Let x= 0xA5 and y= 0xABCDE and string s. 1) Mask the upper 4bits of x 2) clear b
ID: 3931998 • Letter: L
Question
Let x= 0xA5 and y= 0xABCDE and string s.
1) Mask the upper 4bits of x
2) clear bit 6 of x
3) set bit 9 of y
4) interchange x and low 8 bits of y without using a third variable or register
5) invert x.
6) shift left x four bits and shift right y two bits.
7) check if bit 4 of x is 0.
8) check if bit 10 of y is 1.
9) suppose the length s is 5. convert y into decimal ASCII string and put it into s.
10) suppose s-"ABCEXY". write a c function to covert s as "yxecba"
(4) intercbange x dad low 8 bits of y without using a tont variabl (6) Shift left four hits and shift ti (7) Check if bit 4 of s is 0 (s) Check bit 10 of y is Suppose the length s is 5. Convert y into decimal ASCIl string and put it into s. Write a C function to covert s as "yxecba" ABCEXY"Explanation / Answer
Hello Thanks for giving clear description and Typing Question from Image
x= 0xA5 and y= 0xABCDE (Hexa Decimal Numbers. In Decimal(10) X=165 and 703710)
1) Mask the upper 4bits of x
----------------------------------------------------
unsigned char value = 0xa5;
unsigned char mask = 0x0f;
unsigned char mask_result = (value & mask);
2) clear bit 6 of x :-
----------------------------------------
unsigned char value = 0xa5;
unsigned char clear = 0xfb;
unsigned char clear_result = (value & clear);
3) set bit 9 of y :-
-----------------------------
unsigned int value = 0xabcde;
unsigned int set = 0x800;
unsigned int set_result = (value | set);
4) interchange x and low 8 bits of y without using a third variable or register :-
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This is Achieved in 2 steps without using any 3rd variable
1) Clear Low 8 bits of Y (set Low 8 bits to "00000000"
2)Now do unary or(|) with X
unsigned int x = 0x00a5;
unsigned int y = 0xabcde;
y = y &(1111111100000000);
y = (y | x);
5) invert x :-
------------------------
C Code
x = ~x;
6) shift left x four bits and shift right y two bits :-
------------------------------------------------------------------------
x = x <<4;
y = y >>2;
7) check if bit 4 of x is 0.
8) check if bit 10 of y is 1.
------------------------------------------------
int pos = 3 // Postion at which we have to check
int value = 10; //Value on which we have to check
if( (value &(1<<pos) ) == 1)
{
print("bit at pos 3 is 1");
}
check if bit 4 of x is 0
if( (x & (1 <<4)) == 0)
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