Why isn't the compiler generating an error "lvalue required"?

Question

As per my understanding, in the line marked as 'line 2' of the below code, the expression (*ptr)++ should generate "lvalue required" error because *ptr evaluates to a constant value of i=1, which is not lvalue?

So why is the program working successfully? Or am I somewhere wrong in my concepts? If yes, please enlighten me on this.

int main(void)
{
    int i;
    int *ptr = (int *) malloc(5 * sizeof(int));

    for (i=0; i<5; i++)
        *(ptr + i) = i;

    printf("%d ", *ptr++);   //line 1
    printf("%d ", (*ptr)++); //line 2 
    printf("%d ", *ptr);     //line 3
    printf("%d ", *++ptr);   //line 4
    printf("%d ", ++*ptr);   //line 5
}

Please [see why not to cast](http://stackoverflow.com/q/605845/2173917) the return value of `malloc()` and family in `C`. — Sourav Ghosh, Jul 14 '15 at 06:14
"*`*ptr` evaluates to a constant value of `i=1`*": I'm not sure I understand this statement. Can you explain? — Cornstalks, Jul 14 '15 at 06:17
@Cornstalks i mean "*ptr" is value at address stored in ptr, which is equal to 1, so is 1++ valid? — codeluv, Jul 14 '15 at 06:30
What about `int i = 0; i++;`. If `i` is 0, then is `0++` valid? The answer is of course no, `0++` (and `1++`) are not valid. `++` is not applied to the *value*, it's applied to the *object* (in this case `i`, or in your case, the object pointed to by `*ptr`). — Cornstalks, Jul 14 '15 at 06:33
@Cornstalks ok , now i got it. i would have green ticked this if this was an answer. :) — codeluv, Jul 14 '15 at 06:35

score 6 · Accepted Answer · answered Jul 14 '15 at 06:19

You're having a misconception. result of (*ptr) is lvalue, upon which the post increment operator can be applied.

So, in your case,

 printf("%d ", (*ptr)++); //line 2

is fine.

To quote C11 standard, chapter §6.5.3.2, Address and indirection operators, ^{(emphasis mine)}

The unary * operator denotes indirection. If the operand points to a function, the result is a function designator; if it points to an object, the result is an lvalue designating the object.

FWIW, if *ptr would not be a lvalue, you would have got error for writing something like *ptr = 5 also, wouldn't it?

score 1 · Answer 2 · answered Jul 14 '15 at 06:36

1

What about int i = 0; i++;. If i is 0, then is 0++ valid? The answer is of course no, 0++ (and 1++) are not valid. ++ is not applied to the value, it's applied to the object (in this case i, or in your case, the object pointed to by *ptr).

answered Jul 14 '15 at 06:36

Cornstalks

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score 1 · Answer 3 · answered Jul 14 '15 at 06:39

1

An lvalue is something that refers to someplace in memory that can/does hold a value.So if *ptr=10; then *ptr is lvalue.

answered Jul 14 '15 at 06:39

ameyCU

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score 0 · Answer 4 · answered Jul 14 '15 at 06:22

0

#include <stdio.h>

int main(void) {
    int i = 42;
    int *ptr = &i;       /* ptr points to i */
    (*ptr)++;            /* you increment the pointed value */
    printf("%d", i);     /* and there is a proof: 43 is displayed */
    return 0;
}

answered Jul 14 '15 at 06:22

dlask

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taking into account how sometimes undefined behavior leads to expected results I wouldn't consider `43 is displayed` a `proof` ;-) – Kuba Wyrostek Jul 14 '15 at 09:56

Why isn't the compiler generating an error "lvalue required"?

4 Answers4