दो बाइनरी नंबरों का गुणनफल खोजने के लिए, पहले उन्हें सेट करें।
val1 = 11100; val2 = 10001; Console.WriteLine("Binary one: "+val1); Console.WriteLine("Binary two: "+val2);
अब उत्पाद प्राप्त करने के लिए लूप करें।
while (val2 != 0) { digit = val2 % 10; if (digit == 1) { val1 = val1 * factor; prod = displayMul(val1, prod); } else val1 = val1 * factor; val2 = val2 / 10; factor = 10; } Console.WriteLine("Product = {0}", prod);
एक विधि के ऊपर displayMul() को पहले बाइनरी नंबर के साथ बुलाया जाता है।
static long displayMul (long val1, long val2) { long i = 0, rem = 0, mul = 0; long[] sum = new long[30]; while (val1 != 0 || val2 != 0) { sum[i++] =(val1 % 10 + val2 % 10 + rem) % 2; rem =(val1 % 10 + val2 % 10 + rem) / 2; val1 = val1 / 10; val2 = val2 / 10; } if (rem != 0) sum[i++] = rem; i = i-1;; while (i >= 0) mul = mul * 10 + sum[i--]; return mul; }
यहाँ पूरा कोड है -
उदाहरण
using System; class Demo { public static void Main(string[] args) { long val1, val2, prod = 0; long digit, factor = 1; val1 = 11100; val2 = 10001; Console.WriteLine("Binary one: "+val1); Console.WriteLine("Binary two: "+val2); while (val2 != 0) { digit = val2 % 10; if (digit == 1) { val1 = val1 * factor; prod = displayMul(val1, prod); } else val1 = val1 * factor; val2 = val2 / 10; factor = 10; } Console.WriteLine("Product = {0}", prod); } static long displayMul (long val1, long val2) { long i = 0, rem = 0, mul = 0; long[] sum = new long[30]; while (val1 != 0 || val2 != 0) { sum[i++] =(val1 % 10 + val2 % 10 + rem) % 2; rem =(val1 % 10 + val2 % 10 + rem) / 2; val1 = val1 / 10; val2 = val2 / 10; } if (rem != 0) sum[i++] = rem; i = i-1;; while (i >= 0) mul = mul * 10 + sum[i--]; return mul; } }
आउटपुट
Binary one: 11100 Binary two: 10001 Product = 111011100