Int i[5]; printf("%p\n", &i[10]); Why doesnt this give me an error...i mean

Question

C / C++

aklil

Int i[5]; printf("%p\n", &i[10]); Why doesnt this give me an error...i mean

it is printing an address how?

0

07.06.2020

2 ответов

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aklil Автор вопроса

This one for example

0

07.06.2020

Cengizhan Pasaoglu · Accepted Answer

Cengizhan Pasaoglu

Why it should give? There is no access violation in here because in printf, you are just calculating start address of i + 10. If you want to access value of this address, it might be. (Still no guarantee that you get seg fault)

0

07.06.2020

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Int i[5]; printf("%p\n", &i[10]); Why doesnt this give me an error...i mean

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