What happens in the following code?

```cpp
void f() {
char *s = "CS 150";
s[0] = 'X';
cout << s << endl;
}
```

A. Prints "XS 150"

B. Most likely crashes when run

C. Code compiles without warnings

D. Code fails to compile because "CS 150" is const

The code given will most likely crash when run. The reason is that the line s[0] = 'X'; tries to modify a string literal, which is not allowed in C++. String literals are treated as const pointers, and any attempt to modify them will result in undefined behavior.

If you want to modify the string, you should use an array of characters, like char s[] = "CS 150"; instead of a pointer to a string literal. This will create a writable copy of the string.

So, in this case, code fails to compile because "CS 150" is const.

What happens in the following code?```cppvoid f() { char *s = "CS 150"; s[0] = 'X'; cout << s << endl;}```A. Prints "XS 150"B. Most likely crashes when runC. Code compiles without warningsD. Code fails to compile because "CS 150" is const

Answer :

Final answer:

Explanation:

Other Questions

What happens in the following code?

```cpp
void f() {
char *s = "CS 150";
s[0] = 'X';
cout << s << endl;
}
```

A. Prints "XS 150"

B. Most likely crashes when run

C. Code compiles without warnings

D. Code fails to compile because "CS 150" is const