C 4 weeks). Can you explain how to alter the other lines of my code so that the program works correctlly. Thanks in advance, Dave

Feb 21st, 2005, 5:17 PM	#4
Dizzutch Professional Programmer Join Date: Dec 2004 Location: Worcester, MA Posts: 441 Rep Power: 4	you can't return an array in C you'll have to return a pointer to the array. __________________ naked pictures of you \| PFO F@H stats

Feb 21st, 2005, 6:45 PM	#6
Xero Hobbyist Programmer Join Date: Dec 2004 Location: a cardboard box Posts: 118 Rep Power: 4	Edit: Actually Im not sure if this applies for C. Works for C++. Umm Arrays automatically pass by reference (theres another term used for arrays, cant remember). Theres no need for a pointer. __________________ ... Last edited by Xero; Feb 21st, 2005 at 6:48 PM.

Feb 21st, 2005, 6:01 PM	#5
ViZioN Newbie Join Date: Feb 2005 Posts: 3 Rep Power: 0	Yeah Dizzutch i've been told that already today but i'm not too sure how to do that I've only been learning C for 4 weeks Are you able to help me out?

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Programming Forums > Application Development > C++

Returning An Array From a Function