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  1. PO
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    <p>You won't find complete C code for your problem at SO. Your first approach isn't that bad. You could do the following:</p> <ol> <li>Multiply n1 and n2, conversion is done by mulitplication and addition, i. e. <code>a{1,2,3} -&gt; 1*100 + 2*10 + 3*1</code>, easy to implement</li> <li>Count the digits of your multiplication result (use division inside a loop)</li> <li>While looping through the digits you can store them back into another array</li> </ol> <p>If you can't or if you don't want to deal with dynamic array allocation, then think about how big your array for storage must be beforehand and perform a static allocation.</p> <p><strong>Edit</strong></p> <p>Based on the discussion another approach:</p> <p>Suppose, that <code>r = n1 * n2</code></p> <ol> <li>Create a n*m 2D array, where <ul> <li>n = number of digits in n2</li> <li>m = number of digits in n1 + 1</li> </ul></li> <li>Within a loop multiply each digit of n1 with one of the elements of n2, store the result in the array, store the result per-digit in the 2D-array, don't forget to add the carry to each digit</li> <li>Repeat 2 with all other digits of n2</li> <li>Now the array is filled and you'll have to add each digits like you would do it on paper, store each result within a target array, take care of the carry again</li> </ol> <p>There is one thing left in the algorithm: Determine the size of the target array, based on the informations within the intermediate array, you can think about this by using pencil and paper ;)</p>
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    1. POmultiply two arrays with int values
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    1. USSebastian Dressler
    1. USSebastian Dressler
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    1. COYou are right but suppose I want to get 20 digit number and multiply it in 30 digit number it will overflow by multiplying in Int Format
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    2. COI see, now i fully understand the approach of your algorithm. Then my answer isn't what you expected :D But i'll give it another thought.
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    3. COI've made an edit with my additional thoughts, hope that helps.
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