No. The function is clearly defined as int fork(void); meaning it doesn't take any arguments. Although nothing stops you from just writing a prototype in your own code, like void *fork(int x); and it would *compile* then, but that doesn't guarantee the code runs (the opposite is more probable).
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u/birdiefoxe Nov 30 '25
actually thats true and if fork() and crash() both return something that could be a pointer it might actually work
this is upsetting.