Question

width specifier in printf

Asked by: tittu

the following code display output as 0xffffffab  instead 0xab

how to give width specifier in printf to display only 0xab.

i tried all but none of them worked for me.



#include <stdio.h>
int main() {
 
    char m;
    m = 0xab;
    printf("m = %x \n",m);
 
 
}

                                  
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Asked On
2009-09-22 at 22:51:59ID24753943
Topics

C Programming Language

,

Open Source Programming

Participating Experts
3
Points
500
Comments
6

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Answers

 

by: pgnatyukPosted on 2009-09-22 at 23:17:52ID: 25400418

Like that?

#include <stdio.h>
int main() 
{
	char m;
	m = 0xab;
	unsigned char x = 0xAB;
	printf("m = %x \n",m);
	printf("x = 0x%x \n",x);
}
                                              
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by: efnPosted on 2009-09-23 at 02:27:31ID: 25401167

A width specifier is not going to do it, because it can only be used to pad, not to truncate, the char gets promoted to an int when you pass it to a function, and printf expects an unsigned int for %x.

pgnatyuk showed one way to fix it.  Here's another:

printf("m = %x \n", m & 0xFF);

This works because it masks out the leading FFs you get from sign extension.

 

by: tittuPosted on 2009-09-23 at 06:22:08ID: 25402739

is there any other way without doing any mask operations.

 

by: Infinity08Posted on 2009-09-23 at 06:52:25ID: 25403030

You could always use unsigned char ...

 

by: Infinity08Posted on 2009-09-23 at 06:53:12ID: 25403033

Oh, but I see pgnatyuk already suggested that. Sorry. Ignore my previous post, please.

 

by: tittuPosted on 2009-09-24 at 01:44:38ID: 31632309

it worked with unsigned char thanks.
but what difference it makes with signed char ? does
the sign bit is shown as fffffff

20120131-EE-VQP-002

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