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Hello,
I am using Slackware 13.0 64bits in a AMD Athlon 4400 64 x2
and, i found something interesting with C and GCC, look at this:
file1.c
/**********/
int main(void)
{
char *p = 0;
p = (char *)getpointer();
return 0;
}
/***********/
file2.c
/***********/
void *getpointer(void)
{
return 0x1234567887654321;
}
/***********/
can you imagine that if you compile those 2 files
gcc -c file1.c
gcc -c file2.c
gcc file1.o file2.o
and run it, the valued assigned to p is DIFFERENT from 0x1234567887654321? actually, in my
machine p gets the value 0xffffffff87654321
why? because getpointer() was NOT declared in file1.c, you fix this "issue" by adding
void *getpointer(void);
at the beginning of file1.c
why does this happen? because when gcc doesn't find the function getpointer(), I think it
"declares" it something like this
int getpointer(...);
and you are NOT warned that you forgot to declare the function (well, it actually tells
you, if you compile with the -Wall option)
so, the value returned by getpointer() is casted to int, losing 32bits, and then assigned
to p. With a 32bits architecture, ints and pointers are both 32bits long, so there is no
problem. Also, the same happens with long, double, and any other bigger-than-int.
if the getpointer() function returned, for example, char *, and you don't use the (char *)
while calling it, gcc warns you that you are assigning an int value to a char *.
The question is... is this a 100% programmer problem? or should gcc warn you also when you
don't use the -Wall option?
Thanks for reading,
Saludos!
- --
"Do you pine for the nice days of minix-1.1, when men were men and wrote their own device
drivers?" -- Linus Torvalds
"If you want to go somewhere, goto is the best way to get there." -- Ken Thompson
"Value your freedom or you will lose it" -- Richard Stallman
GNU/Linux Registered User #460377
Slackware 13.0 64 bits
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