The actual expression is more like
__internal_argp = (void *) &args;
foo = *((short *) __internal_argp)++;
so while you get the correct value on little-endian, the next value will be
taken from the wrong address.
-- brandon s. allbery [os/2][linux][solaris][japh] allbery@kf8nh.apk.net system administrator [WAY too many hats] allbery@ece.cmu.edu carnegie mellon / electrical and computer engineering KF8NH Kiss my bits, Billy-boy.
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