Reputation: 39
The code seems to be on the right track however it skips out on a row and disregards the first element [0,0] pushing everything completely back by an element.
#include <stdio.h>
#include <stdlib.h>
int main()
{
char **nums;
int i,j,k, num_cases;
int rows, cols;
printf("Enter the number of grids.\n");
scanf("%d", &num_cases);
for(k=0;k<num_cases;k++){
printf("Test Case #%d\n", k+1);
printf("Enter the # of rows & columns separated by a space.\n");
scanf("%d%d", &rows, &cols);
nums = (char**)malloc(rows*sizeof(char*));
for(i=0; i<rows; i++){
nums[i] = (char*)malloc(cols*sizeof(char));
}
printf("Enter your %dx%d grid of letters.\n", rows, cols);
for(i=0; i<rows; i++){
for(j=0; j<cols; j++){
scanf("%c", &nums[i][j]);
}
}
}
for(i=0; i<rows; i++){
for(j=0; j<cols; j++){
printf("[%d][%d] = %c\n", i, j, nums[i][j]);
}
}
for(i=0; i<rows; i++){
free(nums[i]);
}
free(nums);
return 0;
}
Upvotes: 1
Views: 71
Reputation: 34585
This line
scanf("%d%d", &rows, &cols);
is leaving a newline
in the input buffer, because "%d"
format consumes leading whitespace but not trailing whitespace. So when this line is executed
scanf("%c", &nums[i][j]);
It reads the newline
that was left. You can deal with this by adding a space in front of the "%c"
like this
scanf(" %c", &nums[i][j]);
which will consume any leading whitespace before the char you want to read.
Upvotes: 1