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Typecast of one argument to printf impacts other ???

When the below program is compiled and executed,

#include <stdio.h>

#define MASK 0xFFFFFFULL

main() {
unsigned long long a;
unsigned long b, c;

a = 0x12345678ULL;
b = a & MASK;
c = 0x00345678UL;

printf("%d %d\n", sizeof(unsigned long), sizeof(unsigned long
long));
printf("0x%x %lu \t 0x%x %lu \t 0x%x %lu \n",
a & MASK, a & MASK, b, b, c, c);

}

the result from a BigEndian machine is

4 8
0x0 3430008 0x0 3430008 0x345678 3430008

Surprisingly, the result of 'a & MASK' and 'b' are 0x0, when 0x%x
format is used.

And the result from a Little Endian machine is

4 8
0x345678 0 0x345678 0 0x345678 3430008

Again more surprisingly, the results are 0 when %lu is used. Why these
unexpected results ?

Also, when the printf is changed like this,

printf("0x%x %lu \t 0x%x %lu \t 0x%x %lu \n",
(unsigned long)(a & MASK), a & MASK, b, b, c, c);

the result is

0x345678 0 0x345678 3430008 0x345678 3430008

in a Big Endian machine and

0x345678 3430008 0x0 3430008 0x345678 3430008

in a little endian.

And with the below

printf("0x%x %lu \t 0x%x %lu \t 0x%x %lu \n",
(unsigned long)(a & MASK), (unsigned long)(a & MASK), b, b,
c, c);

the results are as expected

0x345678 3430008 0x345678 3430008 0x345678 3430008

in both the machines.

How does typecasting of one argument affect printed values of others ?
Or is there anyother reason behind it and what I see is a misconception
of some hidden fact ?

Nov 15 '05 #1
3 2008

na******@yahoo. co.in wrote:
When the below program is compiled and executed,

#include <stdio.h>

#define MASK 0xFFFFFFULL

main() {
unsigned long long a;
unsigned long b, c;

a = 0x12345678ULL;
b = a & MASK;
c = 0x00345678UL;

printf("%d %d\n", sizeof(unsigned long), sizeof(unsigned long
long));
printf("0x%x %lu \t 0x%x %lu \t 0x%x %lu \n",
a & MASK, a & MASK, b, b, c, c);
For the first vararg parameter, you're passing an unsigned long long
when the corresponding format specifier, %x, expects an unsigned int.
For the second vararg parameter, you're passing an unsigned long long
when %lu expects an unsigned long. Mismatching format specifiers with
the parameter types is *undefined behavior*. The compiler is free to do
ANYTHING IT WISHES.

However, I suppose you want a practical, "what is most-likely happening
behind the scenes" answer. Okay, here is the most-likely reason:

In a typical implementation, when you call printf, the parameters are
pushed onto the stack from right to left, and so, your stack looks like
this:

(separated into 4-byte groups -- big endian)
00345678 c -- bottom of stack
00345678 c
00345678 b
00345678 b
00345678 a & MASK (least significant 4 bytes)
00000000 (most significant 4 bytes)
00345678 a & MASK
00000000
ptr_to_format_s tring -- top of stack (popped first)

And so, printf pops off the pointer to the format string and starts
parsing the format string. It encounters an %x, which means to display
the *unsigned int* parameter in hexadecimal format. And so, since an
unsigned int is likely to be 4 bytes on your platform, 4 bytes are
popped off the stack, and so 00000000 is popped and represented in
hexadecimal. Then %lu is encountered and since it expects an unsigned
long, which is 4 bytes on your platform, 4 bytes are popped, which is
00345678 in hex, or 3430008 in decimal. etc.

Solution: use the right format specifiers: %llx and %llu
}

the result from a BigEndian machine is

4 8
0x0 3430008 0x0 3430008 0x345678 3430008

Surprisingly, the result of 'a & MASK' and 'b' are 0x0, when 0x%x
format is used.

And the result from a Little Endian machine is

4 8
0x345678 0 0x345678 0 0x345678 3430008

Again more surprisingly, the results are 0 when %lu is used. Why these
unexpected results ?

Also, when the printf is changed like this,

printf("0x%x %lu \t 0x%x %lu \t 0x%x %lu \n",
(unsigned long)(a & MASK), a & MASK, b, b, c, c);

the result is

0x345678 0 0x345678 3430008 0x345678 3430008

in a Big Endian machine and

0x345678 3430008 0x0 3430008 0x345678 3430008

in a little endian.

And with the below

printf("0x%x %lu \t 0x%x %lu \t 0x%x %lu \n",
(unsigned long)(a & MASK), (unsigned long)(a & MASK), b, b,
c, c);

the results are as expected

0x345678 3430008 0x345678 3430008 0x345678 3430008

in both the machines.

How does typecasting of one argument affect printed values of others ?
Or is there anyother reason behind it and what I see is a misconception
of some hidden fact ?


Nov 15 '05 #2
On 15 Oct 2005 05:19:02 -0700, na******@yahoo. co.in wrote:
When the below program is compiled and executed,

#include <stdio.h>

#define MASK 0xFFFFFFULL

main() {
unsigned long long a;
unsigned long b, c;

a = 0x12345678ULL;
b = a & MASK;
c = 0x00345678UL;

printf("%d %d\n", sizeof(unsigned long), sizeof(unsigned long
long));
sizeof evaluates to a size_t which is an unsigned integer of some
type. %d requires an int so you have potential undefined behavior
here. A common recommendation is to use %lu and cast the sizeof to
unsigned long.
printf("0x%x %lu \t 0x%x %lu \t 0x%x %lu \n",
a & MASK, a & MASK, b, b, c, c);
Here have more undefined behavior.

%x requires an unsigned int. a & MASK is an unsigned long long.

%lu requires an unsigned long. a & MASK is still an unsigned long
long.

%x requires an unsigned int. b is an unsigned long. So is c.

}

the result from a BigEndian machine is

4 8
0x0 3430008 0x0 3430008 0x345678 3430008

Surprisingly , the result of 'a & MASK' and 'b' are 0x0, when 0x%x
format is used.
You cannot lie to printf and expect things to work, especially since
you know that unsigned long and unsigned long long have different
sizes.

And the result from a Little Endian machine is

4 8
0x345678 0 0x345678 0 0x345678 3430008

Again more surprisingly, the results are 0 when %lu is used. Why these
unexpected results ?
You cannot lie to printf and expect things to work, especially since
you know that unsigned long and unsigned long long have different
sizes.

Also, when the printf is changed like this,

printf("0x%x %lu \t 0x%x %lu \t 0x%x %lu \n",
(unsigned long)(a & MASK), a & MASK, b, b, c, c);

the result is

0x345678 0 0x345678 3430008 0x345678 3430008

in a Big Endian machine and

0x345678 3430008 0x0 3430008 0x345678 3430008

in a little endian.

And with the below

printf("0x%x %lu \t 0x%x %lu \t 0x%x %lu \n",
(unsigned long)(a & MASK), (unsigned long)(a & MASK), b, b,
c, c);

the results are as expected

0x345678 3430008 0x345678 3430008 0x345678 3430008

in both the machines.

How does typecasting of one argument affect printed values of others ?
Or is there anyother reason behind it and what I see is a misconception
of some hidden fact ?


Welcome to the wonderful world of undefined behavior. One
manifestation is "do something to really confuse the user."
<<Remove the del for email>>
Nov 15 '05 #3
na******@yahoo. co.in wrote:
When the below program is compiled and executed,

/* Your program is hopelessly broken. Learn to use specifiers that
match your data types. Also learn that sizeof() returns an unsigned
value. Note the changes below. As a style isssue, using tabs for
output meant to be human-readable is a mistake. All of your
questions are related to your errors. */
#include <stdio.h>

#define MASK 0xFFFFFFULL

int main(void)
{
unsigned long long a;
unsigned long b, c;

a = 0x12345678ULL;
b = a & MASK;
c = 0x00345678UL;

printf("%d %d\n", (int) sizeof(unsigned long),
(int) sizeof(unsigned long long));
printf("%#llx %llu; %#lx %lu; %#lx %lu\n", a & MASK, a & MASK, b,
b, c, c);
return 0;

}
[output]

4 8
0x345678 3430008; 0x345678 3430008; 0x345678 3430008

#include <stdio.h>

#define MASK 0xFFFFFFULL

main() {
unsigned long long a;
unsigned long b, c;

a = 0x12345678ULL;
b = a & MASK;
c = 0x00345678UL;

printf("%d %d\n", sizeof(unsigned long), sizeof(unsigned long
long));
printf("0x%x %lu \t 0x%x %lu \t 0x%x %lu \n",
a & MASK, a & MASK, b, b, c, c);

}

Nov 15 '05 #4

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