3 ms·
"Correct" does not mean "reproducible" just because you think lowly of irreproducible builds. A binary consisting of foo.o and bar.o is correct whether foo.o w
by mike_hock 3y ago
"Correct" does not mean "reproducible" just because you think lowly of irreproducible builds.
A binary consisting of foo.o and bar.o is correct whether foo.o was linked before bar.o or vice versa, provided that both foo.o and bar.o were compiled correctly.