Hi all,
I have a question regarding the gcc behavior (gcc version 3.3.4).
On the following test program it emits a warning:
#include <stdio.h>
int aInt2[6] = {0,1,2,4,9,16};
int aInt3[5] = {0,1,2,4,9};
void print1 (const int* p, size_t cnt)
{
while (cnt--)
printf ("%d\n", *p++);
}
void print2 (const int (*p)[5])
{
size_t cnt;
#if 0
// prohibited:
(*p)[0]=0;
#endif
for (cnt=0; cnt<5; cnt++)
printf ("%d\n", (*p)[cnt]);
}
int main()
{
printf ("test begin\n");
print1 (aInt2, sizeof(aInt2)/sizeof(aInt2[0]));
print2 (&aInt3); // <-- warns here
printf ("test end\n");
return 0;
}
The warning is:
MOD.c: In function `main':
MOD.c:: warning: passing arg 1 of `print2' from incompatible pointer type
Why is that? How is this different from using print1 (aInt2, ...)? All I
want to do is to explicitly show that the pointer is to an array of 5
entries and ensure that it's not modified in anyway inside print2(). However
the #if 0'd assignment operator inside print2() is correctly handled by
gcc -- an error is generated:
MOD.c: In function `print2':
MOD.c:: error: assignment of read-only location
What's wrong with gcc?
There's another thing about the pointer to the array is... The compiler
doesn't generate a warning if I change in print2()
for (cnt=0; cnt<5; cnt++)
to
for (cnt=0; cnt<6; cnt++)
and lets me access (*p)[5], though I think at least a warning should be
generated here.
It doesn't warn me if I put
aInt3[6] = 0;
into main(). But in both cases the compiler "knows" the real size of the
array, still no warning...
Is this OK???
Thanks,
Alex
P.S. I compile it as gcc MOD.c -Wall -oMOD.exe
