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Old 12-14-2006, 08:21 PM   #3 (permalink)
zero2
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I came up with something, but there's probably a way to simplify it even more. Here's my code:

Code:
#include <stdio.h>

void displayIntegerDigit(char[]);

int main ( void ) 
{
	char cAry1[81];
	
	printf( "\nEnter an integer: " );
	gets(cAry1);
	
  displayIntegerDigit(cAry1);
	
	
	return 0;
}
	
	
void displayIntegerDigit(char cTest[])
{
	int iA;
	int i;
	
	printf("\nThe sign: +\n");
	printf("The counts of digits:\n");
	
	iA = 0;
	for ( i = 0; i < 81; i++ )
		{
			if (* ( cTest + i ) == '0' )
				{
					iA++;
				}
		}
	printf("\t\t0 and %d\n", iA);
	
	iA = 0;
	
	for ( i = 0; i < 81; i++ )
		{
			if (* ( cTest + i ) == '1' )
				{
					iA++;
				}
		}
	printf("\t\t1 and %d\n", iA);
	
	iA = 0;
	
	for ( i = 0; i < 81; i++ )
		{
			if (* ( cTest + i ) == '2' )
				{
					iA++;
				}
		}
	printf("\t\t2 and %d\n", iA);
	
	iA = 0;
	
	for ( i = 0; i < 81; i++ )
		{
			if (* ( cTest + i ) == '3' )
				{
					iA++;
				}
		}
	printf("\t\t3 and %d\n", iA);
	
	iA = 0;
	
	for ( i = 0; i < 81; i++ )
		{
			if (* ( cTest + i ) == '4' )
				{
					iA++;
				}
		}
	printf("\t\t4 and %d\n", iA);

	iA = 0;
	
	for ( i = 0; i < 81; i++ )
		{
			if (* ( cTest + i ) == '5' )
				{
					iA++;
				}
		}
	printf("\t\t5 and %d\n", iA);
	
	iA = 0;
	
	for ( i = 0; i < 81; i++ )
		{
			if (* ( cTest + i ) == '6' )
				{
					iA++;
				}
		}
	printf("\t\t6 and %d\n", iA);
	
	iA = 0;
	
	for ( i = 0; i < 81; i++ )
		{
			if (* ( cTest + i ) == '7' )
				{
					iA++;
				}
		}
	printf("\t\t7 and %d\n", iA);
	
	iA = 0;
	
	for ( i = 0; i < 81; i++ )
		{
			if (* ( cTest + i ) == '8' )
				{
					iA++;
				}
		}
	printf("\t\t8 and %d\n", iA);
	
	iA = 0;
	
	for ( i = 0; i < 81; i++ )
		{
			if (* ( cTest + i ) == '9' )
				{
					iA++;
				}
		}
	printf("\t\t9 and %d\n", iA);
	
  return;
}
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