Why this code compiles and returns 1?
int main(int ac, char* av[])
{
return ac[av] == 0;
}
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Radosław Panuszewski
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You sure it's not av[ac] == 0 ? – Tilak Madichetti Apr 09 '17 at 13:18
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4@TilakMaddy Both are correct. – DeiDei Apr 09 '17 at 13:19
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2There must be another question answering this, anyhow (for the short space allowed in a comment): `pointer[i]` is the same than `*(pointer + i)` which is the same than `*(i + pointer)` which is the same than `i[pointer]`. – peppe Apr 09 '17 at 13:20
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@TidyMaddy Yes, I'm sure – Radosław Panuszewski Apr 09 '17 at 13:20
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3`ac[av]` is equivalent to `av[ac]`. Since `ac` (or `argc`) is the number of arguments passed in `av` array, you'll always get out-of-bounds access which is _undefined behavior_. However, it is likely that first pointer past `av` (again, it should be called `argv`) is null pointer, comparison returns true value which is coded as `1`. – myaut Apr 09 '17 at 13:20
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@myaut, `av[ac]` is defined to be 0/null. – chris Apr 09 '17 at 13:23
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1@myaut: absolutely not, `argv[argc]` is guaranteed to be 0. See http://eel.is/c++draft/basic.start.main#2 – peppe Apr 09 '17 at 13:23