Smaller and faster way to calculate modulus of 2 using bit manipulation

algorithms/bit-manipulation/addition.c

@@ -0,0 +1,20 @@
+#include <stdio.h>
+
+int add(int x, int y)
+{
+	while(x > 0)
+	{
+		unsigned int carry = x & y;
+		y = x ^ y;
+		x = carry << 1;
+	}
+	return y;
+}
+
+int main()
+{
+	printf("sum of %d and %d is %d\n", 3, 4, add(3, 4));
+	return 0;
+}
+
+

algorithms/bit-manipulation/check_power_of_2.c

@@ -0,0 +1,15 @@
+#include <stdio.h>
+
+#define check_power_of_2(x) (!( x & (x - 1)) && (x > 1))
+
+int main()
+{
+    for(int i = 1; i < 100; i++)
+    {
+        if(check_power_of_2(i))
+            printf("True! %d is a power of 2\n", i);
+        else
+            printf("False! %d is not a power of 2\n", i);
+    }
+    printf("Hello World\n");
+}

algorithms/bit-manipulation/modulus_with_power_of_2.c

@@ -0,0 +1,20 @@
+/*
+    1. Here is an example of the modulus operation
+    using bit manipulation
+    2. It is supposed to be much faster and more efficient
+    The only condition is that n should be a power of 2
+*/
+
+#include <stdio.h>
+
+#define modulo(x,n) ((x) & (n - 1))
+
+int main()
+{
+    printf("Hello World\n");
+
+    for(int i = 0; i < 100; i++)
+        printf("value of %d modulo 8 is: %d\n", i, modulo(i,10));
+
+    return 0;
+}

algorithms/bit-manipulation/subtraction.c

@@ -0,0 +1,18 @@
+#include <stdio.h>
+
+int subtract(int x, int y)
+{
+	while(y != 0)
+	{
+		unsigned int borrow = y & ~x;
+		x = y ^ x;
+		y = borrow << 1;
+	}
+	return x;
+}
+
+int main()
+{
+	printf("diff of %d and %d is %d\n", 4, 3, subtract(4, 3));
+	return 0;
+}