Comments on: a query for int overflow https://www.emblogic.com/blog/07/a-query-for-int-overflow/ Embedded System and ARM Training Mon, 07 Oct 2013 04:00:03 +0000 hourly 1 http://wordpress.org/?v=3.9.1 By: ashish.dahiya https://www.emblogic.com/blog/07/a-query-for-int-overflow/#comment-906 Mon, 02 Jul 2012 12:12:02 +0000 http://emblogic.org/blog/?p=3733#comment-906 it is something like this
-2147483648 to 0 to 2147483647

overflow means when its value is taken 2147483648 then compiler will take it as -2147473648.

]]> By: ashish.dahiya https://www.emblogic.com/blog/07/a-query-for-int-overflow/#comment-905 Mon, 02 Jul 2012 12:08:33 +0000 http://emblogic.org/blog/?p=3733#comment-905 int (Integer. ) size (4bytes) and range is
signed: -2147483648 to 2147483647
unsigned: 0 to 4294967295.

that,s why when it is 2147483647 then it will be printed same
but when it becomes 2147483648 the it will overflow and then it will move towars negative side and print -2147483648,,,and when it is
2147483649 then it will print -2147483647.and so on………..

]]> By: Kuldeep Dhanotia https://www.emblogic.com/blog/07/a-query-for-int-overflow/#comment-904 Mon, 02 Jul 2012 12:04:09 +0000 http://emblogic.org/blog/?p=3733#comment-904 overflow is nothing its warning to u if u r exceeding its value from upper range side….

]]> By: sumit.ahuja https://www.emblogic.com/blog/07/a-query-for-int-overflow/#comment-899 Mon, 02 Jul 2012 07:39:03 +0000 http://emblogic.org/blog/?p=3733#comment-899 but how is it an overflow case..according to the compiler,
the values keep on repeating between (-)ve tp (+) range of signed..

]]> By: m siddarth https://www.emblogic.com/blog/07/a-query-for-int-overflow/#comment-897 Mon, 02 Jul 2012 06:08:53 +0000 http://emblogic.org/blog/?p=3733#comment-897 Internally the compiler declare the integer variable as signed int hence the range of integer on 32 bit platform -2147483648 to 2147483647.

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