The following code seems to execute fine but how is it valid and whats happening here?
int i;
printf("%d",i["11"]);
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8UB: `i` is an uninitialized array index. – nneonneo Aug 20 '13 at 05:20
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5There are many questions that this is a duplicate of. The difficulty is going to be finding them. – Jonathan Leffler Aug 20 '13 at 05:22
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3What is with C programmers nowadays? The first thing they are trying to learn is how to abuse the system. – Krishnabhadra Aug 20 '13 at 05:24
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What is the output..? @manoj – Kira Aug 20 '13 at 05:24
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3@JonathanLeffler Here is one of the dupes http://stackoverflow.com/questions/381542/in-c-arrays-why-is-this-true-a5-5a -- the question got 550 upvotes. – Ray Toal Aug 20 '13 at 05:28
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1possible duplicate of [Why does i\[arr\] work as well as arr\[i\] in C with larger data types?](http://stackoverflow.com/questions/7181504/why-does-iarr-work-as-well-as-arri-in-c-with-larger-data-types) – nneonneo Aug 20 '13 at 05:28
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Thanks Ray and nneonneo. Either of the suggested duplicates would be a better way of closing this question than 'off topic'. However, as long as it is closed in some way, I'm not going to get too fussed. – Jonathan Leffler Aug 20 '13 at 05:30
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thanks for the link @Ray Toal – manojadams Aug 20 '13 at 05:35
2 Answers
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when you do i["11"] what you do is say to the compiler to take the value of i and add to it the value of "11" and take the value on the address that is the sum of them
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