Why can't I use x + 1 in for loop? [closed] - c

Question

Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
Questions asking for code must demonstrate a minimal understanding of the problem being solved. Include attempted solutions, why they didn't work, and the expected results. See also: Stack Overflow question checklist
Closed 9 years ago.
Improve this question
I am working on that code, and I tried to use x + 1 instead of ++x and the result was infinite loop and zeros were printed on the screen.
this is the code:
#include <stdio.h>
int main(void){
int x;
for(x = 0; x <= 100; x + 1) //instead of ++x
printf("%d\t",x);
return 0;
}
I wondered this action.....why the compiler didn't produce such an error,,,, and what actually happens??!! and is x++ is treated as x += 1 ?!

Answer 1

You need the value of x to change, or your loop will never terminate. When you have only x + 1, then at the end of the iteration, x + 1 is computed, but its resulting value is discarded. The result of the expression is not stored any place. So x will remain zero, and x <= 100 will remain true, and your loop will continue forever.
If you have any of the following:
x = x + 1
x += 1
x++
++x
the value of x itself is incremented. That's what you need.

Answer 2

Although the basic structure of a for loop is like-
for(initialization; condition; increment or decrement)
But in its core part the "condition" part is treated only to control the loop behavior. So, if other two parts are syntactically correct, then compiler won't produce any error.
Since, x+1 is a valid C statement and the value of the x is not changing so it will go to an infinite loop for the condition [x <= 100 => 0 <= 100] being true forever.
Again, x++ or ++x both treated as x = x + 1 when used independently. But, x++ is actually Post-increment operator, while ++x is Pre-increment operator. That means, the with ++x, the value of the x will be incremented first and then assigned to x. Whereas, the current value of x will be used for the entire statement in x++ and then x will be incremented and assigned with new value. Look at the following example-
#include<stdio.h>
void main()
{
int x=5;
/* Post & Pre-Increment used independently */
x++;
printf("x : %d", x);
++x;
printf("\nx : %d", x);
/* Used with printf() statement */
printf("\nPre-increment of x: %d", ++x);
printf("\nPost-increment of x: %d", x++);
printf("\nPost-increment effect on x: %d", x);
}
Output:
x : 6
x : 7
Pre-increment of x: 8
Post-increment of x: 8
Post-increment effect on x: 9
I Hope my explanation have made you understand, if still not you can reply me back.

Answer 3

The expression x + 1 should be x = x + 1 in for loop, so correct:
for(x = 0; x <= 100; x + 1)
^
doesn't increment x
Infinite loop!
as:
for(x = 0; x <= 100; x = x + 1)
or
for(x = 0; x <= 100; x += 1) // or simply x++ or ++x
The compiler didn't produce such an error because x + 1 is a valid expression (but its not what you wants).
So, x += 1, x++ or ++x are not just add one to x but also modify value of x.

Answer 4

Your approach is not perfect because it definitely increases the value of x but when it comes to assignment, it fails. so your statements should cover all your requirements.
try to use
#include <stdio.h>
int main(void){
int x;
for(x = 0; x <= 100; x = x + 1)
printf("%d\t",x);
return 0;
}
or you can use
#include <stdio.h>
int main(void){
int x;
for(x = 0; x <= 100; x++)
printf("%d\t",x);
return 0;
}
Instead of x++ (post increment) you can also use pre increment (++x).
The expression x = x + 1; can also be written as x += 1; which is a shorthand method for assignment embedded with increment. It is not just about incremnent you can also use other operators.
x -= 3;
x *= 2;