can I change "a" to "b"?

时间: 2023-11-05 20:43:00 浏览: 32
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I mean I'm quite new to system verilog, so I need you to show me the concrete code after modification, can you show me the code of the 10 ways of modification you just suggested? thanks a lot

Certainly! Here's the code for the 10 modifications I suggested: 1. Modify the values in both `t1` and `t2` class constructors to be random numbers instead of hard-coded values. ``` interface Int(); class t1; integer a[string]; function new; a["a"] =$random; a["b"] =$random; a["c"] =$random; endfunction endclass class t2; integer b[string]; function new; b["aa"] =$random; b["bb"] =$random; b["cc"] =$random; endfunction endclass t1 inst1 =new; t2 inst2 =new; endinterface ``` 2. Add an additional class that extends `t1` and `t2` and overrides specific values with its own. ``` interface Int(); class t1; integer a[string]; function new; a["a"] =40; a["b"] =60; a["c"] =80; endfunction endclass class t2; integer b[string]; function new; b["aa"] =44; b["bb"] =66; b["cc"] =88; endfunction endclass class t3 extends t1,t2; function new; super.new(); a["a"] =100; b["bb"] =200; endfunction endclass t1 inst1 =new; t2 inst2 =new; t3 inst3 =new; endinterface ``` 3. Use a different data type for the `integer array[string]` variable, such as `real array[string]`. ``` interface Int(); class t1; real a[string]; function new; a["a"] =40; a["b"] =60; a["c"] =80; endfunction endclass class t2; real b[string]; function new; b["aa"] =44; b["bb"] =66; b["cc"] =88; endfunction endclass t1 inst1 =new; t2 inst2 =new; endinterface ``` 4. Change the `virtual` keyword to `pure virtual` in the `Int` interface declaration. ``` interface Int(); pure virtual class t1; integer a[string]; function new; a["a"] =40; a["b"] =60; a["c"] =80; endfunction endclass pure virtual class t2; integer b[string]; function new; b["aa"] =44; b["bb"] =66; b["cc"] =88; endfunction endclass t1 inst1 =new; t2 inst2 =new; endinterface ``` 5. Replace the `initial` block with a `task` or `function` that takes in the `Int` interface as an argument and returns the `array` variable. ``` interface Int(); class t1; integer a[string]; function new; a["a"] =40; a["b"] =60; a["c"] =80; endfunction endclass class t2; integer b[string]; function new; b["aa"] =44; b["bb"] =66; b["cc"] =88; endfunction endclass task test; input Int in; integer array[string]; begin #1; array =in.inst1.a; $display("testint.inst1.a =%p", in.inst1.a); $display("array = %p", array); #2; array =in.inst2.b; $display("testint.inst2.b =%p", in.inst2.b); $display("array =%p", array); end endtask t1 inst1 =new; t2 inst2 =new; endinterface module top; Int in(); initial begin test(in); end endmodule ``` 6. Add a `bit [31:0]` variable to both `t1` and `t2` classes and modify the constructor functions to set it to the sum of the other values. ``` interface Int(); class t1; integer a[string]; bit [31:0] sum; function new; a["a"] =40; a["b"] =60; a["c"] =80; sum =a["a"] +a["b"] +a["c"]; endfunction endclass class t2; integer b[string]; bit [31:0] sum; function new; b["aa"] =44; b["bb"] =66; b["cc"] =88; sum =b["aa"] +b["bb"] +b["cc"]; endfunction endclass t1 inst1 =new; t2 inst2 =new; endinterface ``` 7. Create a new interface that extends `Int` and adds additional methods or variables. ``` interface Int(); class t1; integer a[string]; function new; a["a"] =40; a["b"] =60; a["c"] =80; endfunction endclass class t2; integer b[string]; function new; b["aa"] =44; b["bb"] =66; b["cc"] =88; endfunction endclass t1 inst1 =new; t2 inst2 =new; endinterface interface Int2 extends Int(); parameter int MAX_ARRAY_SIZE =100; function void print_array(); for (int i =0; i < MAX_ARRAY_SIZE; i++) begin $display("a[%0d] = %0d", i, inst1.a[i]); end endfunction endinterface module top; Int2 in(); initial begin in.print_array(); end endmodule ``` 8. Modify the `testint` variable to be a `typedef` of the `Int` interface instead of a `virtual` instance. ``` typedef Int testint_t; testint_t in =new(); ``` 9. Change the `integer` data type to `bit [7:0]` for all instances of `integer` in the code. ``` interface Int(); class t1; bit [7:0] a[string]; function new; a["a"] =40; a["b"] =60; a["c"] =80; endfunction endclass class t2; bit [7:0] b[string]; function new; b["aa"] =44; b["bb"] =66; b["cc"] =88; endfunction endclass t1 inst1 =new; t2 inst2 =new; endinterface ``` 10. Add assertions to the code to ensure that the values in `testint.inst1.a` and `testint.inst2.b` are always greater than zero. ``` interface Int(); class t1; integer a[string]; function new; a["a"] =40; a["b"] =60; a["c"] =80; assert(a["a"] > 0) else $error("Value of a[a] is not greater than zero"); assert(a["b"] > 0) else $error("Value of a[b] is not greater than zero"); assert(a["c"] > 0) else $error("Value of a[c] is not greater than zero"); endfunction endclass class t2; integer b[string]; function new; b["aa"] =44; b["bb"] =66; b["cc"] =88; assert(b["aa"] > 0) else $error("Value of b[aa] is not greater than zero"); assert(b["bb"] > 0) else $error("Value of b[bb] is not greater than zero"); assert(b["cc"] > 0) else $error("Value of b[cc] is not greater than zero"); endfunction endclass t1 inst1 =new; t2 inst2 =new; endinterface ```

There are three types of filter responses: impulse response, step response, and frequency response. Suppose I want to know how the filter reacts to a sudden change in the input, state which type of filter response I should compute? Then, plot the corresponding filter response (first 100 samples) for the two filters separately.

If you want to know how the filter reacts to a sudden change in the input, you should compute the step response. To plot the step response for each filter, we first need to obtain the transfer function for each filter. Once we have the transfer function, we can use it to compute the step response. Assuming we have the transfer functions for the two filters, we can plot the step response in Python using the following code: ```python import numpy as np import matplotlib.pyplot as plt from scipy.signal import lfilter # Define the transfer function for filter 1 b1 = np.array([1, 0.5, 0.1]) a1 = np.array([1, -0.9, 0.2]) # Define the transfer function for filter 2 b2 = np.array([1, -0.8, 0.2]) a2 = np.array([1, -0.5, 0.1]) # Compute the step response for filter 1 zi = lfilter_zi(b1, a1) y1, _ = lfilter(b1, a1, np.ones(100), zi=zi) # Compute the step response for filter 2 zi = lfilter_zi(b2, a2) y2, _ = lfilter(b2, a2, np.ones(100), zi=zi) # Plot the step response for filter 1 plt.figure() plt.plot(y1[:100], 'b') plt.title('Step Response for Filter 1') plt.xlabel('Sample') plt.ylabel('Output') plt.grid(True) # Plot the step response for filter 2 plt.figure() plt.plot(y2[:100], 'r') plt.title('Step Response for Filter 2') plt.xlabel('Sample') plt.ylabel('Output') plt.grid(True) plt.show() ``` Note that we use the `lfilter()` function from the `scipy.signal` library to compute the step response. We also use the `lfilter_zi()` function to obtain the initial conditions for the filter. The resulting plots should show the step response for each filter, which indicates how the filter responds to a sudden change in the input signal.

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You have two binary strings � a and � b of length � n. You would like to make all the elements of both strings equal to 0 0. Unfortunately, you can modify the contents of these strings using only the following operation: You choose two indices � l and � r ( 1 ≤ � ≤ � ≤ � 1≤l≤r≤n); For every � i that respects � ≤ � ≤ � l≤i≤r, change � � a i ​ to the opposite. That is, � � : = 1 − � � a i ​ :=1−a i ​ ; For every � i that respects either 1 ≤ � < � 1≤i<l or � < � ≤ � r<i≤n, change � � b i ​ to the opposite. That is, � � : = 1 − � � b i ​ :=1−b i ​ . Your task is to determine if this is possible, and if it is, to find such an appropriate chain of operations. The number of operations should not exceed � + 5 n+5. It can be proven that if such chain of operations exists, one exists with at most � + 5 n+5 operations. Input Each test consists of multiple test cases. The first line contains a single integer � t ( 1 ≤ � ≤ 1 0 5 1≤t≤10 5 ) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer � n ( 2 ≤ � ≤ 2 ⋅ 1 0 5 2≤n≤2⋅10 5 ) — the length of the strings. The second line of each test case contains a binary string � a, consisting only of characters 0 and 1, of length � n. The third line of each test case contains a binary string � b, consisting only of characters 0 and 1, of length � n. It is guaranteed that sum of � n over all test cases doesn't exceed 2 ⋅ 1 0 5 2⋅10 5 . Output For each testcase, print first "YES" if it's possible to make all the elements of both strings equal to 0 0. Otherwise, print "NO". If the answer is "YES", on the next line print a single integer � k ( 0 ≤ � ≤ � + 5 0≤k≤n+5) — the number of operations. Then � k lines follows, each contains two integers � l and � r ( 1 ≤ � ≤ � ≤ � 1≤l≤r≤n) — the description of the operation. If there are several correct answers, print any of them.

A random number of Rabbit images ranging from 1 to 10 are displayed for each operand and the user is expected to enter the values of the two operands and the result of adding the two operands, in the given text fields. When the user clicks on the button ‘Check!’, one of two things can happen: Case 1: all three input values are correct i) the text changes to ‘"Correct! Have another go?"’. ii) the number of Rabbit images displayed for each of the two operands changes. See Figure 2 for an example. iii) the three text fields are reset (i.e. they are made empty). 2/5 Case 2: at least one of the input values entered is incorrect i) the text changes to ‘Wrong! Try again!’. ii) the number of Rabbit images displayed does NOT change. iii) the text fields do NOT change.Implement SumItUp as a Java application. You application must satisfy ALL the specific requirements given below: a) The title of the top-level container must be ‘Welcome to SumItUp!’. b) The initial text should be ‘Enter two operands, result and click on 'Check!'’. See Figure 1. c) There should be no more than 4 Rabbit images per row. See Hint 1. d) The text fields should be wide enough to display at least TWO characters. e) The button ‘Check!’ must not resize when the GUI is resized. See Hint 2 and Figure 3. f) The ‘plus sign’ icon should appear vertically centered between the two sets of Rabbit images and must not resize when the GUI is resized. See Hint 2 and Figure 3. g) When first launched and whenever a correct answer is given, the number of displayed Rabbit images for each operand should change to any number between 1 and 10 (inclusive). See Hint 3 and Hint 4. Note: It is possible for the next number(s) to be the same as the current number(s). h) Nothing should happen if the user clicks the ‘Check!’ button while at least one of the text fields are empty, i.e. no errors should be thrown in this case. Note: You can assume that only a numeric value will be entered into the text fields.

(a) Consider the case of a European Vanilla Call option which is path independent. Examine the convergence of the Monte Carlo Method using the programme given in ‘MC Call.m’. How does the error vary with the number of paths nP aths? The current time is t = 0 and the Expiry date of the option is t = T = 0.5. Suppose that the current value of the underlying asset is S(t = 0) = 100 and the Exercise price is E = 100, with a risk free interest rate of r = 0.04 and a volatility of σ = 0.5. (b) Now repeat part (a) above but assume that the volatility is σ = 0.05. Does the change in the volatility σ influence the convergence of the Monte Carlo Method? (c) Now repeat part (a) but instead of taking one big step from t = 0 to t = T divide the interval into nSteps discrete time steps by using the programme given in ‘MC Call Small Steps.m’. Confirm that for path independent options, the value of nP aths determines the rate of convergence and that the value of nSteps can be set to 1. (d) Now let us consider path dependent options. The programme given in ‘MC Call Small Steps.m’ is the obvious starting point here. We assume that the current time is t = 0 and the expiry date of the option is t = T = 0.5. The current value of the underlying asset is S(t = 0) = 100 and the risk free interest rate is r = 0.05 and the volatility is σ = 0.3. (i) Use the Monte Carlo Method to estimate the value of an Arithematic Average Asian Strike Call option with Payoff given by max(S(T) − S, ¯ 0). (ii) Use the Monte Carlo Method to estimate the value of an Up and Out Call option with Exercise Price E = 100 and a barrier X = 150. (iii) Comment on the the rate of convergence for part (i) and (ii) above with respect to the parameters nP aths and nP aths使用matlab编程

private final CharSequence getSmallTime() { Context context = getContext(); boolean is24 = DateFormat.is24HourFormat(context); LocaleData d = LocaleData.get(context.getResources().getConfiguration().locale); final char MAGIC1 = '\uEF00'; final char MAGIC2 = '\uEF01'; SimpleDateFormat sdf; String format = is24 ? d.timeFormat24 : d.timeFormat12; // Add seconds to the format string format += ":ss"; if (!format.equals(mClockFormatString)) { /* * Search for an unquoted "a" in the format string, so we can * add dummy characters around it to let us find it again after * formatting and change its size. */ if (AM_PM_STYLE != AM_PM_STYLE_NORMAL) { int a = -1; boolean quoted = false; for (int i = 0; i < format.length(); i++) { char c = format.charAt(i); if (c == ''') { quoted = !quoted; } if (!quoted && c == 'a') { a = i; break; } } if (a >= 0) { // Move a back so any whitespace before AM/PM is also in the alternate size. final int b = a; while (a > 0 && Character.isWhitespace(format.charAt(a-1))) { a--; } format = format.substring(0, a) + MAGIC1 + format.substring(a, b) + "a" + MAGIC2 + format.substring(b + 1); } } mClockFormat = sdf = new SimpleDateFormat(format); mClockFormatString = format; } else { sdf = mClockFormat; } String result = sdf.format(mCalendar.getTime()); if (AM_PM_STYLE != AM_PM_STYLE_NORMAL) { int magic1 = result.indexOf(MAGIC1); int magic2 = result.indexOf(MAGIC2); if (magic1 >= 0 && magic2 > magic1) { SpannableStringBuilder formatted = new SpannableStringBuilder(result); if (AM_PM_STYLE == AM_PM_STYLE_GONE) { formatted.delete(magic1, magic2+1); } else { if (AM_PM_STYLE == AM_PM_STYLE_SMALL) { CharacterStyle style = new RelativeSizeSpan(0.7f); formatted.setSpan(style, magic1, magic2, Spannable.SPAN_EXCLUSIVE_INCLUSIVE); } formatted.delete(magic2, magic2 + 1); formatted.delete(magic1, magic1 + 1); } return formatted; } } return result;}如何实时更新秒

private final CharSequence getSmallTime() { Context context = getContext(); boolean is24 = DateFormat.is24HourFormat(context); LocaleData d = LocaleData.get(context.getResources().getConfiguration().locale); final char MAGIC1 = '\uEF00'; final char MAGIC2 = '\uEF01'; SimpleDateFormat sdf; String format = is24 ? d.timeFormat24 : d.timeFormat12; if (!format.equals(mClockFormatString)) { /* * Search for an unquoted "a" in the format string, so we can * add dummy characters around it to let us find it again after * formatting and change its size. */ if (AM_PM_STYLE != AM_PM_STYLE_NORMAL) { int a = -1; boolean quoted = false; for (int i = 0; i < format.length(); i++) { char c = format.charAt(i); if (c == ''') { quoted = !quoted; } if (!quoted && c == 'a') { a = i; break; } } if (a >= 0) { // Move a back so any whitespace before AM/PM is also in the alternate size. final int b = a; while (a > 0 && Character.isWhitespace(format.charAt(a-1))) { a--; } format = format.substring(0, a) + MAGIC1 + format.substring(a, b) + "a" + MAGIC2 + format.substring(b + 1); } } mClockFormat = sdf = new SimpleDateFormat(format); mClockFormatString = format; } else { sdf = mClockFormat; } String result = sdf.format(mCalendar.getTime()); if (AM_PM_STYLE != AM_PM_STYLE_NORMAL) { int magic1 = result.indexOf(MAGIC1); int magic2 = result.indexOf(MAGIC2); if (magic1 >= 0 && magic2 > magic1) { SpannableStringBuilder formatted = new SpannableStringBuilder(result); if (AM_PM_STYLE == AM_PM_STYLE_GONE) { formatted.delete(magic1, magic2+1); } else { if (AM_PM_STYLE == AM_PM_STYLE_SMALL) { CharacterStyle style = new RelativeSizeSpan(0.7f); formatted.setSpan(style, magic1, magic2, Spannable.SPAN_EXCLUSIVE_INCLUSIVE); } formatted.delete(magic2, magic2 + 1); formatted.delete(magic1, magic1 + 1); } return formatted; } } return result; }添加秒

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