Creating a matrix of strings in C (without malloc and without initializing)

Question

I want to create a matrix of weekly timetable which must have 8 rows and 5 rows. And each string has 10 chars. It's possible to do that with malloc like that:

char ***Schedule = (char ***)malloc(ROW*sizeof(char **));

for (int i = 0; i < ROW; i++)
{
    Schedule[i] = (char **)malloc(COL*sizeof(char *));

    for (int j = 0; j < COL; j++)
        Schedule[i][j] = (char *)malloc(SIZE*sizeof(char));
}

But I want to not use malloc, because it seems too much lines of code and size of matrix is very little already (8*5*10 = 400 bytes).

char Schedule[8][5][10];

The above code doesn't work.

char *Schedule[8][5];

This one works but i don't trust it, because size of strings are uncertain. It may crash.

post some code that doesn't work using `char Schedule[8][5][10];` ([mcve]) — Jean-François Fabre, Nov 01 '17 at 15:48
Note that 10 displayable chars means you need space for 11 bytes. — Martin James, Nov 01 '17 at 15:48
writing in strings in `char *Schedule[8][5];` without allocating is just undefined behaviour — Jean-François Fabre, Nov 01 '17 at 15:49
I would have gone with 'Schedule[8][5][128];' anyway, just in case someone wants to schedule some new subject with more chars. — Martin James, Nov 01 '17 at 15:51
https://codeshare.io/2K6Z4E doesn't compile. "error: assignment to expression with array type Schedule[0][0] = "sdwdwd";" — Atreidex, Nov 01 '17 at 15:51

user2736738 · Accepted Answer · 2017-11-01T16:31:34.527

Maybe you are using the char Schedule[8][5][10] in wrong way, The correct way would be to do like this (an example)

scanf("%s",Schedule[i][j]);

char *Schedule[8][5] can be trust if you know that it is basically the 40 char* to which you can allocate memory as you want.

For the second case you can allocate in Schedule[i][j]=malloc(sizeof(char)*n)

These are the common ways to using char arrays in your code. These are the ways to create a char array and use them. You wouldn't find other ways to do the same.

OP asked in comment:

Scanning the values worked, thanks. But why " Schedule[i][j] = "abc" " doesn't work?

Ans: You cannot use assignment on an array. Arrays are not assignable in C. Here you can use strcpy(Schedule[0][0] ,"sdwdwd"); ¹as you know that destination is capable of holding the string lliteral.

^{1. Use of strcpy is not advised. Here shown as an example to give user a quick idea. For safer strcpy use strcat like this}

Scanning the values worked, thanks. But why " Schedule[i][j] = "abc" " doesn't work? — Atreidex, Nov 01 '17 at 15:57
@Atreidex.: You cannot use assignment on an array. Arrays are not assignable in C. Here you can use `strcpy(Schedule[0][0] ,"sdwdwd");` as you know that destination is capable of holding the string lliteral. — user2736738, Nov 01 '17 at 16:01

Creating a matrix of strings in C (without malloc and without initializing)

1 Answers1