c problem

[saurabh jain]

i tried to run this prog
main()
{
unsigned char m=0x80;
printf("%d",m<<1);
}

what should be the output of above prog.

 

[Emmanuel Delahaye]

saurabh jain vient de nous annoncer :

i tried to run this prog
main()
{
unsigned char m=0x80;
printf("%d",m<<1);
}

what should be the output of above prog.

Nothing sensible unless you fix it to avoid invoking undefined
behaviours.

Fixes and enhancements:

#include <stdio.h>

int main (void)
{
unsigned char m = 0x80;

printf("%u\n", (unsigned) (m << 1u));

return 0;
}

I would have said that m will be converted to an int (or unsigned int,
I'm not sure. True C-gurus will answer that), then the shift occurs.
The new value is now 0x100. This value is converted to unsigned int and
passed to printf() that displays it as an unsigned decimal value (256).

Let's test:

D:\CLC\S\SAURABH>bc proj.prj
256

Of course, it proves nothing, but it lets me a chance that the fixes
and explanation was good...

 

[Darrell Grainger]

i tried to run this prog
main()
{
unsigned char m=0x80;
printf("%d",m<<1);
}

what should be the output of above prog.

Undefined.

 

[Old Wolf]

#include <stdio.h>

int main (void)
{
unsigned char m = 0x80;
printf("%u\n", (unsigned) (m << 1u));
return 0;
}

I would have said that m will be converted to an int (or unsigned int,
I'm not sure. True C-gurus will answer that), then the shift occurs.

Converted to int if INT_MAX >= UCHAR_MAX, otherwise, to unsigned int.

Also the 'u' on '1u' makes no difference (the type of the right-hand
operand of the shift operators doesn't affect the type of the result)