分配2D字符数组

Question

I am making a program to allocate a 20x20 array of characters. Here is what I did:

#include<stdio.h>
#include<stdlib.h>
int main()
{
char *a=(char*) calloc(20,sizeof(char[20]));
a[0]="abcd";
printf("%s\n",a[0]);
return 0;
}

Link to ideone .

The output of the above code is (null) . Can anybody please explain this? According to me, I am allocating a pointer a 20 spaces of size 20 each. So a[0] technically has enough memory to store "abcd", yet the output is null .

Answer 1

The type of variable a that you have is incorrect: it should be char (*a)[20] (yes, with parentheses).

This line is also incorrect:

a[0]="abcd";

you cannot assign C strings like that, because a[0] is not a pointer: it is an array of 20 characters, so you need to use strcpy instead:

#include<stdio.h>
#include<stdlib.h>
int main()
{
    char (*a)[20]=calloc(20, sizeof(char[20]));
    strcpy(a[0], "abcd");
    strcpy(a[1], "wxyz");
    printf("%s\n",a[0]);
    printf("%s\n",a[1]);
    return 0;
}

See the corrected program running here .

Note: Unlike C++, C does not require type casting of void pointers. It is typical for C programs to omit the cast of results returned from malloc , calloc , and realloc , because the type is already known from the type of the variable being assigned.