datatype

Question

{

signed int i =-13;

unsigned int k= i%2;

printf("%d\n",k);

}

a. -13

b.compilation error

c.-1

d.1

Comments

D should be ans

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c is ans

Answer 1

Suppose integers are stored in 2 bytes. Signed int has range -32768 to 32767 AND Unsigned integers has range 0 to 65535.

signed int i =-13; 

Declares i to be signed integer and stores -13 in it.

unsigned int k= i%2;

Declares k to be unsigned integer and stores 65535 in it. 

i % 2 = (-13) % 2 = -1 and -1 is equal to 65535 in unsigned type.

printf("%d\n",k);

k is of unsigned type while format specifier is %d (not %u) therefore berfore printing k is again typecast to signed i.e. -1

hence answer C 
 

Comments

% must give positive value. How it works with C language. Is it modulo operator right?

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Modulo operator gives remainder with sign of dividend in C

(-13) % 2 = -1

(13) % 2 = 1

https://www.lemoda.net/c/modulo-operator/

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Remainder always positive right ? Is it compiler specific or given in standard?

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The C programming Language Dennis Ritchie

If both operands are nonnegative then the remainder is nonnegative and smaller than the divisor; if not, it is guaranteed only that the absolute value of the remainder is smaller than the absolute value of the divisor.

So standard doesn't guarantee correct value for negative values.

Above link shows correct use of Mod operator