subject

The value of esp is 2000 00f8h. the stack has the following values: | 2 b | 2000 00f4| 0 0 | 2000 00f5| 1 f | 2000 00f6| 2 a | 2000 00f7| f f | 2000 00f8 < -esp| 1 0 | 2000 00f9| 5 5 | 2000 00fa| a 7 | 2000 00fb| b 8 | 2000 00fc| 8 b | 2000 00fd| e 0 | 2000 00fe| e 2 | 2000 00ff| 0 a | 2000 0100||after the following instructions are performed, what is the value in ax and in esp? (a) pop ax(b) pop ax(c) pop ax(d) push ax

ansver
Answers: 2

Another question on Computers and Technology

question
Computers and Technology, 21.06.2019 15:30
Some of the items below indicate the steps required to move a slide to a different location in a presentation. select those steps and indicate the order in which they should be performed to move the slide. (for example, if an item describes what you do first, select 1.) included in the list are items that are not part of the process. for these items, select n/a, an abbreviation for not applicable. 1 — first 2 — second 3 — third 4 — fourth 5 — fifth 6 — sixth n/a — not applicable drag the slide to the desired place. enter the slide position desired. hold the mouse button down. select move from the tools menu. select the slide. switch to the notes view.
Answers: 3
question
Computers and Technology, 22.06.2019 15:00
The three logical operators used to write compound conditions are "and," "or," and "not." a: true b: false
Answers: 2
question
Computers and Technology, 22.06.2019 21:00
Which of these is most responsible for differences between the twentieth century to the twenty-first century?
Answers: 2
question
Computers and Technology, 23.06.2019 00:40
Consider the following statements: struct nametype{string first; string last; }; struct coursetype{string name; int callnum; int credits; char grade; }; struct studenttype{nametype name; double gpa; coursetype course; }; studenttype student; studenttype classlist[100]; coursetype course; nametype name; mark the following statements as valid or invalid. if a statement is invalid, explain why.a.) student.course.callnum = "csc230"; b.) cin > > student.name; c.) classlist[0] = name; d.) classlist[1].gpa = 3.45; e.) name = classlist[15].name; f.) student.name = name; g.) cout < < classlist[10] < < endl; h.) for (int j = 0; j < 100; j++)classlist[j].name = name; i.) classlist.course.credits = 3; j.) course = studenttype.course;
Answers: 1
You know the right answer?
The value of esp is 2000 00f8h. the stack has the following values: | 2 b | 2000 00f4| 0 0 | 2000 00...
Questions
question
English, 27.06.2019 21:00
Questions on the website: 13722361