[c] Multiple conditions in a C 'for' loop

I came across this piece of code. I generally use '&&' or '||' to separate multiple conditions in a for loop, but this code uses commas to do that.

Surprisingly, if I change the order of the conditions the output varies.

#include<stdio.h>

int main() {
    int i, j=2;

    for(i=0; j>=0,i<=5; i++)
    {
         printf("%d ", i+j);
         j--;
    }
    return 0;
}

Output = 2 2 2 2 2 2

#include<stdio.h>

int main(){
    int i, j=2;

    for(i=0; i<=5,j>=0; i++)
    {
         printf("%d ", i+j);
         j--;
    }
    return 0;
}

Output = 2 2 2

Can somebody explain the reason? It seems to be checking only the last comma-separated condition.

Do not use this code; whoever wrote it clearly has a fundamental misunderstanding of the language and is not trustworthy. The expression:

j >= 0, i <= 5

evaluates "j >= 0", then throws it away and does nothing with it. Then it evaluates "i <= 5" and uses that, and only that, as the condition for ending the loop. The comma operator can be used meaningfully in a loop condition when the left operand has side effects; you'll often see things like:

for (i = 0, j = 0; i < 10; ++i, ++j) . . .

in which the comma is used to sneak in extra initialization and increment statements. But the code shown is not doing that, or anything else meaningful.

Of course it is right what you say at the beginning, and C logical operator && and || are what you usually use to "connect" conditions (expressions that can be evaluated as true or false); the comma operator is not a logical operator and its use in that example makes no sense, as explained by other users. You can use it e.g. to "concatenate" statements in the for itself: you can initialize and update j altogether with i; or use the comma operator in other ways

#include <stdio.h>

int main(void)  // as std wants
{
  int i, j;

  // init both i and j; condition, we suppose && is the "original"
  // intention; update i and j
  for(i=0, j=2; j>=0 && i<=5; i++, j--)
  {
       printf("%d ", i+j);
  }
  return 0;        
}

Wikipedia tells what comma operator does:

"In the C and C++ programming languages, the comma operator (represented by the token ,) is a binary operator that evaluates its first operand and discards the result, and then evaluates the second operand and returns this value (and type)."

The comma expression takes on the value of the last (eg. right-most) expression.

So in your first loop, the only controlling expression is i<=5; and j>=0 is ignored.

In the second loop, j>=0 controls the loop, and i<=5 is ignored.

As for a reason... there is no reason. This code is just wrong. The first part of the comma-expressions does nothing except confuse programmers. If a serious programmer wrote this, they should be ashamed of themselves and have their keyboard revoked.

There is an operator in C called the comma operator. It executes each expression in order and returns the value of the last statement. It's also a sequence point, meaning each expression is guaranteed to execute in completely and in order before the next expression in the series executes, similar to && or ||.

Completing Mr. Crocker's answer, be careful about ++ or -- operators or I don't know maybe other operators. They can affect the loop. for example I saw a code similar to this one in a course:

for(int i=0; i++*i<-1, i<3; printf(" %d", i));

The result would be $ 1 2$. So the first statement has affected the loop while the outcome of the following is lots of zeros.

for(int i=0; i<3; printf(" %d", i));