Reputation: 70
I am trying to make an X[N][M] array with malloc, and I am having this error. Can anybody help me with that ?
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char argv[])
{
int N,M; // N=megethos rows , M=megethos columns
int **X; // pinakas
int size,i,j;
printf("Give me the rows of the table: ");
scanf("%d",&N);
printf("Give me the columns: ");
scanf("%d",&M);
size = N*sizeof(int*);
X = malloc(size);
size = M * sizeof(int*);
for(i=0; i<N;i++){
X =malloc(size);
}
for(i=0;i<N;i++){
for(j=0;j<M;j++){
printf("Eisagwgh toy %d - %d stoixeioy: ",i+1,j+1);
scanf("%d",&X[i][j]);
}
}
for(i=0;i<N;i++){
printf("\n");
for(j=0;j<M;j++){
printf("%d",X[i][j]);
}
}
free(X);
}
I have already try to change this line:
size = N*sizeof(int*);
to this: size = N*sizeof(int);
but still same error appear.
Upvotes: 0
Views: 64
Reputation: 1885
I suggest this way of allocating memory:
X = malloc(N * sizeof(int*));
for (int i = 0; i < N; i++)
X[i] = (int*)malloc(M * sizeof(int));
instead of this wrong allocating:
size = N*sizeof(int*);
X = malloc(size);
size = M * sizeof(int*);
for(i=0; i<N;i++){
X =malloc(size);
}
Upvotes: 0
Reputation: 760
Please see the below code, I just made a very small change in your code, must see the comment.
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char argv[])
{
int N,M; // N=megethos rows , M=megethos columns
int **X; // pinakas
int size,i,j;
printf("Give me the rows of the table: ");
scanf("%d",&N);
printf("Give me the columns: ");
scanf("%d",&M);
size = N*sizeof(int*);
X = malloc(size);
size = M * sizeof(int*);
for(i=0; i<N;i++){
//Here it intialize each row with size
X[i] =malloc(size);
}
for(i=0;i<N;i++){
for(j=0;j<M;j++){
printf("Eisagwgh toy %d - %d stoixeioy: ",i+1,j+1);
scanf("%d",&X[i][j]);
}
}
for(i=0;i<N;i++){
printf("\n");
for(j=0;j<M;j++){
printf("%d",X[i][j]);
}
}
free(X);
}
Upvotes: 1