size_t in c++?

[saneman]

In a file I have made:

size_t bb;
bb = 3u;
printf("bb %d\n");

which prints:

bb 2280640

But what is 'u' and why does it print the above number?

[Ismo Salonen]

saneman wrote:

In a file I have made:

size_t bb;
bb = 3u;
printf("bb %d\n");

which prints:

bb 2280640

But what is 'u' and why does it print the above number?

Prints garbage from stack, there is no argument given for %d.
Should be : printf("bb %d\n",bb);

ismo

[Victor Bazarov]

saneman wrote:

In a file I have made:

size_t bb;
bb = 3u;
printf("bb %d\n");

which prints:

bb 2280640

But what is 'u' and why does it print the above number?

'u' is a suffix to make your literal *unsigned*. It prints what you see
by *pure chance*. You provided the format specification (%d) but no
corresponding argument, so the behaviour of your program is undefined,
it can do whatever it wants. Search the web for "nasal demons" (with
quotes).

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[blargg]

In article <8U*************@read4.inet.fi>, Ismo Salonen
<no****@another.invalidwrote:

saneman wrote:

size_t bb;
bb = 3u;
printf("bb %d\n");
...

Prints garbage from stack, there is no argument given for %d.
Should be : printf("bb %d\n",bb);

That prints 0 on one of my machines (big-endian, 16-bit int). %d specifies
an int, but size_t isn't necessarily an (unsigned) int. Since the subject
and group is C++, not C, the way to print it is

std::cout << bb;

If printf must be used for some reason, cast bb to an int, or better, use
unsigned long:

printf( "bb %lu\n", (unsigned long) bb );

[Markus Moll]

Hi

Ismo Salonen wrote:

saneman wrote:

>In a file I have made:

size_t bb;
bb = 3u;
printf("bb %d\n");

Prints garbage from stack, there is no argument given for %d.
Should be : printf("bb %d\n",bb);

Even then, %d and size_t don't match. Use %zu for size_t, as it might be
larger than an int.

Markus

[Pete Becker]

On 2008-08-01 09:17:53 -0400, Markus Moll
<ma*********@esat.kuleuven.ac.besaid:

Hi

Ismo Salonen wrote:

>saneman wrote:

>>In a file I have made:

size_t bb;
bb = 3u;
printf("bb %d\n");

Prints garbage from stack, there is no argument given for %d.
Should be : printf("bb %d\n",bb);

Even then, %d and size_t don't match. Use %zu for size_t, as it might be
larger than an int.

Well, yes, non-portably. "%zu" is new with C99, and the C++ standard is
based on C90. The next version of the standard will provide "%zu".

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

[Anarki]

On Aug 1, 6:17*pm, Markus Moll <[e-mail]>
wrote:

Hi

Ismo Salonen wrote:

saneman wrote:

In a file I have made:

size_t bb;
bb = 3u;
printf("bb %d\n");

Prints garbage from stack, there is no argument given for %d.
Should be : printf("bb %d\n",bb);

Even then, %d and size_t don't match. Use %zu for size_t, as it might be
larger than an int.

Markus

excuse me Markus whats %zu? whats the z stand for?

[Ian Collins]

Anarki wrote:

On Aug 1, 6:17 pm, Markus Moll <[e-mail]>
wrote:

>Hi

Ismo Salonen wrote:

>>saneman wrote:
In a file I have made:
size_t bb;
bb = 3u;
printf("bb %d\n");
Prints garbage from stack, there is no argument given for %d.
Should be : printf("bb %d\n",bb);

Even then, %d and size_t don't match. Use %zu for size_t, as it might be
larger than an int.

Markus

excuse me Markus whats %zu? whats the z stand for?

It's the C99 format specifier for size_t.

--
Ian Collins.