What is the output of this code - >>> x = 5 >>> y = x + 3 >>> y = int (str(y) + "2") >>> print(y)?
Output of this code is 82.Since, you can easily find the output by executing this code in a python interpreter, I am assuming you need the explanation of the output.So, we know that x=3 and y=8. In the next line, what is done is that the integer value of y is typecasted into a string, so str(y) gives “8”. Now, since the operands involved in addition are both string data type, python interprets the addition sign as concatenation operator. So, “2” is concatenated to “8” and it becomes “82”. You typecast this string “82” to integer value and store in y. So, y now holds the value 82. And then you output y which gives us 82 as the output.Thanks.
What is output of this code snippet?
The output(with “\n” for each print) would beSo the linechar *s[] = { “knowledge”, “is”, “power” }; creates a 2d array of characters.and in the line p = s; you are making a double pointer point to that memory(to the start of the memory).I think there are 4 properties of C(in simple terms) that is worth mentioning here.Incrementing p(double ptr) would take you to the next word.Incrementing *p(single ptr) would take you to the next letter.An increment operator(++) gets higher precedence over a dereference operator(*)A pre increment operator(++p) increments before returning the value and a post increment operator(p++) increments after returning the value.I am not going into the details of the above 3.Now keep all the above 3 in mind. Here we go..p is pointing to “knowledge”++*p ==> increment *p ==> move pointer by 1 and hence “nowledge”(Point 2 above)*p++ ==> will not do (*p)++, but will do p++(because of point 3. So p now points to “is”) and then apply “*” operator on it. But remember this is a post increment operator you have used. So the line will not return the new value though it is now actually pointing to “is”.(Point 4 above). Since the old value is returned, it would still print “nowledge”++*p ==> like in point 2 above, it will move pointer by 1 and hence “s” (point 2 above)Hope I was clear enough.
What is the output of this code?
Given the following declaration of a method public int method1( int values[][] ) { int mystery = 1; for ( int i[] : values ) for ( int j : i ) mystery *= j; return mystery; } What is the output of the following code segment? int array[][] = { { 3, 2, 5 }, { 2, 2, 4, 5, 6 }, { 1, 1 } }; System.out.println( mystery( array ) );
What is the output of this code sequence?
This is a linear search for the highest value in the array "a". Temporary variables are absolutely necessary to accomplish tasks while programming. Don't be afraid to make many temporary variables, boolean flags, etc. The problem of sorting and searching in computer programming will never end. They are actually two of the most common, and complex problems that programmers face. Does your program search more? Does it insert/remove entries more? How often do I need to search or sort my data? There is no single correct answer. Another option would be to sort this list of numbers, and then you're guaranteed that the far end number would be the highest: 12.3 48.2 65.8 99.6 In this case, performing the sort would be time consuming, but searching for the highest number would be instantaneous. Good luck! P.S. Do you know why the for-loop has "int i=1" instead of "int i=0"? Hint: The code is correct, its a special case for this particular problem, normally if you want to loop through an entire array, you want to start at index "0".
What is the output of the following code?
No, that is not the correct answer. This line is not valid Java syntax: double [] a = (12.5 48.3 65.0); Array literals are declared with { curly brackets }, not ( parentheses ), and there would need to be commas between the items. If the array were declared correctly, the output would be 3, the length of the array. Edit: You asked what the output was. This is the line that outputs something: System.out.println (a.length); It prints out a.length, which is the number of elements in the array called "a". There are 3 elements. It will print out the number 3. What answer were you expecting?
What is the output of the code void main() int k=2,j=3,p=0; p= (k,j, k); printf ("%d", p)?
the answer is 2., (comma operator) has associativity of right to left and only one value can be assigned to the variable p. The right most variable is k hence it’s value 2 will be assigned to p.ANALOGY :return (a,b,c) ;As only one value can be returned and comma operator has right to left associativity hence value of c will be returned.Visit the following link for more on precedence and associativity -Mahajan - The Best Programming Tutorials
What will be output of this code segment?
The thing to keep in mind here is that the expression:ihas the value of the contents of the variable i, and in C a non-zero value is considered true, while a zero value is considered false. So, when we see an expression like this used as a condition in an if statement, it’s really the same as saying:i != 0In other words, we’re actually checking if the contents of variable i is non-zero. If it is non-zero, then the condition is true. If it is zero, then the condition is false.So, the call to the printf function inside the first if statement will definitely execute, because the value of i is -10, and -10 is certainly not zero, so the condition of that first if statement is true. Because the condition is true, the call to the printf function will executed, so a 1 will be printed.Just before the second if statement, we set i to 0. So, because i is zero, the if statement’s condition is false, and the call to the printf function inside that second if statement will not execute.Finally, just before the third if statement, we set i to 5. Because 5 is non-zero, the if statement’s condition is true, and the call to the printf function inside that third if statement will execute. Thus, a 3 will be printed.Because no spaces or newline characters appear in the printf format strings, the output will just have a 1 immediately followed by a 3 on the same line of output:13