Concept explainers
Explanation of Solution
Given statements:
1 + "Welcome " + 1 + 1
Explanation:
The above statement will print “1Welcome 11”. In java “+” is used to merge the java strings using “+” operator. So while executing the statement, the number “1” is concatenate with string “Welcome” followed by “1+1”.
Given statements:
1 + "Welcome " + (1 + 1)
Explanation:
It force “(1+1)” to be executed first, as it is enclosed within the parenthesis so the value returned is “2”. So while executing the statement, the number “1” is concatenate with string “Welcome” followed by value “2”. So the result will be “1Welcome 2”
Given statements:
1 + "Welcome " + ('\u0001' + 1)
Explanation:
It force “('\u0001+1')” to be executed first, as it is enclosed within the parenthesis so the value returned is “2”. So while executing the statement, the number “1” is concatenate with string “Welcome” followed by value “2”. So the result will be “1Welcome 2”
Given statements:
1 + "Welcome " + 'a' + 1
Explanation:
The above statement will print “1Welcome a1”...
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Introduction to Java Programming and Data Structures, Comprehensive Version (11th Edition)
- "### Exercise 1 ###\n", "The table presented below gives a final grade based on the amount of points a student get at the end of the course. Write a program that asks the student for the number of points he got and, based on the number of points the program should print the student's final grade: \n", "\n", " points grade |\n", 11 ] }, }, { ----\n", less than 500 | 0 |\n", between 500 and 550 | 0.5 \n", between 551 and 640 between 641 and 730 between 731 and 800 between 801 and 850 between 851 and 900 between 901 and 960 above 960 "\n", "Use the if-elif-else consditional to solve this problem. \n" "cell_type": "code", 'execution_count": null, "id": "831e820c", "metadata": {}, "outputs": [], "source": [] 1.0 \n", 1.5 \n", 2.0 \n", 2.5 \n", 3.0 \n", 3.5 \n", 4.0 \n", "cell_type": "markdown", "id": "fa947e7e", "metadata": {}, "source": [arrow_forwardx = 9 ; y = x++ What are the values of x and y after the two statements execute? X = y =arrow_forward31. If originally x=4, what is the value of x after the evaluation of the expression: x++arrow_forward
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