How to fill a 2D array in C with user input values? - c

Question

Note: This is a homework question.
Use FOR construction to fill 2D board with values that were given by
user. The program asks for board size n, m and then it asks for each
board value.
My try
#include <stdio.h>
int main(){
printf("Enter the number of columns");
int i = scanf("%d",&i);
printf("Enter the number of rows");
int y = scanf("%d",&y);
int r[i][y];
int a;
int b;
for (a=0; a<i; a++){
for(b=0; b<y; b++){
int r[a][b] = scanf("%d",&a,&b); //bug
}
}
}
Bug: c:13 variable-sized object may not be initialized
EDIT:
#include <stdio.h>
int main(){
printf("Enter the number of columns");
int i;
scanf("%d", &i);
printf("Enter the number of rows");
int y;
scanf("%d", &y);
int r[i][y];
int a;
int b;
for (a=0; a<i; a++){
for (b=0; b<y; b++){
scanf("%d",&r[a][b]);
}
}
}

Answer 1

scanf takes the address of the variable that is being read and returns the number of items read. It does not return the value read.
Replace
int i = scanf("%d",&i);
int y = scanf("%d",&y);
by
scanf("%d",&i);
scanf("%d",&y);
and
int r[a][b] = scanf("%d",&a,&b);
by
scanf("%d",&r[a][b]);
EDIT:
You are using variable length array (VLA) in your program:
int r[i][y];
as i and y are not constants and are variables. VLA are a C99 standard feature.

Answer 2

you have to allocate the 2D array dynamically cause you don't know it size in compilation.
replace
int r[i][y];
with
int *r = malloc(i*y*sizeof(int));
and when finish, add:
free(r);
*and also the SCANF errors, people already answered here.

Answer 3

First the return value of scanf isn't the value that was read from stdin, instead it is the number of input values scanf read.
Secondly, C does not allow creation of arrays by using variables. You have to first create one array by dynamically allocating it. And for each entry in the first array you have to create yet another array.
Don't forget to free the memory you allocate!

Answer 4

The use of scanf(%d, &var) is incorrect.
scanf reads from console an integer (this type is specified by its first paramenter %d) and stores it in the second parameter.
This second parameter must be a pointer, so an & is needed when your variable is not a pointer yet.
Therefore you should correct your code in this way:
int i;
scanf("%d",&i);
int y;
scanf("%d", &y);
And in your for loop
scanf("%d", &r[a][b]);

Answer 5

It doesn't print the message because of line buffering.
If you add \n to your strings (to start a new line), it might do what you expect:
printf("Enter the number of columns\n");
Alternatively, if you really want the user to type on the same line, you need to flush the buffer manually:
printf("Enter the number of columns");
fflush (stdout);