Programming Forums > Application Development > C

kretchpoof · Feb 3rd, 2006, 7:48 PM

I just started to learn C and I typed this example out of Sams Teach Yourself C in 21 Days book plus i added a bit so that I can keep on trying different problems. Well the program works great except when I enter some numbers where the last two digits ending in 23 and then i get a negative number (e.g. 1523 shows a value of -762324629 or 1323 shows a value of -1979282029). I was wondering if it possibly had to do with the processor in some way because i dont see how the code could screw it up like that.

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//Demonstrates a simple function
#include <stdio.h>
#define NO 0
#define YES 1

long cube(long x);

long input, answer;

int choice = YES;

int main (void)
{
	while(choice != NO)
	{
		printf("Enter an interger value: ");
		scanf("%d", &input);
		answer = cube(input);
		printf("The cube of %ld is %ld.\n", input, answer);
		printf("Would you like to cube another number?(1 for yes 0 for no): ");
		scanf("%d", &choice);
		printf("\n");
	}

	return 0;
}

//Function: cube() - Calculates the cubed value of a variable
long cube(long x)
{
	long x_cubed;

	x_cubed = x*x*x;
	return x_cubed;
}

Navid · Feb 3rd, 2006, 8:15 PM

Welcome to the forums.

Great choice on the book, I read it about 6 months ago. The first thing I noticed is this:

(Toggle Plain Text)

int choice = YES;

The variable choice is declared to be an interger (int). An integer is a whole number, and you are assigning text to it. I suggest assigning a number to your the variable choice to condition your while loop for now.

Another important thing that the book dosen't teach you until later on:

(Toggle Plain Text)

scanf("%d", &input);
rewind(stdin);

Whenever the user is requested for input, the input is kept in the buffer, which will cause problems later on for other additional inputs. So make sure to clear out the stdin with rewind(stdin).

Jessehk · Feb 3rd, 2006, 8:20 PM

Quote:

Originally Posted by Navid

Welcome to the forums.

Great choice on the book, I read it about 6 months ago. The first thing I noticed is this:

(Toggle Plain Text)

int choice = YES;

The variable choice is declared to be an interger (int). An integer is a whole number, and you are assigning text to it. I suggest assigning a number to your the variable choice to condition your while loop for now.

If you look at his code, you can see this:

(Toggle Plain Text)

#define NO 0
#define YES 1

Navid · Feb 3rd, 2006, 8:22 PM

Oh, didn't see that. :o

The Dark · Feb 3rd, 2006, 8:51 PM

A signed long variable in 32 bit C can only hold a value up to 2,147,483,647 (2^31 - 1)
1523^3 is 3,532,642,667, which is too big. This causes the value to overflow and become negative.

To fix this, you could try using a 64 bit integer. Unfortunately, the way to do this is different depending on your compiler (e.g. __int64 for Visual C, long long for gcc).
Alternatively, you could use floating point (e.g. double). This gives you a greater range, but possibly at the expense of accuracy.

kretchpoof · Feb 3rd, 2006, 9:55 PM

Hey thanks for all of the replies. I changed around the variable type and that fixed it.

(Toggle Plain Text)

//Demonstrates a simple function
#include <stdio.h>
#define NO 0
#define YES 1

long long cube(long long x);

long input;
long long answer;

int choice = YES;

int main (void)
{
	while(choice != NO)
	{
		printf("Enter an interger value: ");
		scanf("%d", &input);
		answer = cube(input);
		printf("The cube of %ld is %lld.\n", input, answer);
		printf("Would you like to cube another number?(1 for yes 0 for no): ");
		scanf("%d", &choice);
		printf("\n");
	}

	return 0;
}

//Function: cube() - Calculates the cubed value of a variable
long long cube(long long x)
{
	long long x_cubed;

	x_cubed = x*x*x;
	return x_cubed;
}

DaWei · Feb 4th, 2006, 7:34 AM

Just to highlight for Navid:

Quote:

int choice = YES;

The lack of quotation marks is a hint that there might be a preceding definition of this token as a number.

Quote:

Whenever the user is requested for input, the input is kept in the buffer, which will cause problems later on for other additional inputs.

This is just flat not true. Only input that is not requested or that input which fails to meet the specifications of the specified input mechanism will be left in the buffer. Read the documentation for the particular function to determine what might be automatically discarded, used, or left in the buffer. Incidentally, one can dispense with the buffer in most cases if one desires.

Feb 3rd, 2006, 7:48 PM	#1
kretchpoof Newbie Join Date: Feb 2006 Posts: 2 Rep Power: 0	negative cubed numbers I just started to learn C and I typed this example out of Sams Teach Yourself C in 21 Days book plus i added a bit so that I can keep on trying different problems. Well the program works great except when I enter some numbers where the last two digits ending in 23 and then i get a negative number (e.g. 1523 shows a value of -762324629 or 1323 shows a value of -1979282029). I was wondering if it possibly had to do with the processor in some way because i dont see how the code could screw it up like that. (Toggle Plain Text) //Demonstrates a simple function #include <stdio.h> #define NO 0 #define YES 1 long cube(long x); long input, answer; int choice = YES; int main (void) { while(choice != NO) { printf("Enter an interger value: "); scanf("%d", &input); answer = cube(input); printf("The cube of %ld is %ld.\n", input, answer); printf("Would you like to cube another number?(1 for yes 0 for no): "); scanf("%d", &choice); printf("\n"); } return 0; } //Function: cube() - Calculates the cubed value of a variable long cube(long x) { long x_cubed; x_cubed = xxx; return x_cubed; } //Demonstrates a simple function #include <stdio.h> #define NO 0 #define YES 1 long cube(long x); long input, answer; int choice = YES; int main (void) { while(choice != NO) { printf("Enter an interger value: "); scanf("%d", &input); answer = cube(input); printf("The cube of %ld is %ld.\n", input, answer); printf("Would you like to cube another number?(1 for yes 0 for no): "); scanf("%d", &choice); printf("\n"); } return 0; } //Function: cube() - Calculates the cubed value of a variable long cube(long x) { long x_cubed; x_cubed = xxx; return x_cubed; }

Feb 3rd, 2006, 8:15 PM	#2
Navid Hobbyist Programmer Join Date: Feb 2005 Location: Canada Posts: 187 Rep Power: 4	Welcome to the forums. Great choice on the book, I read it about 6 months ago. The first thing I noticed is this: (Toggle Plain Text) int choice = YES; int choice = YES; The variable choice is declared to be an interger (int). An integer is a whole number, and you are assigning text to it. I suggest assigning a number to your the variable choice to condition your while loop for now. Another important thing that the book dosen't teach you until later on: (Toggle Plain Text) scanf("%d", &input); rewind(stdin); scanf("%d", &input); rewind(stdin); Whenever the user is requested for input, the input is kept in the buffer, which will cause problems later on for other additional inputs. So make sure to clear out the stdin with rewind(stdin).

Feb 3rd, 2006, 9:55 PM	#6
kretchpoof Newbie Join Date: Feb 2006 Posts: 2 Rep Power: 0	Hey thanks for all of the replies. I changed around the variable type and that fixed it. (Toggle Plain Text) //Demonstrates a simple function #include <stdio.h> #define NO 0 #define YES 1 long long cube(long long x); long input; long long answer; int choice = YES; int main (void) { while(choice != NO) { printf("Enter an interger value: "); scanf("%d", &input); answer = cube(input); printf("The cube of %ld is %lld.\n", input, answer); printf("Would you like to cube another number?(1 for yes 0 for no): "); scanf("%d", &choice); printf("\n"); } return 0; } //Function: cube() - Calculates the cubed value of a variable long long cube(long long x) { long long x_cubed; x_cubed = xxx; return x_cubed; } //Demonstrates a simple function #include <stdio.h> #define NO 0 #define YES 1 long long cube(long long x); long input; long long answer; int choice = YES; int main (void) { while(choice != NO) { printf("Enter an interger value: "); scanf("%d", &input); answer = cube(input); printf("The cube of %ld is %lld.\n", input, answer); printf("Would you like to cube another number?(1 for yes 0 for no): "); scanf("%d", &choice); printf("\n"); } return 0; } //Function: cube() - Calculates the cubed value of a variable long long cube(long long x) { long long x_cubed; x_cubed = xxx; return x_cubed; } Last edited by kretchpoof; Feb 3rd, 2006 at 10:10 PM.

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Feb 3rd, 2006, 8:22 PM	#4
Navid Hobbyist Programmer Join Date: Feb 2005 Location: Canada Posts: 187 Rep Power: 4	Oh, didn't see that. :o

Feb 3rd, 2006, 8:51 PM	#5
The Dark Expert Programmer Join Date: Jun 2005 Posts: 825 Rep Power: 4	A signed long variable in 32 bit C can only hold a value up to 2,147,483,647 (2^31 - 1) 1523^3 is 3,532,642,667, which is too big. This causes the value to overflow and become negative. To fix this, you could try using a 64 bit integer. Unfortunately, the way to do this is different depending on your compiler (e.g. __int64 for Visual C, long long for gcc). Alternatively, you could use floating point (e.g. double). This gives you a greater range, but possibly at the expense of accuracy.

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Programming Forums > Application Development > C

negative cubed numbers