new to C

emblem4ever · 01-23-2010, 11:44 AM

hi, I'm new to C programming and I'm still very unsure of many things.

I tried making a decimal to hexadecimal converter, using a very simple method, yet it doesn't seem to work and I can't figure out what syntax errors i have, heres my code

Code:

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

 void main 
{
	Union
	{
		int a;
		float b;
	}myUnion;

	printf("Enter the Decimal number to be converted:\n");

	scanf("%f",&myUnion.b\n);

	printf("The number in Hexadecimal is:\n");

	printf("%x",&myUnion.a);
}

I tried to make it really simple, but whats wrong??

**Salem** · 01-23-2010, 11:47 AM

> scanf("%f",&myUnion.b\n);
Remove the \n here

> printf("%x",&myUnion.a);
Remove the & here

> void main
Remove the void here, and replace it with int

emblem4ever · 01-23-2010, 11:54 AM

oh, thanks a lot, however i still get many errors, but i really don't get why,

like it says syntax error before "{" in line 11
and conflicting types for 'printf'
etc.

but according to my knowlegde and other programs i've seen, it should work fine...?

maybe its my compiler, im using Dev-C 4.9.9.2 because im using vista....
hmmmmm

**Salem** · 01-23-2010, 11:57 AM

> Union
Try union

All C keywords are lowercase.

Memloop · 01-23-2010, 11:59 AM

C is case dependent, and the standard data type for unions is union and not Union.

The %x modifier expects an integer, but you're passing it a float. I have no idea why you're storing the input in an union to begin with, let alone a float.

emblem4ever · 01-23-2010, 12:10 PM

because i want to scan the input as a float, i.e. user enters 26.74628, then using the union, "cast" it as an int, so it can then be modified and displayed as a hexadecimal

anyway, so i got it to compile, but when i run the .exe file, the command prompt opens up and I type in a decimal number but when i press ENTER the program just closes.....why would it do this??

Adak · 01-23-2010, 12:42 PM

This is your program with a few wrinkles. It runs fine, but you have NO logic to change the float, or the integer, into a hexadecimal number.

Code:

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

int main()
{
	union
	{
		int a;
		float b;
	}myUnion;

	printf("Enter the Decimal number to be converted:\n");

	scanf("%f",&myUnion.b);


	printf("The float number in base 10 is: %f, and in Hexadecimal is: %x\n", myUnion.b, myUnion.a);

	printf("%d",(int) myUnion.b);      //NOTE the cast to an int

  while((myUnion.a = getchar()) != '\n'); //these two lines work to hold the console window open
  myUnion.a = getchar();

  return 0;
}

emblem4ever · 01-23-2010, 01:12 PM

thanks for all the help, but i think ur still misunderstanding what i want to do

i want to input a float number, like -27.198EE33 (-27.198 x 10^33)
then using the union to read it as an int (sorry for saying cast before) so it can be outputed in hexadecimal form (as already said the hex modifier only works on int values)
so in this case it sould spit out F8A7EE1

but for some reason its not working properly, like if input 13, which is D in hexadecimal, i get 41500000 as an output

Memloop · 01-23-2010, 01:18 PM

That's not how unions work. I suggest you read this:

Union Declarations (C)

Quote:

If a union of two types is declared and one value is stored, but the union is accessed with the other type, the results are unreliable. For example, a union of float and int is declared. A float value is stored, but the program later accesses the value as an int. In such a situation, the value would depend on the internal storage of float values. The integer value would not be reliable.

Adak · 01-23-2010, 01:18 PM

You're saying that the printf() format specification %x will automatically convert your number from a base 10 number, into a base 16 hex number.

That has not been my experience.

%x will print a hex number in the correct format. That is all it will do, AFAIK.

If you want a number converted from base 10 to base 16, you do the conversion or use a standard function to do it for you. After the conversion, you can use %x or %X to print up the hex number.

Memloop · 01-23-2010, 01:20 PM

C99 has a %A specifier that will print a double as hex.

Adak · 01-23-2010, 01:28 PM

%A eh? C99 is sounding better all the time. Thanks Memloop, for remembering to keep me in the loop.

Did I say that?

kermitaner · 01-23-2010, 01:45 PM

Quote:

Originally Posted by Adak

You're saying that the printf() format specification %x will automatically convert your number from a base 10 number, into a base 16 hex number.

That has not been my experience.

%x will print a hex number in the correct format. That is all it will do, AFAIK.

If you want a number converted from base 10 to base 16, you do the conversion or use a standard function to do it for you. After the conversion, you can use %x or %X to print up the hex number.

no, printf can do this:

Code:

 printf("%x\n",15);

should give you a nice little "f"

Adak · 01-23-2010, 01:48 PM

15 is not a variable. It is a hex number.

Try that with an integer variable (base 10), and see who gets the "f".

I should add that my compiler absolutely won't do it. I've tried several times already.

kermitaner · 01-23-2010, 02:02 PM

Quote:

Originally Posted by Adak

15 is not a variable. It is a hex number.

Try that with an integer variable (base 10), and see who gets the "f".

I should add that my compiler absolutely won't do it. I've tried several times already.

Code:

    int i=255;
  printf("%x\n",i);

even gives me 2 "ff"

maybe u should try another compiler...

01-23-2010, 11:44 AM	#1
emblem4ever Registered User Join Date: Jan 2010 Posts: 9	new to C - decimal to hexadecimal converter hi, I'm new to C programming and I'm still very unsure of many things. I tried making a decimal to hexadecimal converter, using a very simple method, yet it doesn't seem to work and I can't figure out what syntax errors i have, heres my code Code: #include <stdio.h> #include <stdlib.h> #include <string.h> void main { Union { int a; float b; }myUnion; printf("Enter the Decimal number to be converted:\n"); scanf("%f",&myUnion.b\n); printf("The number in Hexadecimal is:\n"); printf("%x",&myUnion.a); } I tried to make it really simple, but whats wrong??
emblem4ever is offline

01-23-2010, 11:47 AM	#2
Salem and the hat of Jobseeking Join Date: Aug 2001 Location: The edge of the known universe Posts: 21,639	> scanf("%f",&myUnion.b\n); Remove the \n here > printf("%x",&myUnion.a); Remove the & here > void main Remove the void here, and replace it with int __________________ If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
Salem is offline

01-23-2010, 11:54 AM	#3
emblem4ever Registered User Join Date: Jan 2010 Posts: 9	oh, thanks a lot, however i still get many errors, but i really don't get why, like it says syntax error before "{" in line 11 and conflicting types for 'printf' etc. but according to my knowlegde and other programs i've seen, it should work fine...? maybe its my compiler, im using Dev-C 4.9.9.2 because im using vista.... hmmmmm
emblem4ever is offline

01-23-2010, 11:57 AM	#4
Salem and the hat of Jobseeking Join Date: Aug 2001 Location: The edge of the known universe Posts: 21,639	> Union Try union All C keywords are lowercase. __________________ If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
Salem is offline

01-23-2010, 11:59 AM	#5
Memloop Robot Join Date: Mar 2009 Posts: 258	C is case dependent, and the standard data type for unions is union and not Union. The %x modifier expects an integer, but you're passing it a float. I have no idea why you're storing the input in an union to begin with, let alone a float.
Memloop is offline

01-23-2010, 12:10 PM	#6
emblem4ever Registered User Join Date: Jan 2010 Posts: 9	because i want to scan the input as a float, i.e. user enters 26.74628, then using the union, "cast" it as an int, so it can then be modified and displayed as a hexadecimal anyway, so i got it to compile, but when i run the .exe file, the command prompt opens up and I type in a decimal number but when i press ENTER the program just closes.....why would it do this?? Last edited by emblem4ever; 01-23-2010 at 12:35 PM.
emblem4ever is offline

01-23-2010, 01:12 PM	#8
emblem4ever Registered User Join Date: Jan 2010 Posts: 9	thanks for all the help, but i think ur still misunderstanding what i want to do i want to input a float number, like -27.198EE33 (-27.198 x 10^33) then using the union to read it as an int (sorry for saying cast before) so it can be outputed in hexadecimal form (as already said the hex modifier only works on int values) so in this case it sould spit out F8A7EE1 but for some reason its not working properly, like if input 13, which is D in hexadecimal, i get 41500000 as an output
emblem4ever is offline

01-23-2010, 01:18 PM	#10
Adak Registered User Join Date: Sep 2006 Posts: 3,125	You're saying that the printf() format specification %x will automatically convert your number from a base 10 number, into a base 16 hex number. That has not been my experience. %x will print a hex number in the correct format. That is all it will do, AFAIK. If you want a number converted from base 10 to base 16, you do the conversion or use a standard function to do it for you. After the conversion, you can use %x or %X to print up the hex number.
Adak is online now

01-23-2010, 01:20 PM	#11
Memloop Robot Join Date: Mar 2009 Posts: 258	C99 has a %A specifier that will print a double as hex.
Memloop is offline

01-23-2010, 01:28 PM	#12
Adak Registered User Join Date: Sep 2006 Posts: 3,125	%A eh? C99 is sounding better all the time. Thanks Memloop, for remembering to keep me in the loop. Did I say that?
Adak is online now

01-23-2010, 01:48 PM	#14
Adak Registered User Join Date: Sep 2006 Posts: 3,125	15 is not a variable. It is a hex number. Try that with an integer variable (base 10), and see who gets the "f". I should add that my compiler absolutely won't do it. I've tried several times already. Last edited by Adak; 01-23-2010 at 01:51 PM.
Adak is online now

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