Simple c programming

[vivekgupta]

Given two programmes are same ???

if yes then try both with these inputs 4 2 3 -1 4 -1 -2 4
i am getting different answers :/

don't know why??
plz help

#include<stdio.h>
 int main()
 {
 int x,i=1,sum=0;
 /*printf("no. of numbers");*/
 /*scanf("%d",&n);*/
 for(i=0;i<5;i++)
 {
 printf("enter the values");
 scanf("%d",&x);
 if(x<0)
 {
 continue;
 }
 sum+=x;

 }
 printf("%d",sum);
 }

----------------------------------------------------------------------------------------------------------------


#include<stdio.h>
 int main()
 {
 int x,i=1,sum=0;
 /*printf("no. of numbers");*/
 /*scanf("%d",&n);*/
 for(i=0;i<5; )
 {
 printf("enter the values");
 scanf("%d",&x);
 if(x<0)
 {
 continue;
 }
 sum+=x;
i+=1;
 }
 printf("%d",sum);
 }

[nonoob]

Because the 'continue' will not execute the i+=1 in the second example.

[vivekgupta]

but both have same meaning means na.... if we define increment i++ in for loop or we define it at the end of the loop body (i=i+1 . Then why it is coming like this??

[std10093]

Let's say that in the second piece of code,the input is negative number.Then x<0 is true,so it will go into the body of the if and continue.It will ignore ALL the rest of the body of the loop.Better now?

[vivekgupta]

hmm.. I know that after continue every statement in body will not be excuted. I am not getting the for loop thing .. means in both case the increment of 'i' will be at the end of the loop body ... so it'll be ignored for the -ve input case and control should be at the begining of for loop till it gets the +ve value . Why it is ignoring increment of 'i' in later case and not in the earlier .
Tanks for reply but still waiting for your reply :P

[Matticus]

To expand on the comment by nonoob:

"continue" makes execution jump back to the beginning of the loop.

In the first case, after "continue" is encountered, execution jumps back to the beginning of the "for()" loop, and 'i' is incremented as defined in the body of the loop.

In the second case, after "continue" is encountered, execution jumps back to the beginning of the "for()" loop skipping the "i+=1;" line below it.

[vivekgupta]

skipping the "i+=1; is logical .. it is jumping back to the starting of the loop .But in the first case
for(i=0;i<5;i++) <<--- here should it think as it is incrementing i's value here (at starting of loop body :O)
so basically when does this i gets incremented ??? in "for(i=0;i<5;i++)" or after this in the loop body (n where )

sorry I am slow .

[std10093]

for( i = 0 ; i < 2 ; i++ )
{
        /*start of body of loop*/
        /*your code that does some work(it does not change value of i)*/
        /*end of body of loop*/
}

How i have it in mind is this :
From the point that the code reaches the for(part1 ; part2 ; part3)
-in our example part1 is i =0
part2 is i < 5
part3 is i++

• Just reached the for loop,so execute part1,thus i is set to zero
• stay there and execute part2,thus checking if the condition i < 2 is true.i=0,so i<2 is true.Because of the true,go into the body
• reach start of body of loop
• execute your code
• reach end of body loop
• go to the for(...) and execute part3,thus i is increased by one,so i equals one now
• stay there and execute part2,thus checking if the condition i < 2 is true.i=1,so i<2 is true.Because of the true,go into the body
• reach start of body of loop
• execute your code
• reach end of body loop
• go to the for(...) and execute part3,thus i is increased by one,so it is set to 2 now
• stay there and execute part2,thus checking if the condition i < 2 is true.i=2,so i<2 is false.Because of the false,we say goodbye to this for loop and go on with the rest of code(after the loop)

[vivekgupta]

Dude you are awesomeeeeeeee. Thank you very much.

[std10093]

Dude you are awesomeeeeeeee. Thank you very much.

Not so much as the other two users posted here for sure You are welcome .Hope i helped

[vivekgupta]

Not so much as the other two users posted here for sure You are welcome .Hope i helped

Thank you all @^ guys