Thread: Simple enough task

Hi all,
1st time poster.

I have a fairly simple task that I need to carry out, however I am not terribly good at C programming, so bare with me!!!!
Basically I need to take in a string of 15 binary bits, from this input I will be adding bits 1,6,11 bits 2,7,12 bits 3,8,13 bits 4,9,14 and bits 5,10,15.
These will all give me parity which will either be a 0 or a 1. I then have to output the original 15 bits along with the 5 new bits, so my output will be 20 bits long.

I hope that all makes sense.......so far I have been trying to do it as such........

Code:

#include <stdio.h>
#include <string.h>
#include <math.h>

int main()
{
	int codeword_array[20];
	int i=0;

	printf("Welcome to the 5-bit parity code calculator\nPlease enter the 1st bit of the 15 bit codeword now: \n");
	
	for (int i>=0; i<16; i++)
	{
		int bit_array[i];	
		scanf("%d", bit_array[i])
			if bit !=(0||1)
			{
				printf("All bits must be 0 or a 1");
			}
			codeword_array[i] = bit[i];
	}	

	if (bit_array[0] + bit_array[5] + bit_array[10] = (0||2))
		{	
			codeword_array[15] = 0;
		}
	else if (bit_array[0] + bit_array[5] + bit_array[10] = (1||3))
		{
			codeword_array[15] = 1;	
		}	
	else
		{
			printf("Error\n");
		}
	system("pause");
	return 0;
}

The thinking behind this is that if I use a 20 bit array, and start off filling the 1st 15 blocks, then do the parity check on them and with my answers fill the final 5.
However I am getting a good few errors on this.

Hope someone can help.

Regards,

Niall.

Firt of all you should use == in your if statements. = is an assignment operator which means all your if tests will be true.

You also use "bit" but never declare it anywhere...

What errors are you getting?

Code:

(14) : error C2143: syntax error : missing ';' before 'type'
(14) : error C2143: syntax error : missing ';' before 'type'
(14) : error C2143: syntax error : missing ')' before 'type'
(14) : error C2143: syntax error : missing ';' before 'type'
(14) : warning C4552: '<' : operator has no effect; expected operator with side-effect
(14) : error C2059: syntax error : ')'
(15) : error C2143: syntax error : missing ';' before '{'
(16) : error C2057: expected constant expression
(16) : error C2466: cannot allocate an array of constant size 0
(16) : error C2133: 'bit_array' : unknown size
(18) : error C2143: syntax error : missing ';' before 'if'
(18) : error C2061: syntax error : identifier 'bit'
(22) : error C2109: subscript requires array or pointer type
(25) : error C2059: syntax error : 'if'
(29) : error C2059: syntax error : 'else'
(33) : error C2059: syntax error : 'else'
(39) : error C2143: syntax error : missing ')' before 'string'
(39) : error C2143: syntax error : missing '{' before 'string'
(39) : error C2059: syntax error : '<Unknown>'
(39) : error C2059: syntax error : ')'
(40) : error C2059: syntax error : 'return'
(41) : error C2059: syntax error : '}'

These are all the errors, (after I changed the = to == and declared bit) A lot of them seem to be linked to the one thing, so I think I might just be missing something simple thats causing a lot of them!!!

P.s. I declared bit outside the for loop, is this correct?

Code:

int codeword_array[20];
	int i=0;
	int bit;

	printf("Welcome to the 5-bit parity code calculator\nPlease enter the 1st bit of the 15 bit codeword now: \n");
	
	for (int i>=0; i<16; i++)

you need to use a & before bit_array[i] in your scanf statement.

And you need a ; after the scanf line too.

compile again and post your errors.

Ok, that got rid of a couple, thanks for that. The next one's are:

Code:

(14) : error C2143: syntax error : missing ';' before 'type'
(14) : error C2143: syntax error : missing ';' before 'type'
(14) : error C2143: syntax error : missing ')' before 'type'
(14) : error C2143: syntax error : missing ';' before 'type'
(14) : warning C4552: '<' : operator has no effect; expected operator with side-effect
(14) : error C2059: syntax error : ')'
(15) : error C2143: syntax error : missing ';' before '{'
(16) : error C2057: expected constant expression
(16) : error C2466: cannot allocate an array of constant size 0
(16) : error C2133: 'bit_array' : unknown size
(18) : error C2061: syntax error : identifier 'bit'
(22) : error C2109: subscript requires array or pointer type
(25) : error C2059: syntax error : 'if'
(29) : error C2059: syntax error : 'else'
(33) : error C2059: syntax error : 'else'
(39) : error C2143: syntax error : missing ')' before 'string'
(39) : error C2143: syntax error : missing '{' before 'string'
(39) : error C2059: syntax error : '<Unknown>'
(39) : error C2059: syntax error : ')'
(40) : error C2059: syntax error : 'return'
(41) : error C2059: syntax error : '}'

There seems to be a good few on line 14. Line 14 is the beginning of the for loop,

Code:

	for (int i>=0; i<16; i++)

Sorry about this, I know its very messy, but just trying to get my head around it!!

the fist part of the for statement sets the intial value of i.

for (int i = 0; i<16; i++) will loop from 0 - 15

Ok, thats grand thanks,

But still giving the errors that i'm missing ;'s around line 14. Dunno where this is happening though,

Any ideas?

post your fixed code

Code:

#include <stdio.h>
#include <string.h>
#include <math.h>

int main()
{
	
	int codeword_array[20];
	int i=0;
	int bit;

	printf("Welcome to the 5-bit parity code calculator\nPlease enter the 1st bit of the 15 bit codeword now: \n");
	
	for (int i = 0; i<16; i++)
	{
		int bit_array[i];	
		scanf("%d", &bit_array[i]);
			if bit !==(0||1)
			{
				printf("All bits must be 0 or a 1");
			}
			codeword_array[i] = bit[i];
	}	

 	if (bit_array[0] + bit_array[5] + bit_array[10] == (0||2))
		{	
			codeword_array[15] == 0;
		}
	else if (bit_array[0] + bit_array[5] + bit_array[10] == (1||3))
		{
			codeword_array[15] == 1;	
		}	
	else
		{
			printf("Error\n");
		}
	
	
	system("pause");
	return 0;
}

You can't declare bit_array inside the for loop. Doing so will declare it each iteration.

You test bit after the scanf, but never assign bit a value. You should do...

Code:

scanf("%d", &bit);
if (bit != 0 && bit !=1)
{
  printf("All bits...");
}
else
{
  codeword_array[i] = bit;
}

You don't need bit_array at all. Change bit_array in your subsequent if statements to codeword_array.

Ok, should it be declared globally? I have done it like this now....

Code:

int codeword_array[20];
	int i=0;
	int bit;
	int bit_array[i];	

	printf("Welcome to the 5-bit parity code calculator\nPlease enter the 1st bit of the 15 bit codeword now: \n");
	
	for (int i = 0; i<16; i++)
	{		
		scanf("%d", &bit_array[i]);
			if bit !==(0||1)
			{
				printf("All bits must be 0 or a 1");
			}
			codeword_array[i] = bit[i];
	}	

 	if (bit_array[0] + bit_array[5] + bit_array[10] == (0||2))
		{	
			codeword_array[15] == 0;
		}
	else if (bit_array[0] + bit_array[5] + bit_array[10] == (1||3))
		{
			codeword_array[15] == 1;	
		}	
	else
		{
			printf("Error\n");
		}
	
	
	system("pause");
	return 0;
}

With some new errors

Code:

(11) : error C2057: expected constant expression
(11) : error C2466: cannot allocate an array of constant size 0
(11) : error C2133: 'bit_array' : unknown size
(15) : error C2143: syntax error : missing ';' before 'type'
(15) : error C2143: syntax error : missing ';' before 'type'
(15) : error C2143: syntax error : missing ')' before 'type'
(15) : error C2143: syntax error : missing ';' before 'type'
(15) : warning C4552: '<' : operator has no effect; expected operator with side-effect
(15) : error C2059: syntax error : ')'
(16) : error C2143: syntax error : missing ';' before '{'
(18) : error C2061: syntax error : identifier 'bit'
(22) : error C2109: subscript requires array or pointer type
(25) : error C2059: syntax error : 'if'
(29) : error C2059: syntax error : 'else'
(33) : error C2059: syntax error : 'else'
(39) : error C2143: syntax error : missing ')' before 'string'
(39) : error C2143: syntax error : missing '{' before 'string'
(39) : error C2059: syntax error : '<Unknown>'
(39) : error C2059: syntax error : ')'
(40) : error C2059: syntax error : 'return'
(41) : error C2059: syntax error : '}'

Ok, had posted that last one before I realised there was a reply!!!!

Changing bit_array to codeword_array worked for a lot of them, down to 7 errors now.
code is

Code:

#include <stdio.h>
#include <string.h>
#include <math.h>

int main()
{
	
	int codeword_array[20];
	int i=0;
	int bit;
	
	printf("Welcome to the 5-bit parity code calculator\nPlease enter the 1st bit of the 15 bit codeword now: \n");
	
	for (int i = 0; i<16; i++)
	{		
		scanf("%d", &bit);
		if (bit != 0 && bit !=1)
			{
				printf("All bits must be a 0 or a 1/n");
			}
		else
			{
				codeword_array[i] = bit;
			}
		
		
	

 	if (codeword_array[0] + codeword_array[5] + codeword_array[10] == (0||2))
		{	
			codeword_array[15] = 0;
		}
	else if (codeword_array[0] + codeword_array[5] + codeword_array[10] == (1||3))
		{
			codeword_array[15] = 1;	
		}	
	else
		{
			printf("Error\n");
		}
	
	
	system("pause");
	return 0;
}

/*scanf( "%d", &bit1 );
	
	if ( bit1 != (0 || 1))
	{
		printf( "All bits must be either 0 or 1\n" );
	}
	else
	{	
		printf( "You entered %d", bit1);
	}

and the remaining errors are

Code:

(14) : error C2143: syntax error : missing ';' before 'type'
(14) : error C2143: syntax error : missing ';' before 'type'
(14) : error C2143: syntax error : missing ')' before 'type'
(14) : error C2143: syntax error : missing ';' before 'type'
(14) : warning C4552: '<' : operator has no effect; expected operator with side-effect
(14) : error C2059: syntax error : ')'
(15) : error C2143: syntax error : missing ';' before '{'
(43) : warning C4013: 'system' undefined; assuming extern returning int
(94) : fatal error C1075: end of file found before the left brace '{' at 'c:\users\niall\documents\college\4th year\2nd semester\ee402\5bit parity code\5bit parity code\source.c(6)' was matched

You don't need bit_array at all. If you're storing the bit array in codeword_array, why store it in bit_array too?

and you can't use [i] in the declaration.