对C指针的怀疑

Question

I am trying to understand basic concepts in C. I have two questions :

1.Why will the following piece of code work and the other does not?

/* This works fine */
typedef int SortTableRows[20]; 

SortTableRows* SortTableRowsPtr;

/* This will give error --> subscripted value is neither array nor pointer*/
int SortTableRows[20]; 

SortTableRows* SortTableRowsPtr;

1. Second question is what is the third line of code doing?

    typedef int SortTableRows[20]; SortTableRows* SortTableRowsPtr; SortTableRowsPtr[2][3]=2; //Why the compiler doesnt give any error?how can we use a pointer as a 2d array. 

Answer 1

You've defined SortTableRows to be an array of ints with int SortTableRows[20] . That identifier is now fixed as an array, it cannot be used for anything else. You're then trying to use it as a type to declare a pointer (I think). If you want to declare a pointer to an array of ints you would do it like this:

int *SortTableRowsPtr;
SortTablesRowPtr = SortTablesRows;

In the second question, you're assigning a value to a pointer to a pointer that you're treating as a two dimensional array, which makes no real difference to your compiler (though it makes no real sense in this context), as 'multidimensional' arrays are stored in memory in a linear way, the same as single dimension arrays. Your typedef doesn't make any real sense here, as you're defining a type of array, then making an uninitialised pointer to it (this creates an int ** ), and accessing it as if it points to something valid (which will compile, but would surely crash).

To have this work you'd need to define an array of 20 ints and then point to it:

SortTablesRows table[10] = {};  // equivalent to int table[10][20];
SortTablesRowPtr = table;

This works because you're declaring a pointer to an array of 20 ints, ie

int (*SortTablesRowPtr)[20];

I've expanded this answer. It helps to explain it more simply step by step. Take a table of 3 rows and 10 columns. That's defined:

int table[3][10];

You can then set a pointer to this table. Because it's in two dimensions this isn't right:

int *ptr = table; // wrong - incompatible pointer

But this would point to the second row (and the first column):

int *ptr = table[1];

In other words you want a pointer to an array of 10 ints, ie you want a pointer that points to a type of int n[10] . To do this you can write:

int (*ptr2)[10] = table;

Now you can use ptr2 to access the table directly, either as pointers, arrays, or both. As it stands it points to the first row and first column. If you add one to ptr2 it will select the next row.

ptr2[1][3] = 3;  // change row 2, column 4
(*(ptr2 + 1))[4] = 10 // row 2, column 5 (yuck)

Answer 2

Question 1.

In the first you are telling it that SortTablesRow is an array of 20 ints. So SortTablesRow is a type. In the second you are creating a variable called SortTablesRow which is an array of 20 ints. This can't be used as a type because its a variable.

Question 2.

Here you are typedef'ing SortTablesRow to be an array of 20 ints. You then say SortTablesRowPtr is a pointer to it. So now you have a pointer to an array of 20 ints. This is essentially an int**. So the first array index is indexing the first pointer and the second array index is indexing the second pointer (or the array) and thus looking up in the 20 element int array.