hi..everyone.i got a little bit confusion about prefix form
for some book i learn that; for example
and some told me write in this formCode:for (i=0; i < 5; ++i)
and i know that the output is same but doesn't itCode:for (i=0; i<5; i++)
++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??...
Last edited by cBegginer; 03-19-2005 at 12:02 AM.
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:
Hope this helps,Code:for (i=1;i<=5;i++)
Taran
It will be the same because u only use it for loop counter.
Try this:
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 differenceCode:#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; }
The first form,Originally Posted by cBegginer
has an equivalent:Code:for (i=0; i < 5; ++i)The second form,Code:i = 0; while(i < 5) { /* do stuff */ ++i; }has an equivalent:Code:for (i=0; i < 5; i++)Why would you expect the results to differ?Code:i = 0; while(i < 5) { /* do stuff */ i++; }
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
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.here x is assigned the original value of i first and then i is increased by 1.Code:x =i++;
Example from the book that im reading.
Code:#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; }
When no one helps you out. Call google();
When used alone, ++foo and foo++ give the exact same results:
Is exactly the same as:Code:int foo = 0; foo++; printf( "%", foo );
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:int foo = 0; ++foo; printf( "%", foo );
Quzah.Code:#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; }
Hope is the first step on the road to disappointment.
