Thread: need help with c-code for polynomial calculation

made it small test programm for my infinite variables-issue.

instead of

Code:

int *vielfache = new int grad;
    for (int i=0; i-1 <grad; i++)
        {vielfache[i]=0};

i now use

Code:

#include <stdio.h>

main()
    {int grad=0;
    printf ("\n geben sie den grad des polynoms ein!\n");
    scanf ("%d" &grad);
        int vielfache[grad+1];

        for (int i=0; i<grad; i++)
            {printf ("%d", vielfache[i]);
            }
}

but still get this c99 error....

Code:

test.c:6: error: invalid operands to binary & (have ‘char *’ and ‘int’)
test.c:9: error: ‘for’ loop initial declarations are only allowed in C99 mode
test.c:9: note: use option -std=c99 or -std=gnu99 to compile your code

don´t know why i get an error in line 6... "grad" is int....

Originally Posted by dreadkopp

Code:

test.c:6: error: invalid operands to binary & (have ‘char *’ and ‘int’)

Why are you ignoring the advice in message #2 in this thread about line 12 in your original code? This is the line that corresponds to your line 6 in the new code.

Originally Posted by dreadkopp

Code:

test.c:9: error: ‘for’ loop initial declarations are only allowed in C99 mode
test.c:9: note: use option -std=c99 or -std=gnu99 to compile your code

If this error is not clear enough for you, I am very surprised. There are two (and maybe now just recently three) versions of the Programming Language C standard (ANSI C89/ISO C90 and ISO C99). Apparently, your compiler compiles using ISO C90 as its default. But you are using a ISO C99 feature. Now is that enough information for you to find your own problem?

hope so... thx... now using c99 instead of gcc... seems to work

grrrr... thought i got everything correct now..... but it seens like i
don´t...

looks like this

Code:

#include <stdio.h>

main()
{int grad=0;
float ergebnis=0;
float ableitung=0;
int x=0;
int i=0;
int k=0;
int l=0;

	printf ("\n Dieses Programm berechnet den Funktionswert sowie die erste Ableitung eines beliebigen Polynoms aus\n");
	printf ("\n Geben sie den Grad des Polynoms ein!\n");
	scanf ("%d", &grad);
	int vielfache[grad+1];

		for (i<grad; i++;)
			{vielfache[i]=0;}
									i=0;

	while (k-2 < grad; k++; i++;)
		{printf ("\n Geben sie das \n &k \n . Vielfache des Polynoms ein!\n");
		scanf ("%d", &vielfache[i]);}
									i=0;
									k=0;

	printf ("\n Geben sie einen Wert für x ein!\n");
	scanf ("%d", &x);

	int zwischenwert[grad+1];

		for (l<2*grad; l++;)
			{zwischenwert[l]=0;}
									l=0;
			
	while (i-1<grad;)
		{zwischenwert[l]=vielfache[i]*x;
		i++;
		zwischenwert[l+1]=zwischenwert[l]+vielfache[i];
		ergebnis=zwischenwert[l+1];
		l++;
		l++;
		i++;}

	printf ("Der Funktionswert des Polynoms ist \n &ergebnis \n!");

									i=0;
									l=0;

	while (i-1<grad;)
		{zwischenwert[l]=vielfache[i]*grad*x;
		i++;
		zwischenwert[l+1]=zwischenwert[l]+vielfache[i];
		ergebnis=zwischenwert[l+1];
		l++;
		l++;
		i++;}
	
	ableitung=ergebnis/x;

	printf ("Die erste Ableitung des Polynoms ist \n &ableitung \n!");
 
return 0;

}

when i try to compile with c99 i get the following error:

Code:

program.c: In function ‘main’:
program.c:21: error: expected ‘)’ before ‘;’ token
program.c:36: error: expected expression before ‘)’ token
program.c:50: error: expected expression before ‘)’ token

i don´t get it.... call me stupid but please help...

Read about how to write a main function and while loops.

nevermind my last posting.... got another problem....

program compiles without problems and seems to work correctly... but the uotput is totaly wrong most of the times...

the problem is, that the program is not using decimal numbers even if i declared them as float...

Code:

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

int main()
{int grad=0;
double ergebnis=0;
double ableitung=0;
float x=0;
int i=0;
int k=0;
int l=0;

	printf ("Dieses Programm berechnet den Funktionswert sowie die erste Ableitung eines beliebigen Polynoms aus\n");
	printf ("Geben sie den Grad des Polynoms ein!\n");
	scanf ("%d", &grad);
	float vielfache[grad+1];

		for (i<grad; i++;)
			{vielfache[i]=0;}
									i=0;

	while (k < grad+1 )
		{printf ("Geben sie das %i . Vielfache des Polynoms ein!\n" ,k+1);
		scanf ("%f", &vielfache[i]);
		//printf ("%f %d \n" , vielfache[i] , i);
		k++;
		i++;}
									i=0;
									k=0;

	printf ("Geben sie einen Wert für x ein!\n");
	scanf ("%f", &x);

	double zwischenwert[grad+1];

		for (l<2*grad; l++;)
			{zwischenwert[l]=0;}
									l=0;
			
	while (i<grad+1)
		{zwischenwert[l]=vielfache[i]*x;
		i++;
		zwischenwert[l+1]=zwischenwert[l]+vielfache[i];
		ergebnis=zwischenwert[l+1];
		//printf ("%g %g %g \n" , ergebnis , zwischenwert[l], x);
		l++;
		l++;
		i++;}

	printf ("Der Funktionswert des Polynoms ist:  %f \n" ,ergebnis);

									i=0;
									l=0;

	while (i<grad)
		{zwischenwert[l]=vielfache[i]*grad*x;
		i++;
		zwischenwert[l+1]=zwischenwert[l]+vielfache[i];
		ergebnis=zwischenwert[l+1];
		l++;
		l++;
		i++;}
	
	ableitung=ergebnis/x;

	printf ("Die erste Ableitung des Polynoms ist: %f \n" ,ableitung);
 
return 0;

}

can someone explain why this happens and how to fix it? thank u very much

Originally Posted by grumpy

Converting from float to double requires neither typecasting nor compiler warnings. The set of values that a float can represent is a subset of the set of values that a double can represent, so no information can ever be lost converting float to double.

The reverse is not true - converting double to float can lose information, so good quality compilers (can be configured to) give warnings when a double is implicitly converted to a float.

Aye, I think I got reverse-confused :P. Thanks for pointing out my error. Still, I would probably avoid having both floats and doubles(just for less confusion's sake) unless there's a convincing reason to use both.