Question

$$\begin{array}{l} \text { char } \operatorname{si}[50]=" j a c k" \\ \text { char } \operatorname{s2}[5 \theta]=" j i i l^{\prime \prime} \\ \text { char } s 3[50] ; \end{array}$$ What (if anything) prints when each of the following statements is performed? If the statement contains an error, describe the error and indicate how to correct it. Assume the following variable declarations: a. cout \(<<\) strcpy \((s 3, s 2)<<\) end \(l\) \(\mathbf{b}\) cout \(<<\) straat \(\left(\text { strcat }\left(\text { strcpy }(\mathrm{s} 3, \mathrm{s} 1),^{\prime \prime} \text { and }^{\prime \prime}\right), \mathrm{s} 2\right)\) \(<<\) end 1 \(\mathbf{c}\) cout \(<<\operatorname{str} \operatorname{len}(\mathrm{s} 1)+\operatorname{str} \operatorname{len}(\mathrm{s} 2)<<\mathrm{end} \mathrm{l}\) d. cout \(<<\operatorname{str} \operatorname{len}(\mathrm{s} 3)<<\) end 1

Short answer

Part a: Print "jiil"; Part b: Print "jack and jiil"; Part c: Print 8; Part d: Print 13.

Step-by-step solution

Analyze Declaration Statements

The variable declarations prepare strings \(s1\), \(s2\), and \(s3\). \(s1\) is initialized with "jack", \(s2\) with "jiil” but note the potential mistake with \(5\theta\), possibly a typo meant as \([50]\), and \(s3\) is uninitialized. This could lead to undefined behavior when performing operations on \(s3\) before it's assigned a value.

Understand Part (a) Code

The statement `cout << strcpy(s3, s2) << endl;` attempts to copy \(s2\)'s content into \(s3\) and then print \(s3\). Assuming the syntax error \(5\theta\) is corrected to \([50]\), it will copy "jiil" to \(s3\) and print "jiil" followed by a newline.

Understand Part (b) Code

The statement `cout << strcat(strcat(strcpy(s3, s1), " and "), s2) << endl;` combines several string operations. \(s3\) is first assigned "jack", then " and " is concatenated, making it "jack and ", and finally \(s2\) "jiil" is appended. The final result is "jack and jiil" which is printed.

Understand Part (c) Code

The statement `cout << strlen(s1) + strlen(s2) << endl;` calculates the total length of \(s1\) and \(s2\). \(s1\) is "jack" with length 4, and assuming \(s2\) is correctly declared as "jiil" with length 4, the sum of their lengths is 8. Therefore, 8 is printed followed by a newline.

Understand Part (d) Code

The statement `cout << strlen(s3) << endl;` will print the length of \(s3\) from the last modification in part b. \(s3\) is "jack and jiil" (length 13), thus it prints 13 followed by a newline.

Key concepts

Understanding the strcpy Function

In C++, the strcpy function is crucial for string manipulation. It allows us to copy the content from one string to another. This function takes two arguments: the destination string and the source string.
Here’s how it works:

• The function copies the contents of the source string into the destination.
• The destination string must be large enough to hold the source string to avoid buffer overflows.
• Once the operation is complete, strcpy returns a pointer to the destination string.

Remember: strcpy does not check for buffer overflows, so always ensure your destination string is sized appropriately to store the copied data. For example, if you have `char s1[50] = "hello";`, using `strcpy(s2, s1);` will result in `s2` containing "hello".

Mastering the strcat Function

The strcat function is used to concatenate, or join, two strings in C++. It appends the contents of the source string to the end of the destination string. This makes it a powerful tool for string manipulation.

• Firstly, both strings must be properly null-terminated.
• The destination string must have enough space to accommodate the new combined string.
• The function returns a pointer to the destination string after successfully appending the source to it.

Example Use: Consider you have `char s1[50] = "Hello, "` and `char s2[] = "World!";`. When you perform `strcat(s1, s2);`, the result in `s1` would be "Hello, World!".

Harnessing the Power of strlen Function

The strlen function is employed to determine the length of a string. Unlike other functions, it doesn't modify the original string, but instead helps you understand the size of the string content.

• It takes a single argument: a pointer to the string.
• The function returns the total number of characters in the string, excluding the null terminator.

For instance, if you have `char s1[] = "Programming";`, `strlen(s1);` will return 11 since the word "Programming" has 11 letters. Knowing the length helps in intelligently managing memory and ensuring safe manipulation or copying of strings.

Importance of Variable Declaration

In C++, declaring variables is a vital step before using them. When dealing with strings, it's essential to declare them with enough allocated space for the intended content to prevent errors.

• Declaring a char array, like `char str[50];`, reserves a block of memory to hold up to 49 characters (considering the null terminator).
• Failing to declare a variable properly can lead to undefined behavior or program crashes.

Adequate variable declaration also helps you align your data structures with the logic of your program, ensuring cleaner, more readable code. Always visualize the maximum size your data might require to allocate sufficient memory.

Avoiding Syntax Errors in C++

Syntax errors in C++ are mistakes in the way the code is written. These can prevent your code from compiling and running, usually due to typographical mistakes or incorrect use of functions.

• Common syntax errors include misspelled keywords or functions, such as using "strcpyi" instead of "strcpy".
• Another classic error is mismatched brackets or using incorrect operators.
• Proper indentation and code structure also help in avoiding such mistakes.

To minimize syntax errors, you can use IDE features like syntax highlighting and auto-completion. Consistently test small blocks of code to catch errors early. Always ensure variable names and function calls are accurate and logical.