confusion with increment and decrement operators

hi..everyone.i got a little bit confusion about prefix form
for some book i learn that; for example

Code:

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for (i=0; i < 5; ++i)

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and some told me write in this form

Code:

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for (i=0; i<5; i++)

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and i know that the output is same but doesn't it
++i will increment the value first and if i++ the value will increment after the loop.I really doesn't understand that why the for loop come out with the same answer??...

hi,
the difference between i++ and ++i is that if the i++ (or ++i) is used in an expression, i++ will cause i to be used in the expression then incremented, wheras ++i will increment i and then use it in the expression. As i isn't being used in an expression in your code, there is no difference between i++ and ++i
If you want to have i going from 1 to 5 instead of 0 to 4, simply use the following code instead:

Code:

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for (i=1;i<=5;i++)

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Hope this helps,
Taran

It will be the same because u only use it for loop counter.

Try this:

Code:

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#include <stdio.h>

int main()
{
        int i = 0;
        int j = 0;
        printf("%d\n", i++);//result 0
        printf("%d\n", i++);//result 1
        printf("%d\n", ++j);//result 1
        printf("%d\n", ++j);//result 2
        return 0;
}

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Remember, the difference between the two forms is what value (either the old or the new) is passed on to the surrounding expression. If there is no surrounding expression, if the ++i or i++ appears all by itself, to increment i and do nothing else, you can use either form; it makes no difference

Quote:

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Originally Posted by cBegginer

and i know that the output is same but doesn't it ++i will increment the value first and if i++ the value will increment after the loop.I really doesn't understand that why the for loop come out with the same answer??...

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The first form,

Code:

---

for (i=0; i < 5; ++i)

---

has an equivalent:

Code:

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  i = 0;
  while(i < 5)
  {
      /* do stuff */
      ++i;
  }

---

The second form,

Code:

---

for (i=0; i < 5; i++)

---

has an equivalent:

Code:

---

  i = 0;
  while(i < 5)
  {
      /* do stuff */
      i++;
  }

---

Why would you expect the results to differ?

Ic..ok thanks for helping me.I appreciate it alot.

++i and i++ do not really give you the same result everytime.

++i in this case is called the pre-increment operator. This will first add 1 to i and give you the new value of i.

i++ in this case is called the post -decrement operator, because im lacking words to express this i will do it by example.

Code:

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x =i++;

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here x is assigned the original value of i first and then i is increased by 1.

Example from the book that im reading.

Code:

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#include <stdio.h>
main()
{
        int w,x,y,z, result;
       
        w = x = y = z = 1; /*initialize x and y*/
        printf("Given w =%d, x =%d, y=%d, and z =%d,\n", w, x,y,z);

        result = ++w;
        printf("++w evaluates to %d and w is now %d\n", result, w);

        result = x++;
        printf("x++ evaluates to %d and x is now %d\n", result, x);

        result = --y;
        printf("--y evaluates to %d and y is now %d\n", result, y);

        result = z--;
        printf("z-- evaluates to %d and z is now %d\n", result, z);
        return 0;
}

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When used alone, ++foo and foo++ give the exact same results:

Code:

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int foo = 0;

foo++;
printf( "%", foo );

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Is exactly the same as:

Code:

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int foo = 0;

++foo;
printf( "%", foo );

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In a single expression, the results are the same. The difference is when you use them for something other than a single expression. Like so:

Code:

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#include<stdio.h>
void foo( int bar )
{
    printf("bar is %d\n", bar );
}

int main( void )
{
    int baz;

    baz = 0;
    foo( baz++ );

    baz = 0;
    foo( ++baz );

    return 0;
}

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Quzah.