sizeof on a word-oriented machine

[Doug Gwyn]

In article <11135 at riks.csl.sony.co.jp> diamond at ws.sony.junet (Norman Diamond) writes:
-Consider a machine where each 4-byte word has an address.
-  char x[37];
-What is sizeof x?

37.

-If sizeof x is 37 then a user might do:
-  char *two_xs;
-  two_xs = malloc (2 * sizeof x);
-and get screwed because only 76 bytes will be allocated (2 * 37 rounded
-up to a multiple of 4) when 80 are really needed.

No, only 2*37 bytes are needed.  Arrays are not padded.

-I think 40 is the most reasonable value for sizeof x.  Is the standard's
-example wrong?  May it be ignored?

The example is right.  Feel free to ignore it.