Question

Disassembly Practice Answer (9/9) /* naïve multiplication: returns m*n */ int multiply (int m, int n) { int p; /* product */ for (p = 0; n--> 0; p += m) ; return p; 5) Translate into C code • One of many possible ways to write this

Answer 1

This code can be translated into C in many ways, one out of which is as below:

I am writing only the function

In this you are simply running a loop and adding value of "m"  n times.

int multiply(int m, int n)

{

int p=0;

for(;n>0;n--)

{

p=p+m;

}

return p;

}

another way is directly use * (asterisk) operator

int multiply(int m, int n)

{

return (m*n);

}

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