Programming with Microsoft Visual Basic 2015 (MindTap Course List)
7th Edition
ISBN: 9781285860268
Author: Diane Zak
Publisher: Cengage Learning
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Expert Solution & Answer
Chapter 1.LC, Problem 1RQ
Program Description Answer
The timer control’s “Tick” event procedure is processed each time an interval has elapsed.
Hence, correct answer is option “B”.
Expert Solution & Answer
Explanation of Solution
Tick event procedure:
The timer is a control, which is used to process any specific task at regular intervals.
- The state of the timer control is of two types. They are running or stopped state.
- Running state:
- If the timer “Enabled” property is set to “True”, then the timer is running.
- If the interval of the Timer has expired, then the timer Tick event occurs each time.
- Therefore, the code in the Tick event procedure is processed each time.
- Stopped state:
- If the timer “Enabled” property is set to “False”, then the timer is stopped.
- The Tick event does not occur in the stop state.
Therefore, the timer control’s Tick event procedure occurs at the end of each interval of the Timer control.
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Students have asked these similar questions
The next problem concerns the following C code:
/copy input string x to buf */
void foo (char *x) {
char buf [8];
strcpy((char *) buf, x);
}
void callfoo() {
}
foo("ZYXWVUTSRQPONMLKJIHGFEDCBA");
Here is the corresponding machine code on a Linux/x86 machine:
0000000000400530 :
400530:
48 83 ec 18
sub
$0x18,%rsp
400534:
48 89 fe
mov
%rdi, %rsi
400537:
48 89 e7
mov
%rsp,%rdi
40053a:
e8 di fe ff ff
callq
400410
40053f:
48 83 c4 18
add
$0x18,%rsp
400543:
c3
retq
400544:
0000000000400544 :
48 83 ec 08
sub
$0x8,%rsp
400548:
bf 00 06 40 00
mov
$0x400600,%edi
40054d:
e8 de ff ff ff
callq 400530
400552:
48 83 c4 08
add
$0x8,%rsp
400556:
c3
This problem tests your understanding of the program stack. Here are some notes to
help you work the problem:
⚫ strcpy(char *dst, char *src) copies the string at address src (including
the terminating '\0' character) to address dst. It does not check the size of
the destination buffer.
• You will need to know the hex values of the following characters:
1234
3. Which line prevents compiler optimization? Circle one: 1234
Suggested solution:
Store strlen(str) in a variable before the if statement.
⚫ Remove the if statement.
Replace index 0 && index < strlen(str)) {
5 }
}
=
str [index] = val;
Character Hex value | Character Hex value Character Hex value
'A'
0x41
'J'
Ox4a
'S'
0x53
'B'
0x42
'K'
0x4b
"T"
0x54
0x43
'L'
Ox4c
'U'
0x55
0x44
'M'
0x4d
'V'
0x56
0x45
'N'
Ox4e
'W'
0x57
0x46
'0'
Ox4f
'X'
0x58
0x47
'P'
0x50
'Y'
0x59
0x48
'Q'
0x51
'Z'
Ox5a
'T'
0x49
'R'
0x52
'\0'
0x00
Now consider what happens on a Linux/x86 machine when callfoo calls foo with
the input string "ZYXWVUTSRQPONMLKJIHGFEDCBA".
A. On the left draw the state of the stack just before the execution of the instruction
at address Ox40053a; make sure to show the frames for callfoo and foo and
the exact return address, in Hex at the bottom of the callfoo frame.
Then, on the right, draw the state of the stack just after the instruction got
executed; make sure to show where the string "ZYXWVUTSRQPONMLKJIHGFEDCBA"
is placed and what part, if any, of the above return address has been overwritten.
B. Immediately after the ret instruction at address 0x400543 executes, what is
the value of the program counter register %rip?…
Chapter 1 Solutions
Programming with Microsoft Visual Basic 2015 (MindTap Course List)
Ch. 1.LA - Prob. 1RQCh. 1.LA - Which window is used to set the characteristics...Ch. 1.LA - Prob. 3RQCh. 1.LA - Prob. 4RQCh. 1.LA - Prob. 5RQCh. 1.LA - Prob. 6RQCh. 1.LA - Prob. 7RQCh. 1.LA - Which property determines the initial position of...Ch. 1.LA - Prob. 9RQCh. 1.LA - Prob. 10RQ
Ch. 1.LA - Prob. 11RQCh. 1.LA - Prob. 1ECh. 1.LA - Prob. 2ECh. 1.LA - Prob. 3ECh. 1.LB - Prob. 1RQCh. 1.LB - Prob. 2RQCh. 1.LB - Prob. 3RQCh. 1.LB - Prob. 4RQCh. 1.LB - Prob. 5RQCh. 1.LB - Prob. 1ECh. 1.LB - Prob. 2ECh. 1.LB - Prob. 3ECh. 1.LB - Prob. 4ECh. 1.LC - Prob. 1RQCh. 1.LC - Prob. 2RQCh. 1.LC - Prob. 3RQCh. 1.LC - Prob. 4RQCh. 1.LC - Prob. 5RQCh. 1.LC - Prob. 1ECh. 1.LC - Prob. 2ECh. 1.LC - Prob. 3ECh. 1.LC - Prob. 4ECh. 1.LC - Prob. 5ECh. 1.LC - Prob. 6ECh. 1.LC - Prob. 7ECh. 1.LC - Prob. 8E
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- The next problem concerns the following C code: /copy input string x to buf */ void foo (char *x) { char buf [8]; strcpy((char *) buf, x); } void callfoo() { } foo("ZYXWVUTSRQPONMLKJIHGFEDCBA"); Here is the corresponding machine code on a Linux/x86 machine: 0000000000400530 : 400530: 48 83 ec 18 sub $0x18,%rsp 400534: 48 89 fe mov %rdi, %rsi 400537: 48 89 e7 mov %rsp,%rdi 40053a: e8 di fe ff ff callq 400410 40053f: 48 83 c4 18 add $0x18,%rsp 400543: c3 retq 400544: 0000000000400544 : 48 83 ec 08 sub $0x8,%rsp 400548: bf 00 06 40 00 mov $0x400600,%edi 40054d: e8 de ff ff ff callq 400530 400552: 48 83 c4 08 add $0x8,%rsp 400556: c3 This problem tests your understanding of the program stack. Here are some notes to help you work the problem: ⚫ strcpy(char *dst, char *src) copies the string at address src (including the terminating '\0' character) to address dst. It does not check the size of the destination buffer. • You will need to know the hex values of the following characters:arrow_forwardConsider the following assembly code for a C for loop: movl $0, %eax jmp .L2 .L3: addq $1, %rdi addq %rsi, %rax subq $1, %rsi .L2: cmpq %rsi, %rdi jl .L3 addq ret %rdi, %rax Based on the assembly code above, fill in the blanks below in its corresponding C source code. Recall that registers %rdi and %rsi contain the first and second, respectively, argument of a function. (Note: you may only use the symbolic variables x, y, and result in your expressions below do not use register names.) long loop (long x, long y) { long result; } for ( } return result; __; y--) {arrow_forwardIn each of the following C code snippets, there are issues that can prevent the compiler from applying certain optimizations. For each snippet: Circle the line number that contains compiler optimization blocker. ⚫ Select the best modification to improve optimization. 1. Which line prevents compiler optimization? Circle one: 2 3 4 5 6 Suggested solution: ⚫ Remove printf or move it outside the loop. Remove the loop. • Replace arr[i] with a constant value. 1 int sum (int *arr, int n) { 2 int s = 0; 3 for (int i = 0; i < n; i++) { 4 5 6 } 7 8 } s = arr[i]; printf("%d\n", s); return s; 234206 2. Which line prevents compiler optimization? Circle one: 2 3 4 5 6 Suggested solution: Move or eliminate do_extra_work() if it's not necessary inside the loop. Remove the loop (but what about scaling?). ⚫ Replace arr[i] *= factor; with arr[i] = 0; (why would that help?). 1 void scale (int *arr, int n, int factor) { 5 6 } for (int i = 0; i < n; i++) { rr[i] = factor; do_extra_work ();arrow_forward
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