Given the following method: public void printDollarSign( int k) { int j; if (k >

Question

Given the following method: public void printDollarSign( int k) {        int j;        if (k >0)        {            for(j = 1; j <= k; j++)               System.out.print("$");            System.out.println();            printDollarSign(k- 1);        } } What will be the output if the call isprintDolllarSign(5);? ____________________________ ____________________________ ____________________________ ____________________________ ____________________________ Given the following method: public void printDollarSign( int k) {        int j;        if (k >0)        {            for(j = 1; j <= k; j++)               System.out.print("$");            System.out.println();            printDollarSign(k- 1);        } } What will be the output if the call isprintDolllarSign(5);? ____________________________ ____________________________ ____________________________ ____________________________ ____________________________

Explanation / Answer

Dear, output will be
$$$$$ $$$$ $$$ $$ $



Here is the trace of thefunciton, how we get above answer.

CallingprintDollarSign(5) Now k>0 so it will print5$ in loop and will callprintDollarSign(k-1) $$$$$ CallingprintDollarSign(4) Now k>0 so it will print4$ in loop and will callprintDollarSign(k-1) $$$$ CallingprintDollarSign(3) Now k>0 so it will print3$ in loop and will callprintDollarSign(k-1) $$$ CallingprintDollarSign(2) Now k>0 so it will print2$ in loop and will callprintDollarSign(k-1) $$ CallingprintDollarSign(1) Now k>0 so it will print1$ in loop and will callprintDollarSign(k-1) $ CallingprintDollarSign(0) Since k is 0 now, it willexit $$$$$ $$$$ $$$ $$ $



Here is the trace of thefunciton, how we get above answer.

CallingprintDollarSign(5) Now k>0 so it will print5$ in loop and will callprintDollarSign(k-1) $$$$$ CallingprintDollarSign(4) Now k>0 so it will print4$ in loop and will callprintDollarSign(k-1) $$$$ CallingprintDollarSign(3) Now k>0 so it will print3$ in loop and will callprintDollarSign(k-1) $$$ CallingprintDollarSign(2) Now k>0 so it will print2$ in loop and will callprintDollarSign(k-1) $$ CallingprintDollarSign(1) Now k>0 so it will print1$ in loop and will callprintDollarSign(k-1) $ CallingprintDollarSign(0) Since k is 0 now, it willexit $$$$$ $$$$ $$$ $$ $



Here is the trace of thefunciton, how we get above answer.

CallingprintDollarSign(5) Now k>0 so it will print5$ in loop and will callprintDollarSign(k-1) $$$$$ CallingprintDollarSign(4) Now k>0 so it will print4$ in loop and will callprintDollarSign(k-1) $$$$ CallingprintDollarSign(3) Now k>0 so it will print3$ in loop and will callprintDollarSign(k-1) $$$ CallingprintDollarSign(2) Now k>0 so it will print2$ in loop and will callprintDollarSign(k-1) $$ CallingprintDollarSign(1) Now k>0 so it will print1$ in loop and will callprintDollarSign(k-1) $ CallingprintDollarSign(0) Since k is 0 now, it willexit CallingprintDollarSign(5) Now k>0 so it will print5$ in loop and will callprintDollarSign(k-1) $$$$$ CallingprintDollarSign(4) Now k>0 so it will print4$ in loop and will callprintDollarSign(k-1) $$$$ CallingprintDollarSign(3) Now k>0 so it will print3$ in loop and will callprintDollarSign(k-1) $$$ CallingprintDollarSign(2) Now k>0 so it will print2$ in loop and will callprintDollarSign(k-1) $$ CallingprintDollarSign(1) Now k>0 so it will print1$ in loop and will callprintDollarSign(k-1) $ CallingprintDollarSign(0) Since k is 0 now, it willexit

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