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Why can't I compare the results of the random function?

#include<stdio.h>
#include<stdlib.h>
#include<time.h>

int main (){

srand(time(NULL));

/*Player 1*/
int P1d1, P1d2, P1total;
P1d1 = 1 + rand() % 6;  //Rolls Player 1's first die; random # between 1-6
P1d2 = 1 + rand() % 6;  //Rolls Player 1's second die; random # between 1-6
P1total = P1d1 + P1d2;  //Takes total of both rolls

/*Player 2*/
int P2d1, P2d2, P2total;
P2d1 = 1 + rand() % 6; //Rolls Player 2's first die; random # between 1-6
P2d2 = 1 + rand() % 6; //Rolls Player 2's second die; random # between 1-6
P2total = P2d1 + P2d2; //Takes total of both rolls

if (P1d1 == P1d2 && P2d1 =! P2d2){ <--- ERROR HERE
    printf("Player 1 wins!\n");
    }
else if (P2d1 == P2d2 && P1d1 =! P1d2) {
    printf("Player 2 wins!\n");
    }
}

I get the error message error: lvalue required as left operand of assignment.

I understand that this means the left-hand side of the assignment has to be a variable, but the left-hand side is a variable (int P1d1). Any suggestions?

Jonathon Reinhart
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supersaidso
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    Because it should be `==`, not `=`. – Kerrek SB Sep 27 '13 at 22:21
  • It still gives the same error message with == – supersaidso Sep 27 '13 at 22:24
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    It's != not =! And this is long enough – Jonathon Reinhart Sep 27 '13 at 22:25
  • I would vote to close as this has nothing to do with Rand() and is unlikely to help anyone in the future. But I don't know the right close reason. – Jonathon Reinhart Sep 27 '13 at 22:30
  • @JonathonReinhart I'm sure that it's a duplicate of the same stupid lazy mistake, and it also lacks a minimal understanding -- choose your poison. –  Sep 27 '13 at 22:56
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    @JonathonReinhart: this shouldn't be closed because the asker has clearly done there own research and become stuck. I'd vote to modify the question name to be something to do with the problem "Problem comparing values: lvalue required as left operand of assignment" – dave Sep 27 '13 at 23:48

3 Answers3

9

You probably need to use == (equality) rather than = (assignment). I see you've changed that in your question.

The cause of the error message is:

P2d1 =! P2d2

The inequality operator is !=, not =!.

The way you're written it, it's an assignment = followed by a unary "not" !. And since = has very low precedence, this:

(P1d1 == P1d2 && P2d1 =! P2d2)

is equivalent to this:

((P1d1 == P1d2 && P2d1) = !P2d2)

and of course (P1d1 == P1d2 && P2d1) is not a valid target for an assignment.

This is one of the more ahem interesting things about programming in C. In some languages, a minor typo is likely to introduce a syntax error, and the compiler's error message is likely to tell you exactly what the problem is. In C, minor typos can easily result in code that's still syntactically correct, but that has some absurd meaning -- or, in your case, that produces an error message that's not very illuminating.

A good approach, if you don't understand what a compiler error message means, is to ignore what it says and just look carefully at the earliest line of code that the compiler complains about.

Keith Thompson
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0

= is an assignment operator. == is a comparison operator.

nhgrif
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0

Use !=. 

if (P1d1 == P1d2 && P2d1 =! P2d2){
printf("Player 1 wins!\n");
}

is the same as

if ((P1d1 == P1d2 && P2d1) = (!(P2d2))){
printf("Player 1 wins!\n");
}
EKons
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