Using sizeof() on a parameter array

[Robert Earl]

In article <1991May19.135611.6332 at monu0.cc.monash.edu.au> ins845b at monu4.cc.monash.edu.au (mr  k.l. lentin) writes:

|   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));
|   >}
|   >
|   C passes all arrays as pointers and as a stringis an array, it is passed
|   as a pointer. I am not sure but sizeof(*param) MIGHT return 80 but
|   sizeof(param) should return 2 as I can see it.

No, since `param' has decayed to type `char *', sizeof(*param) returns
sizeof(char), which is 1.

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 robert earl		/	"Love is a many splintered thing"
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