Addressing a single dimensional pointer array with two dimensions

Question

I have the following homework question:

Consider the following declarations and answer the question.
char strarr1[10][32];
char *strarr2[10];

Are strarr1[3][4] and strarr2[3][4] both legal references?

I tried compiling the code with gcc to test it. I was fairly sure that the second reference would throw an error, but it didn't. This is what I compiled with gcc:

int main(void){
    char strarr1[10][32];
    char *strarr2[10];

    char x = strarr1[3][4];
    char y = strarr2[3][4];

    return 0;
}

I'm working under the assumption that the test code I used is correct.

How is it possible to reference strarr2[3][4] when strarr2 is a single-dimensional array?

John Knoeller · Accepted Answer · 2010-02-09 02:19:16Z

1

since strarr2 is an array of char*, the second [4] is an index into the char*

it means the same thing as this

char * temp = strarr2[3];
char y = temp[4];

Since I don't see anywhere in your code where strarr2 is being initialized, nor do I see anywhere that strarr2[3] is being allocated, this code will not work as presented. It will either return garbage or segfault.

edited Feb 9, 2010 at 2:19

answered Feb 9, 2010 at 2:12

John Knoeller

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Comments

dmckee --- ex-moderator kitten · Accepted Answer · 2010-02-09 02:15:36Z

1

They are both legal syntax because of the pointer arithmetic indexing convention. However, in your code:

char y = strarr2[3][4];  // <--- NOT SAFE!

is accessing unallocated memory and generates undefined behavior, ergo it is bad.

So quit it.

answered Feb 9, 2010 at 2:15

dmckee --- ex-moderator kitten

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Comments

Khaled Alshaya · Accepted Answer · 2010-02-09 02:11:13Z

0

It is a single dimensional array of pointers. So, you are indexing the pointer at 3 with offset=4:

char y = *(strarr2[3] + 4);

is the same as:

char y = strarr2[3][4];

answered Feb 9, 2010 at 2:11

Khaled Alshaya

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