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About LinkBacksHow the output of this code is 16?Code:main() { double d; printf("%d", (unsigned int)((double *)0+2)); }
This is a discussion on Help:working of Type casting within the C Programming forums, part of the General Programming Boards category; Code: main() { double d; printf("%d", (unsigned int)((double *)0+2)); } How the output of this code is 16?...
Help:working of Type casting How the output of this code is 16?Code:main() { double d; printf("%d", (unsigned int)((double *)0+2)); }
Last edited by GOGO1104; 09-28-2010 at 11:07 AM.
It's 16 due to pointer arithmetic. (double*)0+2 is the same as 0+2*sizeof(double).
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Didn't we just do this question in another thread?
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