error: expected primary-expression before ‘=’ token when using for loops

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lvalue required as left operand of assignment
lvalue required as left operand of assignment (C program)

I am getting this error on my code and I am not sure why. Please help! The error is:

error: expected primary-expression before ‘=’ token

This is the part of my code i'm having the issue with. a and b were already previously declared as ints.

   int i = 0;
   for( i == a; i < = b; i = i + 1)  // ERROR IS IN THIS LINE
   {  int j = 1;
      int N = static_cast<int>(sqrt(i));

      for( j = 1; j < = N; j = j + 1)   // ERROR IS IN THIS LINE
      {  int P = i%j;
         if( P == 0 && j!= 1 && j!= i)
         {  j = N + 1;
         }
         if( P != 0 && j == N)
         {  cout << i << "is prime" << endl;
         }                                         
      }
   }

This has to be a duplicate. That said, it shows poor research effort and I'm not sure why it's getting upvotes. — Wug, Oct 18 '12 at 20:57
This question is not quite exactly the same but has exactly the same answer. [lvalue required as left operand of assignment](http://stackoverflow.com/questions/6162438/lvalue-required-as-left-operand-of-assignment) — Wug, Oct 18 '12 at 21:01
This one is a much closer match: [lvalue required as left operand of assignment (C program)](http://stackoverflow.com/questions/8570020/lvalue-required-as-left-operand-of-assignment-c-program) — Raymond Chen, Oct 18 '12 at 21:03

Anirudh Ramanathan · Answer 1 · 2012-10-18T21:00:40.757

3

I think you meant to do

if( P != 0 && j == sqrt(i))

= is the assignment operator, == tests for equivalence and returns boolean.

The assignment operator has lesser precedence than logical AND (&&), so it being read as

(P != 0 && j) = sqrt(i)

which is why the lvalue error is being reported.

edited Oct 18 '12 at 21:00

answered Oct 18 '12 at 20:55

Anirudh Ramanathan

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Jonathan Leffler · Answer 2 · 2012-10-18T21:35:53.317

2

Because of the low precedence of assignment, what you wrote:

if (P != 0 && j = sqrt(i))

is interpreted as:

if ((P != 0 && j) = sqrt(i))

and you can see that is not an l-value. If you want to assign the square root of i to j, you probably should write:

if (P != 0 && (j = sqrt(i)) != 0.0)

assuming that j is a floating point variable; if it's an integer type, use 0 in place of 0.0. Otherwise, you should write:

if (P != 0 && j == sqrt(i))

which does an explicit equality comparison.

edited Oct 18 '12 at 21:35

answered Oct 18 '12 at 20:58

Jonathan Leffler

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`==` has precedence over `&&`. – Anirudh Ramanathan Oct 18 '12 at 21:02
@Cthulhu: You were right but worse than that, there was a missing close parenthesis. Fixed. Thanks! – Jonathan Leffler Oct 18 '12 at 21:36

score 2 · Answer 3 · edited Oct 18 '12 at 23:25

2

You cannot put spaces in the <= operator :

i < = b;

becomes

i <= b;

edited Oct 18 '12 at 23:25

Anirudh Ramanathan

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answered Oct 18 '12 at 21:10

gogoprog

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It does matter on g++ ... I just tried – gogoprog Oct 18 '12 at 21:31

score 2 · Answer 4 · answered Oct 18 '12 at 21:11

2

I'll play along. The 'error' is the same in both case:

for( i == a; i < = b; i = i + 1)  // ERROR IS IN THIS LINE
...
for( j = 1; j < = N; j = j + 1)   // ERROR IS IN THIS LINE

The error is the '< =' which is not an operator and not valid C or C++. Sidenote, the 'i == a' in the first loop is a no-op.

answered Oct 18 '12 at 21:11

Nik Bougalis

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I mean it has no side-effects, or to put it simply: it does nothing. It compares i to a. If they're the same, it returns 'true', otherwise it returns 'false'. But the return value is discarded and not used. – Nik Bougalis Oct 18 '12 at 21:13
To initialize i in the for loop as the value of a, I should use "=" instead of "=="? – user1757575 Oct 18 '12 at 21:14
Correct. '==' is a check for equality. '=' is an assignment. – Nik Bougalis Oct 18 '12 at 21:15

score 1 · Answer 5 · answered Oct 18 '12 at 20:56

Write the condition as follows.

if( (P != 0) && (j == sqrt(i)) )

instead of

if( P != 0 && j = sqrt(i))

This = sign is assignment operator, not a comparison operator. You need to check if, the left hand side is equal to the right hand side.

error: expected primary-expression before ‘=’ token when using for loops

5 Answers5