Below is my code snippet
int arr[1][1];
arr[0][0]=55;
arr[0][1]=56;
arr[1][0]=57;
arr[1][1]=58;
printf("\n arr[0,0]=%d",arr[0,0]);
printf("\n arr[0,1]=%d",arr[0,1]);
printf("\n arr[0,2]=%d",arr[0,2]);
I get output as -12 -10 -8
Can someone let me know the meaning of arr[0,0]?
Q Can someone let me know the meaning of arr[0,0]?
A arr[0,0] is equivalent to arr[0].
When you have series of comma separated expressions, the value of the last expression is the value of the entire expression. As an extended example,
arr[10, 200, 50, 983, 52] is equivalent to arr[52]
Your program needs to be reworked to print an integer, as you are expecting with your use of %d in the format specifier.
printf("\n arr[0][0]=%d",arr[0][0]);
If you want to print arr[0][1] and arr[0][2], you need to change the definition of arr to
int arr[1][3];
Looking at your new code
Change
int arr[1][1];
to
int arr[2][2];
Otherwise, the following lines will show undefined behavior:
arr[0][1]=56;
arr[1][0]=57;
arr[1][1]=58;
You are using the comma operator. It evaluates to its second operand, so your code is equivalent to:
printf("\n arr[0,0]=%d",arr[0]);
printf("\n arr[0,1]=%d",arr[1]);
printf("\n arr[0,2]=%d",arr[2]);
This is undefined behaviour in three different ways at once:
arr[0] is not an int (it's an array of int), so printing it with %d is not right1, so arr[1] and arr[2] are out-of-bounds accesses.To fix this, change to something like:
int arr[2][2] = { {1, 2}, {3, 4} };
printf("arr[0][0]=%d\n",arr[0][0]);
printf("arr[1][0]=%d\n",arr[1][0]);
printf("arr[0][1]=%d\n",arr[0][1]);
printf("arr[1][1]=%d\n",arr[1][1]);
Firstly about the value of expression (0,1):
int chk = (0,1);
printf("%d",chk);
This is how you can know the value of(0,1). If you want more detail please see: this.
Therefore arr[0,1] is same as arr[1]
