为什么我的循环不会变得无限

Question

#include <stdio.h>
#include <cs50.h>

int main(void)
{

    int n;
    printf("Please give me an integer greater than zero!\n");
    n=GetInt();

    if(n<0)
    {
    printf("You are giving me a bad value!\n");
    return 1;
    }


    for(int i=n-1;i<n;n--)
    printf("%d\n",n);
    return 0;
}

I would like to know why the loop is not going to infinity if the user enters in a number for n . Lets say that the user puts in 40 for n ; wouldn't i always be n-1 , so 39 and n being 40, then i becomes 38 when n becomes 39 and so on — so wouldn't that make an infinite loop?

Answer 1

for(int i=n-1;i<n;n--)

Lets draw a (really short) table for n = 40 :

  i  |  n
-----+-----
 39  | 40    i < n ? true
 39  | 39    i < n ? false

Thus, we'll exit the loop after the 1 ^st iteration.

Clarification :

I guess you're confused because you think that i is updated in each iteration, but that's the point - it doesn't , its value is fixed and only n is changing.

Answer 2

This loop only runs once. Consider:

 for(int i=n-1;i<n;n--)

• With n == 40 , on the first iteration, i = 39 .
• The condition i < n is true ( 39 < 40 == true ), so we go in to the loop for the first time.
• At the end of the first loop, n gets decremented to 39
• The condition i < n is false ( 39 < 39 == false ), so we don't get a second time through the loop.

---

Now, what happens if we make n increase instead of decrease? Will that run forever?

  for(int i=n-1;i<n;n++)

The answer is "maybe, but probably not":

• Eventually, n will reach the largest value that can be stored in an integer, INT_MAX (defined in limits.h , and on my system it is 2,147,483,647).
• Making an integer larger than INT_MAX causes integer overflow .
• The result of integer overflow on a signed integer is undefined , which means the result could be anything (and indeed, your program could crash).

On most systems, however, the value will probably wrap around to INT_MIN , or -2,147,483,648.

• If this happens, i < n will be false, and your loop will terminate.

But, since integer overflow on signed integers is undefined behaviour , you can't be sure that this will happen. It is better to write your program to avoid this situation.

---

If you really want it to run forever - just write:

while(1) { ... }

or

for(;;) { ... }

These two loops have the advantage that they are common ways to write an infinite loop, and so they are easy for other programmers to read.

Answer 3

The reason is that i is never decremented, so it does only 1 loop:

 i=39 n=40
 i=39 n=39  -> stop

In order to decrement also i you should write:

 for(int i = n-1;i<n;n--,i--)

Answer 4

n will underflow somewhen because int is signed and has a value range of -2147483648 to 2147483647 for example (x86). Somewhen n will get more positive than i.

Edit: The loop has at most 1 iteration.

Edit 2: The loop would have no iterations if n would have the value -2147483648 for example because -2147483648 - 1 will make the value positive (two complement integer arithmetic). But this could never the case because the pre condition is that n may not be negative.

Answer 5

i is only ever set once at the start of the loop. For example if the user enters 10 then i is 9 for the 1st iteration. By the 2nd iteration n is decremented by 1 and i is still 9.

Answer 6

Your for loop:

for(int i=n-1;i<n;n--) {
    printf("%d\n",n);
}

Translates into the following while loop:

{
    int i = n - 1;
    while (i < n) {
        printf(%d\n", n);
        n--;
    }
}

The first clause of the for statement performs initialization. It is not repeated at each iteration, but only once. Thus, i never changes value, and so the loop ends after a single iteration.

Answer 7

This happens because you are decrementing n . At the second iteration i < n is false, you are exiting from the loop.

Answer 8

What will happen is:

//Example n = 100
for (int i = 100 - 1; 99 < 100; 100--)
//We now have 99 < 99 on the next loop
//After that you will have 99 < 98 etc.. it will only run once

To make the loop you want use:

for(int i = n-1; i > n ; n--)

To make an endless loop use:

for(;;) //or while(true)

Answer 9

your condition is wrong in the for loop..in your loop i is not changing only n is changing try this

   for(i=n-1;i<n;i--)
   {
   printf("%d\n",i);
   }

Answer 10

you written perfect for loop... for(int i=n-1;i

#include<stdio.h>
#include<conio.h>
int main()
{
    int i;
    for(i=0;;)
    {
        printf("%d",i);
    }
    return 0;
}

or you can do anything else.... to put in infinite simply make ;; in the other two condition in loop.