How to swap the nibbles whit dev++?

[deeyay]

Hello, i've got a project and im missing one of the parts "yes i've just started whit programming"
I'm using Dev++

How to swap the nibbles.

Ect   1010 - 1101
   To 1101 - 1010
"
 
#define _swapNibble(x) ((x<<4)|(x>>4))
 
int main()
{
    // On executing you will get
    // x before swap = 0x1c
    // x after swap = 0xc1
    char x = 0x1c;
    printf("x before swap = %x\n",x);
 
    _swapNibble(x);
    printf("x afer swap = %x",x);
}

""
This is what i've got but got some faults"

Thank you.

[Rogerup]

Change the #define to:

#define _swapNibble(x) (((x<<4)|(x>>4))&0xFF)

or

#define _swapNibble(x) ((char)((x<<4)|(x>>4)))

[deeyay]

In function `int main()':



\c++\Calc\tip\SAH\you.cpp:10: error: `printf' undeclared (first use this function)


: error: (Each undeclared identifier is reported only once for each function it appears in.)

Execution terminated

. i can't see what i've done wrong..

[Rogerup]

At the beginning of your code put:

#include <stdio.h>

[donbock]

Line 12: you declare x as a char. This type should only be used for characters. If you want a small integral type then use either signed char or unsigned char, depending on whether or not you want the value to be signed. (However, in subsequent comments I argue for using type int.)

Line 5: your macro is named _swapNibble. The C Standard reserves all names with a leading underscore just in case they are needed in future versions of the Standard. You should not use a reserved name. I suggest you remove the leading underscore.

Line 13: conversion specifier %x expects an int argument; however you pass x, which is a char. This is a good reason to declare x as an int.

Line 5: the expression that the macro expands to has type int despite the operand being type char. This is another good reason to declare x as an int.

Line 15: this statement evaluates an expression and then discards the result. You need an assignment statement if you want to assign the result to some variable.