Multiplication of two numbers.

Hello , Actaully i am confused that how we can multiply two number without using any operators.? please help me giving yours ideas....

 

Write a function

function multiply($x,$y){
return $x*$y;
}

echo(multiply(5,4)); // Output: 20

 

Hello dear thanks for Your reply but You have use * operator to call a functions.. my question is that witout using any operators how we can multiply two numbers....

 

Use
function multiply($x,$y){
$z = array();

for($i=0; $i<$y; $i++){
$z[$i] = $x;
}

return array_sum($z);
}

echo(multiply(5,4));

and if you cannot use "for" statement write

$x[0] = 5;
$x[1] = 5;
$x[2] = 5;
$x[3] = 5;

echo array_sum($x);

 

Are you referring to signed multiplication x86 ASM instruction which can be called from C++/Java as below?

asm(imul dest, src1, imm_src);

this does dest := src1 * imm_src

 

Without using any operator???

I think there is a way in List where you need to Give

multiply ( 2 3 )

But not sure about that

 

The question is a bit confusing - or at least difficult.

Do you mean is there a quick way of multiplying a number by 2?

In C or C++, you can do this by shifting, e.g. x = x << 1;

 

Thanks for ur reply.dear can You explain me widely how can we use shift for multiplications...

 

#include<stdio.h>
main()
{
int a,b,result;
printf("nEnter the numbers to be multiplied :");
scanf("%d%d",&a,&b);
result=0;
while(b != 0) // Iterate the loop till b==0
{
if (b&01) // Logical ANDing of the value of b with 01
result=result+a; // Update the result with the new value of a.
a<<=1; // Left shifting the value contained in 'a' by 1.
b>>=1; // Right shifting the value contained in 'b' by 1.
}
printf("nResult:%d",result);
}

 

Note that you can only use shift operations for multiply/divide a number with another number if and only if one of them is a power of 2.

That means if p is any number,

p << 1 means p X (2^1), i.e.: p X 2
p << 2 means p X (2^2), i.e.: p X 4
p << 3 means p X (2^3), i.e.: p X 8
p << 4 means p X (2^4), i.e.: p X 16
p << 5 means p X (2^5), i.e.: p X 32
...
...
...
p << n means p X (2^n)

Similarly,

p >> 1 means p / (2^1), i.e.: p / 2
p >> 2 means p / (2^2), i.e.: p / 4
p >> 3 means p / (2^3), i.e.: p / 8
p >> 4 means p / (2^4), i.e.: p / 16
p >> 5 means p / (2^5), i.e.: p / 32
...
...
...
p >> n means p / (2^n)


Now you can understand that if you need to multiply/divide p with q using shift operations, either p or q must be a number with power of 2.

If you want to check whether a number is a power of 2, use following function.

int IsPowerOfTwo(int x)
{
    return (x != 0) && ((x & (x - 1)) == 0);
}

Code (markup):

 

Surely shift is an operator as well?