C
CBFalconer
Old said:
Well, if they are not the same size, it will be hard to stuff both
types into the storage for the smaller type.
I suspect somequoting is missing.
Well, if they are not the same size, it will be hard to stuff both
types into the storage for the smaller type.
I suspect somequoting is missing.
P
Peter Nilsson
Old Wolf said:
If they're different sizes, then they can't be compatible.
O
Old Wolf
OP's trying to assign a pointer-to-double to a double.
Types don't have to be the same size in order for one
to be assignable to the other; for example, you can
assign a char to a double.
M
Mark McIntyre
Peter said:
#include <stdio.h>
int main(void)
{
char c = 'A';
long double x = c;
printf("%d %d %d\n", sizeof 'A', sizeof c, sizeof x);
return 0;
}
$ ./a.out
4 1 12
P
Peter Nilsson
Old Wolf said:
In that case, you may as well ask Ian why he made the first
statement. As it was, it looked like you were questioning
the value of the second statement in the context of the
first.
True, but no one said they did.
P
Peter Nilsson
Is there a point to this broken code?
G
Geoff
What's broken about it? It compiles without errors or warnings of any
kind.
It shows that it is possible to store one arithmetic type in another
without upsetting the compiler even though they are different sizes.
P
Peter Nilsson
Geoff said:
%d expects and int, not size_t.
I hope that's not your criteria for what makes a correct
program. [P.S. gcc -Wall diagnoses the error.]
You mean _a_ compiler.
Which has what to do with my statement above?
[cf my reply to Old Wolf.]
W
Walter Roberson
sizeof returns size_t, but the printf format used is %d (which is for
int).
printf("%d %d %d\n", (int) sizeof 'A', (int) sizeof c, (int) sizeof x);
R
Richard Heathfield
Geoff said:
Type mismatches between format specifier and subsequent arguments to
printf.
All that means is that it doesn't contain any syntax errors or constraint
violations. It doesn't mean it's correct C code.
<snip>
Type mismatches between format specifier and subsequent arguments to
printf.
All that means is that it doesn't contain any syntax errors or constraint
violations. It doesn't mean it's correct C code.
<snip>
G
Geoff
size_t is unsigned int?
Not all versions of gcc. Maybe yours. This would make it
implementation dependent, would it not? Syntactically correct is not
necessarily logically correct, but no one said it was. Nor does
"broken" make any meaningful distinction between logical correctness
and syntactically correct. As a demonstration of the meaninglessness
of saying "different sizes are incompatible" it is a fine example.
Yep. Actually three different ones.
McIntyre can defend his own demonstration if he wants to. I don't see
the relevance of "incompatible size" to the original bug of trying to
modify a pointer from a float.
I
Ian Collins
In might be, it all depends on the system. On most systems I use, it'sGeoff said:
unsigned long. On 64 bit systems with with 32 bit long, it's probably
unsigned long long.
R
Richard Heathfield
Geoff said:
size_t is an unsigned integer type. It could be unsigned char, unsigned
short int, unsigned int, unsigned long int, or (if you have C99) even
unsigned long long int.
The implementation gets the choice of whether to diagnose this (and it does
not have to document its choice). The behaviour of the program, however,
is undefined, whether or not a diagnosis occurs.
<snip>
size_t is an unsigned integer type. It could be unsigned char, unsigned
short int, unsigned int, unsigned long int, or (if you have C99) even
unsigned long long int.
The implementation gets the choice of whether to diagnose this (and it does
not have to document its choice). The behaviour of the program, however,
is undefined, whether or not a diagnosis occurs.
<snip>
G
Geoff
Ah! Well casting makes everything alright then. Why go to all that
trouble? Why not just use %i or even better, %u?
And why has anyone not noticed that sizeof 'A' returned 4, not 1?
J
James Kuyper
Geoff said:
It's an unsigned type; whether or not it's unsigned int is
implementation-defined.
R
Richard Heathfield
Geoff said:
The value yielded by sizeof 'A' depends on your implementation. It may be
1, 2, 3, 4, or indeed any positive integer.
The value yielded by sizeof 'A' depends on your implementation. It may be
1, 2, 3, 4, or indeed any positive integer.
P
Peter Nilsson
[Not really. The program still doesn't bare any relation to
my statement that in order to be compatible types, two
types must have the same size. Note that compatible type is
a concept that is explicitly defined in the C standard.
Suitability of operands in assignment is a separate issue.]
Because they are also wrong. %u will only work if size_t
happens to be an unsigned int, but on the virtual C machine
you cannot assume this.
C99 added a length modifier of z, so size_t can be printed
(and scanned) with %zu. But for C99, a strictly conforming
program must cast the value.
Because there are very few hosted systems where sizeof(int)
is 1. Indeed, it's held that such systems are not conforming.
[In C++, 'A' has character type; in C, it has type int.]
C
CBFalconer
Geoff said:
Because it's wrong.
Because it's right. Most machines will make it 2 or 4.
K
Keith Thompson
Geoff said:
"%i" does the same thing as "%d". "%u" is correct only if size_t
happens to be unsigned int.
I'm sure it was noticed; nobody mentioned it because it's
unremarkable. See question 8.9 in the comp.lang.c FAQ,
<http://www.c-faq.com/>.
S
santosh
Peter said:
But for C89, a strictly...
<snip>