Why For loop in C runs infinitely when only 'equal to' operator is given in condition

rmnature · 02-23-2009, 09:52 AM

Hi,

I am new to c-programming,i noticed that in the For loop whenever i give equal to in condition the loop runs infinitely but if i give <= the loop is ok.

example: for(i=1;i<=5;i++) print("a")

a is printed 5 times but if i give

for(i=1;i=5;i++) the loop runs infinitely.In this case shouldn't the loop exit after(or even before) first iteration?

wje_lq · 02-23-2009, 09:58 AM

These two do not mean the same thing:

Code:

i==5   /* comparison */
i=5    /* assignment */

Taken as an expression, the first will be true (nonzero) only when, well, i==5.

Taken as an expression, the second will be true (nonzero) always, because that assignment itself is an expression and will always be 5.

Hope this helps.

johnsfine · 02-23-2009, 10:02 AM

That's a very common error, but usually once someone sees that the problem is "=" they have the answer (usually people ask about that because they haven't yet narrowed the problem in their to the "=").

In C i=5 is always an assignment (it changes the value of i to be 5). The result is always 5. As a condition, any non zero result acts like true.

In C you must use i==5 to test whether i is equal to 5.

colucix · 02-23-2009, 10:11 AM

This is because the syntax of a for loop is

Code:

for (initialization, condition, operation) {
some code here
}

this means that the variable is initialized, tested against the condition, and then the code between braces is executed. After the code execution has finished, an operation (for example "++", the increment) is performed on the variable, tested again, and if the condition is matched, the program enters another execution cycle. If you put an equal sign in the condition, the result is that the variable is re-initializated to 5 at every cycle and not matched against the condition: this brings to an infinite loop. If you try to put a double equal sign:

Code:

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

you will see that the code within the for loop is not executed at all, since at the first test against the condition, the result is false.

alan43 · 02-23-2009, 10:57 AM

The value of the expression

Code:

i=5

is 5
so your codes become

Code:

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

which is an infinite loop, as 5 is treated as true.

gergely89 · 02-23-2009, 11:01 AM

The

Quote:

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

especially i=5 will give i the value 5, and at the same resolve as a TRUE condition for the simple reason that i has become a non-null value. This way the looping condition is always true and never ends.

linux

Sergei Steshenko · 02-23-2009, 12:24 PM

Quote:

Originally Posted by johnsfine

That's a very common error, but usually once someone sees that the problem is "=" they have the answer (usually people ask about that because they haven't yet narrowed the problem in their to the "=").

In C i=5 is always an assignment (it changes the value of i to be 5). The result is always 5. As a condition, any non zero result acts like true.

In C you must use i==5 to test whether i is equal to 5.

There is an immortal deadly joke regarding the error:

Code:

if(EnemyMissilesLaunchDetected = 1)
  {
  CommenceCounterStrike();
  }

- I do not remember the exact words, but along the above lines.

In order to avoid such errors reverse the order of operands:

Code:

if(1 = EnemyMissilesLaunchDetected)
  {
  CommenceCounterStrike();
  }

- the compiler wouldn't allow assignment to constant.

I.e. meaning to compare with constant using '==' always put the constant to the left of '=='.

rmnature · 02-24-2009, 08:28 AM

Thanks to all of you