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What will be the output of the program?String x = "xyz";x.toUpperCase(); /* Line 2 */String y = x.replace('Y', 'y');y = y + "abc";System.out.println(y);abcXyZabcxyzxyzabcXyZabc

Question

What will be the output of the program?String x = "xyz";x.toUpperCase(); /* Line 2 */String y = x.replace('Y', 'y');y = y + "abc";System.out.println(y);abcXyZabcxyzxyzabcXyZabc

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Solution

The output of the program will be "xyzabc". Here's why:

  1. The program starts by creating a String variable x and assigning it the value "xyz".

  2. The next line x.toUpperCase(); converts the string to uppercase. However, the result of this operation is not stored anywhere, so it has no effect on the program.

  3. The program then creates another String variable y and assigns it the value of x with all occurrences of 'Y' replaced with 'y'. Since x is "xyz" and there is no 'Y' in it, y is also "xyz".

  4. The next line y = y + "abc"; appends "abc" to y, so y becomes "xyzabc".

  5. Finally, the program prints out the value of y, which is "xyzabc".

This problem has been solved

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