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