Help explain the program a few things you don t understand. Very simple procedure

I think questions 1 and 2 should be looked at together, bool assigns a value to inword is false, if(!isspace(c)&&inword), since &&& requires that the expressions on both its left and right sides be true, because the initial assignment to inword is false, then! inword is true (inword is false, false is false, then the relationship inword==false is true,!

inword is also true, so they are equivalent), so that the first non-whitespace letter entered qualifies as if. When the first non-blank letter enters the if statement, and then assign the value of ture to inword, then inword is true, and then add 1 to n-words, and assign this letter to prev.

The second non-blank letter of the first word enters the while loop, in if(!isspace(c)&&inword) condition, because inword is true, then now! inword is false, because && needs the formula to be true in two paragraphs, then the second letter does not meet the if condition, it will enter the next if(isspace(c)&&inword), because it is a non-blank letter, it does not meet the conditions, assign this letter to prev, then the third character of this word enters the while loop.

Assuming the third character is a whitespace character, then it does not conform to if(!).isspace(c)&&inword), then it goes to the next if, if(isspace(c)&&&inword), because the current isspace(c) and inword are true, then it satisfies the condition, enters the if statement, and then assigns false false to inword, and inword becomes false again. Then assign this whitespace character to prev.

At this point, the next word loops on.

Now to the third question, based on the above analysis, we know that every input character will be assigned a value to prev. in the input'|'The preceding character is the last character to enter the while loop. As is customary, if the last line is complete, we type in a line break, then due to the first if statement in the while loop (c==''), then this line will be counted to the number of lines, but if the last sentence entered is not a complete line, we will generally enter a full stop ending (or other non-line breaks), then the last line will not enter a line break, and the last line will not be counted to the number of lines, at this time, because the last character is assigned to prev, after jumping out of the while loop, it satisfies if(prev!).

'), p-line=1, the record is incomplete.

The variable prev is useless and can be removed.

A word is composed of multiple characters, inword is used to determine whether the obtained characters are in the same word, if in the same word, its value is true, if not, it is false.

He's saying there's a string s

Then let you count how many numbers are in the string, i.e. 0 9for loop through the string, from s[0] until the end of the string if s[i] is one of the characters of 0 9, n++, count +1 because you have one in the string'\0'character, so as of there, the s string is over, followed by 8s34. are no longer traversed.

So the s string is actually traversed as 012xy

So there are 3 numbers in total, choose B

It's very logical, and you should be able to get a result by doing the math with a pen on paper.