Here are we use typecasting from pointer to integer, but output of arithmetic operation are different from expected answer, Why?
Source Code :
int main(){
int *p,*q;
p = 1000;
q = 2000;
printf("%d",q-p);
return 0;
}
Output: 250
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Pointer arithmetic is done in units of the base type. In your case in units of `sizeof(int)`. – Some programmer dude Sep 27 '18 at 10:50
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1Also note that doing arithmetic on two unrelated pointers is highly discouraged, and IIRC even [*undefined behavior*](https://en.wikipedia.org/wiki/Undefined_behavior). – Some programmer dude Sep 27 '18 at 10:52
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1Formally, the behaviour is undefined `"%d"` is the wrong specifier and you're doing pointer arithmetic on two locations that are not part of the same array. – George Sep 27 '18 at 10:52
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To printf the difference of pointers you want to use [`t` length modifier](http://port70.net/~nsz/c/c11/n1570.html#7.21.6.1p7). – alk Sep 27 '18 at 10:56
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1Lastly, the result of pointer arithmetic is a value of the type [`ptrdiff_t`](https://en.cppreference.com/w/c/types/ptrdiff_t), which must be printed using the `t` length modifier. As in `"%td"`. See e.g. [this `printf` reference](https://en.cppreference.com/w/c/io/fprintf) for details about format specifiers. – Some programmer dude Sep 27 '18 at 10:56
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No @Karsten Koop I think its different question because of there are output are different after arithmetic operation. – Geshwar Kumar Dhankar Sep 27 '18 at 11:01
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Yes @alk i want – Geshwar Kumar Dhankar Sep 27 '18 at 11:03
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"*typecasting from pointer to integer*" this is *not* a cast, but an "**implicit** conversion". – alk Sep 27 '18 at 11:23
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Yes @alk, Thank you – Geshwar Kumar Dhankar Sep 27 '18 at 11:25
1 Answers
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The same reason due to which p++ will give you 1004. Since sizeof(int) on your machine is 4, hence each operation is shown wrt sizeof(int), even the difference i.e. 1000/4.
Sandy
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