printf bug ??!

[Thomas Kuehne]

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lanael schrieb am 2006-03-16:
> void main() {
> ulong u =0b100000000000000000000000000000000000000000000000; // 1 and 
> 47 zeros
> ulong u1=0b1000000000000000000000000000000000000000000000000; // 1 and 
> 48 zeros
> printf("u  = \t %llx \t %llb\n",u,u);
> printf("u1 = \t %llx \t %llb\n",u1,u1);
> }
>
> output :
> u  =     800000000000    
> 100000000000000000000000000000000000000000000000
> u1 =     1000000000000    
> 000000000000000000000000000000000000000000000000
>
> it looks like a buffer overflow...

The %llb notation isn't part of the standard printf implementation.
(see C99, ISO 9899 7.19.6.1.8)

I'm not saying that this is a bug or not, but the code isn't portable:
> #include <stdio.h>
>
> int main(){
>	unsigned long long i = 9;
>	printf("%llb\n", i);
>	return 0;
> }

prints "%b" under Linux

Thomas


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