What will you see on the terminal?int main(void){ int *ptr; *ptr = 98; printf("%d\n", *ptr); return (0);}It doesn’t compileSegmentation Fault980

The code you provided will likely result in a Segmentation Fault. This is because you're trying to assign a value to a pointer that hasn't been initialized and doesn't point to a valid memory location. Here's a step-by-step explanation:

1. int *ptr; This line declares a pointer to an integer, but it doesn't initialize it. This means that ptr could be pointing anywhere in memory.

2. *ptr = 98; This line tries to write the value 98 to the memory location that ptr is pointing to. Since ptr hasn't been initialized, it's pointing to some random location in memory, and trying to write to it will likely result in a Segmentation Fault.

3. printf("%d\n", *ptr); This line would print the value that ptr is pointing to, if the previous line hadn't caused a Segmentation Fault.

4. return (0); This line would end the program, returning a status code of 0 to indicate that the program has finished successfully. However, because of the Segmentation Fault, the program would likely have terminated before reaching this line.

To fix this code, you would need to allocate memory for ptr to point to before trying to assign a value to it. For example, you could do this by declaring an integer variable and setting ptr to point to it:

int main(void){
    int num;
    int *ptr = #
    *ptr = 98;
    printf("%d\n", *ptr);
    return (0);
}

In this corrected version of the code, ptr is initialized to point to num, so it's safe to assign a value to *ptr. The program will print 98 and then end successfully.