CODEWITHSUNDEEP
cgcc
GintsDev
GintsDev

Reputation: 21

GCC 2.9 and "lvalue required as left operand of assignment"

I have been playing around with void pointers and created this example:

#include <stdio.h>

struct intint {
    int a;
    int b;
};

struct intshortshort {
    int a;
    short b;
    short c;
};

void fill_me(struct intint **pii, void *piss)
{
    (void*)*pii = piss;                // Question about this line?
}

int main()
{
    struct intint *pii = NULL;
    struct intshortshort iss;
    iss.a = iss.b = iss.c = 13;

    fill_me(&pii, &iss);

    printf("%d..%d\n", pii->a, pii->b);


    return 0;
}

Question:

When I'm using gcc version 2.95.4 everything compiles and works as expected, but gcc version 4.7.3 gives me following error:

void_pointer.c:16:17: error: lvalue required as left operand of assignment

Is there a reason why adding (void *) to lvalue is not allowed anymore?

Edit: Thank you for answers, I think I understood the problem, but the question "why it was ok in the first place?" is still interesting.

Upvotes: 2

Views: 835

Answers (5)

Glenn Teitelbaum
Glenn Teitelbaum

Reputation: 10333

The trick is to cast to a void** and then dereference it:

void fill_me(struct intint **pii, void*piss)
{
        *(void **)pii=piss;
}

Output, as expected:

13..851981

13 is obvious
851,981 - 13 = 851,968
851,968 / (256*256) = 13
there are the 2 shorts as an int

Upvotes: 0

notbad
notbad

Reputation: 2887

The old compiler(old version gcc) may allow cast-as-lvalue, but it ha been removed in the new version, you can find it The cast-as-lvalue, conditional-expression-as-lvalue and compound-expression-as-lvalue extensions, which were deprecated in 3.3.4 and 3.4, have been removed.

Upvotes: 2

AleX
AleX

Reputation: 529

This question isn't about the void*, is about typecast and lvalue.

for example:

#include <stdio.h>

int main()
{
    int n;
    (int)n = 100;
    return 0;
}

this code will causes same error: lvalue required as left operand of assignment.

That because the = operator requires a modifiable-lvalue as its left operand.

So, what is the lvalue? and what's the modifiable-lvalue? Please see wikipedia:Value.

Upvotes: 3

Lundin
Lundin

Reputation: 213892

It doesn't work for the same reason as why

  int x, y;
  (int)x = y;

doesn't work. The result of a cast is not a "lvalue", it does not hold a memory location where you can store something inside. Rather regard the result of (int)x as a temporary, invisible variable, which only exists during the execution of this operation.

Upvotes: 2

user1814023
user1814023

Reputation:

You cannot use cast operator on the left operand of the assignment operator in C.

Upvotes: 0

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