pointer increment and dereference (lvalue required error)

Question

I am trying to understand how pointer incrementing and dereferencing go together, and I did this to try it out:

#include <stdio.h>
int main(int argc, char *argv[])
{
    char *words[] = {"word1","word2"};
    printf("%p\n",words);
    printf("%s\n",*words++);
    printf("%p\n",words);
    return 0;
}

I expected this code to do one of these:

First dereference then increase the pointer (printing word1)
First dereference then increase the value (printing ord1)
Dereference pointer + 1 (printing word2)

But compiler won't even compile this, and gives this error: lvalue required as increment operand am I doing something wrong here?

score 2 · Answer 1 · answered Apr 14 '12 at 02:16

2

You need to put braces around the pointer dereference in the second printf, e.g.:printf("%s\n",(*words)++); Also, if you're attempting to get number 2 in your list there, you need to use the prefix increment rather than postfix.

answered Apr 14 '12 at 02:16

thelazydeveloper

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I gave you +1, but it might be helpful to explain the precedence issue, or at least point to a source that illustrates it. – jpm Apr 14 '12 at 02:17

score 2 · Accepted Answer · answered Apr 14 '12 at 02:21

2

You cannot increment an array, but you can increment a pointer. If you convert the array you declare to a pointer, you will get it to work:

#include <stdio.h>
int main(int argc, char *argv[])
{
    const char *ww[] = {"word1","word2"};
    const char **words = ww;
    printf("%p\n",words);
    printf("%s\n",*words++);
    printf("%p\n",words);
    return 0;
}

answered Apr 14 '12 at 02:21

Sergey Kalinichenko

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I thought char * words[], and char ** words meant the same thing. – yasar Apr 14 '12 at 02:24
@yasar11732 No, they are different types. The `char *words[]` *decays* to `char **words` if you pass it to a function, but otherwise they are different. – Sergey Kalinichenko Apr 14 '12 at 02:28

score 1 · Answer 3 · answered Apr 14 '12 at 02:18

words is the name of the array, so ++ makes no sense on it. You can take a pointer to the array elements, though:

for (char ** p = words; p != words + 2; ++p)
{
    printf("Address: %p, value: '%s'\n", (void*)(p), *p);
}

Instead of 2 you can of course use the more generic sizeof(words)/sizeof(*words).

score 0 · Answer 4 · edited May 23 '17 at 12:20

The problem is with this line:

printf("%s\n",*words++);

It is read as *(words++), i.e. increment a block of memory. That doesn't make sense, it is a bit like trying to do:

int a = 1;
(&a)++; // move a so that it points to the next address

which is illegal in C.

The problem is caused by the distinction between arrays and pointers in C: (basically) an array is a block of memory (allocated at compile time), while a pointer is a pointer to a block of memory (not necessarily allocated at compile time). It is a common trip-up when using C, and there are other question on SO about it (e.g. C: differences between char pointer and array).

(The fix is described in other answers, but basically you want to use a pointer to strings rather than an array of strings.)

pointer increment and dereference (lvalue required error)

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