Using sizeof() on a parameter array
Doug Gwyn
gwyn at smoke.brl.mil
Mon May 20 14:13:22 AEST 1991
In article <12151 at jarthur.Claremont.EDU> jseidman at jarthur.Claremont.EDU (Jim Seidman) writes:
-void test(char param[80]) {
- char local[80];
- printf("sizeof(param) = %d\n", sizeof(param));
- printf("sizeof(local) = %d\n", sizeof(local));
-}
-Now, I thought that this should print out the same size for both, namely
-80. But according to MSC, local has a size of 80, and param has a size
-of 2 (the size of a pointer).
MSC is right. The parameter DOES have type pointer-to-char; you've
just written in in an opaque way.
More information about the Comp.lang.c
mailing list