Suppose we have a variable message with the value It's a beautiful day!. If we want to use this variable in a makefile command, we need to escape the single quote character (') so that the command is properly interpreted. We can do this using the subst function like so: 复制 escaped_message = $(subst $(squote),'\$(squote)',$(message)) escaped_message would have the value It'\$(squote)s a beautiful day! Did you made a mistake by "It'\$(squote)s a beautiful day!"?
时间: 2024-02-02 18:02:57 浏览: 32
I apologize for the mistake. The correct value of `escaped_message` after the substitution would be `It\'s a beautiful day!`.
So, the command should be:
```
escaped_message = $(subst $(squote),'\$(squote)',$(message))
```
This will replace any instances of the `$(squote)` variable (which is set to a single quote character) in the `message` variable with `'\$(squote)'`, effectively escaping the single quotes.
In this example, `escaped_message` would have the value `It\'s a beautiful day!`. We can then use this variable in a makefile command without any issues.
相关问题
Suppose we have a variable message with the value It's a beautiful day!. escaped_message = $(subst $(squote),'$(squote)',$(message)) shouldn't the result be It'$(squote)'s a beautiful day! ? If so, the return result will have two single quote, isn't wrong?
Yes, you are correct. The result of the command `escaped_message = $(subst $(squote),'$(squote)',$(message))` will indeed have two single quotes, which is not what we want.
A better way to escape the single quote in `$(message)` is to replace it with a backslash followed by a single quote (`\'`), like this:
```
escaped_message = $(subst ','\'',$(message))
```
This will replace all single quotes in `$(message)` with the escaped sequence `\'`, which represents a single quote. The resulting value of `escaped_message` will be `It\'s a beautiful day!`, which correctly escapes the single quote without adding any extra quotes.
Given a dataset with only two training samples and . Suppose we want to build a Linear Regression model with learning rate . At first, and . 1)Please calculate the cost . 2)Update the parameters and for one time. 3)Please calculate the new cost after updating.
Given:
- Two training samples: `(2, 3)` and `(4, 5)`
- Initial parameters: `theta0 = 0, theta1 = 0`
- Learning rate: `alpha = 0.01`
To build a Linear Regression model, we need to minimize the cost function. The cost function for Linear Regression is:
```
J(theta0, theta1) = (1/2m) * sum(i=1 to m)[hypothesis(xi) - yi]^2
```
where `m` is the total number of training samples, `xi` is the feature of the `i`th training sample, `yi` is the label of the `i`th training sample, and `hypothesis(xi)` is the predicted value for `xi` using the current values of `theta0` and `theta1`.
1) Calculate the initial cost:
```
hypothesis(2) = theta0 + theta1 * 2 = 0 + 0 * 2 = 0
hypothesis(4) = theta0 + theta1 * 4 = 0 + 0 * 4 = 0
J(theta0, theta1) = (1/4) * [(0 - 3)^2 + (0 - 5)^2] = 13/4 = 3.25
```
So the initial cost is `3.25`.
2) Update the parameters for one time:
```
theta0 = theta0 - alpha * (1/m) * sum(i=1 to m)[hypothesis(xi) - yi] = 0 - 0.01 * (1/2) * [(0 - 3) + (0 - 5)] = 0.04
theta1 = theta1 - alpha * (1/m) * sum(i=1 to m)[(hypothesis(xi) - yi) * xi] = 0 - 0.01 * (1/2) * [(0 - 3) * 2 + (0 - 5) * 4] = 0.11
```
So the updated parameters are `theta0 = 0.04` and `theta1 = 0.11`.
3) Calculate the new cost after updating:
```
hypothesis(2) = theta0 + theta1 * 2 = 0.04 + 0.11 * 2 = 0.26
hypothesis(4) = theta0 + theta1 * 4 = 0.04 + 0.11 * 4 = 0.46
J(theta0, theta1) = (1/4) * [(0.26 - 3)^2 + (0.46 - 5)^2] = 2.69/4 = 0.6725
```
So the new cost after updating is `0.6725`.
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