How to solve "subscripted value is neither array nor pointer nor vector"

Question

I'm solving C programming quiz.

The quiz problem was "what is the output of the following code snnipet?"

uint32_t v = 0xdeadbeef;
printf("%02x", (char *) v[0]);

or uint64_t?

Honestly I didn't understand the problem, so I tested on my local machine.

#include<stdio.h>
#include<stdint.h>

int main() {
    uint32_t v = 0xdeadbeef;
    printf("%02x", (char *) v[0]);    /* (1) */

    int64_t w = 0xdeadbeef;
    printf("%02x", (char *) w[0]);    /* (2) */

}

I'm getting compile error on (1) and (2).

Here is the error message

num1.c: In function ‘main’: error: subscripted value is neither array nor pointer nor vector

So for the question on this post, How can I test this code without compile error?

Expected output : de, ad, be, ef, or 00

The code's a little buggy -- you'll need to cast v, not v[0]. That is, ((char*)v)[0]. That might also give you an idea as to the problem. — Charles Duffy
– Charles Duffy, Commented Dec 12, 2017 at 2:40
Did you copy-and-paste the quiz problem? Either the quiz is incorrect, or the quiz is correct and you copied it incorrectly. — Keith Thompson
– Keith Thompson, Commented Dec 12, 2017 at 2:42
This should probably be ((char *) &v)[0]. Then it would output either de or ef depending on endianness. — Cris Luengo
– Cris Luengo, Commented Dec 12, 2017 at 2:50
The compiler is correct. The code in the quiz problem is wrong. The only way it makes sense is to change (char *)v[0] to ((char *)&v)[0] -- or, more likely, ((unsigned char *)&v)[0] — Keith Thompson
– Keith Thompson, Commented Dec 12, 2017 at 2:57
@JohnBaek yes, because you subscripted a value (v) which is neither array nor pointer nor vector. The subscript operator requires one of those things as operand. — M.M
– M.M, Commented Dec 12, 2017 at 3:05

abdullah · Accepted Answer · 2017-12-12 02:53:07Z

1

I think the problem asks about the first byte of four bytes uint32_t arranged in memory layout. That depends on endianness. If you want to find out the output, you may check this code.

#include<stdio.h>
#include<stdint.h>

int main() {
    uint32_t v = 0xdeadbeef;
    char* pv = (char*)&v;
    printf("%02x\n", (uint8_t)pv[0]);
}

answered Dec 12, 2017 at 2:53

abdullah

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7 Comments

merry-go-round Over a year ago

Yeah it's working. it's printing ef. It seems right answer.

merry-go-round Over a year ago

And at the same time, I'm very confused.

abdullah Over a year ago

What is the confusion?

merry-go-round Over a year ago

That depends on endianness. I learned Endianness is the order of memory arrangement (I learned it about 3 years ago so I may misunderstand what it means). And how is it related to your answer?

abdullah Over a year ago

If you machine is little endian, then it will show ef. And if it is big endian, it will show de. Please check endianness

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