Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The backslash is itself a meta-character. Suppose that you want to match a string that contains a backslash character. How do you suppose you would represent the backslash in the regular expression?

Short Answer

Expert verified
To represent a backslash in a regular expression, it should be written as '\\'.

Step by step solution

01

Understanding the nature of a backslash in regex

It's important to note that the backslash is a special character in regular expressions. Being a meta-character, it has special uses such as initiating escape sequences. So, to represent a backslash in a string using regular expression, it can't just be placed directly in the expression.
02

Representing the backslash

A backslash can be represented in a regular expression by using a double backslash, i.e., '\\'. The first backslash acts as an escape character for the second backslash, yielding a single backslash in the resulting string.
03

Confirming the solution

To verify, one could use this form in a regular expression and test it on a string that contains a backslash. The expression should match any instances of a single backslash in the tested string.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You are an early 20th-century experimental physicist and do not know the value of Planck's constant. By a suitable plot of the following data, and using Einstein's explanation of the photoelectic effect \((\mathrm{K} \mathrm{C})=h f-\phi\) where \(h\) is not known), determine Planck's constant. $$\begin{array}{cc}\begin{array}{c}\text { Wavelength of Lighe } \\\\(\text { nm })\end{array} & \begin{array}{c}\text { Stopping Potential } \\\\(\mathrm{V})\end{array} \\\\\hline 550 & 0.060 \\\\\hline 500 & 0.286 \\\\\hline 450 & 0.563 \\\\\hline 400 & 0.908 \\\\\hline\end{array}$$

According to the energy-lime uncertainty principle. the lifetime \(\Delta f\) of a state and the uncertainty \(\Delta E\) in its energy are inversely proportional. Hydrogen's \(656 \mathrm{~nm}\) red spectral line is the result of an electron making a transition "downward" from a quantum state whose lifetime is about \(10^{-8} s\) (a) What inherent uncertainty in the energy of the emitted photon docs this imply? (Note: Unfortunately. we might use the symbol \(\Delta E\) for the energy difference - i.e., the energy of the photon - but here if means the uncertain in that energy difference.) (b) To what range in wavelengths does this correspond? (As noted in Exercise \(2.57\). the uncertainty principle is one contributor to the broadening of spectral lines.) (c) Obtain a general formula relating \(\Delta \lambda\) to \(\Delta t\).

Determine the wavelength of an X-ray photon that can impart, at most, \(80 \mathrm{keV}\) of kinetic energy to a free electron.

A stationary muon \(\mu^{-}\) annihilates with a stationary antimuon \(\mu^{+}\) (same mass. \(1.88 \times 10^{\circ} 28 \mathrm{~kg}\), but opposite charge). The two disappear, replaced by electromagnetic radiation. (a) Why is it not possible for a single photon to result? (b) Suppose two photons result. Describe their possible directions of motion and wavelengths.

What is the range of frequencies in a 1 ns pulse of (a) 1060 nm infrared laser light and (b) \(100 \mathrm{MHz}\) radio waves? (c) For which is the "uncertainty" in frequency, relative to its approximate value, larger?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free