%u printf

[saurabh]

#include<stdio.h>
int main()
{
int i;
printf("\n%u%u",sizeof(i));
}

output 42433316

plz explain the output
with 1 %u o/p =4bytes
with 2 %u o/p=42433316 ????

[Ian Collins]

saurabh wrote:

#include<stdio.h>
int main()
{
int i;
printf("\n%u%u",sizeof(i));

Here you tell printf to expect two more parameters, but you only pass one.

}

output 42433316

plz explain the output

Please use real words.

with 1 %u o/p =4bytes
with 2 %u o/p=42433316 ????

The first 4 is the valid output, the rest is undefined garbage.

--
Ian Collins.

[Giorgio Silvestri]

"Ian Collins" <ia******@hotmail.comha scritto nel messaggio
news:4k************@individual.net...

saurabh wrote:

#include<stdio.h>
int main()
{
int i;
printf("\n%u%u",sizeof(i));

Here you tell printf to expect two more parameters, but you only pass one.

}

output 42433316

plz explain the output

Please use real words.

with 1 %u o/p =4bytes
with 2 %u o/p=42433316 ????

The first 4 is the valid output, the rest is undefined garbage.

Moreover "%u" is valid in order to print unsigned int.

printf("%u\n",(unsigned int)sizeof(i));

size_t can be "bigger" than unsigned int.

In C99 you can use %zu for size_t.

printf("%zu\n",sizeof(i));

--
Giorgio Silvestri
DSP/Embedded/Real Time OS Software Engineer

[Herbert Rosenau]

On Sun, 20 Aug 2006 19:32:14 UTC, "saurabh" <ro***********@gmail.com>
wrote:

#include<stdio.h>
int main()

int main(void)

{
int i;
printf("\n%u%u",sizeof(i));
}

output 42433316

You're multiple times in undefined behavior:
1. You use uninitialised i.
2. You use more conversion flags in the format string as defined in
actual parameters.
3. sifeof returns type size_t not unsigned int.

Another error is that you lets mis the '\n' immediately before the
format string ends.

The output of your program means nothing. It is completely random
based on the errors. Undefined behavior means that all and anything
one can thing about may occure.

I've seen seldom so manny errors in so little number of lines of code.

Fix all the errors and try again.

--
Tschau/Bye
Herbert

Visit http://www.ecomstation.de the home of german eComStation
eComStation 1.2 Deutsch ist da!

[Clark S. Cox III]

Herbert Rosenau wrote:

On Sun, 20 Aug 2006 19:32:14 UTC, "saurabh" <ro***********@gmail.com>
wrote:

>#include<stdio.h>
int main()

int main(void)

>{
int i;
printf("\n%u%u",sizeof(i));
}

output 42433316

You're multiple times in undefined behavior:
1. You use uninitialised i.
2. You use more conversion flags in the format string as defined in
actual parameters.
3. sifeof returns type size_t not unsigned int.

While #2 and #3 are indeed undefined, there is nothing undefined about
#1 (Note, that the OP doesn't use the *value* of i)
--
Clark S. Cox III
cl*******@gmail.com

[Eric Sosman]

Herbert Rosenau wrote:

On Sun, 20 Aug 2006 19:32:14 UTC, "saurabh" <ro***********@gmail.com>
wrote:

>>#include<stdio.h>
int main()

int main(void)

>>{
int i;
printf("\n%u%u",sizeof(i));
}

output 42433316

You're multiple times in undefined behavior:
1. You use uninitialised i.

No; sizeof does not evaluate its operand.

[...]

--
Eric Sosman
es*****@acm-dot-org.invalid

[Jordan Abel]

2006-08-22 <wm***************************@JUPITER1.PC-ROSENAU.DE>,
Herbert Rosenau wrote:

On Sun, 20 Aug 2006 19:32:14 UTC, "saurabh" <ro***********@gmail.com>
wrote:

>#include<stdio.h>
int main()

int main(void)

>{
int i;
printf("\n%u%u",sizeof(i));
}

output 42433316

1. You use uninitialised i.

so?

You're right about the rest, though.