The C masters are in

When i=3, execute the continue statement, re-execute the for loop, i does not change, and also =3, continue to execute the continue statement, so it is an endless loop!

What is for(;;

I can only explain it to you according to the grammar.。。。

main() no parameter input??

int i=0, s=0 Enter two values i, s is equal to 0for(;; Let; =;Or let;?

if(i==3||i==5) continue;If i is equal to 3 and i is equal to 5, then continue??

if (i==6) break;If i is equal to 6, it jumps.

i++;Let the i value be unconditionally increased by 1.

s+=i;Let s be equal to i.

printf("%d",s);Output two a,b variable values.

It should be an endless loop, and there will be no results on the plane.

for(;;It's to make your statement run in a loop, but here you have if(i==3||i==5) continue;

It's going to be an endless loop.

This program will fall into an endless loop, when it grows to 3, it will execute continue, and the following ** will not be executed, i will not change, and it will enter an endless loop.

1) a.There is an error in the number of real types. e.g. float a=; This may be the case inside the system.

2) a. c=(a-=a-5);a-=a-5; c=a;So a=5, c=5; However, the execution of a=b,b+=4 after the hand is lost is a=0, b=4So the values of a, b, and c are 0, 4, and 5

3) d.The value of x++ is the value before x self-increment, for example, x=1; Then abc will output 2 and D will output 1.

In this case, the content in s3 is: jill 0

s1);After that, the content of s3 is jack 0, stract(()."and"Add and to s3, and the result is that s3 is jack and 0, and then stract(()s2) is to add s2 to s3, and then the result is.

jack and jill 0, and return the pointer to s3, and the output is: jack and jill

c.strlen outputs the number of characters before 0.

strlen(s1) +strlen(s2) = 4 + 4 = 8

Output: 8dThe content of s3 is: jack and jill: so output 13