Here's my code
#include <stdio.h>
#include <math.h>
int main()
{
int d, a[3][3] = {{2, 4, 5}, {6, 7, 8}, {9, 10, 11}};
for (d = 0; d < 5; d++)
{
printf("a[%d] = %d\n\n", d, a[d]);
}
return 0;
}
i'm getting the output
a[0] = 6422048
a[1] = 6422060
a[2] = 6422072
a[3] = 6422084
a[4] = 6422096
i'm curious about where did those values come from, and why is it incrementing by 12?
They are pointers. AND the outputs are increased by 12 next because size of int[3] is 12 bytes.
%d to print the pointers isn't correct. %p should be used.a is array of arrays, when you are trying to output you are printing array. Arrays are like pointers, it will give you an address of array. Another problem is, you are accessing your 2D array like 1D array, your array is 3x3 matrix but you are accessing to its 3th and 4th index with d variable. What you want to do is:
for (int i = 0; d < 3; ++j) {
for(int j = 0; j < 3; ++j)
printf("a[%d][%d] = %i\n\n", i, j , a[i][j]);
}
printf, but, I could be wrong. Anyway, +1.
printf has expectations. %d and a pointer ... not a perfect match. In my world, that's undefined behavior. Perhaps not in C.If you compile with -Wall -Wextra (always do that) you'll get the answer:
$ gcc k.c -Wall -Wextra
k.c: In function ‘main’:
k.c:9:26: warning: format ‘%d’ expects argument of type ‘int’, but argument 3 has type ‘int *’ [-Wformat=]
9 | printf("a[%d] = %d\n\n", d, a[d]);
| ~^ ~~~~
| | |
| int int *
| %ls
It says that a[d] has type int* which means that it's a pointer.
d == 4and so on