Write a C program that will swap any two nibbles of a long int x (64-bit integer

Question

Write a C program that will swap any two nibbles of a long int x (64-bit integer). A nibble is a four-bit aggregation. There are two nibbles in a byte. For this problem the index of the least significant nibble is 0, and the index of the most significant nibble is 15 (so 0 <= m, n <= 15). So, if x = 0x0123456789ABCDEF (a 64-bit integer), and if you swap two nibbles say that are at the 0th and 15th index, x would now be x=0xF123456789ABCDE0. You are not allowed to use division, multiplication, or modulus, relative comparisons (<, >, <=, >=), loops, switches, function calls, macros, conditionals (if or ?:) You are allowed to use all bit level and logic operations, left and right shifts, addition and subtraction, equality and inequality tests, integer constants (<=255), INT_MIN and INT_MAX, and casting between data types. (20 points).

Explanation / Answer

// Comment if u have any doubt

int main() {

long long int y, x = 0x0123456789ABCDEF;

int i, j;

printf("Enter 1st position : ");

scanf("%d", &i);

printf(" Enter 2nd position : ");

scanf("%d", &j);

unsigned long long int m = 0xF;

unsigned long long int k = m << (i << 2); // Find 1st nibble position

unsigned long long int l = m << (j << 2); // Find 2nd nibble position

unsigned long long int u = (x & k); // Extract 1st nibble and store in a variable

unsigned long long int v = (x & l); // Extract 2nd nibble and store in a variable

u = (u >> (i << 2)); // Bring that 1st nibble data to the end

v = (v >> (j << 2)); // Bring that 2nd nibble data to the end

k = ~k;

l = ~l;

y = x & k; // Make 1st nibble position bits as zero ; we will replace those zero bits with 2nd nibble bits   

y = y & l; // Make 2nd nibble position bits as zero ; we will replace those zero bits with 1st nibble bits   

y = y | u << (j << 2);

y = y | v << (i << 2);

printf(" Input number : 0x%16llx ",x);

printf("Swapped number :0x%16llx ", y);

}

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