Conversion from hex to dec confusion

Hi,

In the follwing code (simplified bit)

Code:

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unsigned short x = 0x18f;

int z = 8;

int y = (x & 0x0f ) + z;

cout << y ;

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it is displaying, 17.. i was thinking it will be 23..

23 dec is 17 hex. Have you got a previous cout << hex somewhere? - as hex is sticky and stays until another output type is specified. On my system, the code in post #1 displays 23 in decimal.

Try :

Code:

---

unsigned short x = 0x18f;

int z = 8;

int y = (x & 0x0f ) + z;

cout << dec << y ;

---

I tested this code with VS2019.
I see in the debugger window the value of y being
y 0x00000017 int
It is exactly 23 in decimal. :rolleyes:

Thanks a lot kaud, !. yes, i had somewhere else far before the cout << hex !. I also got this, and tried just after the post, with cout << dec. It worked :). Not sure why it should stay on !!!

yes i know, was wondering, why it should display in hex, unless i gave << hex explicitly. But it is confusing, that cout retains the previous << hex !!! Thanks Victor :)

AFAIK, the 'sticky' manipulators are:

Code:

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[no]boolalpha
[no]showbase
[no]showpoint
[no]showpos
[no]skipws
[no]unitbuf
[no]uppercase

dec/ hex/ oct

fixed/ scientific

internal/ left/ right

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Thanks a lot, kaud for the info