Answer / sandeep gupta

A very interesting question where u need knowledge of
Computer architecture also.... :) :) :) :)
The floating point(FP) no. are stored differently in memory
as: mantissa × 2^exponent [M× 2^E]. M is 23 bit long and E
is 8 bit. 1 bit is for sign of number. The format is :
SEEE EEEE EMMM MMMM MMMM MMMM MMMM MMMM
Now binary of 5.375 is 101.011 which can be written in
normalized form(which has single 1 before decimal point) as
1.01011 × 2^2. This 1 and point(.) is always neglected while
storing the no. in registers. so mantissa is 01011 or
010110000000000000000000 × 2^E where E is 8 bit(0 to 255).
The actual value of the exponent is calculated by
subtracting 127 from the stored value (0 to 255) giving a
range of –127 to +128.
So here we need E=2 which we can get from 129(129-127=2)
whose binary is: 10000001.
Now above format becomes:
0100 0000 1010 1100 0000 0000 0000 0000 which is 40ac0000 in
hex. So stored format is: 00->00->ac->40

Answer / govind verma

i think mr sandeep this value stored in this manner
0100 0000 1010 1100 0000 0000 0000 0000 which is 40ac0000


float takn 4 byte in memory in program p=(char*)&a casting this to char type nw p(it char ponter its point to i byte value) point to lower byte of the a(5.375) which is 0000 0000
p[0]-> contain the fist lower byte data 00(bcoz print bye the hexa formate specifier withe 2 width)
p[1]->contain 00
p[2]-> contain ac
p[3]->contain 40

then output becom ...0000ac40

struct Foo { char *pName; }; main() { struct Foo *obj = malloc(sizeof(struct Foo)); clrscr(); strcpy(obj->pName,"Your Name"); printf("%s", obj->pName); } a. Your Name b. compile error c. Name d. Runtime error

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