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)        {            System.out.print("$");            printDollarSign(k- 1);        } } What will be the output if the call isprintDolllarSign(5);? ____________________________ ____________________________ ____________________________ ____________________________ ____________________________ public void printDollarSign( int k) {        int j;        if (k >0)        {            System.out.print("$");            printDollarSign(k- 1);        } } What will be the output if the call isprintDolllarSign(5);? ____________________________ ____________________________ ____________________________ ____________________________ ____________________________

Explanation / Answer

dear, output will be


Here is the trace that show how thisoutput is obtained.
CallingprintDollarSign(5) here k > 0 so it will print one $here and will call printDollarSign(k - 1); OUTPUT :$ CallingprintDollarSign(4) here k > 0 so it will print one $here and will call printDollarSign(k - 1); OUTPUT :$ CallingprintDollarSign(3) here k > 0 so it will print one $here and will call printDollarSign(k - 1); OUTPUT :$ CallingprintDollarSign(2) here k > 0 so it will print one $here and will call printDollarSign(k - 1); OUTPUT :$ CallingprintDollarSign(1) here k > 0 so it will print one $here and will call printDollarSign(k - 1); OUTPUT :$ CallingprintDollarSign(0) k<=0 so return
CallingprintDollarSign(5) here k > 0 so it will print one $here and will call printDollarSign(k - 1); OUTPUT :$ CallingprintDollarSign(4) here k > 0 so it will print one $here and will call printDollarSign(k - 1); OUTPUT :$ CallingprintDollarSign(3) here k > 0 so it will print one $here and will call printDollarSign(k - 1); OUTPUT :$ CallingprintDollarSign(2) here k > 0 so it will print one $here and will call printDollarSign(k - 1); OUTPUT :$ CallingprintDollarSign(1) here k > 0 so it will print one $here and will call printDollarSign(k - 1); OUTPUT :$ CallingprintDollarSign(0) k<=0 so return

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